Instructor Manual - Twelfth Edition Bernard W. Taylor III A-2 Copyr...

Instructor Manual For Introduction to Management Science Twelfth Edition Bernard

• Taylor III 1 / 4

A-2 Copyright © 2016 Pearson Education, Inc.

• maximization; because of “c

j − zj”

• x
• = 12, x 2 = 0, x 3 = 0, x 4 = 15
• s
• = 20, s 2 = 0, s 3 = 0, s 4 = 45

d) If x

• is selected as the entering basic

variable, Z will increase by 20 for every unit of x

• entered into the solution.

e) This solution is not optimal because there

are positive values in the c j − zj row.

cj Basic 60 50 45 50 0 0 0 0 Variables Quantity x 1 x 2 x3 x4 s1 s2 s 3 s 4

• s 1 20 0 1 0 0 1 0 0 0

50 x 4 15 0 0 0 1 0 1 0 0 60 x

1 12 1 1/2 0 0 0 0 1/10 0

45 x

3 45/8 0 0 1 6 0 −3/4 0 1/8

z j 13,785/8 60 30 45 50 0 65/4 6 45/8 c j − zj 0 20 0 0 0 −65/4 −6 −45/8

c j Basic 60 50 45 50 0 0 0 0 Variables Quantity x 1 x 2 x3 x4 s1 s2 s 3 s 4 50 x 2 20 0 1 0 0 1 0 0 0 50 x 4 15 0 0 0 1 0 1 0 0 60 x

1 2 1 0 0 0 −1/2 0 1/10 0

45 x

3 45/8 0 0 1 6 0 −3/4 0 1/8

z j 2,123.125 60 30 45 50 20 65/4 6 45/8 c j − zj 0 0 0 0 20 −65/4 −6 −45/8 Optimal

• Minimization; because “z

j − cj” and a positive “M” in the objective function.

• x
• = 0
• “M/6 − 5/3” has no real meaning since it

includes “M”; it simply represents a large net decrease in cost if s

• is entered into the solution.
• 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.

• no; because there are positive values in the

z j − cj row.

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A-3 Copyright © 2016 Pearson Education, Inc.c j Basic 8 10 4 0 0 0 M Variables Quantity x 1 x2 x3 s1 s2 s 2 A 3

• x 3 30 2/3 0 0 −1 1/6 0 0

10 x 2 10 1/3 1 0 0 −1/6 0 0 M A

3 70 −2/3 0 1 1 −1/6 −1 1

z j 7M + 220 −2M/3 + 18/3 10 4 M −4 −M/6−1 −M M z j − cj −2M/3−2 0 0 M−4 −M/6−1 −M 0

cj Basic 8 10 4 0 0 0 Variables Quantity x 1 x2 x3 s1 s2 s 2

• x 3 100 0 0 1 0 0 −1

10 x 2 10 1/3 1 0 0 −1/6 0

• s

1 70 −2/3 0 0 1 −1/6 −1

z j 500 10/3 10 4 0 −10/6 −4 z j − cj −14/3 0 0 0 −10/6 −4 Optimal

• Minimization; because z

j − cj and “M” are positive in the objective function.

• x
• = 4

x

• = 6
• no; because both constraints included a slack

or surplus variable (s

• or s2) and an equation

would have added only an artificial variable.

• s
• = 0
• no; because there are positive values in the z

j − cj row.

cj Basic 4 6 0 0 Variables Quantit y x1 x2 s1 s2

• s2 4 0 1 −2 1
• x1 8 1 1 −1 0

z j 32 4 4 −4 0 zj − cj 0 −2 −4 0 Optimal

• Maximization; because c

j − zj

• x
• = 0
• no; at this iteration no optimal solution exists.
• the c

j − zj value of “5” means that if s1 was selected as the entering variable, Z would increase by 5 for every unit of s

• entered into the solution.
• no; there are positive values in the c

j − zj row.

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A-4 Copyright © 2016 Pearson Education, Inc.cj Basic 10 5 0 0 Variables Quantit y x1 x2 s1 s2 10 x1 3 1 0 −1/2 0

• x2 4 0 1 0 0
• s

2 1 0 0 1/2 1

z j 50 10 5 −5 0 cj − zj 0 0 5 0

c j Basic 10 5 0 0 Variables Quantit y x1 x2 s1 s2 10 x1 4 1 0 0 1

• x2 4 0 1 0 0
• s

1 2 0 0 1 2

z j 60 10 5 0 10 cj − zj 0 0 0 −10 Optimal Multiple optimal solutions do not exist.

• Minimize Z = .05x
• + .03x 2 (cost, $)

subject to 8x

• + 6x 2 ≥ 48 (vitamin A, mg)

x

• + 2x 2 ≥ 12 (vitamin B, mg)

x 1, x2 ≥ 0

cj Basic .05 .03 0 0M M Variables Quantity x 1 x2 s1 s2 A1 A 2

M A 1 48 8 6 −1 0 1 0

M A 2 12 1 2 0 −1 0 1

zj 60M 9 M 8 M −M −M M M z j − cj 9 M − .05 8M − .03 −M −M 0 0

c j Basic .05 .03 0 0 M Variables Quantity x 1 x2 s1 s2 A 2 .05 x1 6 1 3/4 −1/8 0 0

M A2 6 0 5/4 1/8 −1 1

zj 6M + .30 .05 5 M/4 + .38 M/8 − .006 −M M z j − cj 0 5M/4 + .38 M/8 − .006 −M 0

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