Duals of LP: Consider the linear programming problem given below: max z= x 1 + 2 x 2 Subject to constraints: 3 x 1 + x 2 ≤ 6 2 x 1 + x 2 = 5 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 3 1 ≤ 6 y 2 2 1 = 5 * ≥ 1 ≥ 2 Step 2: In this case, the second primal constraint is equality and therefore dual variable y 2 is now unrestricted in sign(urs). 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 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

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Answer 2

Expert Solution & Answer

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

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Answer 3

Expert Solution & Answer

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

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Answer 4

Expert Solution & Answer

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

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Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

Answer 8

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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

Answer 9

Chapter 6.5, Problem 5P

Explanation of Solution

Duals of LP:

Consider the linear programming problem given below:

max z= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2=5

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 3 1 ≤6
y2 2 1 =5*
≥1 ≥2

Step 2:

In this case, the second primal constraint is equality and therefore dual variable y2 is now unrestricted in sign(urs).

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

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:

There is no ≥ constraint in this LP.

Step 2:

Replace the equality constraint 2x1+x2=5 by two inequality constraints as,

2x1+x2≤5

−2x1−x2≤−5

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= x1+2x2

Subject to constraints:

3x1+x2≤6

2x1+x2≤5

−2x1−x2≤−5

x1,x2≥0

Construct the following table according to the given LP,

Min W (x1≥0) (x2≥0)
x1

Answer 12

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

Answer 13

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

Answer 14

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Answer 15

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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 that the second primal constraint is equality, so the dual variable y_2 is now unrestricted.

Concept explainers

Explanation of Solution

Explanation of Solution

Explanation of Solution

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2

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Chapter 6 Solutions

Max z
min w	(x1≥0)	(x2≥0)
	x1	x2
(y1≥0)	y1	3	1	≤6
	y2	2	1	=5*
	≥1	≥2

Solution Summary: The author explains that the second primal constraint is equality, so the dual variable y_2 is now unrestricted.

Concept explainers

Explanation of Solution

Explanation of Solution

Explanation of Solution

c. Now put y2=y2'−y2'' in the part (b) answer. So that, min w= 6y1+5y2 Subject to constraints 3y1+2y2≥1 y1+y2≥2 y1≥0,y2 urs Since, y2

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Chapter 6 Solutions

c.

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

min w= 6y1+5y2

Subject to constraints

3y1+2y2≥1

y1+y2≥2

y1≥0,y2 urs

Since,

y2