Resource Allocation and Decision Analysis (ECON 8010) – Spring 2014
“Linear Programming”
Reading: “Note on Linear Programming” (ECON 8010 Coursepak, Page 7) and “Merton Truck
Company” (ECON 8010 Coursepak, Page 19)
Definitions and Concepts:

Linear Programming – mathematical techniques used for solving constrained optimization
problems in which the objective function and each of the constraints can be stated as linear
functions of the choice variables

General ingredients of a constrained optimization problem:
1. Decision Variables – levels of various activities over which the decision maker has a
choice
2. Objective Function – a mathematical function describing the desirability of all the
different possible outcomes that could be realized (i.e., a measure of the desirability
resulting from of all the different possible choices of the decision variables)
3. Constraints – restrictions on the acceptable choices of the decision variables by the
decision maker

Redundant Constraint – a constraint is redundant if the restriction imposed by the
constraint is satisfied at all combinations of the choice variables collectively satisfying all of
the other constraints

Slack – a measure of the amount of each available scarce resource that remains unused at the
solution to the linear programming problem

Shadow Price – the benefit of increasing the available amount of a scarce resource,
measured by the resulting increase in the value of the objective function at the solution to the
linear programming problem

Lower Range – maximum decrease in the available amount of a scarce resource for which
the calculated shadow price is valid

Upper Range – maximum increase in the available amount of a scarce resource for which
the calculated shadow price is valid

Simplex Method – an algorithmic procedure developed by the late mathematician George
Dantzig for solving Linear Programming problems of any size
“Cost Minimization” as a Constrained Optimization Problem:
 suppose a firm produces output (denoted by q ) by hiring two different inputs (denoted by x1
and x2 ) according to the production function F ( x1 , x2 )
 inputs are hired in competitive factor markets, at constant per unit prices of w1 and w2
 firm wants to produce some target level of output (denoted q ) as inexpensively as possible
 We can frame this as a Constrained Optimization Problem with
1. Decision Variables of: x1 and x2
2. an Objective Function of: w1 x1  w2 x2
3. a Constraint of: F ( x1 , x2 )  q
Q:
A:
Is this a constrained optimization problem which “fits the mold of Linear Programming”?
It depends upon the functional form of F ( x1 , x2 ) …
 recall, for Linear Programming we need the objective function to be a linear function
of the choice variables (which w1 x1  w2 x2 clearly is)
 we also need the constraints to be linear functions of the choice variables => whether
this is the case depends upon the functional form of F ( x1 , x2 )  q
To see that our objective function is indeed linear, let’s illustrate it graphically…
 helpful to draw the objective function for varying realized values of production costs
(e.g.,
c24 , c36 , c60 , and c84 )
x2 “ c84 isocost line”  Combinations of the
two inputs which cost exactly $8,400 dollars
“ c60 isocost line”  Combinations of the
two inputs which cost exactly $6,000 dollars
“ c24 isocost line”  Combinations of the
two inputs which cost exactly $2,400 dollars
x1
0 0 
note: the slope of each isocost curve is simply   w1 


so, again, the Objective Function for this problem is clearly linear
the goal of the firm is to Minimize Production Costs => they want to be on the isocost
line closest to the origin, while still producing the target level of output
w 
 2
To illustrate combinations of x1 and x2 satisfying the constraint, we must assume a functional
form of F ( x1 , x2 ) .
(i)
Perfect Substitutes => F ( x1 , x2 )  x1  x2 => the constraint becomes x1  x2  q
x2
Constraint is a linear
function => Linear
Programming is
applicable
x1 0 0
(ii)
Perfect Complements => F ( x1 , x2 )  min{ x1 , x2 } => the constraint of min{x1 , x2 }  q
actually becomes a dual restriction of x1  q and x2  q
x2
Constraints are linear
functions => Linear
Programming is
applicable
Region in
which “both
restrictions”
are satisfied
x1 0 0
(iii)
Cobb-Douglas Production Function => F ( x1 , x2 )  Ax1 x2 => the constraint becomes
x2
Ax1 x2  q
Constraints are not
linear functions =>
Linear Programming
is NOT applicable
x1 0 0
Example 1 – Insulation Production (Coursepak, Page 7):
1.
Specify the objective function of the plant manager.
2.
State equations for all of the constraints facing the plant manager.
3.
Formally state the Linear Programming problem.
4.
Graphically illustrate the “feasible set” of the decision variables.
5.
Determine the solution to this Linear Programming problem.
1.


2.

Specify the objective function of the plant manager.
First recognize that the decision variables can be thought of as “the number of truckloads
of ‘Type B’ insulation” (denoted B ) and “the number of truckloads of ‘Type R’
insulation” (denoted R )
 Often, we have some flexibility in “how we specify the problem” => we could have
thought of our decision variables as “tons of ‘Type B’ insulation” and “tons of ‘Type
R’ insulation”
 In this case, there is no “right” or “wrong” approach => either specification will give
the correct insights (so long as we are consistent throughout the analysis)
Thus, the objective is to maximize contribution, which is equal to: 950B  1,200R
State equations for all of the constraints facing the plant manager.
The plant manager faces several restrictions on her choice
(i)
“…can produce any mix of output, as long as the total weight is no more than 70
tons per day…One truckload of “Type B” insulation weighs 1.4 tons; one truckload
of “Type R” weighs 2.8 tons.” => (1.4) B  (2.8) R  70
(ii)
“…loading facilities can handle up to 30 trucks per day” => B  R  30
(iii) “…can obtain at most 65 canisters of the (flame retarding) agent per day. One
truckload of (finished) “Type B” insulation requires an input of three canisters of
the agent, but one truckload of “Type R” insulation requires only one canister.” =>
3B  R  65
(iv)
3.

implicit assumptions of B  0 and R  0
Formally state the Linear Programming problem.
The plant manager faces the Linear Programming problem of:
Maximize 950 B  1,200 R
Subject to:
(1.4) B  (2.8) R  70
B  R  30
3B  R  65
B0
and R  0
4.
Graphically illustrate the “feasible set” of the decision variables.

R
 1.4
70
B
 (.5) B  25
2.8
2.8
R Slope of line is
equal to (-.5)
“Machine
Constraint”
25 0 B
50
0 
R  (1) B  30
R Slope of line is
equal to (-1)
30 0 B
0 
“Loading
Constraint”
30
R  (3) B  65
R 65 “Flame
Retarding
Agent
Constraint”
Slope of line is
equal to (-3)
0 B
0 65/3=21.6666…
(?)


What does the picture look like when we draw all three constraints in the same graph?
“horizontal intercepts” can be ranked from smallest to largest as “blue < red < green”…
“vertical intercepts” can be ranked from smallest to largest as “green < red < blue”…
But, which of the two following cases do we have?
R
R or
0 B
0 0 B
0 To figure this out, start by determining the intersection of “blue” and “green”…
 at this intersection R  (.5) B  25 and R  (3) B  65 must both be satisfied
simultaneously
 ( 3) B  65  ( .5) B  25  40  (2.5) B  B  240.5  16
 From which it follows R  (.5)(16)  25  (3)(16)  65  17
And then figure out if this point is “above” or “below” “red”…
 recall, the equation for “red” is R  (1) B  30


thus, if B  16 , then R  ( 1)(16)  30  14
since 14  17 it follows that “red” passes below the point of intersection between “blue”
and “green” (i.e., the correct picture is the one above on the left)
From here we can determine the point of intersection between…
 “red” and “green” as: (1) B  30  ( .5) B  25  (.5) B  5  B  10 and R  20
 “red” and “blue” as: (1) B  30  (3) B  65  2 B  35  B  17.5 and R  12.5
Thus, the feasible set is:
R 25 20 12.5 0 B 0 10 17.5 21. 6
Note that if we instead had…
R 0 B
0 …then the constraint corresponding to “red” would be redundant
5.

(i)
Determine the solution to this Linear Programming problem.
From the graphical illustration of the feasible set, it follows that the solution must be (for
a linear objective function with any arbitrary slope) at one of the four “corner points”
Vertical Intercept is optimal (if “orange” is flatter than “green”)
R 0 B 0 (ii)
Intersection of “green” and “red” is optimal (if “orange” is steeper than “green” but
flatter than “red”)
R 0 B 0 (iii)
Intersection of “red” and “blue” is optimal (if “orange” is steeper than “red” but flatter
than “blue”)
R 0 B 0 (iv)
Horizontal Intercept is optimal (if “orange” is steeper than “blue”)
R 0 B 0 note:
 mathematically, it is also possible to have “multiple solutions”
 for example, if slope of “orange” had been exactly equal to slope of “red”...
R 0 
B 0 then any point along “red” would be best => multiple solutions (each yielding the same
maximal value of the objective function)
For the current problem, recall that the objective function of this firm is 950B  1,200R
1
 950
B
v
 Setting v  950 B  1,200 R and solving for R , we obtain R 
1,200
1,200
 slope of the objective function is equal to  950   19 
1,200


 24 
Recall, slope of “green” is    , slope of “red” is  1 , and slope of “blue” is  3
Thus, the objective function is steeper than “green” but flatter than “red” => the
*
solution to the problem is B  10 and R*  20
1
2
R R*  20 0 0 B  10
*
B Further Insights on the Solution:

Again, observe that the solution occurs at the intersection of “red” and “green”
R R*  20 0 B 0 

B*  10
Thus, the constraints illustrated by “green” (i.e., the “machine constraint” of
(1.4) B  (2.8) R  70 ) and “red” (i.e., the “loading constraint” of B  R  30 ) are
both binding => there is “zero slack” for the associated scarce resources at the
solution => these two inputs are the “bottlenecks” in the operation
However, the constraint illustrated by “blue” (i.e., the “flame retarding agent constraint”
of R  3 B  65 ) is NOT BINDING => at the optimal solution, the firm is not using all
of the available “flame retarding agent”
 More precisely, they are only using 20  3(10)  50 units of the “flame retarding
agent”
 Slack of “15” (i.e., 65  50  15 , difference between the available amount and the
amount used) for this resource
 The amount of available flame retarding agent could decrease to 50 units without
altering the solution
(?)
How much would this firm value “additional loading capacity”?
 That is, what if the “loading constraint” (of B  R  30 ) could be altered by
increasing the value of the term on the right-hand side? By how much would the firm
value such an increase?
 Visually:
New “red”
R constraint
New
“optimal
choice”
0 B 0 
As “loading capacity” is increased…
 some points that were not initially feasible are now feasible (illustrated by the “purple
shaded area”)
 the intersection of “green” and “red” move down (“southeast”) along the “green” line
 so long as the increase in machine capacity is sufficiently small, the solution still
occurs at the intersection of “green” and “red” => the new optimal choice is as
illustrated above
 note: this occurs along an “orange” line that is further from the origin => value of the
objective function at the solution is indeed larger
Algebraically:
 state the “loading constraint” in more general terms as: B  R  L => R   B  L
 note, the intersection of “red” and “green” now occurs at:
 B  L   12 B  25 => 12 B  L  25

=> B  2 L  50 and R  2 L  50  L  50  L
so long as the increase in loading capacity is sufficiently small, the solution is now
B*  2 L  50 and R*  50  L => the value of the objective function at the solution is
now
v*  9502 L  50  1,20050  L 
 1,900 L  47,500  60,000  1,200 L
 700 L  12,500


*
as L is increased, v * (i.e., optimal value of objective function) increases by dv
dL  700
the “shadow price” of “loading capacity” is $700 => the firm would be willing to pay up
to $700 to increase “loading capacity” from 30 up to 31
Note: in the discussion above we assumed that L was “sufficiently small”…
 if loading capacity becomes “too big,” then we eventually have…
R 0 B
0 

…in which case the solution is at the intersection of “green” and “blue”
this “cutoff” value of L is the unique value for which all three constraints intersect at the
same point
 for “general L ,” the intersection between “red” and “green” occurs at B  2 L  50
 recall, the intersection between “green” and “blue” occurs at B  16
 thus, the previous discussion applies only if
2 L  50  16
L  33
But, we can also see that the feasible set (and therefore the solution) is qualitatively different if
L is “too small.” For “very small L ” we would have…
R 0 B
0 …in which case the solution is at the vertical intercept of “red”
 this will occur for values of L smaller than the unique value for which the intersection of
“red” and “green” occurs at B  0
 that is, for
0  2 L  50
L  25
Shadow Price of $700 for Loading Capacity is valid for (all other factors fixed) 25  L  33
 since we are starting with a value of L  30 , we say that the
 Lower Range for Loading Capacity is equal to 5
 Upper Range for Loading Capacity is equal to 3
Similarly, for “machine capacity”
Algebraically:
 state “machine constraint” in more general terms as: R   12 B 
 note, the intersection of “red” and “green” now occurs at:
 B  30   12 B  21.8 M => 12 B  30  21.8 M

1
2.8
M
=> B  60  1M.4 and R  30  60  1M.4   1M.4  30
so long as the increase in machine capacity is sufficiently small, the solution is now
B*  60  1M.4 and R *  1M.4  30 => value of the objective function at the solution is now
v*  95060  1M.4   1,2001M.4  30
 21,000  250
1.4 M


thus, as M is increased, v * increases by dM  1.4  178.57
the “shadow price” of “machine capacity” is $178.57 => the firm would be willing to pay
up to $178.57 to increase “machine capacity” from 70 up to 71
dv*
250
Further, it can be shown that
 the Lower Range for Machine Capacity is 10.50
 the Upper Range for Machine Capacity is 14.00
(?)
(A)




What about the Shadow Price for the “flame retarding agent”?
Start by using “economic intuition”…
recall the definition of “Shadow Price” (benefit of increasing the available amount of a
scarce resource, measured by the resulting increase in the value of the objective function
at the solution to the linear programming problem)
the constraint on the “flame retarding agent” was NOT BINDING => at the solution, the
firm was not using all of the available units of this input
thus, the solution would not change if the available amount of “flame retarding agent”
were increased => Shadow Price is equal to zero
further, it follows that the Upper Range for this Shadow Price is infinity (i.e., the shadow
price will still equal zero even as the available amount of the “flame retarding agent” is
increased to infinity)
Recognize that in general:
 if Slack is equal to zero, then Shadow Price will be positive
 if Slack is positive, then Shadow Price is equal to zero
“Merton Truck Company”: (Page 19 of Coursepak)

Choose levels of production for two different types of trucks: x1  (units of Model 101) and
x2  (units of Model 102)

Constraints of:
Engine Assembly constraint:
Metal Stamping constraint:
Model 101 Assembly constraint:
Model 102 Assembly constraint:
x1  2 x2  4,000
2 x1  2 x2  6,000
2 x1  5,000
3 x2  4,500

Objective: maximize contribution
 Per Unit Costs for Model 101 of:
(materials) + (labor) + (variable overhead)
= $24,000 + $4,000 + $8,000 = $36,000
 Per Unit Revenue for Model 101 of $39,000
 Per Unit Costs for Model 102 of:
(materials)+(labor)+(variable overhead)
= $20,000 + $4,500 + $8,500 = $33,000
 Per Unit Revenue for Model 102 of $38,000

Thus, the Objective Function is: v  3,000x1  5,000x2
Questions:
1a. Find the best product mix for Merton.
1b. Determine the Shadow Price of engine assembly capacity.
1d. Determine the Upper Range for the shadow price of engine assembly capacity
determined in (1b).
2.
Merton’s production manager suggests purchasing Model 101 or Model 102 engines
form an outside supplier in order to relieve the capacity problem in the engine assembly
department. If Merton decides to pursue this alternative, it will be effectively “renting”
capacity: furnishing the necessary materials and engine components, and reimbursing
the outside supplier for labor and overhead. Should the company adopt this alternative?
If so, what is the maximum rent it should be willing to pay for a machine-hour of engine
assembly capacity? What is the maximum number of machine hours it should rent?
Note that the constraints can be expressed as x2 
1a.
1
2
x1  2,000 , x2   x1  3,000 ,
x1  2,500 , and x2  1,500 => the feasible set can be illustrated as:
x2 3,000 2,000 1,500 1,000 x1
0 0 
2,000 2,500
1
Slope of “green” is 2 ; slope of “blue” is
3,000
4,000 1.
3
Slope of the objective function ( v  3,000 x1  5,000 x2 ) is 5
 Optimal output combination is x1*  2,000 and x2*  1,000

1b.

To determine the Shadow Price of “engine assembly capacity,” start by stating the Engine
Assembly constraint in more general terms as x1  2 x2  E => x2  21 x1  E2
from here, solve for the point of intersection between “green” and “blue” as:
1
2
x1  E2   x1  3,000

 x1  6,000  E and x2  E  3,000
at this point, the value of the objective function is

thus, the Shadow Price for engine assembly capacity is equal to
v*  3,000(6,000  E )  5,000( E  3,000)
 3,000,000  2,000E
dv*
dE
 2,000
1d.




2.
As “engine assembly capacity” is increased, “green” shifts outward…
the solution is still at the intersection of “green” and “blue” until “green” shifts out
beyond the intersection of “blue” and “purple”
the intersection of “blue” and “purple” occurs at
1,500   x1  3,000 => x1  1,500
thus, the solution is still at the intersection of “green” and “blue” so long as
6,000  E  1,500  E  4,500
starting at E  4,000 , the value of E could increase by 500 and the Shadow Price of
2,000 still applies (i.e., the Upper Range for the Shadow Price is 500)
Based upon the answers to (1b) and (1d), Merton should be willing to pay up to $2,000
per unit to rent additional “engine assembly capacity,” up to a maximum of 500 units of
engine assembly capacity.
Mathematical possibility of an “empty feasible set” (in which case the problem cannot be solved)
 for example, suppose we had constraints of
1. x1  50 (“blue”)
2. x2  40 (“yellow”)
3. x1  x2  70 (“red”)
x2
x2 0 x1
0 0 x1
0  No points for which all three constraints are satisfied!
 “Feasible Set” is an “Empty Set”!
 When we have more than two choice variables (in which case it is not so easy to
graphically illustrate the “feasible set”), then this possibility is not so easy to detect
Multiple Choice Questions:
1.
Which of the following is a valid Objective Function for a Linear Programming problem?
x1 x2
A.
B.
C.
D.
500 x1  4,000 x2  10,000 x3
( x1  x2 )( x3  x4 )
More than one (perhaps all) of the above answers is correct.
2.
_________________ refers to a measure of the amount of each available scarce resource
that remains unused at the solution to a Linear Programming problem.
A.
Slack
B.
Shadow Price
C.
A Redundant Constraint
D.
Lower Range
3.
Consider a Linear Programming problem in which a decision-maker wants to maximize
the function 5 x1  50 x 2  10 x3 . Suppose the “feasible set” is defined by the five
“corner points,”
( x1 , x 2 , x3 ) :
X A  (0,10,1) ,
X B  (6,0,5) ,
X C  (5,8,4) ,
X D  ( 2,4,0) , and X E  (0,0,0) . It follows that the solution to this problem is
X A  (0,10,1) .
A.
X B  (6,0,5) .
B.
C.
D.
4.
X C  (5,8,4) .
More than one of the above answers is correct (i.e., the problem has multiple
solutions).
Consider a Linear Programming problem with the following three constraints:
4 x1  8 x 2  120 , 10 x2  20 x3  200 , and 12 x1  6 x2  24 x3  180 . Which of the
following combinations of ( x1 , x 2 , x3 ) are feasible?
A.
B.
C.
D.
5.
(2,7,5)
(3,8,4)
(7,12,1)
More than one of the above choices is feasible.
Which of the following is NOT one of the three general ingredients which must be
specified when stating a constrained optimization problem?
A.
Decision Variables.
B.
The Objective Function.
C.
Slack.
D.
More than one of the above answers is correct (i.e., more than one of the above is
not one of the “three general ingredients”).
6.
_____________________ developed an algorithmic procedure (known as the Simplex
Method) for solving Linear Programming problems of any size.
A.
John von Neumann
B.
John Glenn
C.
Glenn Danzig
D.
George Dantzig
7.
Which of the following statements is correct?
A.
“At the solution to any Linear Programming problem, all of the constraints must
be binding.”
B.
“Linear Programming can be used to solve ‘maximization’ problems, but not
‘minimization’ problems.”
C.
“For a constraint, if ‘slack’ is equal to zero, then the ‘shadow price’ must be
positive.”
D.
More than one (perhaps all) of the above answers is correct.
8.
Consider a Linear Programming problem with the following two constraints:
2 x1  3 x 2  6,000 and x1  5 x 2  7,900 . Which of the following ( x1 , x 2 ) is a
“corner point” of the feasible set?
x1  750 and x 2  1,500 .
A.
B.
C.
D.
x1  900 and x 2  1,400 .
x1  2,000 and x 2  500 .
More than one (perhaps all) of the above answers is correct.
Problem Solving or Short Answer Questions:
1.
William runs a cattle ranch. He must choose levels of two different types of cattle feed
(denoted x and y ) in order to satisfy multiple minimal nutritional standards, while
minimizing the costs of purchasing cattle feed. Each pound of each type of feed contains
levels of three different nutrients ( A , B , and C ) as summarized in the table below:
Feed
x
y
Nutrient A
2 units
2 units
Nutrient B
5 units
1 unit
Nutrient C
4 units
2 units
Each pound of x costs p x  .60 ; and each pound of y costs p y  .20 . Each day
each cow must be feed at least 60 units of Nutrient A, 50 units of Nutrient B, and 80 units
of Nutrient C.
A.
What are the decision variables for William? What is his Objective Function?
B.
State inequalities which summarize the restrictions on his choice of the decision
variables. Graphically illustrate the “feasible set,” clearly labeling all relevant
intercepts and points of intersection.
C.
Determine and graphically illustrate the solution to his Linear Programming
problem. Determine the total expenditures on cattle feed at this optimal choice.
D.
E.
2.
A new study has been released which states that cows really only need “72 units
of Nutrient C per day.” How would your answer to part (C) change in light of this
new information? Explain.
Suppose instead that the study mentioned in part (D) had stated that cows only
need “68 units of Nutrient C per day.” How would your answer to part (C)
change in light of this new information? Explain.
Bob must choose output levels for three different products ( x , y , and z ) in order to
maximize total contribution, subject to three constraints (an “assembly line constraint,” a
“warehouse space constraint,” and a “loading dock constraint”). He has formulated and
solved his problem as a Linear Programming problem. Based upon his analysis, he has
determined that the optimal levels of output are x*  435 , y*  280 , and z*  115 ,
resulting in a total contribution of $128,560. The values of slack, shadow price, lower
range, and upper range for each of the three constraints are reported below:
Assembly Line Capacity
Warehouse Space Capacity
Loading Dock Capacity
Slack
0
200
0
Shadow Price
4,210
0
890
Lower Range
10.25
200
30.10
Upper Range
8.70
Infinity
16.50
An independent business located in the same industrial park as Bob’s business is
temporarily in need of additional warehouse space.
A.
They approach Bob with an offer to pay $5,000 to rent 150 units of warehouse
space from Bob’s company. Should Bob accept this offer? Explain why or why
not.
B.
Suppose that they had instead offered to pay only $2,000 to rent 150 units of
warehouse space. Should Bob accept this offer? Explain why or why not.
3.
Laura must choose values of x and y to maximize 1,200 x  800 y . Her choice is
restricted by the following constraints: (i) 3 x  4 y  480 , (ii) 2 x  y  200 , (iii)
y  90 , (iv) x  0 , and (v) y  0 .
A.
B.
C.
4.
Graphically illustrate the “feasible set,” clearly labeling all relevant intercepts and
points of intersection.
Determine and graphically illustrate the solution to his Linear Programming
problem. Determine the value of the objective function at this optimal choice.
For constraint (i), determine the value of slack, shadow price, upper range, and
lower range.
Hillary works for a public school district in central Texas. She is in charge of
determining which students will attend the two different existing high schools in the
district. “High School Alpha” has a capacity of 1,250 students; “High School Beta” has a
capacity of 900 students. She has divided the school district into four geographic areas
(“Region A,” “Region B,” “Region C,” and “Region D”). “High School Alpha” is
located within “Region A,” and “High School Beta” is located within “Region B.” The
table below reports the number of students residing in each of the four areas, as well as
the average distance (in miles) that a student in each area must travel to each school:
Region A
Region B
Region C
Region D
Number of Students
600
800
200
400
Distance to Alpha
1
7
10
9
Distance to Beta
7
1
14
2
Her goal is to assign students in each region to the two high schools in order to minimize
total transportation costs over the entire school year. There are a total of 180 school days
in her district. The daily cost of transporting each student to and from school is (.45) per
mile that the student resides from their assigned school.
Let a denote the number of students from “Region A” assigned to “High School
Alpha” (recognize that this implies 600  a students from “Region A” are assigned to
“High School Beta”). Likewise, let b denote the number of students from “Region B”
assigned to “High School Alpha,” let c denote the number of students from “Region C”
assigned to “High School Alpha,” and let d denote the number of students from “Region
D” assigned to “High School Alpha.”
A.
State Hillary’s Objective Function.
B.
State the two restrictions that her choices of a , b , c , and d must satisfy.
C.
Determine the optimal allocation of students across the two schools. Determine
the total annual transportation costs resulting from this choice.
D.
Determine the value of slack at each of the two schools at this optimal allocation.
E.
If the capacity of “High School Alpha” had been 1,350 students, how would the
solution and total annual transportation costs have differed? Explain.
F.
If the capacity of “High School Beta” had been 1,000 students, how would the
solution and total annual transportation costs have differed? Explain.
5.
Jaideep has solved a Linear Programming problem in which he had to choose the optimal
levels of two variables ( x1 and x 2 ), subject to three different constraints (“Constraint I,”
“Constraint II,” and “Constraint III”). He reports values for slack, shadow price, lower
range, and upper range for each of the three constraints as follows:
Constraint I
Constraint II
Constraint III
Slack
0
0
12
Shadow Price
35
16
8
Lower Range
4.50
12.75
3.20
Do you have any concerns with these reported values? Explain.
Upper Range
–5.75
4.25
12.50
Answers to Multiple Choice Questions:
1.
2.
3.
4.
5.
6.
7.
8.
B
A
A
B
C
D
C
B
Answers to Problem Solving or Short Answer Questions:
1A.
His decision variables are x (pounds of “feed x”) and y (pounds of “feed y”). His
Objective Function is: p x x  p y y  (.6) x  (.2) y (the value of which he aims to
1B.
minimize).
His choice is restricted by having to choose feed combinations for which the cattle get at
least 60 units of Nutrient A, at least 50 units of Nutrient B, and at least 80 units of
Nutrient C. Mathematically these constraints are: 2 x  2 y  60 , 5 x  y  50 , and
4 x  2 y  80 . Graphically:
y 50 40 33. 3 30 25 20 x
0 0 5 10
3. 3
20
30 1C.
Recognize that the Objective Function has a slope of
slopes of the constraints are
 p x  .6

 3 , while the
py
.2
 50
 40
 5 (“blue”),
 2 (“green”), and
20
10
 30
 1 (“red”). Thus, the solution is at the intersection of “blue” and “green.”
30
Algebraically, the optimal choice of x must satisfy  5 x  50  2 x  40 =>
x*  3. 3 . It follows that the optimal y is y*  33. 3 . Graphically:
y 50 33. 3 20 x
0 0 3. 3 1D.
2A.
30
If instead this final constraint were 4 x  2 y  72 , then the “red” line still passes below
the intersection of “green” and “blue.” Thus, the solution is still at the intersection of
“green” and “blue.” However, this intersection now occurs at:  5 x  50  2 x  36
=> x* 
1E.
10  4. 6 (with a corresponding value of y of y*  803  26 . 6 ).
For the final constraint of 4 x  2 y  Z , once the value of Z is dropped below
Z  70 , then the intersection of “blue” and “green” occurs below the “red” line, the
solution now occurs at the intersection of “red” and “blue” (i.e., x*  5 and y*  25 ).
14
3
Note that there is slack of 200 for the Warehouse Constraint. Thus, available warehouse
space could be decreased by 150 units (which is less than 200) without altering Bob’s
2B.
3A.
optimal choice. Because of this, he should be willing to accept $5,000 for these unused
150 units of warehouse space.
Based upon the logic above, if he were made a “take-it-or-leave-it” offer for any amount
of warehouse space up to 200 units for any positive payment, then he should accept the
offer. So, he should be willing to accept this even lower amount of only $2,000 for these
150 units of warehouse space.
Given the constraints of: (i) 3 x  4 y  480 , (ii) 2 x  y  200 , (iii) y  90 , (iv)
x  0 , and (v) y  0 , her feasible set is:
y 200 120 90 72 x
0 0 3B.
40 55 64
160 100
Recognize that the slope of the Objective function is
1, 200
800

3
2
, while the slopes of the
200
 43 (“red”), and 100
 2 (“blue”). Thus, the
optimal choice is x*  64 and y*  72 (i.e., at the intersection of “red” and “blue”).
This choice results in v*  1,200 (64 )  800 (72)  134,400 .
constraints are 0 (“green”),
120
160
y 90 72 0 x
0 40 64
100
3C.
Constraint (i) ( 3 x  4 y  480 , illustrated by the “red” line) is binding. Thus, slack is
equal to 0. The solution will still occur at the intersection of “red” and “blue” so long as
“red” intersects “blue” at a point with 55  x  100 . Expressing this constraint in more
general terms as 3 x  4 y  Z i (or equivalently as y 
x
Zi
4
), the intersection
 2 x  200 => x  160  Z5i . Thus,
this intersection occurs at 55  x  100 so long as 300  Z i  525 . Starting with
Z i  480 , it follows that the Upper Range is 45 and the Lower Range is 180. Finally,
between “red” and “blue” occurs at
3
4
x
3
4
Zi
4
for 300  Z i  525 , the solution is x*  160 


v*  1,200 160  Z5i  800
dv*
 80 .
is equal to dZ
i
4A.
4B.

2 Zi
5

Zi
5
and y* 
2 Zi
5
 120 , for which
 120  80 Z i  96,000 . Thus, the Shadow Price
Hillary’s Objective Function is
(180 )(. 45)1a  7 (600  a )  7b  1(800  b )  10 c  14 ( 200  c )  9 d  2( 400  d ) 
 81a  4,200  7 a  7b  800  b  10 c  2,800  14 c  9 d  800  2 d 
 818,600  6 a  6b  4c  7 d 
She aims to minimize the value of this function.
Her choice is restricted by the capacity of each school. For chosen values of a , b , c ,
and d , the total number of students attending “High School alpha” is simply
a  b  c  d . Recall, the capacity of this school is 1,250 students. Thus, she faces a
constraint of a  b  c  d  1,250 . Similarly, the total number of students attending
“High School Beta” is (600  a )  (800  b )  ( 200  c )  ( 400  d ) , which can
be simplified to 2,000  ( a  b  c  d ) . Since the capacity of “High School Beta” is
900 students, this implies a restriction of 2,000  ( a  b  c  d )  900 =>
1,100  a  b  c  d .
4C.
If the only restrictions on her choice were that 0  a  600 , 0  b  800 ,
0  c  200 , and 0  d  400 , then she would clearly minimize the value of the
objective function by choosing that a  600 , b  0 , c  200 , and d  0 . However,
this choice would violate the constraint of 1,100  a  b  c  d (since it would assign
1,200 students to “High School Beta,” a school with a capacity of only 900 students).
From here, in order to identify the solution to the constrained optimization problem, she
must reassign 300 students from “High School Beta” to “High School Alpha.” The two
types of students that should could potentially reassign are either “Region B” students or
“Region D” students. Reassigning one “Region D” student will increase the value of the
Objective Function by ($81)(7)=$567, whereas reassigning one “Region B” student will
increase the value of the Objective Function by only ($81)(6)=$486. Since she wants to
minimize the value of the Objective Function the best choice is to reassign 300 “Region
B” students from “High School Beta” to “High School Alpha.” Thus, her optimal choice
4D.
4E.
4F.
5.
is a  600 , b  300 , c  200 , and d  0 , which gives rise to annual transportation
costs of 818,600  3,600  1,800  800  0   81(6,000 )  486 ,000 .
At the optimal choice 1,100 students are assigned to “High School Alpha.” Since this
school has a capacity of 1,250 students, slack is equal to 150. At the optimal choice 900
students are assigned to “High School Beta.” Since this school has a capacity of 900
students, slack is equal to 0.
Increasing the availability of a scarce resource for which slack is positive at the solution
to the Linear Programming problem will in general not alter the solution. Thus, if the
capacity of “High School Alpha” had been 1,350 students (instead of 1,250 students), the
solution would not differ at all.
At the initial solution, “High School Beta” is filled to capacity. Thus, increasing capacity
of this school should alter the solution to the problem (and ultimately lead to a lower
value of the objective function). If the capacity had been 1,000 instead of only 900, then
the second constraint is 2,000  ( a  b  c  d )  1,000 => 1,000  a  b  c  d .
Thus (applying the same logic used in answering part (4C) above), the optimal choice is
a  600 , b  200 , c  200 , and d  0 . This choice leads to annual transportation
costs of 818,600  3,600  1,200  800  0   81(5, 400 )  437 , 400 . Thus,
having this extra capacity at “High School Beta” would allow the district to realize
transportation costs that are $48,600 less per year.
Based upon the reported values, there appear to be some mistakes in his results. First,
“Upper Range” specifies the maximum increase in the available amount of a scarce
resource for which the calculated shadow price is valid. The value of “Upper Range”
should always be greater than or equal to zero. Thus, the reported value of (–5.75) for
“Constraint I” is not an acceptable value for “Upper Range.” Second, it must always be
true that if “slack” is positive, then the corresponding shadow price is equal to zero. This
is not true for the values reported above corresponding to “Constraint III” (suggesting an
additional error).