Basic Components of a Queuing System Arrivals
Service facilities
Arrival rate is known and constant. e.g.: 3 customers arrive every 15 minutes at a restaurant.
Service rate is known and constant. e.g.: A counter at a bank can serve 1 person in 5 minutes.
Arrival rate follows Poisson distribution. Arrivals (customers) are patient.
Service rate follows exponential distribution. Service systems are classified based on number of channels and number of phases.
Waiting line (queue) Customers are served on FIFO basis.
Example: Burger kiosk
Example: Drive-thru
Example: Bank, post office
Example: School registration
Queuing Cost Service Cost
Waiting Cost
The cost that we have to pay for providing the service.
The cost that we have to pay for making the customers wait.
đ?‘ đ?‘ đ??śđ??śđ?‘ đ?‘
đ?œ†đ?œ†đ?œ†đ?œ†đ??śđ??śđ?‘¤đ?‘¤
Number of server
Service cost per server
Arrival rate
Average waiting time
Waiting cost at a given time
Objective: To determine level of services that can minimize total queuing cost. Total queuing cost = Total service cost + Total waiting cost = đ?‘ đ?‘ đ??śđ??śđ?‘ đ?‘ + đ?œ†đ?œ†đ?œ†đ?œ†đ??śđ??śđ?‘¤đ?‘¤
Assumptions: 1. Arrival rate is known and constant. 2. Arrival rate follows Poisson distribution. 3. Arrivals are patient. 4. Arrivals are served on FIFO basis. 5. Service rate is known and constant. 6. Service rate follows exponential distribution. 7. Service rate is greater than arrival rate (đ?œ‡đ?œ‡ > đ?œ†đ?œ†).
Number of customers in the system:
Average time a customer spends in the queue:
đ?œ†đ?œ† đ??żđ??ż = đ?œ‡đ?œ‡ − đ?œ†đ?œ†
đ?œ†đ?œ† đ?‘Šđ?‘Šđ?‘žđ?‘ž = đ?œ‡đ?œ‡(đ?œ‡đ?œ‡ − đ?œ†đ?œ†)
Number of customers in the queue:
Probability that the service facility is being used (busy):
đ?œ†đ?œ†2 đ??żđ??żđ?‘žđ?‘ž = đ?œ‡đ?œ‡(đ?œ‡đ?œ‡ − đ?œ†đ?œ†)
đ?œ†đ?œ† đ?œŒđ?œŒ = đ?œ‡đ?œ‡
1 đ?‘Šđ?‘Š = đ?œ‡đ?œ‡ − đ?œ†đ?œ†
đ?œ†đ?œ† đ?‘ƒđ?‘ƒ0 = 1 − đ?œ‡đ?œ‡
Average time a customer spends in the system:
Probability that no one is in the system:
Aman’s Barber Shop has one barber named Raju. Customers arrive at the shop at the rate of two per hour. Raju takes 15 minutes to perform a haircut and he is paid by the shop owner RM5 per hour for his service. The shop owner estimates that the cost of customer waiting time in term of customer dissatisfaction and lost goodwill is RM10 per hour. a)
Calculate: i) The average number of customers at the shop. ii) The average number of customers waiting for a haircut. iii) The average time a customer spends at the shop. iv) The average time a customer spends waiting before getting a haircut. v) The probability that Raju is free. vi) Total cost per hour spent by Aman’s Barber Shop.
b) Suppose that the shop owner knows another barber named Gopal who can gives a haircut in 10 minutes. The shop owner wants to hire Gopal. Gopal agrees to work at the shop only if he were paid RM8 per hour. Determine whether it is worthwhile to replace Raju with Gopal.
Single channel λ = 2 customers per 1 hour
μ = 1 haircut per 15 minutes x4
x4
= 4 haircuts per 60 minutes 𝜆𝜆 𝐿𝐿 = 𝜇𝜇 − 𝜆𝜆
2 = 4−2
= 1 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜆𝜆2 𝐿𝐿𝑞𝑞 = 𝜇𝜇(𝜇𝜇 − 𝜆𝜆) 22 = 4(4 − 2)
1 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2
1 𝑊𝑊 = 𝜇𝜇 − 𝜆𝜆
1 = 4−2
1 = ℎ𝑜𝑜𝑜𝑜𝑜𝑜 2
𝜆𝜆 𝑊𝑊𝑞𝑞 = 𝜇𝜇(𝜇𝜇 − 𝜆𝜆) 2 = 4(4 − 2) 1 = ℎ𝑜𝑜𝑜𝑜𝑜𝑜 4
𝜆𝜆 𝑃𝑃0 = 1 − 𝜇𝜇
2 =1− 4 1 = 2
𝐶𝐶𝑠𝑠 = 5
𝐶𝐶𝑤𝑤 = 10 (Based on time spent at the shop) 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝐶𝐶𝑠𝑠 + 𝜆𝜆𝜆𝜆𝐶𝐶𝑤𝑤
1 =1 5 +2 (10) 2 = 𝑅𝑅𝑅𝑅𝑅𝑅
Gopal đ??śđ??śđ?‘ đ?‘ = 8
đ??śđ??śđ?‘¤đ?‘¤ = 10 (Based on time spent at the shop) đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡đ?‘‡ đ?‘?đ?‘?đ?‘?đ?‘?đ?‘?đ?‘?đ?‘?đ?‘? = đ?‘ đ?‘ đ??śđ??śđ?‘ đ?‘ + đ?œ†đ?œ†đ?œ†đ?œ†đ??śđ??śđ?‘¤đ?‘¤
1 =1 8 +2 (10) 4 = đ?‘…đ?‘…đ?‘…đ?‘…đ?‘…đ?‘…
It is more worthwhile to hire Gopal because total cost for hiring Gopal is lower than total cost for hiring Raju.
Îť = 2 customers per 1 hour Îź = 1 haircut per 10 minutes x6
x6
= 6 haircuts per 60 minutes 1 đ?‘Šđ?‘Š = đ?œ‡đ?œ‡ − đ?œ†đ?œ†
1 = 6−2
1 = â„Žđ?‘œđ?‘œđ?‘œđ?‘œđ?‘œđ?‘œ 4
Assumptions: 1. Arrival rate is known and constant. 2. Arrival rate follows Poisson distribution. 3. Arrivals are patient. 4. Arrivals are served on FIFO basis. 5. Service rate is known and constant. 6. Service rate follows exponential distribution. 7. All servers perform at the same rate. 8. Number of server times service rate is greater than arrival rate (đ?‘ đ?‘ đ?œ‡đ?œ‡ > đ?œ†đ?œ†).
Probability that the service facility is being used (busy):
Number of customers in the queue:
đ?œ†đ?œ† đ?œŒđ?œŒ = đ?‘ đ?‘ đ?œ‡đ?œ‡
đ?œ†đ?œ† đ?‘ đ?‘ đ?œŒđ?œŒ đ?‘ƒđ?‘ƒđ?‘œđ?‘œ đ?œ‡đ?œ‡ đ??żđ??żđ?‘žđ?‘ž = đ?‘ đ?‘ ! 1 − đ?œŒđ?œŒ 2
Number of customers in the system:
Probability that no one is in the system: đ?‘ƒđ?‘ƒ0 =
1 đ?œ†đ?œ† ∑đ?‘›đ?‘›=đ?‘ đ?‘ −1 đ?‘›đ?‘›=0 đ?‘›đ?‘›! đ?œ‡đ?œ‡
��
1
1 đ?œ†đ?œ† + đ?‘ đ?‘ ! đ?œ‡đ?œ‡
đ?‘ đ?‘
1 1 − đ?œŒđ?œŒ
đ?œ†đ?œ† đ??żđ??ż = đ??żđ??żđ?‘žđ?‘ž + đ?œ‡đ?œ‡
Average time a customer spends in the system: đ??żđ??ż đ?‘Šđ?‘Š = đ?œ†đ?œ†
Average time a customer spends in the queue: đ??żđ??żđ?‘žđ?‘ž đ?‘Šđ?‘Šđ?‘žđ?‘ž = đ?œ†đ?œ†
A bank has four tellers counter with one officer at each counter. The services at these tellers are exponentially distributed with mean of 5 minutes per customer. Arriving customers form a single line and proceeds to whichever counter that is free. The arrival of customers is Poisson distributed with mean of 36 per hour. Suppose that each officer is paid RM20 per hour and the cost that the bank has to pay for each hour a customer spends at the bank is RM6. Calculate a) b) c) d) e) f)
the probability that there are no customers in the system. the average number of customer in the queue. the average number of customers at the bank. the average time a customer spends at the bank. the average time a customer spends in the queue. the total queuing cost for the bank.
Multi channel, s = 4 Îź = 1 customer per 5 minutes
Îť = 36 customers per 1 hour
x 12
x 12
= 12 customers per 60 minutes đ?‘ƒđ?‘ƒ0 = = =
∑đ?‘›đ?‘›=đ?‘ đ?‘ −1 đ?‘›đ?‘›=0
1 đ?œ†đ?œ† đ?‘›đ?‘›! đ?œ‡đ?œ‡
��
đ?‘›đ?‘›=4−1 1 36 ∑đ?‘›đ?‘›=0 đ?‘›đ?‘›! 12
1 ��
1 đ?œ†đ?œ† + đ?‘ đ?‘ ! đ?œ‡đ?œ‡ 1
đ?‘ đ?‘
1 1 − đ?œŒđ?œŒ
1 36 + 4! 12
1
4
1
3 1−4
1 0 1 1 1 2 1 3 3 + 3 + 3 + 3 + 13.5 0! 1! 2! 3!
đ?œ†đ?œ† 36 3 = = đ?œŒđ?œŒ = đ?‘ đ?‘ đ?œ‡đ?œ‡ 4(12) 4
1 = 51 2
2 = 51
4 𝜆𝜆 𝑠𝑠 3 36 2 81 𝜌𝜌 𝜇𝜇 𝑃𝑃𝑜𝑜 27 4 12 51 34 𝐿𝐿𝑞𝑞 = = = = 1.5882 = 2 2 3 𝑠𝑠! 1 − 𝜌𝜌 17 3 4! 1 − 2 4
𝜆𝜆 27 36 78 + = = 4.5882 𝐿𝐿 = 𝐿𝐿𝑞𝑞 + = 𝜇𝜇 17 12 17
78 𝐿𝐿 17 13 𝑊𝑊 = = = = 0.1275 𝜆𝜆 36 102
27 𝐿𝐿𝑞𝑞 17 3 𝑊𝑊𝑞𝑞 = = = = 0.0441 𝜆𝜆 36 68
𝐶𝐶𝑠𝑠 =20
𝐶𝐶𝑤𝑤 = 6 (Based on time spent at the bank) 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝐶𝐶𝑠𝑠 + 𝜆𝜆𝜆𝜆𝐶𝐶𝑤𝑤
13 = 4 20 + 36 (6) 102 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅.53