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Exam 3: Linear Programming: Sensitivity Analysis and Interpretation of Solution

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Question 21

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If the optimal value of a decision variable is zero and its reduced cost is zero, this indicates that alternative optimal solutions exist.

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Question 22

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Sensitivity analysis information in computer output is based on the assumption of

A) no coefficient change.
B) one coefficient change.
C) two coefficient change.
D) all coefficients change.

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Question 23

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A negative dual price for a constraint in a minimization problem means

Decreasing the objective function coefficient of a variable to its lower limit will create a revised problem that is unbounded.

The dual price associated with a constraint is the improvement in the value of the solution per unit decrease in the right-hand side of the constraint.

An objective function reflects the relevant cost of labor hours used in production rather than treating them as a sunk cost. The correct interpretation of the dual price associated with the labor hours constraint is

In a linear programming problem, the binding constraints for the optimal solution are 5X + 3Y $\le$ 30 2X + 5Y $\le$ 20 a.Fill in the blanks in the following sentence: As long as the slope of the objective function stays between _______ and _______, the current optimal solution point will remain optimal. b.Which of these objective functions will lead to the same optimal solution? 1) 2X + 1Y 2) 7X + 8Y 3) 80X + 60Y 4) 25X + 35Y

A constraint with a positive slack value

Eight of the entries have been deleted from the LINDO output that follows. Use what you know about linear programming to find values for the blanks. MIN 6 X1 + 7.5 X2 + 10 X3 SUBJECT TO 2) 25 X1 + 35 X2 + 30 X3 >= 2400 3) 2 X1 + 4 X2 + 8 X3 >= 400 END LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) 612.50000 $\begin{array}{ccc}\text { VARIABLE } & \text { VALUE } & \text { REDUCED COST } \\\mathrm{X} 1 &-&1.312500\\\mathrm{X}_{2} &-&-\\\mathrm{X} 3 &27.500000&-\end{array}$ $\begin{array}{ccc}\text { ROW }& \text { SLACK OR SURPLUS } & \text { DUAL PRICE } \\\hline2)&-&-.125000\\3)&-&-781250\end{array}$ NO. ITERATIONS= 2 RANGES IN WHICH THE BASIS IS UNCHANGED: $\begin{array}{lrrr}&&&\text { OBJ. COEFFICIENT RANGES }\\& \text { CURRENT}& \text {ALLOWABLE }& \text {ALLOWABLE }\\\text { VARIABLE }&\text { COEFFICIENT } & \text { INCREASE } & \text { DECREASE }\\\hline\mathrm{X} 1 &6.000000&-&-\\\mathrm{X} 2 & 7.500000 & 1.500000 & 2.500000 \\\mathrm{X} 3 & 10.000000 & 5.000000 & 3.571429\end{array}$ $\begin{array}{cccc}&&&\text { RIGHT HAND SIDE RANGES }\\&\text { CURRENT } & \text { ALLOWABLE }& \text { ALLOWABLE }\\\text { ROW } & \text { RHS } & \text { INCREASE } & \text { DECREASE } \\\hline2 & 2400.000000 & 1100.000000 & 900.000000 \\3 & 400.000000 & 240.000000 & 125.714300\end{array}$

Decision variables must be clearly defined before constraints can be written.

A section of output from The Management Scientist is shown here. $\begin{array} { c c c c } \frac { \text { Variable } } { 1 } & \frac { \text { Lower Limit } } { 60 } & \frac { \text { Current Value } } { 100 } & \frac { \text { Upper Limit } } { 120 }\end{array}$ What will happen to the solution if the objective function coefficient for variable 1 decreases by 20?

A section of output from The Management Scientist is shown here. $\frac { \text { Constraint } } { 2 }$$\frac { \text { Lower Limit } } { 240 }$$\frac { \text { Current Value } } { 300 }$$\frac { \text { Upper Limit } } { 420 }$ What will happen if the right-hand side for constraint 2 increases by 200?

When the right-hand sides of two constraints are each increased by one unit, the objective function value will be adjusted by the sum of the constraints' dual prices.

A negative dual price indicates that increasing the right-hand side of the associated constraint would be detrimental to the objective.

Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a minimization objective function and all $\ge$ constraints. $\text { Input Section }$ $\begin{array}{l}\text { Objective Function Coefficients }\\\begin{array}{c|c}\hline X & Y \\\hline 5 & 4\end{array}\end{array}$ $\begin{array}{|c|c|c|c|}\hline \text { Constraints } & & & \text { Req'd. } \\\hline \# 1 & 4 & 3 & 60 \\\hline \# 2 & 2 & 5 & 50 \\\hline \# 3 & 9 & 8 & 144 \\\hline\end{array}$ $\text { Output Section }$ $\begin{array}{|l|c|c|c|}\hline \text { Variables } & 9.6 & 7.2 & \\\hline \text { Profit } & 48 & 28.8 & 76.8 \\\hline\end{array}$ $\begin{array}{|c|c|c|}\hline \text { Constraint } & \text { Usage } & \text { Slack } \\\hline \# 1 & 60 & 1.35 \mathrm{E}-11 \\\hline \# 2 & 55.2 & -5.2 \\\hline \# 3 & 144 & -2.62 \mathrm{E}-11 \\\hline\end{array}$ a. Give the original linear programming problem. b. Give the complete optimal solution.

The reduced cost for a positive decision variable is 0.

Which of the following is not a question answered by sensitivity analysis?

How is sensitivity analysis used in linear programming? Given an example of what type of questions that can be answered.

The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the

Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a maximization objective function and all $\le$ constraints. Input Section $\begin{array}{l}\text { Objective Function Coefficients }\\ X & Y \\ 4 & 6\end{array}$ $\begin{array}{|c|c|c|c|}\hline \text { Constraints } & & & \text { Avail. } \\\hline \# 1 & 3 & 5 & 60 \\\hline \# 2 & 3 & 2 & 48 \\\hline \# 3 & 1 & 1 & 20 \\\hline\end{array}$ $\text { Output Section }$ $\begin{array}{|l|c|c|c|}\hline \text { Variables } & 13.333333 & 4 & \\\hline \text { Profit } & 53.333333 & 24 & 77.333333 \\\hline\end{array}$ $\begin{array}{|c|c|c|}\hline \text { Constraint } & \text { Usage } & \text { Slack } \\\hline \# 1 & 60 & 1.789 \mathrm{E}-11 \\\hline \# 2 & 48 & -2.69 \mathrm{E}-11 \\\hline \# 3 & 17.333333 & 2.6666667 \\\hline\end{array}$ a.Give the original linear programming problem. b.Give the complete optimal solution.