Duals of LP: Consider the linear programming problem given below: max z= 3 x 1 + x 2 Subject to constraints: x 1 + x 2 ≤ 1 − x 1 + x 2 ≥ 2 x 1 , x 2 ≥ 0 a. To find the dual of the above LP proceed as follows: Step 1: Max z min w ( x 1 ≥ 0 ) ( x 2 ≥ 0 ) x 1 x 2 ( y 1 ≥ 0 ) y 1 1 1 ≤ 1 y 2 -1 1 ≥ 2 * ≥ 3 ≥ 1 Step 2: In this case, the second primal constraint is ≥ and therefore dual variable y 2 must satisfy y 2 ≤ 0 . The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

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Answer 1

Expert Solution & Answer

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

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Answer 2

Expert Solution & Answer

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

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Answer 3

Expert Solution & Answer

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

Answer 8

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Answer 9

Chapter 6.5, Problem 6P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

−x1+x2≥2

x1,x2≥0

a.

To find the dual of the above LP proceed as follows:

Step 1:

Max z
min w (x1≥0) (x2≥0)
x1 x2
(y1≥0) y1 1 1 ≤1
y2 -1 1 ≥2*
≥3 ≥1

Step 2:

In this case, the second primal constraint is ≥ and therefore dual variable y2 must satisfy y2≤0.

The asterisk (*) sign represents the places where rules must be applied to determine part of the dual

Answer 10

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Answer 11

Explanation of Solution

b.

As the above linear programming problem is not a normal max problem, therefore the following steps are to be performed to convert it into a normal form.

Step 1: Multiply each ≥ constraint by -1 as mentioned below:

−x1+x2≥2

−1(−x1+x2)≤−2

x1−x2≤−2

Step 2:

Replace the equality constraint by two inequality constraints as,

There is no equality constraint in this LP.

Step 3:

Now replace each urs variable x_i by xi=xi'−xi'' as there is no urs variable in the given LP.

Now write the transformed LP as,

max z= 3x1+x2

Subject to constraints:

x1+x2≤1

x1−x2≤−2

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0

Answer 12

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

Answer 13

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 6.1 - Prob. 1P Ch. 6.1 - Prob. 2P Ch. 6.1 - Prob. 3P Ch. 6.1 - Prob. 4P Ch. 6.1 - Prob. 5P Ch. 6.2 - Prob. 1P Ch. 6.2 - Prob. 2P Ch. 6.3 - Prob. 1P Ch. 6.3 - Prob. 2P Ch. 6.3 - Prob. 3P

Solution Summary: The author explains how to find the dual of the linear programming problem. The asterisk (*) sign represents the places where rules must be applied.

Explanation of Solution

Explanation of Solution

Explanation of Solution

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2

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Chapter 6 Solutions

Max z
min w	(x1≥0)	(x2≥0)
	x1	x2
(y1≥0)	y1	1	1	≤1
	y2	-1	1	≥2*
	≥3	≥1

Solution Summary: The author explains how to find the dual of the linear programming problem. The asterisk (*) sign represents the places where rules must be applied.

Explanation of Solution

Explanation of Solution

Explanation of Solution

c. Now put y¯2=−y2 in the part (b) answer. So that, min w= y1+2y2.....(2) Subject to constraints y1−y2≥3 y1+y2

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Chapter 6 Solutions

c.

Now put y¯2=−y2 in the part (b) answer. So that,

min w= y1+2y2.....(2)

Subject to constraints

y1−y2≥3

y1+y2