The exam is reasonably straight forward. There are 40 multiple choice questions worth 1.5 points each. Please clearly mark your solutions on the answer sheet on page 9.
There are two LP formulations worth 20 points each. You do NOT need to solve the problem. It is just a formulation.
 Decision alternatives
 should be identified before decision criteria are established.
 are limited to quantitative solutions
 are evaluated as a part of the problem definition stage.
 are best generated by brainstorming.
 Decision criteria
 are the choices faced by the decision maker.
 are the problems faced by the decision maker.
 are the ways to evaluate the choices faced by the decision maker.
 must be unique for a problem.
 Problem definition
 includes specific objectives and operating constraints.
 must occur after to the quantitative analysis process.
 must involve the analyst and but not the user of the results.
 must involve the user of the results and but not the analyst.
 A model that uses a system of symbols to represent a problem is called
 mathematical.
 iconic.
 analog.
 constrained.
 The maximization or minimization of a quantity is the
 goal of management science.
 decision for decision analysis.
 constraint of operations research.
 objective of linear programming.
 Decision variables
 tell how much or how many of something to produce, invest, purchase, hire, etc.
 represent the values of the constraints.
 measure the objective function.
 must exist for each constraint.
 Which of the following is a valid objective function for a linear programming problem?
 Max 5xy
 Min 4x + 3y + (2/3)z
 Max 5x^{2 } + 6y^{2}
 Min (x_{1} + x_{2})/x_{3}
 Which of the following statements is NOT true?
 A feasible solution satisfies all constraints.
 An optimal solution satisfies all constraints.
 An infeasible solution violates all constraints.
 A feasible solution point does not have to lie on the boundary of the feasible region.
 Slack
 is the difference between the left and right sides of a constraint.
 is the amount by which the left side of a < constraint is smaller than the right side.
 is the amount by which the left side of a > constraint is larger than the right side.
 exists for each variable in a linear programming problem.
 The improvement in the value of the objective function per unit increase in a righthand side is the
 sensitivity value.
 dual price.
 constraint coefficient.
 slack value.
 As long as the slope of the objective function stays between the slopes of the binding constraints
 the value of the objective function won’t change.
 there will be alternative optimal solutions.
 the values of the dual variables won’t change.
 there will be no slack in the solution.
 A constraint that does not affect the feasible region is a
 nonnegativity constraint.
 redundant constraint.
 standard constraint.
 slack constraint.
 Whenever all the constraints in a linear program are expressed as equalities, the linear program is said to be written in
 standard form.
 bounded form.
 feasible form.
 alternative form.
 All of the following statements about a redundant constraint are correct EXCEPT
 A redundant constraint does not affect the optimal solution.
 A redundant constraint does not affect the feasible region.
 Recognizing a redundant constraint is easy with the graphical solution method.
 At the optimal solution, a redundant constraint will have zero slack.
 All linear programming problems have all of the following properties EXCEPT
 a linear objective function that is to be maximized or minimized.
 a set of linear constraints.
 alternative optimal solutions.
 variables that are all restricted to nonnegative values.
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Questions 16 – 19
Use this graph to answer the questions._{ }
Max 20X + 10Y_{ }
s.t. 12X + 15Y < 180
15X + 10Y < 150
3X – 8Y < 0
X, Y > 0
 Which area (I, II, III, IV, or V) forms the feasible region?
 Area I is the feasible region
 Area II is the feasible region
 Area III is the feasible region
 Area IV is the feasible region
 Which point (A, B, C, or D) is optimal?
 A
 B
 C
 D
 Which constraints are binding?
 Contraint 1
 Constraints 1 and 2
 Constraints 2 and 3
 Constraint 3
 Which slack variables are zero?
 S_{1} and S_{2}
 S_{1} and S_{3}
 S_{3}
 S_{2} and S_{3}
 In a linear programming problem,
 the objective function and the constraints must be quadratic functions of the decision variables.
 the objective function and the constraints must be nonlinear functions of the decision variables.
 the objective function must be linear functions of the decision variables but and the constraints may be nonlinear functions.
 the objective function and the constraints must be linear functions of the decision variables.
 A negative dual price for a constraint in a minimization problem means
 as the righthand side increases, the objective function value will increase.
 as the righthand side decreases, the objective function value will increase.
 as the righthand side increases, the objective function value will decrease.
 as the righthand side decreases, the objective function value will decrease.
 If a decision variable is not positive in the optimal solution, its reduced cost is
 what its objective function value would need to be before it could become positive.
 the amount its objective function value would need to improve before it could become positive.
 zero.
 its dual price.
 A constraint with a positive slack value
 will have a positive dual price.
 will have a negative dual price.
 will have a dual price of zero.
 has no restrictions for its dual price.
 The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the
 optimal solution.
 dual solution.
 range of optimality.
 range of feasibility.
 The range of feasibility measures
 the righthandside values for which the objective function value will not change.
 the righthandside values for which the values of the decision variables will not change.
 the righthandside values for which the dual prices will not change.
 the lefthandside values for which the values of the decision variables will not change.
 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
 the maximum premium (say for overtime) over the normal price that the company would be willing to pay.
 the upper limit on the total hourly wage the company would pay.
 the reduction in hours that could be sustained before the solution would change.
 the number of hours by which the righthand side can change before there is a change in the solution point.
 A section of output is shown here.
Variable

Current Coefficient

Allowable Increase

Allowable Decrease

X1

100.000000

20.000000

40.000000

What will happen to the solution if the objective function coefficient for variable 1 decreases by 20?
 Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same.
 The value of the objective function will change, but the values of the decision variables and the dual prices will remain the same.
 The same decision variables will be positive, but their values, the objective function value, and the dual prices will change.
 The problem will need to be resolved to find the new optimal solution and dual price.
 A section of output is shown here.
Row

Current RHS

Allowable Increase

Allowable Decrease

X1

100.000000

20.000000

40.000000

What will happen if the righthandside for constraint 2 increases by 200?
 Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same.
 The value of the objective function will change, but the values of the decision variables and the dual prices will remain the same.
 The same decision variables will be positive, but their values, the objective function value, and the dual prices will change.
 The problem will need to be resolved to find the new optimal solution and dual price.
 The amount that the objective function coefficient of a decision variable would have to improve before that variable would have a positive value in the solution is the
 dual price.
 surplus variable.
 reduced cost.
 upper limit.
 The dual price measures, per unit increase in the right hand side,
 the increase in the value of the optimal solution.
 the decrease in the value of the optimal solution.
 the improvement in the value of the optimal solution.
 the change in the value of the optimal solution.
 Sensitivity analysis information in computer output is based on the assumption of
 no coefficient change.
 one coefficient change.
 two coefficient change.
 all coefficients change.
 The amount by which an objective function coefficient would have to improve before it would be possible for the corresponding variable to assume a positive value in the optimal solution is called the
 reduced cost.
 relevant cost.
 sunk cost.
 dual price.
 Which of the following is not a question answered by sensitivity analysis?
 If the righthand side value of a constraint changes, will the objective function value change?
 Over what range can a constraint’s righthand side value without the constraint’s dual price possibly changing?
 By how much will the objective function value change if the righthand side value of a constraint changes beyond the range of feasibility?
 By how much will the objective function value change if a decision variable’s coefficient in the objective function changes within the range of optimality?
LINDO output is given for the following linear programming problem.
MIN 12 X1 + 10 X2 + 9 X3
SUBJECT TO
2) 5 X1 + 8 X2 + 5 X3 >= 60
3) 8 X1 + 10 X2 + 5 X3 >= 80
END
LP OPTIMUM FOUND AT STEP 1
OBJECTIVE FUNCTION VALUE
1) 80.000000
Variable

Value

Reduced Cost

X1

.000000

4.000000

X2

8.000000

.000000

X3

.000000

4.000000

Row

Slack or Surplus

Dual Price

2)

4.000000

.000000

3)

.000000

1.000000

RANGES IN WHICH THE BASIS IS UNCHANGED:


Obj. Coefficient Ranges

Variable

Current
Coefficient

Allowable
Increase

Allowable
Decrease

X1

12.000000

INFINITY

4.000000

X2

10.000000

5.000000

10.000000

X3

9.000000

INFINITY

4.000000



Righthand Side Ranges

Row

Current
RHS

Allowable
Increase

Allowable
Decrease

2

60.000000

4.000000

INFINITY

3

80.000000

Infinity

5.000000

 What is the solution to the problem?
 x_{1} = 4, x_{2} = 0, x_{3} = 4, s_{1} = 4, s_{2} = 0, z = 800
 x_{1} = 0, x_{2} = 8, x_{3} = 0, s_{1} = 4, s_{2} = 0, z = 8
 x_{1} = 0, x_{2} = 8, x_{3} = 0, s_{1} = 4, s_{2} = 0, z = 80
 x_{1} = 12, x_{2} = 10, x_{3} = 9, s_{1} = 0, s_{2} = 0, z = 80
 Which constraints are binding?
 Constraint 1
 Constraint 2
 Constraint 3
 Constraint 4
 Interpret the reduced cost for x_{1}.
 c_{1} would have to decrease by 4 or more for x_{1} to become positive.
 c_{1} would have to increase by 4 or more for x_{1} to become positive.
 c_{1} would have to decrease by 4 or more for x_{1} to become negative.
 c_{1} would have to increase by 4 or more for x_{1} to become negative.
 Interpret the dual price for constraint 2.
 Increasing the righthand side by 1 will cause a negative improvement, or decrease, of 1 in this minimization objective function.
 Increasing the righthand side by 1 will cause a negative improvement, or increase, of 1 in this minimization objective function.
 Increasing the righthand side by 1 will cause an improvement, or increase, of 1 in this minimization objective function.
 Increasing the righthand side by 1 will have no effect in this minimization objective function.
 What would happen if the cost of x_{1} dropped to 10 and the cost of x_{2} increased to 12?
 The solution would change.
 The value of the objective function would not change..
 Constraint 1 would become binding.
 The solution would not change.
.
 The reduced cost
 for a positive decision variable is 1.
 for a negative decision variable is 0.
 for a positive decision variable is 0.
 for a positive decision variable is 1.
 The constraint 5x_{1} – 2x_{2}< 0
 passes through the point (50, 20).
 passes through the point (20, 50).
 passes through the point (10, 50).
 passes through the point (50, 50).
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PART II – Formulate the following linear program (20 points)
CLEARLY IDENTIFY YOUR VARIABLES. THE FORMULATION SHOULD BE WRITTEN IN A FORM READY FOR INPUT (i.e. variables on the left and constants on the right).
A company wants a high level, aggregate production plan for the next 6 months. Projected orders for the company’s product are listed in the table. Over the 6month period, units may be produced in one month and stored in inventory to meet some later month’s demand. Because of seasonal factors, the cost of production is not constant, as shown in the table.
The cost of holding an item in inventory for 1 month is $4/unit/mo. Items produced and sold in the same month are not put in inventory. The maximum number of units that can be held in inventory is 250. The initial inventory level at the beginning of the planning horizon is 200 units; the final inventory level at the end of the planning horizon is to be 100. The problem is to determine the optimal amount to produce in each month so that demand is met while minimizing the total cost of production and inventory. Shortages are not permitted.

Demand

Production

Month

(units)

cost ($/unit)

1

1300

100

2

1400

105

3

1000

110

4

800

115

5

1700

110

6

1900

110

PART III – Formulate the following linear program (20 points)
CLEARLY IDENTIFY YOUR VARIABLES. THE FORMULATION SHOULD BE WRITTEN IN A FORM READY FOR INPUT (i.e. variables on the left and constants on the right).
CSL is a chain of computer service stores. The number of hours of skilled repair time that CSL requires during the next five months is as follows:
January 
6,000 hours 
February 
7,000 hours 
March 
8,000 hours 
April 
9,500 hours 
May 
11,000 hours 
At the beginning of January, 50 skilled technicians work for CSL. Each skilled technician can work up to 160 hours per month. To meet future demand, new technicians must be trained. It takes one month to train a new technician. During the month of training, a trainee must be supervised for 50 hours by an experienced technician. Each experienced technician is paid $2,000 a month (even if he or she does not work the full 160 hours). During the month of training, a trainee is paid $1,000 a month. At the end of each month 5% of CSL’s experienced technicians quit to join Plum Computers. Formulate (do not solve) a linear program whose solution will enable CSL to minimize labor cost incurred in meeting the service requirements for the next five months
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