1.
a) x1 = 10, x2 = 40, s3 = 30, z = 420 b) Yes; all cj – zj row values are zero or negative. c) x3 = 0; s2 = 0 d) Maximize Z = 10x1 + 2x2 + 6x3 e) 3 f) Since there are three decision variables, a three-dimensional graph is required.
2.
a) minimization; because zj – cj is being calculated on the bottom row and not cj – zj b) x1 = 20, x3 = 10, s1 = 10, z = 240 c) Yes; all zj – cj values are negative or zero. d) Minimize Z = 6x1 + 20x2 + 12x3 e) 3 constraints f) Yes, one; because there are 3 constraints but only 2 surplus variables. Since both ≤ and ≥ constraints have slack or surplus variables, and an equality does not, then one of the three constraints was an =. g) x2 = 0
3.
a) maximization; because of “cj-zj” b) x1 = 12, x2 = 0, x3 = 0, x4 = 15 c) s1 = 20, s2 = 0, s3 = 0, s4 = 45 d) If x2 is selected as the entering basic variable, Z will increase by 20 for every unit of x2 entered into the solution. e) This solution is not optimal because there are positive values in the cj - zj row.
cj
Basic Variables
Quantity
60 x1
50 x2
45 x3
50 x4
0 s1
0 s2
0 s3
0 s4
0 50 60 45
s1 x4 x1 x3
20 15 12 45/8
0 0 1 0
1 0 1/2 0
0 0 0 1
0 1 0 6
1 0 0 0
0 1 0 -3/4
0 0 1/10 0
0 0 0 1/8
zj
13,785/8
60
30
45
50
0
65/4
6
45/8
0
20
0
0
0
-65/4
-6
-45/8
cj – zj
cj
Basic Variables
Quantity
60 x1
50 x2
45 x3
50 x4
0 s1
0 s2
0 s3
0 s4
50 50 60 45
x2 x4 x1 x3
20 15 2 45/8
0 0 1 0
1 0 0 0
0 0 0 1
0 1 0 6
1 0 -1/2 0
0 1 0 -3/4
0 0 1/10 0
0 0 0 1/8
zj
2,123.125
60
30
45
50
20
65/4
6
45/8
0
0
0
0
20
-65/4
-6
-45/8
cj-zj Optimal 4.
a) minimization; because “zj – cj ” and a positive “M ” in the objective function. b) x3 = 0 c) “M/6 - 5/3” has no real meaning since it includes “M ”; it simply represents a large net decrease in cost if s2 is entered into the solution. d) two; since there are two artificial variables remaining in the tableau, it will take at least two more tableaus to eliminate them, and, they must be eliminated to insure a feasible solution. e) no; because there are positive values in the zj – cj row.
cj
Basic Variables
Quantity
8 x1
10 x2
4 x3
0 s1
0 s2
0 s2
M A3
4 10 M
x3 x2 A3
30 10 70
2/3 1/3 -2/3
0 1 0
0 0 1
-1 0 1
1/6 -1/6 -1/6
0 0 -1
0 0 1
zj
7M+220
-2M/3 + 18/3
10
4
M-4
-M/6-1
-M
M
-2M/3-2
0
0
M-4
-M/6-1
-M
0
zj-cj
cj
Basic Variables
Quantity
8 x1
10 x2
4 x3
0 s1
0 s2
0 s2
4 10 0
x3 x2 s1
100 10 70
0 1/3 -2/3
0 1 0
1 0 0
0 0 1
0 -1/6 -1/6
-1 0 -1
zj
500
10/3
10
4
0
-10/6
-4
-14/3
0
0
0
-10/6
-4
zj-cj Optimal 5.
a) minimization; because zj – cj and “M” is positive in the objective function. b) x3 = 4 x2 = 6 c) no; because both constraints included a slack or surplus variable (s1 or s2) and an equation would have added only an artificial variable. d) s2 = 0 e) no; because there are positive values in the zj – cj row.
cj
Basic Variables
Quantity
4 x1
6 x2
0 s1
0 s2
0 4
s2 x1
4 8
0 1
1 1
-2 -1
1 0
zj
32
4
4
-4
0
0
-2
-4
0
zj-cj Optimal 6.
a) maximization; because cj – zj b) x2 = 0 c) no; at this iteration no optimal solution exists. d) the cj – zj value of “5” means that if s1 was selected as the entering variable, Z would increase by 5 for every unit of s1 entered into the solution. e) no; there are positive values in the cj – zj row.
7.
Minimize Z = .05x1 + .03x2 (cost, $) subject to 8x1 + 6x2 ≥ 48 (vitamin A, mg) x1 + 2x2 ≥ 12 (vitamin B, mg) x1, x2 ≥ 0
cj
Basic Variables
.05
.03
0
0
M
M
Quantity
x1
x2
s1
s2
A1
A2
M
A1
48
8
6
–1
0
1
0
M
A2
12
1
2
0
–1
0
1
60M
9M
8M
–M
–M
M
M
9M – .05 8M – .03 –M
–M
0
0
zj zj – cj cj
Basic Variables
Quantity
10 x1
5 x2
0 s1
0 s2
10 5 0
x1 x2 s2
3 4 1
1 0 0
0 1 0
-1/2 0 1/2
0 0 1
zj
50
10
5
-5
0
0
0
5
0
cj-zj
cj
Basic Variables
Quantity
10 x1
5 x2
0 s1
0 s2
10 5 0
x1 x2 s1
4 4 2
1 0 0
0 1 0
0 0 1
1 0 2
zj
60
10
5
0
10
0
0
0
-10
cj-zj
Optimal Multiple optimal solutions do not exist. 8. a) Minimize Zd = 60y1 + 40y2 subject to 12y1 + 4y2 ≥ 9 4y1 + 8y2 ≥ 7 y1, y2 ≥ 0 b) y1 = the marginal value of one additional hr of process 1; y2 = the marginal value of one additional hr of process 2 For the s1 column, the cj – zj value of $.55 is the marginal value of 1 hr of process 1 production time. For the s2 column, the cj – zj value of $0.60 is the marginal value of 1 hr of process 2 production time. c) cj, basic: cj
Basic Variables
9+∆
7
0
0
Quantity
x1
x2
s1
s2
9+∆
x1
4
1
0
1/10
–1/20
7
x2
3
0
1
–1/20
3/20
zj
57 + 4∆
9+∆
7
cj – zj
0
11/20 – ∆/10
12/20 + ∆/20
0 –11/20 – ∆/10 –12/20 + ∆/20
Since c2 = 7 + ∆; ∆ = c2 – 7. Thus
Solving for the cj – zj inequalities:
c2 – 7 ≥ –4 c2 ≥ 3
–11/20 – ∆/10 ≤ 0 –∆/10 ≤ 11/20 –∆ ≤ 11/2 ∆ ≥ –11/2
Summarizing, 3 ≤ c2 ≤ 18.
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
d) q1:
c1 – 9 ≥ 11/2 c1 ≥ 7/2 –12/20 + ∆/20 ≤ 0 ∆/20 ≤ 12/20 ∆ ≤ 12
x1: 4 + ∆/10 ≥ 0 ∆/10 ≥ –4 –∆ ≥ –80 ∆ ≥ –40
x2: 3 – ∆/20 ≥ 0 –∆/20 ≥ –3 –∆ ≥ –60 ∆ ≤ 60
Therefore, –40 ≤ ∆ ≤60. Since
Since c1 = 9 + ∆ , ∆ = c1 – 9. Thus
q1 = 60 + ∆ ∆ = q1 – 60 –40 ≤ q1 – 60 ≤ 60 20 ≤ q1 ≤ 120
c1 – 9 ≤ 12 c1 ≤ 12 Summarizing, 7/2 ≤ c1 ≤ 12. c2, basic: cj
Basic Variables
9
7+∆
0
0
Quantity
x1
x2
s1
s2
9
x1
4
1
0
1/10
–1/20
7+∆
x2
3
0
1
–1/20
3/20
zj
57 + 3∆
9
7+∆
11/20 – ∆/20
12/20 + 3∆/20
0
0
–11/20 + ∆/20
–12/20 – 3∆/20
cj – zj
Solving for the cj – zj inequalities: –11/20 + ∆/20 ≤ 0 ∆/20 ≤ 11/20 ∆ ≤ 11 Since c2 = 7 + ∆; ∆ = c2 – 7. Thus c2 – 7 ≤ 11 c2 ≤ 18 –12/20 – 3∆/20 ≤ 0 –3∆/20 ≤ 12/20 ∆ ≥ –4
e) The marginal value of 1 hr of process 1 production time is $.55. The sensitivity range for q1, production hours, is 20 ≤ q1 ≤ 120. Thus, the company would purchase up to 120 hr at the marginal value price.
q2: x1: 4 – ∆/20 ≥ 0 –∆/20 ≥ –4 –∆ ≥ –80 ∆ ≤ 80
x2: 3 +3 ∆/20 ≥ 0 3∆/20 ≥ –3 ∆ ≥ –20
Therefore, –20 ≤ ∆ ≤ 80. Since q2 = 40 + ∆ ∆ = q2 – 40 –20 ≤ q2 – 40 ≤ 80 20 ≤ q2 ≤ 120