Calculating the extra payment: Let x 1 is the number of motorcycle produced at plant 1, x 2 is the number of motorcycle produced at plant 2 and x 3 is the number of motorcycle produced at plant 2. In order to minimize the cost of meeting these demands, the following Linear Programming has to be solved: min z= 300 x 1 + 280 x 2 + 225 x 3 Such that 20 x 1 + 16 x 2 + 10 x 3 ≤ 21 , 000 5 x 1 + 8 x 2 + 7 x 3 ≤ 9 , 400 x 1 ≤ 750 x 2 ≤ 750 x 3 ≤ 750 x 1 + x 2 + x 3 ≥ 1 , 400 All variables ≥ 0 The output of this Linear Programming is given below: min z= 300 x 1 + 280 x 2 + 225 x 3 Subject to constraint 20 x 1 + 16 x 2 + 10 x 3 ≤ 21 , 000 5 x 1 + 8 x 2 + 7 x 3 ≤ 9 , 400 x 1 ≤ 750 x 2 ≤ 750 x 3 ≤ 750 x 1 + x 2 + x 3 ≥ 1 , 400 End LP optimum is found at step 3

Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms

4th Edition

ISBN: 9780534423551

Author: Wayne L. Winston

Publisher: Cengage Learning

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Chapter 5.3, Problem 5P

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Calculating the extra payment:

Let x₁ is the number of motorcycle produced at plant 1, x₂ is the number of motorcycle produced at plant 2 and x₃ is the number of motorcycle produced at plant 2.

In order to minimize the cost of meeting these demands, the following Linear Programming has to be solved:

min z= 300x1+280x2+225x3

Such that

20x1+16x2+10x3≤21,000

5x1+8x2+7x3≤9,400

x1≤750

x2≤750

x3≤750

x1+x2+x3≥1,400

All variables ≥ 0

The output of this Linear Programming is given below:

min z= 300x1+280x2+225x3

Subject to constraint

20x1+16x2+10x3≤21,000

5x1+8x2+7x3≤9,400

x1≤750

x2≤750

x3≤750

x1+x2+x3≥1,400

End

LP optimum is found at step 3

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Chapter 5.3, Problem 4P

Chapter 5.3, Problem 6P

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Solution Summary: The author explains that the following Linear Programming has to be solved in order to minimize the cost of meeting these demands.

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