Optimal solution: Consider the linear programming problem given below: max z= 13 x 1 + 16 x 2 + 16 x 3 + 14 x 4 + 39 x 5 Subject to constraints: 11 x 1 + 53 x 2 + 5 x 3 + 5 x 4 + 29 x 5 ≤ 40 3 x 1 + 6 x 2 + 5 x 3 + x 4 + 34 x 5 ≤ 20 x 1 , x 2 , x 3 , x 4 , x 5 ≤ 1 x 1 , x 2 , x 3 , x 4 , x 5 ≥ 0 Here, x i = Fraction of investment I purchased by star oil. The Lindo output of the LP is given in the following figure: LP optimum found at step 5 Objective function value: 57.44902. Variable Value Reduced Cost x 1 1.000000 0.000000 x 2 0.200860 0.000000 x 3 1.000000 0.000000 x 4 1.000000 0.000000 x 5 0.288084 0.000000 Row Slack or Surplus Dual prices 1 0.000000 0.190418 2 0.000000 0.984644 3 0.000000 7.951474 4 0.799140 0.000000 5 0.000000 10.124693 6 0.000000 12.063268 7 0.711916 0.000000 Number of iterations: 5 Ranges in which the basis is unchanged: Obj coefficient ranges Variable Current coefficient Allowable increase Allowable Decrease x 1 13.000000 Infinity 7.951474 x 2 16.000000 45.104527 9.117647 x 3 16.000000 Infinity 10.124693 x 4 14.000000 Infinity 12.063268 x 5 39.000000 51.666668 30.245285 Right-hand side ranges Row Current RHS Allowable increase Allowable Decrease 1 40.000000 38.264706 9.617647 2 20.000000 11.275863 8.849056 3 1.000000 1.139373 1.000000 4 1.000000 Infinity 0.799140 5 1.000000 1.995745 1.000000 6 1.000000 2.319149 1.000000 7 1.000000 Infinity 0.711916

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Answer 1

Question

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

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Answer 2

Question

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Answer 6

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Answer 7

Chapter 6, Problem 29RP

Program Plan Intro

Optimal solution:

Consider the linear programming problem given below:

max z= 13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

Here,

xi= Fraction of investment I purchased by star oil.

The Lindo output of the LP is given in the following figure:

LP optimum found at step 5

Objective function value: 57.44902.

Variable Value Reduced Cost
x1 1.000000 0.000000
x2 0.200860 0.000000
x3 1.000000 0.000000
x4 1.000000 0.000000
x5 0.288084 0.000000

Row Slack or Surplus Dual prices
1 0.000000 0.190418
2 0.000000 0.984644
3 0.000000 7.951474
4 0.799140 0.000000
5 0.000000 10.124693
6 0.000000 12.063268
7 0.711916 0.000000

Number of iterations: 5

Ranges in which the basis is unchanged:

Obj coefficient ranges
Variable Current coefficient Allowable increase Allowable Decrease
x1 13.000000 Infinity 7.951474
x2 16.000000 45.104527 9.117647
x3 16.000000 Infinity 10.124693
x4 14.000000 Infinity 12.063268
x5 39.000000 51.666668 30.245285

Right-hand side ranges
Row Current RHS Allowable increase Allowable Decrease
1 40.000000 38.264706 9.617647
2 20.000000 11.275863 8.849056
3 1.000000 1.139373 1.000000
4 1.000000 Infinity 0.799140
5 1.000000 1.995745 1.000000
6 1.000000 2.319149 1.000000
7 1.000000 Infinity 0.711916

Answer 8

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Answer 9

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Answer 10

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Answer 11

Explanation of Solution

c.

Since both the objective function coefficients have a reduced cost of zero, hence 100% rule can be applied.

A 25% decrease in the NPV from investment 2 implies

Δb1=10I1=38.26

Δc2=4

Allowable decrease corresponding to c2=16, as it can be observed from the Lindo output is D2=7.95

Now applying the formula,

rj=−ΔcjDj

Here,

r2=47.95

r2=0.50

In the same way, a 25% decrease in the NPV from investment 4 implies

Δc4=14(0.25)

Δc4=−3

Answer 12

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Answer 13

Explanation of Solution

d.

Since both the constraints are the binding constraints, hence 100% rule can be applied.

Since the amount available to invest in the time 0 is increased from $40 million to $50 million,

D2=8.85

Δb1=10

Allowable increase corresponding to b1=40, as can be seen from the Lindo output is,

I1=38.26

Now applying the formula

r2=0.56

Here,

r1=1038.26

r1=0.26

Similarly, since the amount available to invest in time 1 is decreased from $20 million to $15 million,

Δb1=15−20

Δb1=−5

Allowable decrease corresponding to b2=20, as can be seen from the Lindo output is D2=8

Answer 14

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Answer 15

Explanation of Solution

e.

Define RM2 to be the pounds of raw material 2 purchased annually. The initial table is now changed by the introduction of the RM2 column. New linear program is,

max z=13x1+16x2+16x3+14x4+39x5

Subject to constraints:

11x1+53x2+5x3+5x4+29x5≤40

3x1+6x2+5x3+x4+34x5≤20

x1,x2,x3,x4,x5≤1

x1,x2,x3,x4,x5≥0

To determine whether a new activity causes the current basis to be optimal or not, price out the new activity. That is calculate c¯j by applying the formula given below:

c¯j=cBVB−1aj−cj

The new column for the investment 6 is

a6=[51000000] and cBVB−1=[0

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

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Answer 19

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 the Lindo output of the LP in the following figure.

Explanation of Solution

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

Explanation of Solution

Explanation of Solution

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Chapter 6 Solutions

Variable	Value	Reduced Cost
x1	1.000000	0.000000
x2	0.200860	0.000000
x3	1.000000	0.000000
x4	1.000000	0.000000
x5	0.288084	0.000000

Row	Slack or Surplus	Dual prices
1	0.000000	0.190418
2	0.000000	0.984644
3	0.000000	7.951474
4	0.799140	0.000000
5	0.000000	10.124693
6	0.000000	12.063268
7	0.711916	0.000000

Obj coefficient ranges
Variable	Current coefficient	Allowable increase	Allowable Decrease
x1	13.000000	Infinity	7.951474
x2	16.000000	45.104527	9.117647
x3	16.000000	Infinity	10.124693
x4	14.000000	Infinity	12.063268
x5	39.000000	51.666668	30.245285

Right-hand side ranges
Row	Current RHS	Allowable increase	Allowable Decrease
1	40.000000	38.264706	9.617647
2	20.000000	11.275863	8.849056
3	1.000000	1.139373	1.000000
4	1.000000	Infinity	0.799140
5	1.000000	1.995745	1.000000
6	1.000000	2.319149	1.000000
7	1.000000	Infinity	0.711916

Solution Summary: The author explains the Lindo output of the LP in the following figure.

Explanation of Solution

b. If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes. As it can be seen from the Lindo output, the Obj coefficient ranges block determines

Explanation of Solution

Explanation of Solution

Explanation of Solution

Want to see the full answer?

Chapter 6 Solutions

b.

If NPV from investment 1 changes to $5 from $13 then the objective function coefficient of x1 changes.

As it can be seen from the Lindo output, the Obj coefficient ranges block determines