SOLUTIONS MANUAL for Operations Management: Sustainability and Supply Chain Management 14th Edition by ACADEMIAMILL

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C H A P T E R

Operations and Productivity

DISCUSSION QUESTIONS 1. The text suggests four reasons to study OM. We want to understand (1) how people organize themselves for productive enterprise, (2) how goods and services are produced, (3) what operations managers do, and (4) this costly part of our economy and most enterprises. LO 1.1: Define operations management AACSB: Application of knowledge 2. With some 40% of all jobs being in the OM field, the career opportunities are prolific. The text suggests many career opportunities. OM students find initial jobs throughout the OM field, including supply chain, logistics, purchasing, production planning and scheduling, plant layout, maintenance, quality control, inventory management, etc. LO 1.3: Identify career opportunities in operations management AACSB: Application of knowledge 3. Possible responses include: Adam Smith (work specialization/ division of labor), Charles Babbage (work specialization/division of labor), Frederick W. Taylor (scientific management), Walter Shewart (statistical sampling and quality control), Henry Ford (moving assembly line), Charles Sorensen (moving assembly line), Frank and Lillian Gilbreth (motion study), and Eli Whitney (standardization). LO 1.1: Define operations management AACSB: Application of knowledge 4.

See references in the answer to Question 3.

LO 1.1: Define operations management AACSB: Application of knowledge 5. The actual charts will differ, depending on the specific organization the student chooses to describe. The important thing is for students to recognize that all organizations require, to a greater or lesser extent, (a) the three primary functions of operations, finance/accounting, and marketing; and (b) that the emphasis or detailed breakdown of these functions is dependent on the specific competitive strategy employed by the firm. LO 1.1: Define operations management AACSB: Application of knowledge 6. The basic functions of a firm are marketing, accounting/ finance, and operations. An interesting class discussion: “Do all firms/organizations (private, government, not-for-profit) perform these three functions?” The authors’ hypothesis is yes, they do. LO 1.1: Define operations management AACSB: Application of knowledge

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OPERATIONS AND PRODUCTIVITY

7. The 10 strategic decisions of operations management are product design, quality, process, location, layout, human resources, supply chain management, inventory, scheduling (intermediate and short-term), and maintenance. We find this structure an excellent way to help students organize and learn the material. LO 1.1: Define operations management AACSB: Application of knowledge 8. The 10 decisions as applied to Amazon: (1) Product design—defining the product may initiate an interesting discussion. Is the product the over 12 million products available, or is the product the ease of order entry, convenience, and home delivery? Probably both. (2) Quality management—quality is built into every aspect of the Amazon culture, from item and order identification, and from the supplier through receipt, process design, human resource training, inventory tracking, etc. Multiple checks of items are standard. Errors in order receipts or shipment are very expensive. (3) Process strategy—receipt, movement to storage, and pulling of product for shipment are all state of the art. Amazon’s process mimics in many ways an assembly line … where the workers stand still and the product is brought to the worker, rather than movement of personnel up and down aisles to “pull” merchandise. (4) Location strategy—Amazon’s facilities are strategically located worldwide to facilitate rapid delivery. (5) Layout strategy—facilities are designed and redesigned to meet the changing state-of-the-art innovations, from Kivas (moving storage bins) to “pull to light.” (6) Human resources—a non-union workforce has allowed continuing innovation in job design, work assignments, flexible short-term and long-term (seasonal) work schedules, and benefits. (7) Supply chain management—volume has given Amazon substantial negotiating flexibility. Amazon has also been willing to design and produce its own “copycat” products when it seems advantageous. (8) Inventory management—sophisticated item receipt and storage over multiple facilities, along with customer order tracking that facilitates shipment of orders from multiple locations to ensure complete orders, gives Amazon a huge advantage. (9) Scheduling—the combination of inventory management, superior software, and a great process allows Amazon to ship in 15 minutes … the time from the customer’s keyboard order click to shipment is 15 minutes. (10) Maintenance—the facility maintenance is rather simple as factories go … it is complex, but they are not making rockets either. However, the software for customer order entry, Kiva control, inventory, and order management—over multiple facilities—is complex and requires ongoing development, backups, and updating. LO 1.1: Define operations management AACSB: Application of knowledge 9. Four areas that are important to improving labor productivity are (1) basic education (basic reading and math skills), (2) diet of the labor force, (3) social overhead that makes labor available (water, sanitation, transportation, etc.), and (4) maintaining and expanding the skills necessary for changing technology and knowledge, as well as for teamwork and motivation. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Application of knowledge 10. Productivity is harder to measure when the task becomes more intellectual. A knowledge society implies that work is more intellectual and therefore harder to measure. Because the U.S. and many other countries are increasingly “knowledge” societies, productivity is harder to measure. Using labor-hours as a measure of productivity for a postindustrial society versus an industrial or agriculture society is very different. For example, decades spent developing a marvelous new drug or winning a very difficult legal case on intellectual property rights may be significant for postindustrial societies, but not show much in the way of productivity improvement measured in labor-hours. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 11. Productivity is difficult to measure because precise units of measure may be lacking, quality may not be consistent, and exogenous variables may change. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Reflective thinking 12. Mass customization is the flexibility to produce to meet specific customer demands, without sacrificing the low cost of a product-oriented process. Rapid product development is a source of competitive advantage. Both rely on agility within the organization. LO 1.1: Define operations management AACSB: Application of knowledge Copyright ©2023 Pearson Education, Inc.

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13. Labor productivity in the service sector is hard to improve because (1) many services are labor intensive and (2) they are individually (personally) processed (the customer is paying for that service—the haircut), (3) it may be an intellectual task performed by professionals, (4) it is often difficult to mechanize and automate, and (5) it is often difficult to evaluate for quality. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Reflective thinking 14. Taco Bell designed meals that were easy to prepare; with actual cooking and food preparation done elsewhere; automation to save preparation time; reduced floor space; manager training to increase span of control. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Application of knowledge 15. Bureau of Labor Statistics (stats.bls.gov) is a good place to start. Results will vary for each year, but overall data for the economy will range from 0.9% to 4.8%, and mfg. could be as high as 5% and services between 1% and 2%. The data will vary even more for months or quarters. The data are frequently revised, often substantially. LO 1.7: Compute multifactor productivity AACSB: Application of knowledge

ETHICAL DILEMMA AMERICAN CAR BATTERY INDUSTRY You may want to begin the discussion by asking how ethical it is for you to be in the lead battery business when you know that any batteries you recycle will very likely find their way to an overseas facility (probably Mexico) with, at best, marginal pollution containment. Then after a likely conclusion of “Well someone has to provide batteries,” you can move to the following discussion. (a) As owner of an independent auto repair shop trying to dispose of a few old batteries each week, your options may be limited. But as an ethical operator, your first option is to put pressure on your battery supplier to take your old batteries. Alternatively, shop for a battery supplier who wants your business enough to dispose of your old batteries. Third, because there is obviously a market for the lead in old batteries, some aggressive digging may uncover an imaginative recycler who can work out an economical arrangement for pickup or delivery of your old batteries. Another option is, of course, to discontinue the sale of batteries. (This is a problem for many small businesses; ethical decisions and regulation may be such that they often place an expensive and disproportionate burden on a small firm.) (b) As manager of a large retailer responsible for disposal of thousands of used batteries each week, you should have little trouble finding a battery supplier with a reverse supply chain suitable for disposal of old batteries. Indeed, a sophisticated retailer, early on in any supply-chain development process, includes responsible disposal of environmentally dangerous material as part of the negotiations. Disposal of old batteries should be a minor issue for a large retailer. (c) For both a small and large retailer, the solution is to find a “sustainable” solution or get out of the battery business. Burying the batteries behind the store is not an option. Supplement 5: Sustainability in the Supply Chain provides some guidelines for a deeper class discussion.

END-OF-CHAPTER PROBLEMS 1.1 (a)

(b)

120 boxes = 3.0 boxes/hour 40 hours 125 boxes = 3.125 boxes/hour 40 hours

1.2

0.125 box = 4.167% 3.0

(a) Labor productivity is 160 valves/80 hours = 2 valves per hour (b) New labor productivity = 180 valves/80 hours = 2.25 valves per hour (c) Percentage change in productivity = .25 valve/2 valves = 12.5%

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1.3 0.15 =

57,600 , where L = number of laborers (160)(12)(L) employed at the plant

So, L =

1.4 (a)

(b)

OPERATIONS AND PRODUCTIVITY

57,600 = 200 laborers employed (160)(12)(0.15)

Units produced 100 pkgs = = 20 pkgs/hour Input 5 133 pkgs = 26.6 pkgs per hour 5

1.5

6.6 = 33.0% 20

Resource

Last Year

This Year

Change

Percentage Change

Labor

1,000 = 3.33 300

1,000 = 3.64 275

0.31

0.31 = 9.3% 3.33

Resin

1,000 = 20 50

1,000 = 22.22 45

2.22

2.22 = 11.1% 20

Capital

1,000 = 0.1 10,000

1,000 = 0.09 11,000

–0.01

−0.01 = −10.0% 0.1

Energy

1,000 = 0.33 3,000

1,000 = 0.35 2,850

0.02

0.02 = 6.1% 0.33

1.6 Production Labor-hour @ $10 Resin @ $5 Capital cost/month Energy

Last Year

This Year

1,000 $3,000 250 100 1,500 $4,850

1,000 $2,750 225 110 1,425 $4,510

[(1,000 / 4,510) − (1,000 / 4,850)] = (1,000 / 4,850) 0.222 − 0.206 0.016 = = 7.8% improvement* 0.206 0.206 *With rounding to 3 decimal places.

1.7 Productivity =

Output Input 65 65 = (520 × 13) $6,760 = .0096 rug per labor $

(a) Labor productivity =

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65 (b) Multifactor = productivity (520 × $13) + (100 × $5) + (20 × $50)

65 = .00787 rug per $ $8, 260

1.8 (a) Labor productivity = 1,000 tires/400 hours = 2.5 tires/hour. (b) Multifactor productivity is 1,000 tires/(400 × $12.50 + 20,000 × $1 + $5,000 + $10,000) = 1,000 tires/$40,000 = 0.025 tire/dollar. (c) Multifactor productivity changes from 1,000/40,000 to 1,000/39,000, or from 0.025 to 0.02564; the ratio is 1.0256, so the change is a 2.56% increase. Last Year

This Year

Change

Percentage Change

Labor hours

1,500 = 4.29 350

1,500 = 4.62 325

0.33 4.29

= 7.7%

Capital invested

1,500 = 0.10 15,000

1,500 = 0.08 18,000

− 0.02 0.1

= –20%

Energy (btu)

1,500 = 0.50 3,000

1,500 = 0.55 2,750

0.05 0.50

= 10%

1.9

Productivity of capital did drop; labor productivity increased as did energy, but by less than the anticipated 15%.

1.10

Multifactor productivity is: 375 autos/[($20 × 10,000) + ($1,000 × 500) + ($3 × 100,000)] = 375/(200,000 + 500,000 +300,000) = 375/1,000,000 = .000375 auto per dollar of inputs

1.11

(a) Before: 500/20 = 25 boxes per hour;

After: 650/24 = 27.08 (b) 27.08/25 = 1.083, or an increase of 8.3% in productivity (c) New labor productivity = 700/24 = 29.167 boxes per hour 1.12

1,500 × 1.25 = 1,875 (new demand)

Outputs = Productivity Inputs 1,875 = 2.344 Labor-hours 1,875 New process = ≅ 800 labor-hours 2.344 800 = 5 workers 160 1,500 Current process = = 2.344 Labor-hours 1,500 = labor-hours ≅ 640 2.344 640 = 4 workers 160 Add one worker. Copyright ©2023 Pearson Education, Inc.

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1.13

OPERATIONS AND PRODUCTIVITY

(a) Labor change: 1,500 1,500 = = .293 loaf/$ (640 × $8) 5,120 1,875 = 0.293 loaf/$ (800 × $8) (b) Investment change:

1,500 1,500 = = .293 loaf/$ (640 × $8) 5,120 1,875 1,875 = = .359 loaf/$ (640 × 8) + (100) 5,220 .293 – .293 = 0 (labor) .293 .359 – .293 Percentage change : = .225 .293 = 22.5% (investment)

The better option is to purchase a new blender because it generates more loaves per dollar.

1,500 (640 × 8) + 500 + (1,500 × 0.35) 1,500 = = 0.244 loaf/$ 6,145 1,875 New process = (800 × 8) + 500 + (1,875 × 0.35) 1,875 = = 0.248 loaf/$ 7,556.25 0.248 – 0.244 Percentage change = = 1.6% 0.244

1.14

1.15

Old process =

(a)

6,600 vans = 0.10 x labor-hours x = 66,000 labor-hours

There are 300 laborers. So,

66,000 labor-hours = 220 labor-hours/laborer 300 laborers on average, per month 6,600 vans = 0.11, so x = 60,000 labor-hours x labor-hours 60,000 labor-hours so, = 200 labor-hours/laborer 300 laborers on average, per month

(b) Now

1.16

$ output 52($90) + 80($198) = Labor-hours 8(45) $20,520 = = $57.00 per labor-hour 360

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1.17

(a) Last year =

1,500 (350 × 8) + (15,000 × 0.0083) + (3,000 × 0.6)

1,500 2,800 + 124.50 + 1,800 1,500 = = 0.317 doz / $ 4,724.5 1500 (b) This year = (325 × 8) + (18,000 × 0.0083) + (2,750 × 0.6) =

= 0.341 doz / $

0.341 − 0.317 0.317 = 0.076, or 7.6% increase

CASE STUDY UBER TECHNOLOGIES, INC. 1. First, some drivers (maybe most) may not require a wage that equals those fully engaged in the “taxi” business. It truly could be a supplemental income. . . . “I’m going that way anyhow so let’s make a few dollars while on the way.” Similarly, the capital investment cost approaches zero as the car is going that direction anyhow. These are idle or underutilized resources. From society’s perspective, Uber and its like competitors are desirable because both idle or wasted labor and capital resources are being utilized. At the same time, as a bonus, Uber is reducing traffic and auto pollution while speeding up the transport of individuals and local commerce. As a competitor for the traditional taxi service, Uber seems to be an enhancement in efficiency. For those faculty who what to spend some time on the larger productivity message, this case provides such an opportunity. Uber, as Joseph Schumpeter would suggest, has developed a disruptive technology (creative destruction, in a Schumpeterian translation). Innovations such as this are exactly how economic efficiency is enhanced. The traditional taxi services, with some imagination, could have developed and adopted this technology, but most were ensconced in their own regulatory cocoon. As is often the case, it takes an outsider, such as Uber et al. to be creative by putting unused resources to use and providing society greater efficiency. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 2. Perhaps a business model similar to Uber’s can be applied to the trucking industry. And, indeed, Uber has established an Uber app for the trucking industry. An estimated 30% of trucking backhauls are empty. However, the number of independent truckers or truckers with the latitude to alter their route may be very small. And this number must be a tiny fraction of independent automobile drivers. So, the ability to “Uberize” trucking may be very difficult, but utilizing that idle 30% would be huge benefit to society. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 3. Perhaps the Uber model can be used for package delivery, documents, and everything from flowers to groceries. Airbnb (www.airbnb.com) is applying a similar model to short-term rentals of rooms, apartments, and homes—competing with more traditional bed and breakfast facilities and hotels. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 4. As Uber expands into food delivery, package delivery, courier service, etc., many of the disadvantages of the Uber model are slowly being overcome. However, a major ongoing disadvantage is the proliferation of regulations, with some countries and airports even banning ride-sharing companies. Additionally, some jurisdictions, namely California, require ride-sharing companies to treat drivers as employees rather than independent contractors. This is a significant issue as the Uber business model does not hire drivers. This and related liability disputes are still being defined. The issue is compounded by the need to have enough drivers on hand when needed. The other major issue is Uber’s very extensive, sophisticated, necessarily complex, and fast software network with its inherent vulnerabilities, from “bugs” to network downtime and hacks. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking Copyright ©2023 Pearson Education, Inc.

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VIDEO CASE STUDIES FRITO-LAY: OPERATIONS MANAGEMENT IN MANUFACTURING

This case provides a great opportunity for an instructor to stimulate a class discussion early in the course about the pervasiveness of the 10 decisions of OM with this case alone or in conjunction with the Hard Rock Cafe case. There is a short video (7 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. Product design: Each of Frito-Lay’s 40-plus products must be conceived, formulated (designed), tested (market studies, focus groups, etc.), and evaluated for profitability.  Quality: The standards for each ingredient, including its purity and quality, must be determined.  Process: The process that is necessary to produce the product and the tolerance that must be maintained for each ingredient by each piece of equipment must be specified and procured.  Location: The fixed and variable costs of the facility, as well as the transportation costs and the delivery distance, given the freshness, must be determined.  Layout: The Frito-Lay facility would be a process facility, with great care given to reducing movement of material within the facility.  Human resources: Machine operators may not have inherently enriched jobs, so special consideration must be given to developing empowerment and enriched jobs.  Supply chain management: Frito-Lay, like all other producers of food products, must focus on developing and auditing raw material from the farm to delivery.  Inventory: Freshness and spoilage require constant effort to drive down inventories.  Scheduling: The demand for high utilization of a capital-intensive facility means effective scheduling will be important.  Maintenance: High utilization requires good maintenance, from machine operator to the maintenance department and depot service. 

LO 1.1: Define operations management AACSB: Reflective thinking 2.

Determining output (in some standard measure, perhaps pounds) and labor-hours would be a good start for single-factor productivity.

For multifactor productivity, we would need to develop and understand capital investment and energy, as well as labor, and then translate those into a standard, such as dollars. LO 1.6: Compute single-factor productivity LO 1.7: Computer multifactor productivity AACSB: Reflective thinking 3. Hard Rock performs all 10 of the decisions as well, only with a more service-sector orientation. Each of these is discussed in the solution to the Hard Rock Cafe case. LO 1.8: Identify the critical variables in enhancing productivity AACSB: Reflective thinking 2

HARD ROCK CAFE: OPERATIONS MANAGEMENT IN SERVICES

There is a short video (7 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. Hard Rock’s 10 decisions: This is early in the course to discuss these in depth, but still a good time to get the students engaged in the 10 OM decisions around which the text is structured.  Product design: Hard Rock’s tangible product is food and like any tangible product it must be designed, tested, and “costed out.” The intangible product includes the music, memorabilia, and service.  Quality: The case mentions the quality survey as an overt quality measure, but quality can be discussed from a variety of perspectives— hiring the right people, food ingredients, good suppliers, speed of service, friendliness, etc.  Process: The process can be discussed from many perspectives: (a) the process of processing a guest, to their seat, taking the order, order processing, delivery of the meal, payment, etc.; (b) the process of how a meal is prepared (see, for instance, how one would make a Hard Rock Hickory BBQ Bacon Cheeseburger (Figure 5.9) or a Buffalo Chicken Mac & Cheese (Figure 14.9) or use the Method Analysis tool discussed in Chapter 10; or (c) some subset of any of these. Copyright ©2023 Pearson Education, Inc.

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Location: Hard Rock Cafes have traditionally been located in tourist locations, but that is beginning to change. Layout: Little discussion in the case, but students may be very aware that a kitchen layout is critical to efficient food preparation and that a bar is critical in many food establishments for profitability. The retail shop in relation to the restaurant and its layout is a critical ingredient for profitability at Hard Rock.  Human resources: Jim Knight, VP for Human Resources at Hard Rock, seeks people who are passionate about music, love to serve, and can tell a story. This OM decision is a critical ingredient for success of a Hard Rock Cafe and an integral part of the Hard Rock dining experience.  Supply chain management: Although not discussed in the case, students should appreciate the importance of the supply chain in any food service operation. Some items like leather jackets have a 9-month lead time. Contracts for meat and poultry are signed 8 months in advance.  Inventory: Hard Rock, like any restaurant, has a critical inventory issue that requires that food be turned over rapidly and that food in inventory be maintained at the appropriate and often critical temperatures. But the interesting thing about Hard Rock’s inventory is that they maintain $40 million of memorabilia with all sorts of special care, tracking, and storage issues.  Scheduling: Because most Hard Rock Cafe’s sales are driven by tourists, the fluctuations in seasonal, daily, and hourly demands for food are huge. This creates a very interesting and challenging task for the operations managers at Hard Rock. (Not mentioned in the case, linear programming is actually used in some cafes to schedule the waitstaff.)  Maintenance/reliability: The Hard Rock Cafe doors must open every day for business. Whatever it takes to provide a reliable kitchen with hot food served hot and cold food served cold must be done. Bar equipment and point-of-sale equipment must also work.  

LO 1.1: Define operations management AACSB: Reflective thinking 2. Productivity of kitchen staff is simply the output (number of meals) over the input (hours worked). The calculation is how many meals prepared over how many hours spent preparing them. The same kind of calculation can be done for the waitstaff. In fact, Hard Rock managers begin with productivity standards and staff to achieve those levels. (You may want to revisit this issue when you get to Chapter 10 and Supplement 10 on labor standards and discuss how labor can be allocated on a per-item basis with more precision.) LO 1.6: Compute single-factor productivity AACSB: Analytical thinking 3. Each of the 10 decisions discussed in Question 1 can be addressed with a tangible product like an automobile.  Product design: The car must be designed, tested, and costed out. The talents may be those of an engineer or operations manager rather than a chef, but the task is the same.  Quality: At an auto plant, quality may take the form of measuring tolerances or wear of bearings, but there is still a quality issue.  Process: With an auto, the process is more likely to be an assembly-line process.  Location: Hard Rock Cafe may want to locate at tourist destinations, but an auto manufacturer may want to go to a location that will yield low fixed or variable cost.  Layout: An automobile assembly plant is going to be organized on an assembly line criterion.  Human resources: An auto assembly plant will be more focused on hiring factory skills rather than a passion for music or personality.  Supply chain management: The ability of suppliers to contribute to design and low cost may be a critical factor in the modern auto plant.  Inventory: The inventory issues are entirely different—tracking memorabilia at Hard Rock, but an auto plant requires tracking a lot of expensive inventory that must move fast.  Scheduling: The auto plant is going to be most concerned with scheduling material, not people.  Maintenance: Maintenance may be even more critical in an auto plant as there is often little alternate routing, and downtime is very expensive because of high fixed and variable cost. LO 1.4: Explain the distinction between goods and services AACSB: Reflective thinking

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OPERATIONS AND PRODUCTIVITY

There is a short video (6.5 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. Celebrity’s 10 decisions: It is early in the course to discuss these in depth, but still a good time to get the students engaged in the 10 OM decisions around which the text is structured.  Product design: Celebrity’s product consists of a complete “premium” vacation/holiday experience. It includes accommodations, ports-of-call, shipboard facilities, food, service, etc. Students should appreciate the full scope of how Celebrity Cruises designs all of the many attributes of its “product.”  Quality: The case mentions the quality survey as an overt quality measure, but quality can be discussed from a variety of perspectives—hiring the right people, food ingredients, good suppliers, speed of service, cleanliness, friendliness, etc.  Process: Operation of a successful cruise line consists of many processes. The process can be discussed from various perspectives: (a) the process of welcoming a guest aboard, (b) bill and payment processing, (c) delivery of meals, (d) supply chain, (e) off ship excursions, etc. The methods analysis tools discussed in Chapter 10 provide a way for students to address and analyze these processes.  Location: Celebrity Cruises provides a unique opportunity for students to address the many aspects of the location decision. First, where in the world are the customers? Second, from what home ports will Celebrity operate? Third, where are the locations of the ports-of-call for the ship?  Layout: How should the ship itself be designed … how many restaurants, how many kitchens, what other amenities (i.e. gym, spa, theater, shops, library, etc.)? What shipboard features will distinguish differences in pricing?  Human resources: The unique international flavor of the crew on cruise ships generates a wide variety of special recruiting, motivational, and teamwork issues. A service-oriented staff, carefully recruited and well trained, is a critical ingredient for success of a “hotel at sea” and an integral part of the premium Celebrity Cruises experience.  Supply chain management: Students should appreciate the importance of the supply chain for a floating hotel that is going to be at sea for days or even weeks at a time.  Inventory: Because there is seldom resupply once at sea, inventory, but particularly food inventory for hundreds of people, is a critical issue. Food requirements must be accurately forecasted and be maintained at the appropriate and often critical temperatures. Food is only one of the many inventory items to be maintained: water, fuel, cleaning supplies, clothes, and memorabilia require all sorts of special care, tracking, and storage issues. 



Scheduling: Fluctuations in location and season create a very interesting and challenging task for the operations managers. Not only the ships and port access and excursions, but also food deliveries and crews, must all be scheduled. Maintenance/reliability: The ship is open every day for business. Minor maintenance is performed while the ship is operating, with more significant maintenance performed annually and major long-term maintenance conducted in dry dock every 5 years.

LO 1.2: Identify the 10 strategic decisions of operations management AACSB: Reflective thinking 2. Celebrity’s 10 OM decisions are also executed by a manufacturing firm. See, for instance, the Frito-Lay case discussed earlier in this chapter. Indeed, the theme of the text is that these 10 decisions are pervasive in OM. It matters little if the product is a Frito-Lay product, an iPhone, or a premium vacation with Celebrity Cruises; all of these 10 decisions are going to be made. The distinction is the implementation and emphasis placed on each. For instance, product design at Frito-Lay may begin with selecting the proper potatoes, cooking oils, and temperature. Celebrity, as noted above, has a very different product design task. Similarly, quality of Frito-Lay chips may be dependent on precise cutting blades and processing temperature, while Celebrity’s quality manifests itself in accommodations, food, and service. Students should be challenged to recognize that the 10 decisions are made, albeit with distinctions dependent upon the product and strategy. LO 1.2: Identify the 10 strategic decisions of operations management ACSB: Reflective thinking 3

CELEBRITY CRUISES: OPERATIONS MANAGEMENT AT SEA

3. Celebrity’s 10 OM decisions are also executed by a retail firm. Indeed, the theme of the text is that these 10 decisions are pervasive in OM. It matters little if the product is a retail firm or a restaurant (such as Hard Rock, discussed in the prior case) or a premium vacation with Celebrity Cruises; all of these 10 decisions are going to be made. Perhaps in a different way and with different emphasis, but they will be made. For instance, Hard Rock’s product is a unique memorabilia-filled dining experience. Celebrity’s product is a holiday with premium accommodations, food, and service. Students should be challenged to recognize that the 10 decisions are made, albeit with distinctions dependent upon the product and strategy. LO 1.2: Identify the 10 strategic decisions of operations management ACSB: Reflective thinking

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4. The differences between a land-based hotel and the “hotel at sea” may be very small in terms of guest expectations and the quality decision. However, the emphasis on various aspects of the other decisions can be expected to change. For instance, for the “hotel at sea” the location decision changes as a function of the season, port-of-call performance, and even weather. A hotel may or may not include dining excellence a part of its product, but for most cruise lines, a premium dining experience is critical. In the case of supply chain, logistics, and inventory, for the ship there is often no resupply; therefore, there is an added emphasis on forecasts, logistics, and inventory. Forecasts must be accurate, suppliers punctual, and inventory counts precise. Similarly, maintenance onboard ship must remove all variability; the emergency backup may be days away. Most hotels will very likely have little in common with the implementation of the human resource function at an international cruise line with employees from dozens of countries. But they both must be successful at the HR decision. LO 1.2: Identify the 10 strategic decisions of operations management ACSB: Reflective thinking

ADDITIONAL CASE STUDIES (available in MyLab Operations Management) 1

NATIONAL AIR EXPRESS

This case can be used to introduce the issue of productivity and how to improve it, as well as the difficulty of good consistent measures of productivity. This case can also be used to introduce some of the techniques and concepts of OM. 1. The number of stops per driver is certainly a good place to start. However, mileage and number of shipments will probably be good additional variables. (Regression techniques, addressed in Chapter 4, can be addressed here.) LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 2. Customer service should be based on an analysis of customer requirements. Document requirements in terms of services desired (supply needs, preprinted waybills, package weights, pickup and drop-off requirements) should all be considered. (The house of quality technique discussed in Chapter 5 is one approach for such an analysis.) LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 3. Other companies in the industry do an effective job of establishing very good labor standards for their drivers, sorters, and phone personnel. Difficult perhaps, but doable. (Work measurement in Chapter 10 addresses labor standards.) LO 1.8: Identify the critical variables in enhancing productivity AACSB: Analytical thinking 2

ZYCHOL CHEMICALS CORPORATION

1. The analysis of the productivity data is shown on the next page. Both labor and material productivity increased, but capital equipment productivity did not. The net result is a large negative change in productivity. If this is a one-time change in the accounting procedures, this negative change should also be a one-time anomaly. The effect of accounting procedures is often beyond the control of managers. For example, perhaps the capital allocation is based on an accelerated allocation of depreciation of newly installed technology. This accounting practice will seriously impact near-term productivity and then later years’ productivity figures will benefit from the reduced depreciation flows. This highlights the difficulty in accounting for costs in an effective managerial manner. Decisions and evaluation of operating results should be based on sound managerial accounting practices and not necessarily generally accepted financial accounting principles. LO 1.6: Compute single-factor productivity LO 1.7: Compute multifactor productivity AACSB: Analytical thinking

CHAPTER 1

OPERATIONS AND PRODUCTIVITY

2. An analysis of adjusted results reduces the negative impact on the capital allocation but there is still a negative growth in multifactor productivity. After adjustment for inflation, the material costs are still higher in 2022. Yet, one must be aware of the extra volatility of the cost of petroleum-based products. Did the manager have control over his price increases? One should look at the changes in a petroleumbased price index, including the cost of oil, over the last two years in order to gain a better understanding of the degree to which the manager had control over these costs. The increase in wages was beyond the manager’s control, and a constant rate should be used for comparing both years’ results. Yet a negative result still remains. Even when material costs in 2022 are converted to the original cost of $320, a negative 5% growth in productivity remains. The increase in the capital base is responsible yet should not persist in future years if the increase was the result of an adoption of new technology. LO 1.6: Compute single-factor productivity LO 1.7: Compute multifactor productivity AACSB: Analytical thinking 3. The manager did not reach the goal. An analysis of the changes in capital costs is warranted. Even after adjusting for inflation, multifactor productivity was not positive. However, labor and materials productivity were favorable. The capital investment cost (as figured by the accounting department) was so large as to make his multifactor productivity negative. Multifactor productivity has fallen by 11.61% before adjustment and by 7.87% after the adjustment for inflation. LO 1.7: Compute multifactor productivity AACSB: Application of knowledge Single-Factor Productivity Analysis Production (Units) Material Used (Barrels) Material Cost per Barrel

4,500 700 $320.00

Labor-Hours Compensation Rate

22,000 $20.00

2021

Capital Applied ($) Producer Price Index (PPI)

2022

$224,000

6,000 900 $360.00

$440,000

28,000 $22.00

$375,000 120

Adjusted Cost*

Adjusted Total Cost

$324,000

$345.60 (360/1.04167)

$311,040 (900 × 345.60)

$616,000

$21.12 (22/1.04167) $595,200 (620,000)/1.04167)

$591,360 (28,000 × $21.12) $595,200

$620,000 125

*Change in PPI = 4.167% = (125/120 − 1) = 0.04167 Total Cost

Multifactor Productivity (MFP) Analysis Labor Productivity (Units per hour) Material Productivity (Units per barrel) Capital Productivity (Units per $)

$1,039,000

$1,560,000

$1,497,600 (Adjusted)

2021

2022

4,500/22,000 = 0.2045

6,000/28,000 = 0.2143

4.79%

Nearly reached the goal

4,500/700 = 6.4286

6,000/900 = 6.6667

3.70%

Positive change

4,500/375,000 = 0.0120

6,000/620,000 = 0.0097

−19.17%

Large negative change

MFP Before Adjustment per $) MFP After Adjustment (per $)

2021 0.00508 0.00508

2022 0.00449 0.00468

% Change

(0.00449 − 0.00508)/0.00508 = −11.61% (0.00468 – 0.00508)/0.00508 = −7.88%

C H A P T E R

Operations Strategy in a Global Environment

DISCUSSION QUESTIONS 1. Global seems the better label for Boeing because authority and responsibility reside in the U.S.—the home country. LO 2.5: Identify and explain four global operations strategy options AACSB: Application of knowledge 2. Six reasons to internationalize: Reduce costs, improve supply chain, provide better goods and services, attract new markets, learn to improve operations, attract and retain global talent. LO 2.5: Identify and explain four global operations strategy options AACSB: Analytical thinking 3. No. Sweetness at Coca-Cola is adjusted for the tastes of individual countries. LO 2.1: Define mission and strategy AACSB: Application of knowledge 4. A mission is an organization’s purpose—what good or service it will contribute to society. LO 2.1: Define mission and strategy AACSB: Application of knowledge 5. Strategy is an organization’s action plan—how it is going to achieve its purpose. LO 2.1: Define Mission and strategy AACSB: Application of knowledge 6. A mission specifies where the organization is going and a strategy specifies how it is going to get there. LO 2.1: Define mission and strategy AACSB: Application of knowledge 7. The answer to this question will depend on the establishment studied, but should probably include some of the following considerations: The mission: diagnose automobile problems and make the necessary repair at a fair price for the local customer. Points to consider, or options, within the 10 decision areas are Decision:

Option:

Product

Repair work of American and/or foreign vehicles; specialized (tune-ups, lubrication, wheel alignment, etc.) versus general repair; frame and body repair versus engine and power train repair; repair and maintenance only versus repair, maintenance, and sales of fuel; professional staffing versus rental of tools and space for do-it-yourself repair work Appropriate level of quality; warranty; method of measuring and maintaining quality (customer complaints, inspection by supervising mechanic, etc.) Use of general versus special-purpose diagnostic and repair equipment (in particular, the degree to which computer controlled diagnostic equipment is employed) In-town, shopping mall, highway

Quality Process Location

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Layout

Single bay/multibay; general-purpose bay versus special-purpose bay (lubrication/tire repairs and installation/wheel alignment/ engine and power train repair, etc.) Human Resources Employment of certified versus noncertified repairpersons; employment of specialists versus general mechanics Supply Chain Choice of supplier(s) for both general and original manufacturer parts and supplies Scheduling Hours of operation (8:00 A.M.–5:00 P.M.; 24-hour towing; weekends/holidays), repairs versus motor vehicle safety inspections, etc.; service by appointment versus walk-in (or drive-up) service Inventory Quantity and variety of repair parts (fan belts, filters, mufflers, headlights, etc.) to stock; whether to stock generic or original manufacturer parts Maintenance Bays with hydraulic lifts versus easier-to-maintain “basement” work areas; preventive maintenance of equipment versus breakdown LO 2.1: Define mission and strategy AACSB: Application of knowledge 8. Library or Internet assignment: Student is to identify a mission and strategy for a firm. BusinessWeek, Fortune, The Wall Street Journal, and Forbes all have appropriate articles. LO 2.1: Define mission and strategy AACSB: Application of knowledge 9. OM strategy changes during a product’s life cycle: During the introduction stage, issues such as product design and development are critical, then during the growth stage the emphasis changes to product and process reliability; from there we move to concern for increasing the stability of the manufacturing process and cost cutting; and finally, in the decline stage pruning the line to eliminate items not returning good margin becomes important. Figure 2.5 provides a more expansive list. LO 2.1: Define mission and strategy AACSB: Application of knowledge 10. The text focuses on three conceptual strategies—cost leadership, differentiation, and response. Cost leadership by Walmart—via low overhead, vicious cost reduction in the supply chain; differentiation, certainly any premium product—all fine dining restaurants, upscale autos—Lexus, etc.; response, your local pizza delivery service, FedEx, etc. LO 2.2: Identify and explain three strategic approaches to competitive advantage AACSB: Reflective thinking 11. An operations strategy statement for Southwest Airlines would include a focus on efficient, low-cost service with high capital utilization (high aircraft and gate utilization), flexible non-union empowered employees, low administrative overhead, etc. Southwest’s strategy was complicated by the purchase of AirTran. First, there was a major organizational culture issue. Southwest’s culture is unique. The company really does think of itself as a family, with a fun culture. AirTran’s culture was different. Integrating the two cultures was a challenge. Related to this were human resources issues such as seniority, pay rate, and promotion policies, all of which were complicated by union issues. On the tangible side, Southwest’s use of just Boeing 737s was complicated by AirTran’s use of other types of planes. To maintain the “one plane” efficiency (pilot training maintenance, inventory, etc.), Southwest leased the planes that were not Boeing 737s to Delta. The merger can be considered a success, providing Southwest with a 20%–25% increase in capacity, 14 new domestic cities, and 7 international cities. The success speaks well of the capability of Southwest management. LO 2.1: Define mission and strategy AACSB: Application of knowledge 12. The integration of OM with marketing and accounting is pervasive. You might want to cite examples such as developing new products. (Marketing must help with the design, the forecast and target costs; accounting must ensure adequate cash for development and the necessary capital equipment.) Similarly, new technology or new processes emanating from operations must meet the approval of marketing and the capital constraints imposed by the accounting department. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking

13. To summarize outsourcing trends:  Not everyone who outsources is 100% satisfied, and future arrangements may be revised or insourced.  IT will be a major expansion area, according to Gartner, Inc.  More laws may be passed to protect U.S. jobs. Copyright ©2023 Pearson Education, Inc.

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  

Foreign firms will increase their outsourcing to the U.S. Outsourcing will continue to grow. Current practices will be improved.

LO 2.3: Understand the significance of key success factors and core competencies AACSB: Reflective thinking

14. Cost savings in recent years from outsourcing has been significant. It may be possible to reduce labor costs by as much 75%. But more realistically, this figure is in the 20%–40% range. Overall savings in the 10%–30% range are possible. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 15. Internal issues include the following:  Employment—morale may drop, and employees may lose their jobs.  Facilities—may need to be changed if components arrive in different stages of assembly.  Logistics—now includes customs, timing, and insurance. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 16. The company should identify its own core competencies and then consider a list of candidate activities and firms for outsourcing. The factor-rating method can be used to compare various companies on a set of factors that management considers important. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 17. Bad outsourcing decisions may result in the following outcomes:  Higher transportation cost  Loss of control  Future competition from the provider  Negative impact on employees  Quick gains at the expense of long-term objectives LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 18. McDonald’s fits the categorization in the text as a multidomestic, as opposed to international, global, or transnational. This is the concept of exporting the management talent and process allowing flexibility in the product itself. In the case of McDonald’s, this export is operations management expertise, which it has implemented world-wide. Interestingly, McDonald’s likes to call itself multilocal. LO 2.5: Identify and explain four global operations strategy options AACSB: Application of knowledge

ETHICAL DILEMMA Here is an interesting scenario. A firm can save $10 million in production costs per year. All it has to do is locate manufacturing in Vietnam, which is not a democracy, where sustainability is not an issue, and where some employees are exploited. Nike faced a similar dilemma in Vietnam, where it was accused of paying less than a livable wage ($1.60 per day). Students may (or may not) be prepared to discuss this current and sensitive subject. AACSB: Reflective thinking

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END-OF-CHAPTER PROBLEMS 2.1

Arrow; Bidermann International, France Braun Household Appliances; Procter & Gamble, U.S. Volvo Autos; Geely, China Firestone Tires; Bridgestone, Japan Godiva Chocolate; Campbell Soup, U.S. Haagen-Dazs Ice Cream; great globalization discussion example: Haagen-Dazs was established in New York City; now owned by Pillsbury (U.S.A.), which is owned by General Mills (U.S.A.), but Nestlé SA (Switzerland) is licensed to sell Haagen-Dazs in the U.S. Jaguar Autos; Tata, India GE Appliances; Haier, China Lamborghini; Volkswagen, Germany Goodrich; Michelin, France Alpo Pet Foods; Nestlé, Switzerland

2.2 The corruption perception index maintained by Transparency International (www.transparency.org) gives a 1-to-100 scale (100 being least corrupt to 1 being most corrupt). Also see Chapter 8, Table 8.2. For 2020 most corrupt: Venezuela (#176), China (#78), USA (#25), Switzerland (#3), Denmark (#1). A lively class discussion can also take place regarding who pays bribes, as shown on the same Web site 2.3 The World Bank’s “Ease of Doing Business” rankings can be found at www.doingbusiness.org/en/rankings. (Also see Table 8.1 in the text.) For 2019, easiest-to-hardest: USA #6, China #31, Switzerland #36, Mexico #60. These rankings change—and sometimes quickly—based on world events (such as China’s new influence over Hong Kong). 2.4 The three methods are cost leadership, differentiation, and response. Cost leadership can be illustrated by Walmart and Dell, with low overhead and huge buying power to pressure its suppliers into concessions. Differentiation can be illustrated by almost any restaurant or restaurant chain, such as Red Lobster, which offers a distinct menu and style of service than others. Response can be illustrated by a courier service such as FedEx, that guarantees specific delivery schedules; or by a custom tailor, who will hand-make a suit specifically for the customer. 2.5 Cost leadership: institutional food services, such as Sodexho, provide meal service to college campuses and similar institutions. Such firms often get their contracts by being low bidder to provide service. Response: a catering firm (the customer picks the menu, time, and date). Differentiation: virtually all restaurants seek differentiation in menu, in taste, in service. This is particularly true of fine dining restaurants, but also true of fast food restaurants. For instance, Burger King likes to talk about meals “anyway you want them,” and McDonald’s has a playground or seating area for children. 2.6

(a) The maturing of a product may move the OM function to focus on more standardization, make fewer product changes, find optimum capacity, stabilize the manufacturing process, lower labor skills, use longer production runs, and institute cost cutting and design compromises. (b) Technological innovation in the manufacturing process may mean new human resources skills (either new personnel and/or training of existing personnel), and added capital investment for new equipment or processes. Product design, layout, maintenance procedures, purchasing, inventory, quality standards, and procedures may all need to be revised. (c) A design change will, at least potentially, require the same changes as noted in (b).

2.7

Specific answers to this question depend on the organization considered. Some general thoughts follow: (a) For a producer with high energy costs, major oil prices change the cost structure, result in higher selling prices, and, if the company is energy inefficient compared to other producers, result in a change in competitive position. Conversely, when oil prices drop it is a bonanza for heavy fuel users such as airlines. (b) More restrictive quality of water and air legislation increases the cost of production and may, in some cases, prohibit the use of specific technologies. The high cost of process modification to meet more rigid standards has resulted in the closing of numerous plants including paper mills and steel mills. (c) A decrease in the number of young prospective employees entering the U.S. labor market can contribute to a tighter job market. High unemployment rates can have the opposite effect.

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(d) Inflation, especially at high or uncertain rates, makes it more difficult to predict both the cost of production and the market demand. (e) Legislation moving health insurance from a before-tax benefit to taxable income will reduce the take-home pay of employees by the amount of the taxes. This could have a significant effect on the income of employees in the lower pay classifications, putting substantial pressure on operations managers to increase wages in these classifications. (This does not mean that it is not a good idea for society—i.e., to make employees more sensitive to the cost of health insurance.) 2.8 (a) Using the weighted model, with the four weights totaling 1.0, England has a risk of 2.3 and Canada a risk of 1.7. Now Canada is selected. England = .1(2) + .6(3) + .2(1) + .1(1) = 2.3 Canada = .1(3) + .6(1) + .2(3) + .1(2) = 1.7 (b) When each of the weights is doubled, the selection stays the same: Canada. 2.9 With weights given, the results are Mexico

= 3.3 = [.4(1) + .2(7) + .1(3) + .1(5) + .1(4) + .1(3)]

Panama

= 4.1

Costa Rica = 4.4 Peru

= 4.2

Mexico is the lowest-risk country for the firm to outsource to. 2.10

(a) The results of the factor rating method are as follows: Overnight Shipping

Worldwide Delivery

United Freight

Weighted total

800

815

775

Weighted average

81.5

77.5

The best outsource provider is Worldwide Delivery. (b) Nothing changes in the weighted averages if every one of the weights is doubled. The weighted totals will double. (c) If the three Overnight Shipping ratings increase by 10%, to 99, 77, and 77, respectively, the new weighted average is 88, and the weighted sum is 880. So Overnight is now the preferred logistics provider. 2.11 Selection Criteria

Criterion Weight

Computations for Manila

Computations for Delhi

1. Flexibility

0.5

0.5 × 5 = 2.5

0.5 × 1 = 0.5

0.5 × 9 = 4.5

2. Trustworthiness

0.1

0.1 × 5 = 0.5

0.1 × 2 = 0.2

3. Price

0.2

0.2 × 4 = 0.8

0.2 × 3 = 0.6

0.2 × 6 = 1.2

4. Delivery

0.2

0.2 × 5 = 1.0

0.2 × 6 = 1.2

Total score

1.0

4.8

2.8

7.1

Moscow Bell is clearly the highest rated for Walker’s help desk.

2.12 Provider

Score

5W + 320 = (60 + 15 + 125 + 15 + 30 + 75)

4W + 330

3W + 370

5W + 255

Computations for Moscow

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Find all w from 1 to 30 so that: 3w + 370 ≥ 5w + 320, or 50 ≥ 2w, or w ≤ 25 3w + 370 ≥ 4w + 330, or 40 ≥ w, or w ≤ 40 3w + 370 ≥ 5w + 255, or 115 ≥ 2w, or w ≤ 57.5 Company C is recommended for all w such that 1.0 ≤ w ≤ 25.0 2.13 Global. Its level of integration goes beyond multinational. The collection of parts and subassemblies coming from other countries is carefully orchestrated. It is not transnational because its “home” is clearly the U.S., and there is little sense of “local responsiveness.”

CASE STUDY RAPID-LUBE 1. To provide economical preventative maintenance and interior auto cleaning, primarily to vehicles owned by individuals (as opposed to businesses), in the U.S. LO 2.1: Define mission and strategy AACSB: Reflective thinking 2. This case is a good way to get the student thinking about the 10 decisions around which the text is organized. Rapid-Lube’s approach to these 10 decisions includes the following:  Product design: A narrow product strategy could be defined as “lubricating automobiles” (more in Chapter 5).  Quality strategy: Because of limited task variety, high repetition, good training, and good manuals, quality should be relatively easy to maintain.  Process strategy: The process strategy allows employees and capital investment to focus on doing this mission well, rather than trying to be a “general-purpose” garage or gas station.  Location strategy: Facilities are usually located near residential areas.  Layout strategy: The three bays are designed specifically for the lubrication and vacuuming tasks to minimize wasted movement on the part of the employees and to contribute to the speedier service.  Supply chain management: Purchasing is facilitated by negotiation of large purchases and custom packaging.  Human resources strategy: Human resources strategy focuses on hiring a few employees with limited skills and training them in a limited number of tasks during the performance of which they can be closely supervised.  Inventory: Inventory investment should be relatively low, and they should expect a high turnover.  Scheduling: Scheduling is quite straightforward with similar times for most cars. Once volume and fluctuation in volume are determined, scheduling should be very direct—assisting both staffing and customer relations.  Maintenance: There is relatively little equipment to be maintained, and therefore little preventive maintenance is required. With three bays and three systems, there is backup available in the case of failure. LO 2.2: Identify and explain three strategic approaches to competitive advantage AACSB: Reflective thinking 3. Specialization of personnel and facilities should make Rapid-Lube more efficient. Jobs/tasks accomplished per man hour would be a good place to start. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Reflective thinking

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VIDEO CASE STUDIES 1

STRATEGY AT NAUTIQUE BOAT COMPANY

There is a short video (8 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. See attached figure for one possible activity mapping for Nautique. Less ambitious students may simply connect the six ovals from material noted in the case.

Activity Mapping for Nautique LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 2. The strengths of Nautique include international name recognition as a product of quality, innovation, and prestige. The firm offers a constant stream of new high-tech products from a world-class product development team. It has dedicated employees and a management team with low turnover. Weaknesses include limited production space and facilities for current demand. It can be difficult to maintain an effective, well-trained, and loyal workforce in a tight Florida labor market. Also, factory jobs can be exhausting (10-hour days), hot (no air conditioning), and assigned under challenging working conditions (sprayed epoxy and paint chemicals in the air). The opportunities for Nautique include an increase in boat sales corresponding to an affluent society across global markets. The introduction of electric boats opens new doors for ecoconscious buyers. The threats to Nautique include a large number of bigger players who provide lower priced products. Heavy investment in electric boats may not necessarily yield success. Overseas tariffs/political issues can have a major impact on foreign sales, which have become a bigger part of the revenue stream in recent years. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Application of knowledge Copyright ©2023 Pearson Education, Inc.

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3. This nearly century-old company constantly innovates with upgraded products. Early products were wooden outboards, then mahogany inboards, and now fiberglass inboards—and soon electric inboards. The labor force is increasingly educated (managers with master’s degrees) and now has career paths for factory workers and managers. The Ski Nautique brand was introduced in 1961, and by the 1990s the Nautique name became more well-known than “Correct Craft” (the parent company brand). Nautique started expanding from water skiing to wake-boarding in the 1990s, which became the driving force in sales in the 2010s. “Wake Surfing” is now becoming the main sales driver. The move to electric boats in the 2020s is the possible future of the company. The Nautique Regatta brings owners of boats to company sponsored events to enhance customer loyalty. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Application of knowledge 4. The impact of electric boats is likely parallel to the impact of electric cars and trucks on the auto industry. It is a revolution, but possibly a 10-year transitionary one. Nautique is known as a tech leader in boating and strives to keep that edge. This means major changes in supply chains, away from expensive internal combustion engines to electric motors and battery suppliers. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Application of knowledge 2

HARD ROCK CAFE’S GLOBAL STRATEGY

There is a short video (9 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. Identify the strategic changes that have taken place at Hard Rock Cafe. What we want to do here is help the student understand that an optimum mix of internal strengths and opportunities drives strategies in a changing environment.  Initially, Hard Rock was a London cafe serving classic American food.  Then it became a “theme” chain with memorabilia in tourist destinations.  Then it added stores.  Then it added live music and a rock concert.  Then it became an established name and began opening hotels and casinos.  Then it upgraded its menu.  Then it moved into cities that are not the typical tourist destination. LO 2.1: Define mission and strategy AACSB: Reflective thinking 2.

As these strategic changes have taken place—the 10 decisions of OM change:  Location: From a London cafe, to tourist destinations, to non-tourist locations.  Product design: New menu items  Quality: The entire evaluation of quality and quality control got much more complex.  Process: The kitchen process changed when Hard Rock went from hamburgers to lobster, and additional changes were made as the

firm moved to retail merchandising.  Layout: Added retail stores, added live music facilities.  Supply chain management: Purchase memorabilia and lobsters—new expectations of the supply chain.  Inventory: From food to clothing to memorabilia, to expanded food items in inventory—how do you keep lobsters alive and how

long?  Human resources: The range of talents needed keeps expanding; from cooks of classic American fare and waitstaff and bartenders, to

merchandisers, to cooks for a wider more expensive menu, to coordinators and performers for the live music facilities. The case says little about scheduling and maintenance, but every change in product (food or merchandise) and every change in equipment and processes changes scheduling and maintenance. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Application of knowledge Copyright ©2023 Pearson Education, Inc.

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Hard Rock fits in the multidomestic strategy, which uses the existing domestic model globally.

LO 2.5: Identify and explain four global operations strategy options AACSB: Reflective thinking 3

OUTSOURCING OFFSHORE AT DARDEN

There is a short video (10 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. Since this case was written and filmed, Darden spun off in 2015 Red Lobster into a separate company. 1. There are numerous outsourcing opportunities available to a restaurant, including food supplies, all other supplies, janitorial, data processing, benefits, marketing, and bookkeeping. Darden outsources the seafood and produce part of its supply chain, but maintains tight quality standards “from farm to fork.” LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 2. When a giant like Darden procures supplies in 35 countries, it needs to have a large staff “on the ground” to arrange for training, quality control, contracts, expediting, language/cultural issues, and so on. With very tight standards, it will not use a supplier until all its expectations for reliability/quality are met. Once trained, a supplier need not be managed as closely, freeing Darden supply-chain personnel to seek out the next provider. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Analytical thinking 3. In other industries, perhaps where 48-hour freshness is not a critical issue, supply chains may differ. Challenges come from culture, communications, distance, and documents. Companies like Walmart have used alliances. P&G reorganized along product lines instead of geography to increase coordination. Mercedes decided to build some models in the U.S. to get closer to customers. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Application of knowledge 4. Darden outsources seafood harvesting and preparation offshore because (a) it may not legally own/control the catch in foreign waters; (b) labor intensity of food preparation means it is cheaper for that work to be done offshore; (c) bulk food purchases are capital intensive and not part of Darden’s core competence. Darden has recently spun off its Red Lobster restaurants in part because of the special challenges and costs connected with this part of its business. LO 2.3: Understand the significance of key success factors and core competencies AACSB: Reflective thinking

C H A P T E R

Project Management

DISCUSSION QUESTIONS 1. There are many possible answers. Project management is needed in large construction jobs, in implementing new information systems, in new product development/marketing, in creating a new assembly line, and so on. LO 3.1: Use a Gantt chart for scheduling AACSB: Application of knowledge

2. Project organizations make sure existing programs continue to run smoothly while new projects are successfully completed. LO 3.1: Use a Gantt chart for scheduling AACSB: Written and oral communication

The three phases involved in managing a large project are planning, scheduling, and controlling.

LO 3.1: Use a Gantt chart for scheduling AACSB: Application of knowledge

4. PERT and CPM help answer questions relating to which task elements are on (or likely to be on) the critical path and to probable completion times for the overall project. Some specific questions include:  When will the entire project be completed?  Which are the critical activities or tasks in the project; that is, the activities that will delay the entire project if completed behind schedule?  Which are the noncritical activities; that is, those that can run behind schedule without delaying the whole project? How far behind schedule can these activities run without disrupting the completion time?  What is the probability that the project will be completed by a specific date?  At any particular date, is the project on schedule, behind schedule, or ahead of schedule?  On any given date, is the money spent equal to, less than, or greater than the budgeted amount?  Are there enough resources available to finish the project on time?  If the project is required to be finished in a shorter amount of time, what is the least-cost way to accomplish this? LO 3.2: Draw AOA and AON networks AACSB: Analytical thinking

5. WBS is a hierarchical subdivision of effort required to achieve an objective. It defines a project by breaking it down into manageable parts and even finer subdivisions. LO 3.1: Use a Gantt chart for scheduling AACSB: Analytical thinking

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Project Management

6. A Gantt chart is a visual device that shows the duration of tasks in a project. It is a low-cost means of ensuring that (1) all activities are planned, (2) their order of performance is documented, (3) the activity times are recorded, and (4) the overall project time is developed. LO 3.1: Use a Gantt chart for scheduling AACSB: Analytical thinking

7. The difference between AOA and AON is that activities are shown on arrows in the former and on the node in the latter. We primarily use AON in this chapter. LO 3.2: Draw AOA and AON networks AACSB: Application of knowledge

Any late start or extension of an activity on the critical path will delay the completion of the project.

LO 3.4: Determine a critical path AACSB: Reflective thinking

To crash an activity, the project manager would pay money to add resources (overtime, extra help, etc.).

LO 3.6: Crash a project AACSB: Application of knowledge

10. Activity times used in PERT are assumed to be described by a beta probability distribution. Given optimistic (a), pessimistic (b), and most likely (m) completion times, average or expected time is given by: t=

a + 4m + b 6

and the variance by:

 (b − a )  Variance =   6 

LO 3.5: Calculate the variance of activity times AACSB: Analytical thinking

11. Earliest start (ES) of an activity is the latest of the earliest finish times of all its predecessors. Earliest finish (EF) is the earliest start of an activity plus its duration. Latest finish (LF) of an activity is the earliest of the latest start times of all successor activities. Latest start (LS) of an activity is its latest finish less its duration. LO 3.3: Complete forward and backward passes for a project AACSB: Application of knowledge

12. The critical path is the shortest time possible for the completion of a series of activities, but that shortest time is the longest path through the network. Only the longest path allows time for all activities in the series; any smaller amount will leave activities unfinished. LO 3.4: Determine a critical path AACSB: Written and oral communication

13. Dummy activities have no time duration. They are inserted into an AOA network to maintain the logic of the network, such as when two activities have exactly the same beginning and ending events. A dummy activity is inserted with one of them so that the computer software can handle the problem. LO 3.2: Draw AOA and AON networks AACSB: Reflective thinking Copyright ©2023 Pearson Education, Inc.

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PROJECT MANAGEMENT

14. They are (1) optimistic time (a), an estimate of the minimum time an activity will require; (2) most likely time (m), an estimate of the normal time an activity will require; and (3) pessimistic time (b), an estimate of the maximum time an activity will require. LO 3.5: Calculate the variance of activity times AACSB: Analytical thinking 15. No. In networks, there is no possibility that crashing a non-critical task can reduce the project duration. Only critical tasks offer the possibility of reducing path length. However, other criteria for crashing may exist; for instance, skills required in one of the activities may also be needed elsewhere. LO 3.6: Crash a project AACSB: Written and oral communication 16.

Total PERT project variance is computed as the sum of the variances of all activities on the critical path.

LO 3.5: Calculate the variance of activity times AACSB: Application of knowledge 17. Slack: the amount of time an activity can be delayed and not affect the overall completion time of the whole project. Slack can be determined by finding the difference between the earliest start time and the latest start time, or the earliest finish time and the latest finish time for a given activity. LO 3.4: Determine a critical path AACSB: Application of knowledge 18. If there are a sufficient number of tasks along the critical path, we can assume that project completion time is described by a normal probability distribution with mean equal to the sum of the expected times of all activities on the critical path and variance equal to the sum of the variances of all activities on the critical path. The fundamental assumption required is that the number of activities on the critical path is large enough that the mean of the sum of the beta distributions is distributed approximately as the normal distribution.

LO 3.4: Determine a critical path AACSB: Analytical thinking

19. Widely used project management software includes MS Project, Oracle Primavera, Mind View, HP Project, and Fast Track. LO 3.1: Use a Gantt chart for scheduling AACSB: Application of knowledge 20. Waterfall approach: the project progresses smoothly, in a step-by-step manner, through each phase to completion. This usually applies to well-established projects. Agile project management: ill-defined projects that require collaboration and constant feedback to adjust to the many unknowns of the evolving technology and project specifications. AACSB: Reflective thinking

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Project Management

ETHICAL DILEMMA Large projects with time/cost overruns are not uncommon situations in the world of project management. Why do MIS projects commonly sport 200%–300% cost overruns and completion times twice those projected? Why do massive construction projects run so late and so overbudgeted? Students are expected to read about such projects and come up with explanations, especially related to ethics. In the case of MIS projects, long software development tasks are almost doomed to failure because of the changes in technology and staff that take place. It’s a necessity to break large projects down into smaller 3- to 6-month modules or pieces that are self-contained. This protects the organization from a total loss should the massive project never be completed. In every case, quality project management means open communication, realistic timetables, good staff, and use of software like MS Project to build and maintain a schedule. Bidding on a contract with a schedule that is not feasible may be unethical as well as poor business.

AACSB: Application of knowledge

ACTIVE MODEL EXERCISE** ACTIVE MODEL 3.1: Gantt Chart 1.

Both A and H are critical activities. Describe the difference between what happens on the graph when you increase A vs. increasing H.

When you increase H, it is the only task to change on the chart. However, when you increase A, then all critical tasks move to the right, and the slack for the noncritical tasks increases. 2.

Activity F is not critical. By how many weeks can you increase activity F until it becomes critical? 6 weeks

3. Activity B is not critical. By how many weeks can you increase activity B until it becomes critical? What happens when B becomes critical? 1 week. Activity D also becomes critical. 4.

What happens when you increase B by 1 more week after it becomes critical?

Activities A, C, and E become noncritical, and the project takes 1 additional week. 5. Suppose that building codes may change and, as a result, activity B would have to be completed before activity C could be started. How would this affect the project? Activity B becomes critical, and the project takes 1 additional week.

Active Model 3.1 appears in MyLab Operations Management.

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PROJECT MANAGEMENT

END-OF-CHAPTER PROBLEMS 3.1

Some possible Level 3[(a)] and Level 4[(b)] activities for the house appear for each Level 2 activity below.

3.2 Here are some detailed activities to add to Lawson’s WBS:** 1.1.1

Set initial goals for fund-raising

1.1.2

Set strategy including identifying sources and solicitation

1.1.3

Raise the funds

1.2.1

Identify voters’ concerns

1.2.2

Analyze competitor’s voting record

1.2.3

CHAPTER 3

1.3.1

Hire campaign manager and political advisor

1.3.2

Get volunteers

1.3.3

Hire a staff

1.3.4

Hire media consultants

1.4.1

Identify filing deadlines

1.4.2

File for candidacy

1.5.1

Train staff for audit planning

Project Management

Students could make many other choices. **

Source: Modified from an example found in M. Hanna and W. Newman, Operations Management: Prentice Hall, Upper Saddle River, NJ (2001): p. 722.

3.3

Hours

3.4

(a) AON network:

(b) AOA network:

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PROJECT MANAGEMENT

3.5

Critical path is A–C–F–H. Time = 21 days. This is an AON network.

3.6

Critical path is A–C–F–G–I. Time = 21 days. This is an AOA network.

3.7 The paths through this network are J–L–O, J–M–P, J–M–N–O, K–P, and K–N–O. Their path durations are 23, 18, 22, 13, and 17. J–L–O is the critical path; its completion time is 23 weeks.

3.8 (a)

(b) Critical path is B–D–E–G (c) Total project takes 26 days (d) Activity

Time

Slack

Critical

A B C D E F G

2 5 1 10 3 6 8

0 0 0 5 15 1 18

2 5 1 15 18 7 26

13 0 11 5 15 12 18

15 5 12 15 18 18 26

13 0 11 0 0 11 0

No Yes No Yes Yes No Yes

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3.9 (a)

(b, c) There are four paths:

3.10

Path

Time (hours)

A–C–E–G B–D–F–G A–C–D–F–G B–E–G

19.5 24.9 28.7 (critical) 15.7

(a)

(b, c) Task A B C D E F G H I

Time

Slack

9 7 3 6 9 4 6 5 3

0 9 9 16 16 12 25 22 31

9 16 12 22 25 16 31 27 34

0 9 18 20 16 21 25 26 31

9 16 21 26 25 25 31 31 34

0 0 9 4 0 9 0 4 0

Activities on the critical path: A, B, E, G, I Project completion time = 34

3.11

(a) AOA diagram of the project:

(b)

The critical path, listing all critical activities in chronological order: A→B→E→ F A→C→F

1 + 1 + 2 + 2 = 6 ( not CP ) 1 + 4 + 2 = 7. This is the CP.

Project Management

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(c)

PROJECT MANAGEMENT

The project completion time (in weeks): 7 (This is the length of CP.)

(d)

The slack (in weeks) associated with any and all non-critical paths through the project: Look at the paths that aren’t critical— only 1 here—so from above:

A→B→E→F

7 – 6 = 1 week slack.

3.12

Note: Activity times are shown as an aid for Problem 3.15. They are not required in the solution to Problem 3.12.

3.13

(a)

(b) Critical path is B–E–F–H. (c) Time = 16 weeks

3.14

(a)

(b) Activity

Time

Slack

Critical

S T U V W X Y Z

4 6 3 4 8 7 2 1

0 0 4 4 7 6 15 13

4 6 7 8 15 13 17 14

0 1 4 11 7 9 15 16

4 7 7 15 15 16 17 17

0 1 0 7 0 3 0 3

Y N Y N Y N Y N

Activities on the critical path: S, U, W, Y Project completion time = 4 + 3 + 8 + 2 = 17 weeks

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Project Management

3.15 Activity

Time

Slack

Critical

A B C D E F G H

6 7 3 2 4 6 10 7

0 0 6 6 7 7 11 13

6 7 9 8 11 13 21 20

2 0 8 12 7 8 11 14

8 7 11 14 11 14 21 21

2 0 2 6 0 1 0 1

No Yes No No Yes No Yes No

The critical path is given for activities B, E, G. Total project completion time is 21 weeks.

3.16

(a)

(b) Task A B C D E F G H I J K L M N O P Q R S T U **

Time

Slack

0.0 0.0 8.0 8.0 8.0 9.0 8.0 8.0 10.0 8.0 12.0 14.0 15.0 11.0 13.0 13.0 8.1 15.5 16.5 17.0 18.0

0.0 8.0 8.1 9.0 9.0 10.0 10.0 11.0 11.0 12.0 14.0 15.0 15.5 13.0 14.0 14.5 13.1 16.5 17.0 18.0 18.0

0.0 0.0 10.4 12.0 10.0 13.0 9.0 11.0 11.0 8.0 12.0 14.0 15.0 12.0 14.5 14.0 10.5 15.5 16.5 17.0 18.0

0.0 8.0 10.5 13.0 11.0 14.0 11.0 14.0 12.0 12.0 14.0 15.0 15.5 14.0 15.5 15.5 15.5 16.5 17.0 18.0 18.0

0.0 0.0 2.4 4.0 2.0 4.0 1.0 3.0 1.0 0.0 0.0 0.0 0.0 1.0 1.5 1.0 2.4 0.0 0.0 0.0 0.0

0.0 8.0 0.1 1.0 1.0 1.0 2.0 3.0 1.0 4.0 2.0 1.0 0.5 2.0 1.0 1.5 5.0 1.0 0.5 1.0 ** 0.0

Note: Start (A) and Finish (U) are assigned times of zero.

Critical path is A–B–J–K–L–M–R–S–T–U, for 18 days. (c)

(i) No, transmissions and drivetrains are not on the critical path. (ii) No, halving engine building time will reduce the critical path by only one day. (iii) No, it is not on the critical path.

(d) Reallocating workers not involved with critical path activities to activities along the critical path will reduce the critical path length.

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PROJECT MANAGEMENT

3.17

(a) Estimated (expected) time for C = [8 + (4 × 12) + 16]/6 = 72/6 = 12 weeks 2

16  (16 − 8)  (b) Variance for C is   = 9 = 1.78  6  (c) Critical path is A–C–F–H–J–K. (d) Time on critical path = 7.67 + 12 + 9.67 + 2 + 6.67 + 2.17 = 40.18 weeks (rounded). (e) Variance on critical path = 1 + 1.78 + 5.44 + 0 + 1.78 + 0.03 = 10.03.

36 − 40.18 = –1.32, which is about 9.3% chance 3.17 (.093 probability) of completing project before week 36.

(f) Z =

Note that based on possible rounding in part (d)—where time on critical path could be 40.3—the probability can be as low as 8.7%. So a student answer between 8.7% and 9.3% is valid. Summary table for Problem 3.17 follows: Activity

Activity Time Early Start

Early Finish

Late Start

Late Finish

Slack

Variance

A B C D E F G H I J K

7.67 9.67 12 6.33 2 9.67 3 2 6 6.67 2.17

7.67 17.34 19.67 14 19.34 29.34 22.67 31.34 35.34 38.01 40.18

0.0 8 7.67 25.01 17.67 19.67 28.34 29.34 32.01 31.34 38.01

7.67 17.67 19.67 31.34 19.67 29.34 31.34 31.34 38.01 38.01 40.18

0 0.33 0 17.34 0.33 0 8.67 0 2.67 0 0

1 13.44 1.78 1 0.11 5.44 0.11 0 0 1.78 0.03

0 7.67 7.67 7.67 17.34 19.67 19.67 29.34 29.34 31.34 38.01

3.18 Activity

A B C D E F

11 27 18 8 17 16

15 31 18 13 18 19

19 41 18 19 20 22

t =

a + 4m + b 6 15 32 18 13.17 18.17 19

σ =

b −a 6

1.33 2.33 0 1.83 0.5 1

Variance 1.77 5.44 0 3.36 0.25 1

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3.19

Project Management

(a)

Activity

Expected

Variance

A B C D E F G H I J K

3 2 1 6 2 6 1 3 10 14 2

6 4 2 7 4 10 2 6 11 16 8

8 4 3 8 6 14 4 9 12 20 10

5.83 3.67 2.00 7.00 4.00 10.00 2.17 6.00 11.00 16.33 7.33

0.69 0.11 0.11 0.11 0.44 1.78 0.25 1.00 0.11 1.00 1.78

LS 7.17 5.33 0.00 2.00 9.00 13.00 15.83 23.00 18.00 20.00 29.00

LF Slack Critical 13.00 7.17 No 9.00 5.33 No 2.00 0.00 Yes 9.00 0.00 Yes 13.00 0.00 Yes 23.00 0.00 Yes 18.00 2.83 No 29.00 0.00 Yes 29.00 2.83 No 36.33 18.00 No 36.33 0.00 Yes

(b, c) Activity A B C D E F G H I J K

Time 5.83 3.67 2.00 7.00 4.00 10.00 2.17 6.00 11.00 16.33 7.33

ES 0.00 0.00 0.00 2.00 9.00 13.00 13.00 23.00 15.17 2.00 29.00

EF 5.83 3.67 2.00 9.00 13.00 23.00 15.17 29.00 26.17 18.33 36.33

The critical path is given by activities C, D, E, F, H, K. Average project completion time is 36.33 days.

Expected completion time for the project is 36.33 days. Project variance = Sum of variances of activities on critical path = 0.11 + 0.11 + 0.44 + 1.78 + 1.00 + 1.78 = 5.22. Standard deviation = 2.28.

(d)

40 − 36.33   P (t ≤ 40) = P  z ≤ = P [ z ≤ 1.61] = 0.946 2.28  

3.20 (a)

Activity

Variance

A B C D

9 4 9 5

10 10 10 8

11 16 11 11

10 10 10 8

0.11 4 0.11 1

(b) Critical path is A– C with mean (te) completion time of 20 weeks. The other path is B–D, with mean completion time of 18 weeks. (c) Variance of A– C = (Variance of A) + (Variance of C) = 0.11 + 0.11 = 0.22 Variance of B–D = (Variance of B) + (Variance of D) =4+1=5 (d) Probability A–C is finished in 22 weeks or less =

22 − 20   PZ ≤  = P( Z ≤ 4.26) ≅ 1.00 0.22  

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PROJECT MANAGEMENT

(e) Probability B–D is finished in 22 weeks or less = 22 − 18   P Z ≤  = P ( Z ≤ 1.79) = 0.963  5  (f) The critical path has a relatively small variance and will almost certainly be finished in 22 weeks or less. Path B–D has a relatively high variance. Due to this, the probability B–D is finished in 22 weeks or less is only about 0.96. Since the project is not finished until all activities (and paths) are finished, the probability that the project will be finished in 22 weeks or less is not 1.00 but is approximately 0.96.

3.21

(a)

Activity

Expected Time

Variance

A B C D E F G H I J K L M N

4 1 6 5 1 2 1 4 1 2 8 2 1 6

6 2 6 8 9 3 7 4 6 5 9 4 2 8

7 3 6 11 18 6 8 6 8 7 11 6 3 10

5.83 2.00 6.00 8.00 9.17 3.33 6.17 4.33 5.50 4.83 9.17 4.00 2.00 8.00

0.25 0.11 ** 0.00 1.00 ** 8.03 0.44 1.36 ** 0.11 ** 1.36 0.69 ** 0.25 0.44 ** 0.11 ** 0.44

Activity A B C D E F G H I J K L M N

Time 5.83 2.00 6.00 8.00 9.17 3.33 6.17 4.33 5.50 4.83 9.17 4.00 2.00 8.00

ES 0.00 0.00 5.83 5.83 11.83 13.83 13.83 21.00 25.33 30.83 30.83 35.66 40.00 42.00

EF 5.83 2.00 11.83 13.83 21.00 17.16 20.00 25.33 30.83 35.66 40.00 39.66 42.00 50.00

LS 0.00 9.83 5.83 9.67 11.83 17.67 19.16 21.00 25.33 33.17 30.83 38.00 40.00 42.00

LF 5.83 11.83 11.83 17.67 21.00 21.00 25.33 25.33 30.83 38.00 40.00 42.00 42.00 50.00

Slack 0.00 9.83 0.00 3.84 0.00 3.84 5.33 0.00 0.00 2.34 0.00 2.34 0.00 0.00

Critical Yes No Yes No Yes No No Yes Yes No Yes No Yes Yes

The critical path is given by activities A, C, E, H, I, K, M, N. Expected completion time is 50 days. P(Completion in 53 days). Σ Variances on critical path = 10.55 so, σcp = 3.25.

53 − 50   P (t ≤ 53) = P  z ≤ = P [ z ≤ 0.92 ] = 0.821 = 82.1% 3.25   (b) x − 50   P z ≤  3.25   where z = 2.33 for 99% probability. x − 50 . Then 3.25 x = 50 + (2.33)(3.25) = 57.57 ≅ 58 days

So 2.33 =

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Project Management

3.22 (a) This project management problem can be solved using PERT. The results are below. As you can see, the total project completion time is about 32 weeks. The critical path consists of activities C, H, M, and O. Project completion time = 32.05, so the marketing campaign should start 32 weeks before the film release. 2

9  13 − 10   3  Variance (C) =   =  =  6   6  36 2

 9 − 5  16 Variance (H) =   =  6  36 2

2.25  6.5 − 5  Variance (M) =   = 6 36   2

9 8−5 Variance (O) =   = 6 36   9 16 2.25 9 Project variance = + + + » 1.00 36 36 36 36 Project standard deviation = 1.00 Activity A B C D E F G H I J K L M N O P

Activity Time Early Start Early Finish 2.17 3.5 11.83 5.17 3.83 7 3.92 7.47 10.32 3.83 4 4 5.92 1.23 6.83 7

0 0 0 0 0 2.17 3.5 11.83 11.83 11.83 5.17 3.83 19.3 15.67 25.22 16.9

2.17 3.5 11.83 5.17 3.83 9.17 7.42 19.3 22.15 15.66 9.17 7.83 25.22 16.9 32.05 23.9

Late Start

Late Finish

10.13 11.88 0 14.65 15.98 12.3 15.38 11.83 14.9 19.98 19.82 19.82 19.3 23.82 25.22 25.05

12.3 15.38 11.83 19.82 19.82 19.3 19.3 19.3 25.22 23.82 23.82 23.82 25.22 22.05 32.05 32.05

Slack 10.13 11.88 0 14.65 15.98 10.13 11.88 0 3.07 8.15 14.65 15.98 0 8.15 0 8.15

(b) P (completing campaign in <= 32 weeks) is 50%. There is a 50% chance of taking longer and a 50% chance of finishing earlier. (c) As can be seen in the following analysis, the changes do not have any impact on the critical path or the total project completion time. A summary of the analysis is below. Project completion time = 32.05 Project standard deviation = 1.00

CHAPTER 3

Activity

Activity Time

A B C D E F G H I J K L M N O P

2.17 3.5 11.83 5.17 3.83 7 3.92 7.47 0 0 4 4 5.92 1.23 6.83 7

3.23

PROJECT MANAGEMENT

Early Start

Early Finish

Late Start

Late Finish

Slack

0 0 0 0 0 2.17 3.5 11.83 11.83 11.83 5.17 3.83 19.3 11.83 25.22 13.07

2.17 3.5 11.83 5.17 3.83 9.17 7.42 19.3 11.83 11.83 9.17 7.83 25.22 13.07 32.05 20.07

10.13 11.88 0 14.65 15.98 12.3 15.38 11.83 25.22 23.82 19.82 19.82 19.3 23.82 25.22 25.05

12.3 15.38 11.83 19.82 19.82 19.3 19.3 19.3 25.22 23.82 23.82 23.82 25.22 22.05 32.05 32.05

10.13 11.88 0 14.65 15.98 10.13 11.88 0 13.38 11.98 14.65 15.98 0 11.98 0 11.98

(a) Probability of completion is 17 months or less: 17 − 21   = P [ z ≤ − 2.0 ] P (t ≤ 17) = P  z ≤ 2   = 1 − P [ z ≥ 2.0 ] = 1 − 0.97725 = 0.0228 (b) Probability of completion in 20 months or less:

20 − 21   = P [ z ≤ − 0.5] P (t ≤ 20 ) = P  z ≤ 2   = 1 − P [ z ≥ 0.5] = 1 − 0.69146 = 0.3085 (c) Probability of completion in 23 months or less:

23 − 21   P (t ≤ 23) = P  z ≤ = P [ z ≤ 1.0 ] = 0.84134 2   (d) Probability of completion in 25 months or less: 25 − 21   P (t ≤ 25) = P  z ≤ = P [ z ≤ 2.0 ] = 0.97725 2   (e)

x − 21   = 1.645 for a 95% chance of completion P z ≤ 2   by the x date. Then x = 21 + 2(1.645) = 24.29, or 24 months.

3.24

(a) Expected times for individual activities (using [(a + 4m + b)/6]). A = 5, B = 6, C = 7, D = 6, E = 3. Expected project completion time = 15 (Activities A–C–E). (b) Variance for individual activities (using [(b – a)/6]2). A = 1; B = 1; C = 1; D = 4; E = 0. Project variance = Σ variances on critical path = 1 + 1 + 0 = 2.

CHAPTER 3

Project Management

3.25 (a)

(b) Activity

Time

Slack

σ2

A B C D E F G H

7 3 9 4 5 8 8 6

0 7 7 16 16 21 29 37

7 10 16 20 21 29 37 43

0 13 7 25 16 21 29 37

7 16 16 29 21 29 37 43

0 6 0 9 0 0 0 0

2 1 3 1 1 2 1 2

4 1* ** 9 1* ** 1 ** 4 ** 1 ** 4

Activities on the critical path: A, C, E, F, G, H. Project Completion time = 43.

σ = 4.8

(c)

49 − 43 = 1.25 4.8 P(t ≤ 49) = .89435 z=

P(t ≥ 49) = (1 − .89435) = 0.10565 3.26 (a) Activity A B C D

Expected Time t

Variance

8 1 9 2

10 6 12 5

12 23 15 8

[8 + 4(10) + 12] / 6 = 10 [1 + 4(6) + 23] / 6 = 8 [9 + 4(12) + 15] / 6 = 12 [2 + 4(5) + 8] / 6 = 5

[(12 – 8)/6] = 0.44 2 [(23 – 1)/6] = 13.44 2 [(15 – 9)/6] = 1.00 2 [(8 – 2)/6] = 1.00

Critical path is A– C – D with mean (t) completion time of 10 + 12 + 5 = 27 days. The other path is B–C– D, with mean completion time of 25 days. (b) Variance of A– C – D = (Variance of A) + (Variance of C) + (Variance of D) = 0.44 + 1.00 + 1.00 = 2.44 Standard deviation of A– C – D =

2.44 = 1.56

Probability A– C – D is finished in 29 days or less =

29 − 27   P Z ≤  = P ( Z ≤ 1.28) = 89.97% 1.56   (c) Variance of B– C – D = (Variance of B) + (Variance of C) + (Variance of D) = 13.44 + 1.00 + 1.00 = 15.44

Standard deviation of B– C – D =

15.44 = 3.93

Probability B– C – D is finished in 29 days or less =

29 − 25   P Z ≤  = P( Z ≤ 1.02) = 84.61% 3.93   (d) No. The non-critical path has an even lower probability of finishing within 29 days than the critical path. 3.27 (a) Activity Q R S T

Expected Time t

Variance

7 1 10 1

10 6 12 5

12 23 16 7

[7 + 4(10) + 12] / 6 = 9.83 [1 + 4(6) + 23] / 6 = 8 [10 + 4(12) + 16] / 6 = 12.33 [1 + 4(5) + 7] / 6 = 4.67

[(12 – 7)/6] = .69 2 [(23 – 1)/6] = 13.44 2 [(16 – 10)/6] = 1.00 2 [(7 – 1)/6] = 1.00

CHAPTER 3

PROJECT MANAGEMENT

The expected completion time (t) is 9.83 + 8.00 + 12.33 + 4.67 = 34.83 days. (b) Variance of Q–R–S–T = (Variance of Q) + (Variance of R) + (Variance of S) + (Variance of T) = 0.69 + 13.44 + 1 + 1 = 16.13 Standard deviation of Q–R – S – T =

16.33 = 4.02

Probability Q– R – S – T is finished in 38 days or less =

38 − 34.83   P Z ≤  = P ( Z ≤ 0.79) = 78.52% 4.02  

3.28

Helps to modify the AON with the lowest costs to crash. This problem is based on the data in Problem 3.11. 1. The critical path is A → C → F; C is cheapest to crash, so take it to 3 weeks at $200 (and $200 < $250) 2. Now both paths through are critical. We would need to shorten A or F, or shorten C and either B/E. This is not worth it, so we would not bother to crash any further.

3.29

Critical path C–E at 12 days. Activity A B C D E

Daily Crash Costs $100 50 100 150 200

Maximum Crash 1 day 2 days 1 day 2 days 3 days

To crash by 4 days, from 12 days to 8 days: • Crash C by 1 day ($100) to 11 days total. • Now crash E by 1 day ($200) and A by 1 day ($100) to 10 days total. • Now crash E by 2 days ($400) and D by 2 days ($300) to 8 days total. • Total additional cost to crash 4 days = $1,100.

3.30 Crash costs per unit time are $600 for A, $900 for B, and $1,000 for C. (a) A offers the cheapest path to a single-day reduction. (b) A cannot supply a second reduction, so the next best choice is B, which adds $900. (c) The total for both days is $1,500.

3.31

(a) Project completion time = 16 (Activities A–D–G) Figure for Problem 3.31

(b)

Total cost = $12,300.

(c)

Crash D 1 week at an additional cost of $75.

CHAPTER 3

Activity

Norm. Time–Crash Time

Project Management

Crash $–Normal $ $/time

A B C D E F G

3.32

1 1 0 4 3 1 2

$600 600 0 300 300 1,200 600

(d) Activity

Crash

Cost

D G A E

4 2 1 1

$300 600 600 100

7 weeks

$1,600

$600 600 — 75 100 1,200 300

(a)

Project completion time = 14 weeks Task A B C D E F G

Time 3 2 1 7 6 2 4

EF 0 0 0 3 2 1 10

LS 3 2 1 10 8 3 14

LF 0 2 11 3 4 12 10

Slack 3 4 12 10 10 14 14

0 2 11 0 2 11 0

(b) To crash to 10 weeks, we follow 2 steps: 1. Crash D by 2 weeks ($150). 2. Crash D and E by 2 weeks each ($150 + $100). Total crash cost = $400 (c) Using POM for Windows software, minimum project completion time = 7. Crashing cost = $1,550. Normal Time (days)

Crash Time (days)

Normal Cost ($)

A B C D E F G

3 2 1 7 6 2 4

2 1 1 3 3 1 2

1,000 2,000 300 1,300 850 4,000 1,500

3.33

AON network

Crash Cost Crash Cost/Pd Crash By ($) (days) 1,600 2,700 300 1,600 1,000 5,000 2,000

$600 700 0 75 50 1,000 250

1 0 0 4 3 0 2

6 Design 5 Software Start

8 Wiring 4 Chip Install

End 8 Testing

Crashing Cost ($) 600 0 0 300 150 0 500

CHAPTER 3

PROJECT MANAGEMENT

Activity

Daily Crash Costs

Maximum Crash

Design Wiring Chip Install Software Testing

$100 50 100 150 200

1 day 2 days 1 day 2 days 3 days

To crash by 4 days, from 13 days to 9 days, Crash Wiring by 1 day ($50) to reach 12 days. Crash Wiring by a second day ($50) and Chip Install by 1 day ($100) to reach 11 days.  Crash Software by 2 days ($300) and Testing by 2 days ($400) to reach 9 days total.  Total cost to crash 4 days = $900.  

VIDEO CASE STUDIES 1

PROJECT MANAGEMENT AT ARNOLD PALMER HOSPITAL

There is a short video (8 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. Also note that the Global Company Profile in Chapter 6 highlights this hospital.

1. Construction project network:

LO 3.2: Draw AOA and AON networks AACSB: Analytical thinking 2. The critical path is Activities 1–3–5–6–8–10–11–12–14–16–17–19–20–21. The project length is 47 months—about four years from start to finish. LO 3.4: Determine a critical path AACSB: Analytical thinking 3. Building a hospital is much more complex than an office building for several reasons. In this case, hundreds of “users” of the new building had extensive input. Second, the design of the new layout (circular, pod design) is somewhat radical compared to traditional “linear” hospitals. Third, the hospital was built with future expansion in mind. Fourth, the guiding principles impacted on design/construction. Fifth, hospitals, by their very nature, are more complex from a safety, health hazard, security, quiet, serenity perspective than an office building. LO 3.4: Determine a critical path AACSB: Application of knowledge 4. Since there were 13 months of planning prior to the proposal/ review stage (listed as Activity 1) and the project then took 47 months (for a total of 60 months), 22% of the time was spent in planning. Construction itself started with activity 14 (19.75 months); there were 13 + 19.75 months of work (= 32.75 months) out of 60 months, or 55% of the time was spent prior to building beginning. These figures reflect the importance of preplanning placed on the project. Rather than having to redo walls or rooms, mockups and detailed planning meetings saved time on the back end and resulted in a building that met the needs of patients and staff alike. It always pays to spend extra time on the front end of a large project.

CHAPTER 3

Project Management

MANAGING HARD ROCK’S ROCKFEST

There is a short video (9 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. Network diagram

Hard Rock’s ES, EF, LS, LF, and slack: Activity time

Early Start

Early Late Start Late Finish Finish

A B C D E F G H I J K L M N O P Q R

7 3 3 5 6 4 2 4 4 10 2 3 8 6 7 20 4 4

0 7 7 10 15 21 25 10 14 18 10 15 18 10 27 10 10 14

7 10 10 15 21 25 27 14 18 28 12 18 26 16 34 30 14 18

0 7 18 10 15 21 25 16 20 24 21 23 26 28 27 14 19 23

7 10 21 15 21 25 27 20 24 34 23 26 34 34 34 34 23 27

S T U V W X Y Z

3 4 6 7 4 8 6 6

25 7 11 11 18 17 10 16

28 11 17 18 22 25 16 22

31 13 17 23 30 23 22 28

34 17 23 30 34 31 28 34

Slack 0 0 11 0 0 0 0 6 6 6 11 8 8 18 0 4 9 9 6 6 6 12 12 6 12 12

CHAPTER 3

PROJECT MANAGEMENT

The critical path is A–B–D–E–F–G–O, with a timeline of 34 weeks.

LO 3.4: Determine a critical path AACSB: Analytical thinking 2.

Activities C, K, L, M, N, Q, R, V, W, Y, and Z have slacks of 8 or more weeks.

LO 3.4: Determine a critical path AACSB: Analytical thinking 3.

Major challenges a project manager faces in a project like Rockfest: (1) keeping in touch with each person responsible for major activities, (2) last-minute surprises, (3) cost overruns, (4) too many fans trying to enter the venue, (5) resignations of one of the key managers or promoters, and many others.

LO 3.4: Determine a critical path AACSB: Analytical thinking 4. Work breakdown structure, with example of Level 1, 2, 3, 4 tasks: 1.0 Rockfest event, with site selected (A)

[Level 1]

1.1 Select local promoter (B)

[Level 2]

1.1.1 Web site (D)

[Level 3]

1.1.2 TV deal (E) 1.1.3 Hire director (F) 1.1.3.1 Camera placement (G)

[Level 4]

1.1.4 Headline entertainers (H) 1.1.4.1 Support entertainers (I) 1.1.4.2 Travel for talent (J) 1.1.4.3 Passes/stage credentials (O) 1.1.5 Staff travel (P) 1.1.6 Merchandise deals (Y) 1.1.6.1 Online sales of merchandise (Z) 1.1.7 Hire sponsor coordinator (Q) 1.1.7.1 Finalize sponsors (R) 1.1.7.2 Signage for sponsors (S) 1.2 Hire production manager (C) 1.2.1 Sound/staging (N) 1.2.2 Venue capacity (K) 1.2.2.1 TicketMaster contract (L) 1.2.2.2 Onsite ticketing (M) 1.3 Hire operations manager (T) 1.3.1 Site plan (U) 1.3.1.1 Power, etc. (X) 1.3.2 Security director (V) 1.3.2.1 Set police/fire plan (W) Answers may vary somewhat at Level 3 and Level 4. Level 2 activities should be activities B, C, and T. Copyright ©2023 Pearson Education, Inc.

CHAPTER 3

Project Management

LO 3.4: Determine a critical path AACSB: Analytical thinking

ADDITIONAL CASE STUDIES (available in MyLab Operations Management) 1 1.

SHALE OIL COMPANY See table below.

LO 3.4: Determine a critical path AACSB: Analytical thinking 2.

From the precedence data supplied in the problem, we can develop the following AON network:

The following table indicates the expected times, variances, and slacks needed to complete the rest of the case: Activity

Opt

Most Likely

Pess

E(t)

Slack

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

1.0 1.5 2.0 1.0 1.0 2.0 2.0 1.0 1.0 1.0 2.0 15.0 1.0 3.0 3.0 14.0 1.0 2.0 5.0 10.0 4.0 1.0 1.0 1.0 1.0 2.0 1.5 1.0 3.0

2.0 2.0 3.0 2.0 2.0 2.5 4.0 2.0 1.5 1.5 2.5 20.0 1.5 5.0 8.0 21.0 5.0 5.0 10.0 15.0 5.0 2.0 2.0 2.0 2.0 4.0 2.0 3.0 5.0

2.5 2.5 4.0 3.0 4.0 3.0 5.0 3.0 2.0 2.0 3.0 30.0 2.0 8.0 15.0 28.0 10.0 10.0 20.0 25.0 8.0 3.0 2.5 3.0 3.0 6.0 2.5 5.0 10.0

1.92 2.00 3.00 2.00 2.17 2.50 3.83 2.00 1.50 1.50 2.50 20.83 1.50 5.17 8.33 21.00 5.17 5.33 10.83 15.83 5.33 2.00 1.92 2.00 2.00 4.00 2.00 3.00 5.50

0.25 0.17 0.33 0.33 0.50 0.17 0.50 0.33 0.17 0.17 0.17 2.50 0.17 0.83 2.00 2.33 1.50 1.33 2.50 2.50 0.67 0.33 0.25 0.33 0.33 0.67 0.17 0.67 1.17

0.00 1.92 3.92 3.92 3.92 3.92 3.92 6.92 5.92 5.92 6.08 6.08 6.42 6.42 7.75 8.92 7.42 7.42 8.58 7.92 29.92 12.75 14.75 26.92 23.75 16.08 35.25 37.25 40.25

1.92 3.92 6.92 5.92 6.08 6.42 7.75 8.92 7.42 7.42 8.58 26.92 7.92 11.58 16.08 29.92 12.58 12.75 19.42 23.75 35.25 14.75 16.67 28.92 25.75 20.08 37.25 40.25 45.75

0.00 1.92 3.92 22.50 10.25 13.42 29.58 6.92 26.67 24.50 19.92 12.42 15.92 28.08 33.42 8.92 28.17 26.00 22.42 17.42 29.92 31.33 33.33 33.25 33.25 41.75 35.25 37.25 40.25

1.92 3.92 6.92 24.50 12.42 15.92 33.42 8.92 28.17 26.00 22.42 33.25 17.42 33.25 41.75 29.92 33.33 31.33 33.25 33.25 35.25 33.33 35.25 35.25 35.25 45.75 37.25 40.25 45.75

0.000 0.000 0.000 18.580 6.333 9.500 25.670 0.000 20.750 18.580 13.830 6.330 9.500 21.670 25.670 0.000 20.750 18.580 13.830 9.500 0.000 18.580 18.580 6.330 9.500 25.670 0.000 0.000 0.000

CHAPTER 3

Activity 1 2 3 8 16 21 27 28 29 Variance for critical path:

PROJECT MANAGEMENT

0.25 0.17 0.33 0.33 2.33 0.67 0.17 0.67 1.17

0.0625 0.0289 0.1089 0.1089 5.4289 0.4489 0.0289 0.4489 1.3689 8.0337

Therefore, σ = 8.0337 = 2.834. As an approximation, we can use the customary equation for the normal distribution:

Due date − E (t ) σ

(Note: This might be a good time to discuss the difference between a continuous and a discrete probability distribution and the appropriate procedure for using a continuous distribution as an approximation to a discrete, if you have not already done so.) Finish Time

1 day early 2 days early 3 days early 4 days early 5 days early 6 days early 7 days early

–0.353 –0.706 –1.058 –1.411 –1.764 –2.117 –2.470

Probability

36.3 24.0 14.5 7.9 3.9** 1.7 0.7

The appropriate procedure for using the normal distribution gives 3.0%—roughly a 30% difference.

There is, by the approximate procedure used, a 3.9% probability of finishing 5 days, or 1 week, early.

LO 3.5: Calculate the variance of activity times AACSB: Analytical thinking 3.

In order to shorten the shutdown, Shale Oil would have to determine the costs of decreasing the activities on the critical path. This is the vessel and column branch of the network, which is typically the longest section in a shutdown. The cost of reducing activity time by one time unit for each activity in this branch would have to be calculated. The activity with the lowest of these costs could then be acted upon. Perhaps the repairs to the vessels and columns could be expedited with workers from some of the other branches with high slack time. However, delivery on materials could be an overriding factor.

LO 3.6: Crash a project AACSB: Analytical thinking 2

SOUTHWESTERN UNIVERSITY: A

CHAPTER 3

Activity Mean tA = 30 tB = 60 tC = 65 tD = 55 tE = 30 tF = 0.1 tG = 30 tH = 20 tI = 30 tJ = 10 tK = 0.1 tL = 30 **

S.D. 3.33** 10.00 8.33** 11.66** 1.67 0.00 1.67** 3.33** 6.67** 0.67 0.00 6.67**

Project Management

Variance 11.11 100.00 69.44 136.11 2.78 0.00 2.78 11.11 44.44 0.44 0.00 44.44

Critical path

= A − C − D − G − H − I − L (260 days)

Critical path

Variance of critical path = 11.11 + 69.44 + 136.11 + 2.78 + 11.11 + 44.44 + 44.44 = 319.43 Standard deviation of critical path = 17.87 days LO 3.2: Draw AOA and AON networks LO 3.4: Determine a critical path LO 3.5: Calculate the variance of activity times AACSB: Analytical thinking

P (Completion < 270 days) = P(t ≤ 270)

270 − 260   P Z ≤ = P [ Z ≤ 0.56 ] = 0.712 = 71.2% 17.87   LO 3.5: Calculate the variance of activity times AACSB: Analytical thinking

Crash to 250 days and to 240 days

Activity

Normal Time (days)

Crash Time (days)

Crash Cost/Day

30 60 65 55 30 0** 30 20 30 10 0** 30

20 20 50 30 25 0 25 10 20 8 0 20

$1,500 $3,500 $4,000 $1,900 $9,500 $0 $2,500 $2,000 $2,000 $6,000 $0 $4,500

A B C D E F G H I J K L **

Rounded to zero from 0.1.

To crash to 250 days (from the current 260 days), select A at $1,500/day × 10 days = $15,000. To crash to 240 days now (from the current 250 days), select D at $1,900/day × 10 days = $19,000. Total cost to crash to 240 days = $34,000.

LO 3.6: Crash a project AACSB: Analytical thinking

C H A P T E R

Forecasting

DISCUSSION QUESTIONS 1. Qualitative models incorporate subjective factors into the forecasting model. Qualitative models are useful when subjective factors are important. When quantitative data are difficult to obtain, qualitative models may be appropriate. LO 4.2: Explain when to use each of the four qualitative models AACSB: Application of knowledge 2.

Approaches are qualitative and quantitative. Qualitative is relatively subjective; quantitative uses numeric models.

LO 4.2: Explain when to use each of the four qualitative models AACSB: Application of knowledge 3. Short-range (under 3 months), medium-range (3 months to 3 years), and long-range (over 3 years). LO 4.1: Understand the three time horizons and which models apply for each AACSB: Application of knowledge 4.

The steps that should be used to develop a forecasting system are: (a) Determine the purpose and use of the forecast (b) Select the item or quantities that are to be forecasted (c) Determine the time horizon of the forecast (d) Select the type of forecasting model to be used (e) Gather the necessary data (f) Validate the forecasting model (g) Make the forecast (h) Implement and evaluate the results

LO 4.1: Understand the three time horizons and which models apply for each AACSB: Application of knowledge 5.

Any three of: sales planning, production planning and budgeting, cash budgeting, analyzing various operating plans.

LO 4.1: Understand the three time horizons and which models apply for each AACSB: Application of knowledge 6. There is no mechanism for growth in these models; they are built exclusively from historical demand values. Such methods will always lag trends. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 7. Exponential smoothing is a weighted moving average where all previous values are weighted with a set of weights that decline exponentially. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 46

CHAPTER 4

FORECASTING

8. MAD, MSE, and MAPE are common measures of forecast accuracy. To find the more accurate forecasting model, forecast with each tool for several periods where the demand outcome is known, and calculate MSE, MAPE, or MAD for each. The smaller error indicates the better forecast. LO 4.4: Compute three measures of forecast accuracy AACSB: Analytical thinking 9.

The Delphi method involves the following steps: (a) Assembling a group of experts in such a manner as to preclude direct communication between identifiable members of the group (b) Assembling the responses of each expert to the questions or problems of interest (c) Summarizing these responses (d) Providing each expert with the summary of all responses (e) Asking each expert to study the summary of the responses and respond again to the questions or problems of interest. (f) Repeating steps (b) through (e) several times as necessary to obtain convergence in responses. If convergence has not been obtained by the end of the fourth cycle, the responses at that time should probably be accepted and the process terminated—little additional convergence is likely if the process is continued.

LO 4.2: Explain when to use each of the four qualitative models AACSB: Analytical thinking 10. A time-series model predicts on the basis of the assumption that the future is a function of the past, whereas an associative model incorporates into the model the variables of factors that might influence the quantity being forecast. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking 11. A time series is a sequence of evenly spaced data points with the four components of trend, seasonality, cyclical, and random variation. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 12. When the smoothing constant, α, is large (close to 1.0), more weight is given to recent data; when α is low (close to 0.0), more weight is given to past data. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 13. Seasonal patterns are of fixed duration and repeat regularly. Cycles vary in length and regularity. Seasonal indices allow “generic” forecasts to be made specific to the month, week, etc., of the application. LO 4.5: Develop seasonal indices AACSB: Application of knowledge 14. Exponential smoothing weighs all previous values with a set of weights that decline exponentially. It can place a full weight on the most recent period (with an alpha of 1.0). This, in effect, is the naive approach, which places all its emphasis on last period’s actual demand. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Application of knowledge 15.

Adaptive forecasting refers to computer monitoring of tracking signals and self-adjustment if a signal passes its present limit.

16.

CHAPTER 4 F O R E C A S T I N G

Tracking signals alert the user of a forecasting tool to periods in which the forecast was in significant error.

LO 4.7: Use a tracking signal AACSB: Application of knowledge 17. The correlation coefficient measures the degree to which the independent and dependent variables move together. A negative value would mean that as X increases, Y tends to fall. The variables both move, but move in opposite directions. LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge 18.

Independent variable (x) is said to explain variations in the dependent variable (y).

LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge 19. Nearly every industry has seasonality. The seasonality must be filtered out for good medium-range planning (of production and inventory) and performance evaluation. LO 4.5: Develop seasonal indices AACSB: Analytical thinking 20. There are many examples. The demand for raw materials and component parts such as steel or tires is a function of demand for goods such as automobiles. LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge 21. Obviously, as we go farther into the future, it becomes more difficult to make forecasts, and we must diminish our reliance on the forecasts. LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge 22.

(a) This problem gives students a chance to tackle a realistic problem in business, i.e., not enough data to make a good forecast. As can be seen in the accompanying figure, the data contain both seasonal and trend factors.

Averaging methods are not appropriate with trend, seasonal, or other patterns in the data. Moving averages smooth out seasonality. Exponential smoothing can forecast January next year, but not farther. Because seasonality is strong, a naive model that students create on their own might be best. (b) One model might be: Ft+1 = At–11 That is forecastnext period = actualone year earlier to account for seasonality. But this ignores the trend. One very good approach would be to calculate the increase from each month last year to each month this year, sum all 12 increases, and divide by 12. The forecast for next year would equal the value for the same month this year plus the average increase over the 12 months of last year. Copyright ©2023 Pearson Education, Inc.

CHAPTER 4

FORECASTING

148 = 17 + 12 = 29 12

where 148 = total monthly increases from last year to this year. The forecasts for each of the months of next year then become: Jan. Feb. Mar. Apr. May Jun.

29 26 32 35 42 50

July. Aug. Sep. Oct. Nov. Dec.

56 53 45 35 38 29

LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Reflective thinking

ETHICAL DILEMMA This is an ethical dilemma that will certainly encourage a strong response from your class. Would managers make OM decisions with such weak correlations? Colleges do it all the time, with alternatives that are limited and nonstandard. Should high school GPA and class rank be used? Are they uniform throughout the nation, or even comparable within a country, city, or school district? For instance, would accepting the top 10% of a high school class for admission to college be a better solution? Encourage the students to find quantifiable and equitable alternate predicting factors. It is not an easy task.

ACTIVE MODEL EXERCISES (AVAILABLE IN MYLAB OPERATIONS MANAGEMENT) ACTIVE MODEL 4.1: Moving Averages 1.

What does the graph look like when n = 1? The forecast graph mirrors the data graph but one period later.

What happens to the graph as the number of periods in the moving average increases? The forecast graph becomes shorter and smoother.

What value for n minimizes the MAD for this data? n = 1 (a naïve forecast)

ACTIVE MODEL 4.2: Exponential Smoothing 1.

What happens to the graph when alpha equals zero? The graph is a straight line. The forecast is the same in each period.

What happens to the graph when alpha equals one? The forecast follows the same pattern as the demand (except for the first forecast) but is offset by one period. This is a naive forecast.

Generalize what happens to a forecast as alpha increases. As alpha increases, the forecast is more sensitive to changes in demand.

At what level of alpha is the mean absolute deviation (MAD) minimized? alpha = .16

CHAPTER 4 F O R E C A S T I N G

ACTIVE MODEL 4.3: Exponential Smoothing with Trend Adjustment 1.

Scroll through different values for alpha and beta. Which smoothing constant appears to have the greater effect on the graph? Alpha

2. With beta set to zero, find the best alpha, and observe the MAD. Now find the best beta. Observe the MAD. Does the addition of a trend improve the forecast? alpha = .11, MAD = 2.59; beta above .6 changes the MAD (by a little) to 2.54.

ACTIVE MODEL 4.4: Trend Projections 1.

What is the annual trend in the data? 10.54

2. Use the scrollbars for the slope and intercept to determine the values that minimize the MAD. Are these the same values that regression yields? No, they are not the same values. For example, an intercept of 57.81 with a slope of 9.44 yields a MAD of 7.17.

END-OF-CHAPTER PROBLEMS 4.1 (a)

374 + 368 + 381 = 374.33 pints 3

(b) Week of August 31 September 7 September 14 September 21 September 28 October 5

Weighted Moving Average

Pints Used 360 389 410 381 368 374

381 × .1 = 38.1 368 × .3 = 110.4 374 × .6 = 224.4 372.9 Forecast 372.9

Pints 360 389 410 381 368 374

Forecasting Error 0 29 44.2 6.36 −7.912 −.3296

Error x .20 0 5.8 8.84 1.272 −1.5824 −.06592

Forecast 360 (given) 360 365.8 374.64 375.912 374.3296 374.2636

The forecast for October 12 is 374.26.

4.2

(a) No, the data appear to have no consistent pattern (see part d for graph). Year

Forecast

(b) (c)

Demand 3-year moving 3-year weighted

9.0 7.0 6.4

13.0 7.7 7.8

8.0 9.0 11.0

12.0 10.0 9.6

13.0 11.0 10.9

9.0 11.0 12.2

11.0 11.3 10.5

7.0 11.0 10.6

9.0 8.4

CHAPTER 4

FORECASTING

(d) The 3-year moving average appears to give better results.

4.3

Year

Demand Naive Exp. Smoothing

9.0 7.0 6.4

5.0 9.0 7.4

9.0 5.0 6.5

Forecast

13.0 8.0 12.0 9.0 13.0 8.0 7.5 9.7 9.0

13.0 12.0 10.2

9.0 13.0 11.3

11.0 9.0 10.4

7.0 11.0 10.6

7.0 9.2

Naive tracks the ups and downs best but lags the data by one period. Exponential smoothing is probably better because it smooths the data and does not have as much variation.

4.4 (a) FJuly = FJune + 0.2(Forecasting error) = 42 + 0.2(40 – 42) = 41.6 (b) FAugust = FJuly + 0.2(Forecasting error) = 41.6 + 0.2(45 − 41.6) = 42.3 (c) The banking industry has a great deal of seasonality in its processing requirements

4.5 (a)

3,700 + 3,800 = 3,750 miles in the 6th year. 2

(b) Year

Mileage

2-Year Moving Average

1 2 3 4 5

3,000 4,000 3,400 3,800 3,700

3,500 3,700 3,600 Totals

MAD =

Error

|Error|

−100 100 100 100

100 100 100 300

300 = 100. 3

CHAPTER 4 F O R E C A S T I N G

4.5

Year

Mileage

Forecast

Error

|Error|

1 2 3 4 5

3,000 4,000 3,400 3,800 3,700

3,600 3,640 3,640

−200 160 60

200 160 60 420

Forecast for year 6 is 3,740 miles.

MAD =

4.5

420 = 140 3

(d)

Year

Mileage

Forecast

Forecast Error

Error × α = .50

New Forecast

1 2 3 4 5

3,000 4,000 3,400 3,800 3,700

3,000 3,000 3,500 3,450 3,625 Total

0 1,000 −100 350 75 1,325

0 500 −50 175 38

3,000 3,500 3,450 3,625 3,663

The forecast is 3,663 miles.

4.6 2

Y Sales

X Period

January February March April May June July August September October November December

20 21 15 14 13 16 17 18 20 20 21 23

1 2 3 4 5 6 7 8 9 10 11 12

1 4 9 16 25 36 49 64 81 100 121 144

XY 20 42 45 56 65 96 119 144 180 200 231 276

Sum Average

218 18.2

78 6.5

650

1,474

(a)

(b) i) Naive The coming January = December = 23 ii) 3-month moving (20 + 21 + 23)/3 = 21.33 iii) 6-month weighted [(0.1 × 17) + (.1 × 18) + (0.1 × 20) + (0.2 × 20) + (0.2 × 21) + (0.3 × 23)]/1.0 = 20.6

CHAPTER 4

FORECASTING

iv) Exponential smoothing with alpha = 0.3 FOct = 18 + 0.3(20 − 18) = 18.6

FNov = 18.6 + 0.3(20 − 18.61) = 19.02 FDec = 19.02 + 0.3(21 − 19.02) = 19.61 FJan = 19.61 + 0.3(23 − 19.61) = 20.62 ≈ 21 v) Trend

 x = 78, x = 6.5,  y = 218, y = 18.17

 xy − nx y  x 2 − nx 2 1474 − (12)(6.5)(18.2) 54.4 = = 0.38 b= 650 − 12(6.5)2 143 a = y − bx b=

a = 18.2 − 0.38(6.5) = 15.73 Forecast = 15.73 + .38(13) = 20.67, where next January is the 13th month. (c) Only trend provides an equation that can extend beyond one month

4.7

Present = Period (week) 6.

 1  1 1 1  (a) So: F7 =   A6 +   A5 +   A4 +   A3  1.0 4 4 6   3  1 1 1 1 =   (52) +   (63) +   (48) +   (70) = 56.76 patients, 3 4 4       6 or 57 patients 1 1 1 1 where 1.0 =  weights , , , 3 4 4 6

(b) If the weights are 20, 15, 15, and 10, there will be no change in the forecast because these are the same relative weights as in part (a), i.e., 20/60, 15/60, 15/60, and 10/60. (c) If the weights are 0.4, 0.3, 0.2, and 0.1, then the forecast becomes 56.3, or 56 patients. (96 + 88 + 90) = 91.3 3 (88 + 90) (b) = 89 2 (c)

4.8 (a)

Temperature

2-day M.A.

93 94 93 95 96 88 90

— — 93.5 93.5 94.0 95.5 92.0

|Error|

(Error)

— — 0.5 1.5 2.0 7.5 2.0 13.5

— — 0.25 2.25 4.00 56.25 4.00 66.75

Absolute% Error — — 100(.5/93) 100(1.5/95) 100(2/96) 100(7.5/88) 100(2/90)

= 0.54% = 1.58% = 2.08% = 8.52% = 2.22% 14.94%

MAD = 13.5/5 = 2.7

(d) MSE = 66.75/5 = 13.35 (e) MAPE = 14.94%/5 = 2.99%

4.9

CHAPTER 4 F O R E C A S T I N G

(a, b) The computations for both the 2- and 3-month averages appear in the table; the results appear in the figure below.

(c) MAD (2-month moving average) = .750/10 = .075 MAD (3-month moving average) = .793/9 = .088 Therefore, the 2-month moving average seems to have performed better. Table for Problem 4.9(a, b, c): Forecast

Month January February March April May June July August September October November December

4.9

Price per Chip $1.80 1.67 1.70 1.85 1.90 1.87 1.80 1.83 1.70 1.65 1.70 1.75

2-Month Moving Average

|Error| 3-Month Moving Average

2-Month Moving Average

3-Month Moving Average

1.723 1.740 1.817 1.873 1.857 1.833 1.777 1.727 1.683 Totals

.035 .165 .125 .005 .085 .005 .115 .115 .025 .075 .750

.127 .160 .053 .073 .027 .133 .127 .027 .067 .793

1.735 1.685 1.775 1.875 1.885 1.835 1.815 1.765 1.675 1.675

(d) Table for Problem 4.9(d): α = .1

α = .3

α = .5

Month

Price per Chip Forecast

|Error|

Forecast

|Error|

Forecast

|Error|

January February March April May June July August September October November December

$1.80 1.67 1.70 1.85 1.90 1.87 1.80 1.83 1.70 1.65 1.70 1.75 Totals MAD (total/12)

$.00 .13 .09 .07 .11 .07 .00 .03 .11 .15 .08 .02 $.86 $.072

$1.80 1.80 1.76 1.74 1.77 1.81 1.83 1.82 1.82 1.79 1.75 1.73

$.00 .13 .06 .11 .13 .06 .03 .01 .12 .14 .05 .02 $.86 $.072

$1.80 1.80 1.74 1.72 1.78 1.84 1.86 1.83 1.83 1.76 1.71 1.70

$.00 .13 .04 .13 .12 .03 .06 .00 .13 .11 .01 .05 $.81 $.0675

$1.80 1.80 1.79 1.78 1.79 1.80 1.80 1.80 1.81 1.80 1.78 1.77

CHAPTER 4

FORECASTING

α = .5 is preferable, using MAD, to α = .1 or α = .3. One could also justify excluding the January error and then dividing by n = 11 to compute the MAD. These numbers would be $.078 (for α = .1), $.078 (for α = .3), and $.074 (for α = .5). Note: If table numbers are not rounded, answers would be $0.077 (for alpha = 0.3) and $0.072 (for alpha = 0.5). 4.10

Year Demand (a) 3-year moving (b) 3-year weighted

Forecast

5.0 4.7 4.5

10.0 5.0 5.0

8.0 6.3 7.3

7.0 7.7 7.8

9.0 8.3 8.0

12.0 8.0 8.3

14.0 9.3 10.0

15.0 11.7 12.3

13.7 14.0

Forecast

4 5

6.0 4.7

4.0 5.1

5.0 4.8

10.0 4.8

8.0 6.4

7.0 6.9

9.0 6.9

12.0 7.5

14.0 8.9

15.0 10.4

11.8

2 1.3

3 1.1

4 0.2

5 5.2

|Error| = |Actual – Forecast| Year Exp. Smoothing error

1 1

6 1.6

7 0.1

8 2.1

9 4.5

10 5.1

11 4.6

MAD 2.4

These calculations were completed in Excel. Calculations are slightly different in Excel OM and POM for Windows due to rounding differences.

(b) MSE Year Error2

1 1

2 1.69

3 1.19

4 0.06

MSE =

Σ Errors 2 104.87 = = 9.53 11 years 11

5 26.69

6 2.61

7 0.02

8 4.38

9 19.93

10 26.27

11 21.05

CHAPTER 4 F O R E C A S T I N G

4.12

Day

Monday

Actual Demand

Forecast Demand

Tuesday

Wednesday

Thursday

Friday

← Answer

Ft = Ft–1 + α(At–1 – Ft–1)

Let α = .25. Let Monday forecast demand = 88 F2 = 88 + .25(88 – 88) = 88 + 0 = 88 F3 = 88 + .25(72 – 88) = 88 – 4 = 84 F4 = 84 + .25(68 – 84) = 84 – 4 = 80 F5 = 80 + .25(48 – 80) = 80 – 8 = 72 4.13

(a) Exponential smoothing, α = 0.6:

Year 1 2 3 4 5 6

Demand 45 50 52 56 58 ?

Exponential Smoothing α = 0.6

Absolute Deviation 4.0 6.6 4.6 5.8 4.3

41 41.0 + 0.6(45–41) = 43.4 43.4 + 0.6(50–43.4) = 47.4 47.4 + 0.6(52–47.4) = 50.2 50.2 + 0.6(56–50.2) = 53.7 53.7 + 0.6(58–53.7) = 56.3

Σ = 25.3 MAD = 5.06

Exponential smoothing, α = 0.9: Year 1 2 3 4 5 6

Demand

Exponential Smoothing α = 0.9

Absolute Deviation

45 50 52 56 58 ?

41 41.0 + 0.9(45–41) = 44.6 44.6 + 0.9(50–44.6 ) = 49.5 49.5 + 0.9(52–49.5) = 51.8 51.8 + 0.9(56–51.8) = 55.6 55.6 + 0.9(58–55.6) = 57.8

4.0 5.4 2.5 4.2 2.4

Σ = 18.5 MAD = 3.7

(b) Year 1 2 3 4 5 6

3-year moving average: Demand

3-Year Moving Average

Absolute Deviation

45 50 52 56 58 ?

(45 + 50 + 52)/3 = 49 (50 + 52 + 56)/3 = 52.7 (52 + 56 + 58)/3 = 55.3

7 5.3

Σ = 12.3 MAD = 6.2

CHAPTER 4

(c)

FORECASTING

Trend projection:

Year

Demand

Trend Projection

45 50 52 56 58 ?

42.6 + 3.2 × 1 = 45.8 42.6 + 3.2 × 2 = 49.0 42.6 + 3.2 × 3 = 52.2 42.6 + 3.2 × 4 = 55.4 42.6 + 3.2 × 5 = 58.6 42.6 + 3.2 × 6 = 61.8

1 2 3 4 5 6

Absolute Deviation 0.8 1.0 0.2 0.6 0.6

Σ = 3.2 MAD = 0.64 Y = a + bX b=

 XY – nXY  X 2 – nX 2

a = Y – bX 2

1 2 3 4 5

45 50 52 56 58

45 100 156 224 290

1 4 9 16 25 2

Then: ΣX = 15, ΣY = 261, ΣXY = 815, ΣX = 55, X = 3, Y = 52.2 Therefore: 815 – 5 × 3 × 52.2 = 3.2 55 – 5 × 3 × 3 a = 52.20 – 3.20 × 3 = 42.6 Y6 = 42.6 + 3.2 × 6 = 61.8 b=

(d)

Comparing the results of the forecasting methodologies for parts (a), (b), and (c).

Forecast Methodology

MAD

Exponential smoothing, α = 0.6 Exponential smoothing, α = 0.9 3-year moving average Trend projection

5.06 3.7 6.2 0.64

Based on a mean absolute deviation criterion, the trend projection is to be preferred over the exponential smoothing with α = 0.6, exponential smoothing with α = 0.9, or the 3-year moving average forecast methodologies.

4.14 Method 1:

MAD: (0.20 + 0.05 + 0.05 + 0.20)/4 = .125 ← better MSE: (0.04 + 0.0025 + 0.0025 + 0.04)/4 = .021

Method 2:

MAD: (0.1 + 0.20 + 0.10 + 0.11) / 4 = .1275 MSE: (0.01 + 0.04 + 0.01 + 0.0121) / 4 = .018 ← better

CHAPTER 4 F O R E C A S T I N G

4.15

(a)

Year

Sales

1 2 3 4 5 6

450 495 518 563 584

Forecast 3-Year Moving Average

Absolute Deviation

(450 + 495 + 518)/3 = 487.7 (495 + 518 + 563)/3 = 525.3 (518 + 563 + 584)/3 = 555.0

75.3 58.7

Σ = 134 (b)

134 = 67 MAD = 2

Year

Sales

1 2 3 4 5

450 495 518 563 584

Error2

(75.3)2 = 5,675 (58.7)2 = 3,442

S = 9,117 9,117 = 4,558.5 MSE = 2

4.16

(a)

Year

Sales Y

450 495 518 563 584

1 4 9 16 25

450 990 1,554 2,252 2,920

Σ = 2610

Σ = 55

Σ = 8166

1 2 3 4 5

X = 3, Y = 522 Y = a + bX b=

 XY − nXY  X 2 − nX 2

8166 − (5)(3)(522) 336 = = 33.6 55 − (5)(9) 10

a = Y − bX = 522 − (33.6)(3) = 421.2 y = 421.2 + 33.6 x y = 421.2 + 33.6 × 6 = 622.8 (b) MAD Year

Sales

Forecast Trend

Absolute Deviation

1 2 3 4 5 6

450 495 518 563 584

454.8 488.4 522.0 555.6 589.2 622.8

4.8 6.6 4.0 7.4 5.2

CHAPTER 4

FORECASTING

164.4 = 32.88 5

4.17 Year

Sales

1 2 3 4 5 6

450 495 518 563 584

Forecast Exponential Smoothing α = 0.6

Absolute Deviation

410.0 410 + 0.6(450 − 410) = 434.0 434 + 0.6(495 − 434) = 470.6 470.6 + 0.6(518 − 470.6) = 499.0 499 + 0.6(563 − 499) = 537.4 537.4 + 0.6(584 − 537.4) = 565.6

40.0 61.0 47.4 64.0 46.6

Σ = 259 MAD = 51.8

Year

Sales

1 2 3 4 5 6

450 495 518 563 584

Forecast Exponential Smoothing α = 0.9

Absolute Deviation

410.0 410 + 0.9(450 − 410) = 446.0 446 + 0.9(495 − 446) = 490.1 490.1 + 0.9(518 − 490.1) = 515.2 515.2 + 0.9(563 − 515.2) = 558.2 558.2 + 0.9(584 − 558.2) = 581.4

40.0 49.0 27.9 47.8 25.8

Σ = 190.5 MAD = 38.1

(Refer to Solved Problem 4.1) For α = 0.3, absolute deviations for years 1–5 are 40.0, 73.0, 74.1, 96.9, and 88.8, respectively. So MAD = 372.8/5 = 74.6. MADα= 0.3 = 74.6 MADα= 0.6 = 51.8 MADα= 0.9 = 38.1 Because it gives the lowest MAD, the smoothing constant of α = 0.9 gives the most accurate forecast. 4.18 We need to find the smoothing constant α. We know in general that Ft = Ft–1 + α(At−1 – Ft−1); t = 2, 3, 4. Choose either t = 3 or t = 4 (t = 2 won’t let us find α because F2 = 50 = 50 + α(50 − 50) holds for any α). Let’s pick t = 3. Then F3 = 48 = 50 + α(42 − 50)

48 = 50 + 42α − 50α

−2 = –8α

So,

.25 = α

CHAPTER 4 F O R E C A S T I N G

Now we can find F5: F5 = 50 + α(46 − 50)

F5 = 50 + 46α – 50α = 50 − 4α

α = .25, F5 = 50 − 4(.25) = 49

For

The forecast for time period 5 = 49 units. Trend adjusted exponential smoothing: α = 0.1, β = 0.2

4.19 Month

Unadjusted Forecast

Income

February March April May June July August

70.0 68.5 64.8 71.7 71.3 72.8

Adjusted Forecast

Trend

65.0 65.5 65.89 65.92 66.62 67.31 68.16

0.0 0.1 0.16 0.13 0.25 0.33

|Error|

65 65.6 66.05 66.06 66.87 67.64 68.60

Error

5.0 2.9 1.2 5.6 4.4 5.2 24.3

25.0 8.4 1.6 31.9 19.7 26.6 113.2

MAD = 24.3/6 = 4.05, MSE = 113.2/6 = 18.87. Note that all numbers are rounded.

Note: To use POM for Windows to solve this problem, a period 0, which contains the initial forecast and initial trend, must be added. Trend adjusted exponential smoothing: α = 0.1, β = 0.8

4.20 Month

Demand (y)

February March April May June July Totals Average August forecast

70.0 68.5 64.8 71.7 71.3 72.8 419.1 69.85

Unadjusted Forecast

Trend

Adjusted Forecast

65.0 65.5 66.16 66.57 67.49 68.61

0 0.4 0.61 0.45 0.82 1.06

65.0 65.9 66.77 67.02 68.31 69.68

71.30

Error

|Error|

5.00 2.60 −1.97 4.68 2.99 3.12 16.42 2.74 (Bias)

5.0 2.6 1.97 4.68 2.99 3.12 20.36 3.39 (MAD)

Error

25.00 6.76 3.87 21.89 8.91 9.76 76.19 12.70 (MSE)

Based on the MSE criterion, the exponential smoothing with α = 0.1, β = 0.8 is to be preferred over the exponential smoothing with α = 0.1, β = 0.2. Its MSE of 12.70 is lower. Its MAD of 3.39 is also lower than that in Problem 4.19.

4.21 F5 = α A4 + (1 – α ) ( F4 + T4 ) = ( 0.2 )(19) + ( 0.8)( 20.14 )

= 3.8 + 16.11 = 19.91

T5 = β ( F5 – F4 ) + (1 – β ) T4 = ( 0.4 )(19.91 – 17.82 )

+ ( 0.6)( 2.32 ) = 0.4 ( 2.09) + 1.39 = 0.84 + 1.39 = 2.23

FIT5 = F5 + T5 = 19.91 + 2.23 = 22.14 F6 = αA5 + (1 – α ) ( F5 + T5 ) = ( 0.2 )( 24 ) + ( 0.8)( 22.14 ) = 4.8 + 17.71 = 22.51

CHAPTER 4

FORECASTING

T6 = β ( F6 – F5 ) + (1 – β ) T5 = 0.4 ( 22.51 – 19.91) + 0.6 ( 2.23) = 0.4 ( 2.6 ) + 1.34

= 1.04 + 1.34 = 2.38 FIT6 = F6 + T6 = 22.51 + 2.38 = 24.89

4.22 F7 = αA6 + (1 – α )( F6 + T6 ) = (0.2)(21) + (0.8)(24.89) = 4.2 + 19.91 = 24.11 T7 = β( F7 – F6 ) + (1 – β)T6 = (0.4)(24.11 – 22.51) + (0.6)(2.38) = 2.07 FIT7 = F7 + T7 = 24.11 + 2.07 = 26.18 F8 = αA7 + (1 – α )( F7 + T7 ) = (0.2)(31)

+ (0.8)(26.18) = 27.14 T8 = β ( F8 – F7 ) + (1 – β ) T7 = 0.4 ( 27.14 – 24.11)

+ 0.6 ( 2.07) = 2.45

FIT8 = F8 + T8 = 27.14 + 2.45 = 29.59 F9 = αA8 + (1 – α ) ( F8 + T8 ) = ( 0.2 )( 28)

+ ( 0.8)( 29.59) = 29.28

T9 = β ( F9 – F8 ) + (1 – β ) T8 = ( 0.4 )( 29.28 – 27.14 )

+ ( 0.6 )( 2.45) = 2.32

FIT9 = F9 + T9 = 29.28 + 2.32 = 31.60 4.23

Students must determine the naive forecast for the four months. The naive forecast for March is the February actual of 83, etc.

(a)

Actual Forecast |Error| |% Error| March April May June

101 96 89 108

120 114 110 108

19 18 21 0 58

100 (19/101) = 18.81% 100 (18/96) = 18.75% 100 (21/89) = 23.60% 100 (0/108) = 0% 61.16%

58 = 14.5 4 61.16% MAPE (for management) = = 15.29% 4 MAD (for management) =

Actual Naive

(b) March April May June

101 96 89 108

83 101 96 89

|Error| |% Error| 18 5 7 19 49

100 (18/101) = 17.82% 100 (5/96) = 5.21% 100 (7/89) = 7.87% 100 (19/108) = 17.59% 48.49%

49 = 12.25 4 48.49% MAPE (for naïve) = = 12.12% 4 Naive outperforms management. MAD (for naïve) =

CHAPTER 4 F O R E C A S T I N G

(c) MAD for the manager’s technique is 14.5, while MAD for the naive forecast is only 12.25. MAPEs are 15.29% and 12.12%, respectively. So the naive method is better.

4.24

Month

Number of Accidents (y)

January February March April Totals Averages

30 40 60 90 220 y = 55

 xy − n x y

 x 2 − nx 2 100 = = 20 5 a = y − bx

650 − 4(2.5)(55) 30 − 4(2.5)2

1 2 3 4 10 x = 2.5

30 80 180 360 650

1 4 9 16 30

650 − 550 30 − 25

= 55 − (20)(2.5) =5 The regression line is y = 5 + 20x. The forecast for May (x = 5) is y = 5 + 20(5) = 105. 4.25

Season Fall Winter Spring Summer

Year1 Year2 Demand Demand 200 350 150 300

Average Year1–Year2 Demand

Average Season Demand

Seasonal Index

Year3 Demand

225.0 325.0 157.5 292.5

250 250 250 250

0.90 1.30 0.63 1.17

270 390 189 351

250 300 165 285

Average Yr1 to Yr2  Yr1 Demand + Yr2 Demand  = 2 Demand for season  Sum of Ave Yr1 to Yr2 Demand 4 Average Yr1 to Yr2 Demand Seasonal index = Average Seasonal Demand New Annual Demand Yr3 = × Seasonal index 4 1200 = × Seasonal index 4

Average seasonal demand =

4.26 Year

Winter

Spring

Summer

Fall

1 2 3 4

1,400 1,200 1,000 900 4,500

1,500 1,400 1,600 1,500 6,000

1,000 2,100 2,000 1,900 7,000

600 750 650 500 2,500

CHAPTER 4

FORECASTING

20,000 = 1, 250 16 6,000 Average over spring : = 1,500 4 1,500 Spring index : = 1.2 1, 250 Average over all seasons :

 5,600  Answer :   (1.2) = 1,680 sailboats  4  4.27

Quarter

Year 1

Year 2

Year 3

Average Average Quarterly Seasonal Demand Demand Index

Winter Spring Summer Fall

73 104 168 74

65 82 124 52

89 146 205 98

75.67 110.67 165.67 74.67

4.28

106.67 106.67 106.67 106.67

0.709 1.037 1.553 0.700

The year 26 quarter numbers are 101 through 104.

(1) Quarter

(2) Quarter Number

(3) Forecast (77 + .43Q)

(4) Seasonal Factor

(5) Adjusted Forecast [(3) × (4)]

Winter Spring Summer Fall

101 102 103 104

120.43 120.86 121.29 121.72

.8 1.1 1.4 .7

96.344 132.946 169.806 85.204

4.29

(a) See the table below. 2,880 − 5(3)(180) 55 − 5(3)2

2,880 − 2,700 55 − 45

180 = 18 10 a = 180 − 3(18) = 180 − 54 = 126 =

y = 126 + 18 x For next year (x = 6), the number of disk drives (in millions) is forecast as y = 126 + 18(6) = 126 + 108 = 234. Table for Problem 4.29 Year (x) 1 2 3 4 5 Totals 15 x = 3

Disk Drives (y) 140 160 190 200 210 900 y = 180

126 + 18x

Error

140 320 570 800 1,050 2,800

1 4 9 16 25 55

144 162 180 198 216

–4 –2 10 2 –6

16 4 100 4 36 160

(b) MSE = 160/5 = 32 (c) MAPE = 13.23%/5 = 2.65%

|% Error| 100 (4/140) = 2.86% 100 (2/160) = 1.25% 100 (10/190) = 5.26% 100 (2/200) = 1.00% 100 (6/210) = 2.86% 13.23%

CHAPTER 4 F O R E C A S T I N G

4.30 Year X

Patients Y

36 33 40 41 40 55 60 54 58 61 478

1 4 9 16 25 36 49 64 81 100 385

1,296 1,089 1,600 1,681 1,600 3,025 3,600 2,916 3,364 3,721 23,892

36 66 120 164 200 330 420 432 522 610 2,900

1 2 3 4 5 6 7 8 9 10 55

Given: Y = a + bX where: b=

 XY − nXY  X 2 − nX 2

a = Y − bX 2

and ΣX = 55, ΣY = 478, ΣXY = 2900, ΣX = 385, ΣY = 23892, X = 5.5, Y = 47.8, Then: 2,900 − 2,629 271 = = = 3.28 385 − 302.5 82.5 385 − 10 × 5.52 a = 47.8 − 3.28 × 5.5 = 29.76 b=

2,900 − 10 × 5.5 × 47.8

and Y = 29.76 + 3.28X. For:

X = 11: Y = 29.76 + 3.28 × 11 = 65.8 X = 12: Y = 29.76 + 3.28 × 12 = 69.1 Therefore: Year 11 → 65.8 patients Year 12 → 69.1 patients The model “seems” to fit the data pretty well. One should, however, be more precise in judging the adequacy of the model. Two possible approaches are computation of (a) the correlation coefficient, or (b) the mean absolute deviation. The correlation coefficient: r=

n  XY −  X  Y  n  X 2 − (  X )2   n  Y 2 − (  Y )2      10 × 2,900 − 55 × 478 10 × 385 − 552  10 × 23,892 − 4782     29,000 − 26, 290

238,920 − 228,484 3,850 − 3,025 2,710 2,710 = = = 0.924 825 × 10,436 2,934.3 r 2 = 0.853 The coefficient of determination of 0.853 is quite respectable—indicating our original judgment of a “good” fit was appropriate.

CHAPTER 4

Year X

Patients Y

Trend Forecast

1 2 3 4 5 6 7 8 9 10

36 33 40 41 40 55 60 54 58 61

29.8 + 3.28 × 1 = 33.1 29.8 + 3.28 × 2 = 36.3 29.8 + 3.28 × 3 = 39.6 29.8 + 3.28 × 4 = 42.9 29.8 + 3.28 × 5 = 46.2 29.8 + 3.28 × 6 = 49.4 29.8 + 3.28 × 7 = 52.7 29.8 + 3.28 × 8 = 56.1 29.8 + 3.28 × 9 = 59.3 29.8 + 3.28 × 10 = 62.6

FORECASTING

Absolute Deviation Deviation 2.9 – 3.3 0.4 – 1.9 – 6.2 5.6 7.3 – 2.1 – 1.3 – 1.6

2.9 3.3 0.4 1.9 6.2 5.6 7.3 2.1 1.3 1.6

Σ = 32.6 MAD = 3.26 The MAD is 3.26—this is approximately 7% of the average number of patients and 10% of the minimum number of patients. We also see absolute deviations, for years 5, 6, and 7 in the range 5.6–7.3. The comparison of the MAD with the average and minimum number of patients and the comparatively large deviations during the middle years indicate that the forecast model is not exceptionally accurate. It is more useful for predicting general trends than the actual number of patients to be seen in a specific year. 4.31

(a) Week

Calls

1 2 3 4 5 6 7

50 35 25 40 45 35

Forecast Exponential Smoothing α = 0.20

Deviation

50.00 50.00 + 0.20(50 − 50.00) = 50.00 50.00 + 0.20(35 − 50.00) = 47.00 47.00 + 0.20(25 − 47.00) = 42.60 42.60 + 0.20(40 − 42.60) = 42.08 42.08 + 0.20(45 − 42.08) = 42.66 42.66 + 0.20(35 − 42.66) = 41.13

0.00 −15.00 −22.00 −2.60 2.92 −7.66

(b) Week

Calls

1 2 3 4 5 6 7

50 35 25 40 45 35

Forecast Exponential Smoothing α = 0.60

Deviation

50.00 50.00 + 0.60(50 − 50.00) = 50.00 50.00 + 0.60(35 − 50.00) = 41.00 41.00 + 0.60(25 − 41.00) = 31.40 31.40 + 0.60(40 − 31.40) = 36.56 36.56 + 0.60(45 − 36.56) = 41.62 41.62 + 0.60(35 − 42.63) = 37.65

0.00 −15.00 −16.00 8.60 8.44 −6.62

(c) MAD for α = 20% = (0.00 + 15.00 + 22.00 + 2.60 + 2.92 + 7.66) / 6 = 50.18 / 6 = 8.36 MAD for α = 60% = (0.00 + 15.00 + 16.00 + 8.60 + 8.44 + 6.62) / 6 = 54.66 / 6 = 9.11 Although the MAD is lower for α = 20%, the forecast using α = 60% is closer to the actual. 4.32 Week t

Actual Value At

Smoothed Value Ft (α = 0.30)

Trend Estimate Tt ( β = 0.20)

Forecast FITt

Forecast Error

1 2 3 4 5 6 7

50 35 25 40 45 35

50.00 50.00 45.50 38.72 37.65 38.54 36.55

0.00 0.00 – 0.90 –2.08 –1.88 –1.33 –1.46

50.00 50.00 44.60 36.64 35.77 37.21 35.09

0.00 –15.00 –19.60 3.36 9.23 –2.21

CHAPTER 4 F O R E C A S T I N G

The MAD = (0.00 + 15.00 + 19.60 + 3.36 + 9.23 + 2.21) / 6 = 8.23. To evaluate the trend adjusted exponential smoothing model, actual week 7 calls, which were 33, are compared to the forecast value. The model appears to be producing a MAD close to that given by simple exponential smoothing using α = 0.20 and α = 0.60, but its forecast is closer to actual. 4.33 (a) There is not a strong linear trend in sales over time. (b, c) Bob wants to forecast by exponential smoothing (setting February’s forecast equal to January’s sales) with alpha = 0.1. Sherry wants to use a 3-period moving average. January February March April May

Sales

Bob

Sherry

Bob’s Error

400 380 410 375 405

— 400 398 399.2 396.8

— — — 396.67 388.33 MAD =

— 20.0 12.0 24.2 8.22 16.11

Sherry’s Error — — — 21.67 16.67 19.17

(d) Note that Bob has more forecast observations, while Sherry’s moving average does not start until month 4. Also note that the MAD for Bob is an average of 4 numbers, while Sherry’s is only 2. Bob’s MAD for exponential smoothing (16.11) is lower than that of Sherry’s moving average (19.17). So his forecast seems to be better. 4.34

[(0.4×8) + (0.3×13) + (0.2×9) + (0.1×5)] / (0.4 + 0.3 + 0.2 + 0.1) = (3.2 + 3.9 + 1.8 + 0.5) / 1.0 = 9.4 / 1/0 = 9.4

4.35

(a) (b) (c)

Week

Forecast

Registration Naive 2-week moving 4-week moving

21 22

25 21 21.5

27 25 23

35 29 27 35 26 31 23.75 27

33 29 32 29

37 33 31 31

41 37 35 33.5

37 41 39 35

37 39 37

(d) Based on the MAD, the naive forecast has performed better because it has the smallest MAD of 4.11. Based on the MSE, the naive forecast has performed better because it has the smallest MSE of 20.6. 4.36 Period

Demand

Exponentially Smoothed Forecast

1 2 3 4 5 6 7

7 9 5 9 13 8 Forecast

5 5 + 0.2 × (7 − 5) = 5.4 5.4 + 0.2 × (9 − 5.4) = 6.12 6.12 + 0.2 × (5 − 6.12) = 5.90 5.90 + 0.2 × (9 − 5.90) = 6.52 6.52 + 0.2 × (13 − 6.52) = 7.82 7.82 + 0.2 × (8 − 7.82) = 7.86

CHAPTER 4

FORECASTING

4.37 2

Actual

Forecast

|Error|

Error

95 108 123 130

100 110 120 130

5 2 3 0 10

25 4 9 0 38

MAD = 10/4 = 2.5, MSE = 38/4 = 9.5 4.38 (a) 3-Month Moving Average: Month

3-Month Moving Average

Absolute Deviation

(11 + 14 + 16)/3 = 13.67 (14 + 16 + 10)/3 = 13.33 (16 + 10 + 15)/3 = 13.67 (10 + 15 + 17)/3 = 14.00 (15 + 17 + 11)/3 = 14.33 (17 + 11 + 14)/3 = 14.00 (11 + 14 + 17)/3 = 14.00 (14 + 17 + 12)/3 = 14.33 (17 + 12 + 14)/3 = 14.33 (12 + 14 + 16)/3 = 14.00 (14 + 16 + 11)/3 = 13.67

3.67 1.67 3.33 3.00 0.33 3.00 2.00 0.33 1.67 3.00

Sales

January February March April May June July August September October November December January February

11 14 16 10 15 17 11 14 17 12 14 16 11

Σ = 22.00 MAD = 2.20 (b) 3-Month Weighted Moving Average Month

Sales

January February March April May June July August September October November December January February

11 14 16 10 15 17 11 14 17 12 14 16 11

3-Month Weighted Moving Average

Absolute Deviation

(1 × 11 + 2 × 14 + 3 × 16)/6 = 14.50 (1 × 14 + 2 × 16 + 3 × 10)/6 = 12.67 (1 × 16 + 2 × 10 + 3 × 15)/6 = 13.50 (1 × 10 + 2 × 15 + 3 × 17)/6 = 15.17 (1 × 15 + 2 × 17 + 3 × 11)/6 = 13.67 (1 × 17 + 2 × 11 + 3 × 14)/6 = 13.50 (1 × 11 + 2 × 14 + 3 × 17)/6 = 15.00 (1 × 14 + 2 × 17 + 3 × 12)/6 = 14.00 (1 × 17 + 2 × 12 + 3 × 14)/6 = 13.83 (1 × 12 + 2 × 14 + 3 × 16)/6 = 14.67 (1 × 14 + 2 × 16 + 3 × 11)/6 = 13.17

4.50 2.33 3.50 4.17 0.33 3.50 3.00 0.00 2.17 3.67

Σ = 27.17 MAD = 2.72 (c) Based on a mean absolute deviation criterion, the 3-month moving average with MAD = 2.2 is to be preferred over the 3-month weighted moving average with MAD = 2.72. (d) Interest rates, business cycles, or seasonal factors might also be considered.

4.39

CHAPTER 4 F O R E C A S T I N G

7 9 5 11 10 13 55

1 2 3 4 5 6 21

1 4 9 16 25 36 91

xy 7 18 15 44 50 78 212

y = 9.17 x = 3.5 y = 5.27 + 1.11x Period 7 forecast = 13.07 Period 12 forecast = 18.64, but this is far outside the range of valid data. 4.40 To compute seasonalized or adjusted sales forecast, we just multiply each seasonalized index by the appropriate trend forecast. Yˆ = Index × Yˆ Seasonal

Trend forecast

Hence, for Quarter I: YˆI = 1.25 × 120,000 = 150,000 Quarter II: YˆII = 0.90 × 140,000 = 126,000 Quarter III: Yˆ = 0.75 × 160,000 = 120,000 III

Quarter IV: YˆIV = 1.10 × 180,000 = 198,000 4.41 Week 1 Week 2 Week 3 Week 4 Averages

Mon.

Tue.

Wed.

Thu.

Fri.

Sat.

210 215 220 225 217.5

178 180 176 178 178

250 250 260 260 255

215 213 220 225 218.3

160 165 175 176 169

180 185 190 190 186.3

Overall average = 204

(a) Seasonal indices: 1.066 (Mon) 0.873 (Tue) 1.25 (Wed) 1.07 (Thu) 0.828 (Fri) 0.913 (Sat)

(b) To calculate for Monday of Week 5 = 201.74 + 0.18(25) × 1.066 = 219.9 rounded to 220 Forecast

4.42

220 (Mon) 180 (Tue) 258 (Wed) 221 (Thu) 171 (Fri) 189 (Sat)

(a) $4,000 + $0.20(15,000) = $7,000 (b) $4,000 + $0.20(25,000) = $9,000

4.43

(a) Graph of demand

The observations obviously do not form a straight line but do tend to cluster about a straight line over the range shown.

CHAPTER 4

FORECASTING

(b) Least-squares regression: Y = a + bX b=

 XY − nXY  X 2 − nX 2

a = Y − bX Assume Appearances X

Demand Y

3 4 7 6 8 5 9

3 6 7 5 10 7 ?

9 16 49 36 64 25

9 36 49 25 100 49

XY 9 24 49 30 80 35

ΣX = 33, ΣY = 38, ΣXY = 227, ΣX = 199, X = 5.5, Y = 6.33. Therefore: 227 – 6 × 5.5 × 6.333 = 1.0286 199 – 6 × 5.5 × 5.5 a = 6.333 – 1.0286 × 5.5 = .6762

Y = .676 + 1.03 x (rounded) The following figure shows both the data and the resulting equation:

(c) If there are nine performances by Maroon 5, the estimated sales are: Y9 = .676 + 1.03 × 9 = .676 + 9.27 = 9.93 guitars ≈ 10 guitars 2

(d) R = .82 is the correlation coefficient, and R = .68 means 68% of the variation in sales can be explained by TV appearances.

4.44

Given Y = 36 + 4.3X (a)

Y = 36 + 4.3(70) = 337

(b)

Y = 36 + 4.3(80) = 380

(c)

CHAPTER 4 F O R E C A S T I N G

4.45

Let x1, x2 ,  , x6 be the prices and y1, y2 ,  , y6 be the number sold. 6

 xi

X = i =1 6

= Average price = 3.2583

 yi

Y = i =1 6

= Average number sold = 550

 xi yi = 9,783

i =1 6

 xi = 67.1925

i =1

Then y = a + bx, where y = number sold, x = price, and 6

 xi yi − nx y

b = i =61 = 2 2  x i − nx

(9,783) − 6(3.25833)(550) 67.1925 − 6(3.25833) 2

i =1

−969.489 = = −277.6 3.49222 a = y − bx = 550 − [(−277.6)(3.25)] = 1, 454.6 So at x = 2.80, y = 1,454.6 − 277.6($2.80) = 677.32. Now round to the nearest integer: Answer: 677 lattes.

4.46

(a)

16 12 18 14

330 270 380 300

5,280 3,240 6,840 4,200

256 144 324 196

1,280

19,560

920

60 = 15 4 1,280 y= = 320 4  xy − nx y 19,560 − 4(15)(320) 360 b= = = = 18 20  x 2 − nx 2 920 − 4(15)2 a = y − bx = 320 − 18(15) = 50

Y = 50 + 18 x (b) If the forecast is for 20 guests, the bar sales forecast is 50 + 18(20) = $410. Each guest accounts for an additional $18 in bar sales. 4.47

Y = 7.5 + 3.5X1 + 4.5X2 + 2.5X3 (a) 28 (b) 43 (c) 58

4.48

(a) Ŷ = 13,473 + 37.65(1860) = 83,502 (b) The predicted selling price is $83,502, but this is the average price for a house of this size. There are other factors besides square footage that will impact the selling price of a house. If such a house sold for $95,000, then these other factors could be contributing to the additional value. Copyright ©2023 Pearson Education, Inc.

CHAPTER 4

FORECASTING

(c) Some other quantitative variables would be age of the house, number of bedrooms, size of the lot, size of the garage, etc. 2

(d) Coefficient of determination = (0.63) = 0.397. This means that only about 39.7% of the variability in the sales price of a house is explained by this regression model that only includes square footage as the explanatory variable. 4.49

(a) Given: Y = 90 + 48.5X1 + 0.4X2 where: Y = expected travel cost X1 = number of days on the road X2 = distance traveled, in miles r = 0.68 (coefficient of correlation) If: Number of days on the road → X1 = 5 and distance traveled → X2 = 300 then: Y = 90 + 48.5 × 5 + 0.4 × 300 = 90 + 242.5 + 120 = 452.5 Therefore, the expected cost of the trip is $452.50. (b) The reimbursement request is much higher than predicted by the model. This request should probably be questioned by the accountant. (c) A number of other variables should be included, such as: 1. The type of travel (air or car) 2. Conference fees, if any 3. Costs of entertaining customers 4. Other transportation costs—cab, limousine, special tolls, or parking

In addition, the correlation coefficient of 0.68 is not exceptionally high. It indicates that the model explains approximately 46% of the overall variation in trip cost. This correlation coefficient would suggest that the model is not a particularly good one. 4.50

(a) Least-squares equation: Y = − 0.158 + 0.1308X (b) Y = − 0.158 + 0.1308(22) = 2.719 million (c) Coefficient of correlation = r = 0.966 Coefficient of determination = r2 = 0.934

4.51 Year

Crime Rate X

Patients Y

1 2 3 4 5 6 7 8 9 10 Column totals

58.3 61.1 73.4 75.7 81.1 89.0 101.1 94.8 103.3 116.2 854.0

36 33 40 41 40 55 60 54 58 61 478

3,398.9 1,296 2,098.8 3,733.2 1,089 2,016.3 5,387.6 1,600 2,936.0 5,730.5 1,681 3,103.7 6,577.2 1,600 3,244.0 7,921.0 3,025 4,895.0 10,221.2 3,600 6,066.0 8,987.0 2,916 5,119.2 10,670.9 3,364 5,991.4 13,502.4 3,721 7,088.2 76,129.9 23,892 42,558.6

Given: Y = a + bX where b=

 XY − nXY  X 2 − nX 2

a = Y − bX 2

and ΣX = 854, ΣY = 478, ΣXY = 42558.6, ΣX = 76129.9, ΣY = 23892, X = 85.4, Y = 47.8. Then:

CHAPTER 4 F O R E C A S T I N G

42558.6 − 10 × 85.4 × 47.8 2

76129.9 − 10 × 85.4 1737.4 = = 0.543 3197.3 a = 47.8 − 0.543 × 85.4 = 1.43 and

42558.6 − 40821.2 76129.9 − 72931.6

Y = 1.43 + 0.543X

For: X = 131.2 : Y = 1.43 + 0.543(131.2) = 72.7 X = 90.6 : Y = 1.43 + 0.543(90.6) = 50.6 Therefore: Crime rate = 131.2 → 72.7 patients Crime rate = 90.6 → 50.6 patients Note that rounding differences occur when solving with Excel. Also note that a crime rate of 131.2 is outside the range of the data set used to determine the regression equations, so caution is advised. (a) It appears from the following graph that the points do scatter around a straight line.

4.52

(b) Developing the regression relationship, we have the following: (Summer months) Year 1 2 3 4 5 6 7 8 9 10 11 12

Tourists (Millions) (X)

Ridership (100,000’s) (Y)

7 2 6 4 14 15 16 12 14 20 15 7

1.5 1.0 1.3 1.5 2.5 2.7 2.4 2.0 2.7 4.4 3.4 1.7

49 4 36 16 196 225 256 144 196 400 225 49

2.25 1.00 1.69 2.25 6.25 7.29 5.76 4.00 7.29 19.36 11.56 2.89

10.5 2.0 7.8 6.0 35.0 40.5 38.4 24.0 37.8 88.0 51.0 11.9

Given: Y = a + bX where:

 XY − nXY  X 2 − nX 2

a = Y − bX 2

and ΣX = 132, ΣY = 27.1, ΣXY = 352.9, ΣX = 1,796, ΣY = 71.59, X = 11, Y = 2.26. Then:

352.9 − 12 × 11 × 2.26 2

352.9 − 298.3 54.6 = = 0.159 1,796 − 1, 452 344

1,796 − 12 × 11 a = 2.26 − 0.159 × 11 = 0.511

CHAPTER 4

and

FORECASTING

Y = 0.511 + 0.159X (c) Given a tourist population of 10,000,000, the model predicts a ridership of: Y = 0.511 + 0.159 × 10 = 2.101, or 2,101,000 persons. (d) If there are no tourists at all, the model predicts a ridership of 0.511, or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists (zero) is outside the range of data used to develop the model. (e) The standard error of the estimate is given by: S yx = =

 Y 2 − a  Y − b  XY n−2 71.59 − 0.511 × 27.1 − 0.159 × 352.9 12 − 2

71.59 − 13.85 − 56.11 = .163 10 = .404 (rounded to .407 in POM for Windows software) =

(f) The correlation coefficient and the coefficient of determination are given by: r=

= =

n  XY −  X  Y  n  X 2 − (  X )2   n  Y 2 − (  Y )2      12 × 352.9 − 132 × 27.1 12 × 1796 − 132 2  12 × 71.59 − 27.12     4234.8 − 3577.2  21552 − 17424  859.08 − 734.41 657.6 657.6 = = 0.917 × 11.166 64.25 4128 × 124.67

and r 2 = 0.840 4.53 (a)

421 2.90 377 2.93 585 3.00 690 3.45 608 3.66 390 2.88 415 2.15 481 2.53 729 3.22 501 1.99 613 2.75 709 3.90 366 1.60 Column totals 6,885 36.96

177,241 8.41 142,129 8.58 342,225 9.00 476,100 11.90 369,664 13.40 152,100 8.29 172,225 4.62 231,361 6.40 531,441 10.37 251,001 3.96 375,769 7.56 502,681 15.21 133,956 2.56 3,857,893 110.26

XY 1220.9 1104.6 1755.0 2380.5 2225.3 1123.2 892.3 1216.9 2347.4 997.0 1685.8 2765.1 585.6 20,299.5

Given: Y = a + bX where: b=

 XY − nXY  X 2 − nX 2

a = Y − bX 2

and ΣX = 6885, ΣY = 36.96, ΣXY = 20299.5, ΣX = 3857893, ΣY = 110.26, X = 529.6, Y = 2.843. Then: b=

20, 299.5 − 13 × 529.6 × 2.843

3,857,893 − 13 × 529.62 726 = = 0.0034 211,703 a = 2.84 − 0.0034 × 529.6 = 1.03

20, 299.5 − 19,573.5 3,857,893 − 3,646,190

CHAPTER 4 F O R E C A S T I N G

and Y = 1.03 + 0.0034X As an indication of the usefulness of this relationship, we can calculate the correlation coefficient:

n  XY −  X  Y  n  X 2 − (  X )2   n  Y 2 − (  Y )2      13 × 20, 299.5 − 6,885 × 36.96 13 × 3,857,893 − 6,8852  13 × 110.26 − 36.962     263,893.5 − 254,469.6

1, 433.4 − 1,366.0 50,152,609 − 47, 403, 225 9423.9 = 2,749,384 × 67.0 =

9423.9 = 0.692 1, 658.13 × 8.21

r 2 = 0.479 A correlation coefficient of 0.692 is not particularly high. The coefficient of determination, r2, indicates that the model explains only 47.9% of the overall variation. Therefore, although the model does provide an estimate of GPA, there is considerable variation in GPA, which is as yet unexplained. For (b) X = 350: Y = 1.03 + 0.0034 × 350 = 2.22 (c) X = 800: Y = 1.03 + 0.0034 × 800 = 3.75 Note: When solving this problem, care must be taken to interpret significant digits. Also note that X = 800 is outside the range of the data set used to determine the regression relationship, so caution is advised.

4.54

(a)

Quarter

Contracts X

1 2 3 4 5 6 7 8 Totals Average

153 172 197 178 185 199 205 226 1,515 189.375

Sales Y

8 23,409 10 29,584 15 38,809 9 31,684 12 34,225 13 39,601 12 42,025 16 51,076 95 290,413 11.875

64 100 225 81 144 169 144 256 1,183

XY 1,224 1,720 2,955 1,602 2,220 2,587 2,460 3,616 18,384

b = (18,384 − 8 × 189.375 × 11.875)/(290,413 − 8 × 189.375 × 189.375) = 0.1121 a = 11.875 − 0.1121 × 189.375 = −9.3495 Sales (y) = − 9.349 + 0.1121 (Contracts) (b) r = (8 × 18,384 − 1,515 × 95) ((8 × 290, 413 − 1,5152 )(8 × 1,183 − 952 )) = 0.8963 Sxy = 1,183 − (−9.3495 × 95) − (0.112 × 18,384 / 6) = 1.3408

4.55 (a) 35 + 20(80) + 50(3.0) = 1,785 (b) 35 + 20(70) + 50(2.5) = 1,560

CHAPTER 4

FORECASTING

4.56 Given: ΣX = 15, ΣY = 20, ΣXY = 70, ΣX2 = 55, ΣY2 = 130, X = 3, Y = 4  XY − nXY (a) b=  X 2 − nX 2 a = Y − bX b=

70 − 5 × 3 × 4 55 − 5 × 32

70 − 60 10 = =1 55 − 45 10

a = 4 − 1× 3 = 4 − 3 = 1 Y = 1 + 1X (b) Correlation coefficient:

n  XY −  X  Y  n  X 2 − (  X )2   n  Y 2 − (  Y )2      5 × 70 − 15 × 20

5 × 55 − 152  5 × 130 − 20 2     350 − 300 50 = = 50 × 250  275 − 225 650 − 400  50 = = 0.45 111.80 The correlation coefficient indicates that there is a positive correlation between bank deposits and consumer price indices in Birmingham, Alabama—i.e., as one variable tends to increase (or decrease), the other tends to increase (or decrease). (c) Standard error of the estimate:

 Y 2 − a  Y − b  XY = n−2

S yx =

130 − 20 − 70 = 3

130 − 1 × 20 − 1 × 70 3

40 = 13.3 = 3.65 3

4.57

Column totals

2 1 4 5 3

4 1 4 6 5

4 1 16 25 9

16 1 16 36 25

8 1 16 30 15

Given: Y = a + bX where: b=

 XY − nXY  X 2 − nX 2

a = Y − bX 2

and ΣX = 15, ΣY = 20, ΣXY = 70, ΣX = 55, ΣY = 94, X = 3, Y = 4, then: b=

70 − 5 × 4 × 3

55 − 5 × 32 a = 4 − 1 × 3 = 1.0

70 − 60 = 1.0 55 − 45

and Y = 1.0 + 1.0X. The correlation coefficient:

CHAPTER 4 F O R E C A S T I N G

n  XY −  X  Y  n  X 2 − (  X )2   n  Y 2 − (  Y )2      5 × 70 − 15 × 20 350 − 300 = 2 2 5 × 55 − 15  5 × 94 − 20   275 − 225  470 − 400     50 50 = = 0.845 50 × 70 59.16

Standard error of the estimate:

S yx = =

 Y 2 − a  Y − b  XY = n−2

94 − 1 × 20 − 1 × 70 5−2

94 − 20 − 70 = 1.333 = 1.15 3

4.58

Using software, the regression equation is: Games lost = 6.41 + 0.533 × days rain.

4.59

(a, b)

Period

Demand Forecast

1 2 3 4 5 6 7 8 9 10

20 21 28 37 25 29 36 22 25 28

Error Running Sum

20 20 20.5 24.25 30.63 27.81 28.41 32.20 27.11 26.05

0.00 1.00 7.50 12.75 −5.63 1.19 7.59 −10.20 −2.10 1.95

0.00 1.00 8.50 21.25 15.63 16.82 24.41 14.21 12.10 14.05

|Error|

0.00 1.00 7.50 12.75 5.63 1.19 7.59 10.20 2.10 1.95 MAD ≈ 5.00

 ( At − Ft )

4.60

Tracking signal = t =1

MAD

Month

May June July August September October November December

100 80 110 115 105 110 125 120

100 104 99 101 104 104 105 109

|At – Ft | 0 24 11 14 1 6 20 11 Sum: 87

(At – Ft ) 0 −24 11 14 1 6 20 11 Sum: 39

87 = 10.875 8 39 Tracking signal = = 3.586 10.875

So: MAD:

CHAPTER 4

FORECASTING

4.61 (a) Week 1 2 3 4 5 6 7 8 9 10 11 12

Actual Cumulative Cum. Tracking Miles Forecast Error Error Σ |Error| MAD Signal 17 21 19 23 18 16 20 18 22 20 15 22

17.00 17.00 17.80 18.04 19.03 18.83 18.26 18.61 18.49 19.19 19.35 18.48

0.00 +4.00 +1.20 +4.96 −1.03 –2.83 +1.74 −0.61 +3.51 +0.81 −4.35 +3.52

– 4.00 5.20 10.16 9.13 6.30 8.04 7.43 10.94 11.75 7.40 10.92

0.00 4.00 5.20 10.16 11.19 14.02 15.76 16.37 19.88 20.69 25.04 28.56

0 2 1.73 2.54 2.24 2.34 2.25 2.05 2.21 2.07 2.28 2.38

2 3 4 4 2.7 3.6 3.6 5 5.7 3.2 4.6

(b) The MAD = 28.56/12 = 2.38 (c) The cumulative error and tracking signals appear to be consistently positive, and at week 10, the tracking signal exceeds 5 MADs.

CASE STUDY SOUTHWESTERN UNIVERSITY: (B) This is the second in a series of integrated case studies that run throughout the text. 1. One way to address the case is with separate forecasting models for each game. Clearly, the homecoming game (week 2) and the fourth game (craft festival) are unique attendance situations.

Game

Model

1 2 3 4 5 Total

y = 30,713 + 2,534x y = 37,640 + 2,146x y = 36,940 + 1,560x y = 22,567 + 2,143x y = 30,440 + 3,146x

Forecasts 2022 2023 48,453 52,660 47,860 37,567 52,460 239,000

50,988 54,806 49,420 39,710 55,606 250,530

0.92 0.90 0.91 0.88 0.93

where y = attendance and x = time. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 2.

Revenue in 2022 = (239,000) ($50/ticket) = $11,950,000 Revenue in 2023 = (250,530) ($52.50/ticket) = $13,152,825

LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 3. In games 2 and 5, the forecast for 2023 exceeds stadium capacity. With this appearing to be a continuing trend, the time has come for a new or expanded stadium. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking

CHAPTER 4 F O R E C A S T I N G

VIDEO CASE STUDIES 1

FORECASTING TICKET REVENUE FOR ORLANDO MAGIC BASKETBALL GAMES

There is a short video (10.5 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. The case is a great example of the real-world use of multiple regression in an industry (professional sports) that is of interest to many students. The Orlando Magic team is a leader in using business analytic tools to optimize revenue. 1. Regression model using “day of the week” as independent variable:

Revenue = – $15,944 + $24,334x,

R2 = .23

LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking 2. Regression model using “rating of the opponent” as independent variable:

Revenue = $19,546 + $22,160x,

R2 = .77

LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking 3. Using the multiple regression model in the case:

Revenue = $14,996 + $10,801 (4) + $23,379 (3) + $10,784 (3) = $160,743 where x1 = day of week = 4, x2 = Miami Heat rating = 3, and x3 = Christmas season = 3. (Note: the solution assumes that Thursday was a working day. Some students may consider it a holiday because it was “during the Christmas season.” In that case, x1 would be 3, and the forecast would be $149,942.) LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking 4. Time of day for game, other competing sports events within 100 miles on that date, special half-time or pregame entertainment planned, date set for a special group night (for example, Boy Scouts or Rotary). These may be potential independent variables for Perez’s model. LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge

FORECASTING AT HARD ROCK CAFE

There is a short video (8 minutes) available in MyLab Operations Management that is filmed specifically for this text and supplements this case. 1. Hard Rock case uses forecasting for (1) sales (guest counts) at cafes, (2) retail sales, (3) banquet sales, (4) concert sales, (5) evaluating managers, and (6) menu planning. They could also employ these techniques to forecast: retail store sales of individual (SKU) product demands; sales of each entrée; sales at each workstation, etc. LO 4.1: Understand the three time horizons and which models apply for each AACSB: Application of knowledge 2. The POS system captures all the basic sales data needed to drive individual cafe’s scheduling/ordering. It also is aggregated at corporate HQ. Each entrée sold is counted as one guest at a Hard Rock Cafe. LO 4.1: Understand the three time horizons and which models apply for each AACSB: Information technology

CHAPTER 4

FORECASTING

3. The weighting system is subjective, but is reasonable. More weight is given to each of the past 2 years than to 3 years ago. This system actually protects managers from large sales variations outside their control. One could also justify a 50%–30%–20% model or some other variation. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 4. Other predictors of cafe sales could include season of year (weather); hotel occupancy; spring break from colleges; beef prices; promotional budget; etc. LO 4.6: Conduct a regression and correlation analysis AACSB: Application of knowledge

Y = a + bx

Month 1 2 3 4 5 6 7 8 9 10 Totals Average

Advertising X Guest Count Y 14 17 25 25 35 35 45 50 60 60 366 36.6

21 24 27 32 29 37 43 43 54 66 376 37.6

15,772 − 10 × 36.6 × 37.6 15,910 − 10 × 36.6 2

196 441 294 289 576 408 625 729 675 625 1,024 800 1,225 841 1,015 1,225 1,369 1,295 2,025 1,849 1,935 2,500 1,849 2,150 3,600 2,916 3,240 3,600 4,356 3,960 15,910 15,950 15,772

= 0.7996 ≈ .8

a = 37.6 − 0.7996 × 36.6 = 8.3363 ≈ 8.3 Y = 8.3363 + 0.7996 X At $65,000; y = 8.3 + .8 (65) = 8.3 + 52 = 60.3, or 60,300 guests. For the instructor who asks other questions than this one: 2 r = 0.8869 Std. error = 5.062 LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking

ADDITIONAL CASE STUDIES (available in MyLab Operations Management) 1

THE NORTH-SOUTH AIRLINES Northern Airlines Data

Year

Airframe Cost per Aircraft

Engine Cost per Aircraft

Average Age (hrs)

2015 2016 2017 2018 2019 2020 2021

51.80 54.92 69.70 68.90 63.72 84.73 78.74

43.49 38.58 51.48 58.72 45.47 50.26 79.60

6,512 8,404 11,077 11,717 13,275 15,215 18,390

CHAPTER 4 F O R E C A S T I N G

Southeast Airlines Data Year

Airframe Cost per Aircraft

Engine Cost per Aircraft

Average Age (hrs)

2015 2016 2017 2018 2019 2020 2021

13.29 25.15 32.18 31.78 25.34 32.78 35.56

18.86 31.55 40.43 22.10 19.69 32.58 38.07

5,107 8,145 7,360 5,773 7,150 9,364 8,259

Utilizing the software package provided with this text, we can develop the following regression equations for the variables of interest: Northern Airlines—Airframe Maintenance Cost:  Cost = 36.10 + 0.0026 × Airframe age  Coefficient of determination = 0.7695  Coefficient of correlation = 0.8772 Northern Airlines—Engine Maintenance Cost:  Cost = 20.57 + 0.0026 × Airframe age  Coefficient of determination = 0.6124  Coefficient of correlation = 0.7825 Southeast Airlines—Airframe Maintenance Cost:  Cost = 4.60 + 0.0032 × Airframe age  Coefficient of determination = 0.3905  Coefficient of correlation = 0.6249 Southeast Airlines—Engine Maintenance Cost:  Cost = – 0.67 + 0.0041 × Airframe age  Coefficient of determination = 0.4600  Coefficient of correlation = 0.6782 The following graphs portray both the actual data and the regression lines for airframe and engine maintenance costs for both airlines.

CHAPTER 4

FORECASTING

Note that the two graphs have been drawn to the same scale to facilitate comparisons between the two airlines. Comparison:  Northern Airlines: There seem to be modest correlations between maintenance costs and airframe age for Northern Airlines. There is certainly reason to conclude, however, that airframe age is not the only important factor.  Southeast Airlines: The relationships between maintenance costs and airframe age for Southeast Airlines are much less well defined. It is even more obvious that airframe age is not the only important factor—perhaps not even the most important factor. Overall, it would seem that:  Northern Airlines has the smaller variance in maintenance costs—indicating that its day-to-day management of maintenance is working pretty well.  Maintenance costs seem to be more a function of airline than of airframe age.  The airframe and engine maintenance costs for Southeast Airlines are not only lower, but more similar than those for Northern Airlines. From the graphs, at least, they appear to be rising more sharply with age.  From an overall perspective, it appears that Southeast Airlines may perform more efficiently on sporadic or emergency repairs, and Northern Airlines may place more emphasis on preventive maintenance.

Ms. Young’s report should conclude that:  There is evidence to suggest that maintenance costs could be made to be a function of airframe age by implementing more effective management practices.  The difference between maintenance procedures of the two airlines should be investigated.  The data with which she is currently working does not provide conclusive results. Concluding Comment: The question always arises, with this case, as to whether the data should be merged for the two airlines, resulting in two regressions instead of four. The solution provided is that of the consultant who was hired to analyze the data. The airline’s own internal analysts also conducted regressions, but did merge the data sets. This shows how statisticians can take different views of the same data. LO 4.6: Conduct a regression and correlation analysis AACSB: Analytical thinking

CHAPTER 4 F O R E C A S T I N G

DIGITAL CELL PHONE, INC.

Objectives:  

Selection of an appropriate time-series forecasting model, based on a plot of the data. The importance of combining a qualitative model with a quantitative model in situations where technological change is occurring.

1. A plot of the data indicates a linear trend (least squares) model might be appropriate for forecasting. Using linear trend you obtain the following: 2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

480 436 482 448 458 489 498 430 444 496 487 525 575 527 540 502 508 573 508 498 485 526 552 587 608 597 612 603 628 605 627 578 585 581 632 656

1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1,024 1,089 1,156 1,225 1,296

666

19,366

16,206

378,661

10,558,246

Average 18.5

537.9

450.2

10,518.4

Totals

 xy − nx y 2

480

230,400

872 1,446 1,792 2,290 2,934 3,486 3,440 3,996 4,960 5,357 6,300 7,475 7,378 8,100 8,032 8,636 10,314 9,652 9,960 10,185 11,572 12,696 14,088 15,200 15,522 16,524 16,884 18,212 18,150 19,437 18,496 19,305 19,754 22,120 23,616

190,096 232,324 200,704 209,464 239,121 248,004 184,900 197,136 246,016 237,169 275,625 330,625 277,729 291,600 252,004 258,064 328,329 258,064 248,004 235,225 276,676 304,704 344,569 369,664 356,409 374,544 363,609 394,384 366,025 393,129 334,084 342,225 337,561 399,424 430,336

378,661 − 36 × 18.5 × 537.9 2

 x − nx 16, 206 − (36 × 18.5 ) a = y − bx = 537.9 − 5.25 × 18.5 = 440.85

293,284.6

20390.0 = 5.25 3885.0

CHAPTER 4

r= = =

FORECASTING

n  xy −  x  y 2

[ n  x − (  x )2 ][ n  y 2 − ( y)2 ] (36)(378,661) − (666)(19,366) [(36) × (16, 206) − (666)2 ][(36)(10,558,246) − (19,366)2 ] 13,631,796 − 12,897,756

[(583, 416) − (443,556)][380,096,856) − (375,041,956)] 737,040 734,040 = = [139,860][5,054,900] 706,978,314,000 734,040 = = .873 840,820 r 2 = .76 y = 440.85 + 5.25 (time) r = 0.873, indicating a reasonably good fit The student should report the linear trend results, but deflate the forecast obtained based on qualitative information about industry and technology trends. 2 Because there is limited seasonality in the data, the linear trend analysis above provides a good r of .76. However, a more precise forecast can be developed addressing the seasonality issue, which is done below. Methods a and c yield r2 of .85 2 and .86, respectively, and methods b and d, which also center the seasonal adjustment, yield r of .93 and .94, respectively. LO 4.3: Apply the naive, moving-average, exponential smoothing, and trend methods AACSB: Analytical thinking 2. Four approaches to decomposition of the Digital Cell Phone data can address seasonality, as follows: (a) Multiplicative seasonal model, 2 Cases = 443.87 + 5.08 (time), r = .85, MAD = 20.89 (b) Multiplicative Seasonal Model, with centered moving averages (CMA), which is not covered in our text but can be seen in Render, Stair, Hanna, and Hall’s Quantitative Analysis for Management 2 Cases = 432.28 + 5.73 (time), r = .93, MAD = 12.84 (c) Additive seasonal model, 2 Cases = 444.29 + 5.06 (time), r = .86, MAD = 20.02 (d) Additive seasonal model, with centered moving averages; 2 Cases = 431.31 + 5.72 (time), r = .94, MAD = 12.28

The two methods that use the average of all data have very similar results, and the two CMA methods also look quite close. As suggested with this analysis, CMA is typically the better technique. LO 4.5: Develop seasonal indices AACSB: Analytical thinking

Solutions Manual for Operations Management, 14e

Operations Management Fourteenth Edition Jay Heizer Barry Render Chuck Munson