Consider the following variables: x 1 = items of type 1 x 2 = items of type 2 Weight of item is a 1 lb and item 2 weighs a 2 lb. Therefore, the constraints is a 1 x 1 +a 2 x 2 ≤ b Each type 1 item carries a benefit of c 1 units, each type 2 item carries a benefit of c 2 units...

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Answer 1

Expert Solution & Answer

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 2

Expert Solution & Answer

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Answer 8

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

Answer 9

Chapter 4, Problem 16RP

a.

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

Answer 10

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

Answer 11

b.

Explanation of Solution

from the result of part (a), it can be observed that the constraint in linear programming problem is less than or equal to type.

So, add the slack variable s₁ to the constraint.

Thus, the standard form of the linear programming problem is,

Max z= c₁x₁+c₂x₂

Subject to constraints,

a₁x₁+a₂x₂+s₁ = b

The initial simplex table is shown below:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 z = 0
0 a₁ a₂ 1 b s₁ = b

Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.

Therefore, choose entering variable to be x₂.

The leaving variable is calculated with the use of the following formula,

Leaving variable = {(rhs/x₁), x₁ ≥ 0}

Consider the below table:

z x₁ x₂ s₁ rhs Basic variable
1 -c₁ -c₂ 0 0 -
0 a₁ a₂ 1 b (b/a₂)→

From the above table, the pivot table is intersection of both entering and leaving variables...

Answer 12

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Answer 13

c.

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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See solution

Answer 17

Ch. 4.1 - Prob. 1P Ch. 4.1 - Prob. 2P Ch. 4.1 - Prob. 3P Ch. 4.4 - Prob. 1P Ch. 4.4 - Prob. 2P Ch. 4.4 - Prob. 3P Ch. 4.4 - Prob. 4P Ch. 4.4 - Prob. 5P Ch. 4.4 - Prob. 6P Ch. 4.4 - Prob. 7P

Consider the following variables: x 1 = items of type 1 x 2 = items of type 2 Weight of item is a 1 lb and item 2 weighs a 2 lb. Therefore, the constraints is a 1 x 1 +a 2 x 2 ≤ b Each type 1 item carries a benefit of c 1 units, each type 2 item carries a benefit of c 2 units...

Solution Summary: The author explains that the constraint in linear programming problem is less than or equal to type.

Concept explainers

Explanation of Solution

Consider the following variables:

x₁ = items of type 1

x₂ = items of type 2

Weight of item is a₁ lb and item 2 weighs a₂ lb.

Therefore, the constraints is a₁x₁+a₂x₂ ≤ b

Each type 1 item carries a benefit of c₁ units, each type 2 item carries a benefit of c₂ units...

Explanation of Solution

Explanation of Solution

The following are the assumptions that are violated by this formula:

Deterministic Nature

Additivity

Direct Proportionality

Fractionally

The follo...

Want to see the full answer?

Chapter 4 Solutions

z	x₁	x₂	s₁	rhs	Basic variable
1	-c₁	-c₂	0	0	z = 0
0	a₁	a₂	1	b	s₁ = b

Consider the following variables: x 1 = items of type 1 x 2 = items of type 2 Weight of item is a 1 lb and item 2 weighs a 2 lb. Therefore, the constraints is a 1 x 1 +a 2 x 2 ≤ b Each type 1 item carries a benefit of c 1 units, each type 2 item carries a benefit of c 2 units...

Solution Summary: The author explains that the constraint in linear programming problem is less than or equal to type.

Concept explainers

Explanation of Solution