Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
Chapter 3 – Linear Programming: Computer Solution and Sensitivity Analysis
(#3) Given the following Excel spreadsheet for a linear programming model and Solver window, indicate the formula for
cell B13 and fill in the Solver window with the appropriate information to solve the problem.
Answer: B13 contains: (B10*C4)+(B11*D4)+(B12*E4) = Profit (Z)
Solver Input: Target Cell: (Profit Z) B13; Maximize Z (cell B13); Change Cells: B10:B12
Constraints:
B10:B12 ≥ 0 (x1,x2,x3 ≥ 0); G6 ≤ F6 (Usage ≤ Available for A); G7 ≤ F7 (Usage ≤ Available for B)
Problem # 3
Products:
Profit per Unit:
(4)
Column C
Column D
Column E
1 (x1)
2 (x2)
3 (x3)
115.00
90.00
130.00
Resources:
Column F
Column G
Column H
Available
Usage
Left Over
Resource A
(6)
6.3
4.7
5.8
345
0.00
345.00
Resource B
(7)
18.1
11.8
14.6
710
0.00
710
Production
Column B
Product 1 =
(10)
Product 2 =
(11)
Product 3 =
(12)
Profit =
0.00
(13)
#5) The following model was solved graphically
in Chapter 2 (Problem 20):
Maximize Z = 5x1 + 8x2 ($, Profit)
Subject to the following constraints:
 3x1 + 5x2 ≤ 50 (units, resource 1)
 2x1+4x2 ≤ 40 (units, resource 2)
 1x1 + 0x2 ≤ 8 (units, resource 3)
 0x1 + 1x2 ≤ 10 (units, resource 4)
and also x1, x2 ≥ 0
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
Given the following Excel spreadsheet for this model, indicate the formulas in cells F6, F7, F8, F9, G6, G7, G8, G9, and
B14, and fill in the Solver window with the necessary information to solve the model. Solve the model using Excel.
Answer: Column F (Usage) = SUM {constraint coefficients (x1, x2) * Optimal objective variable’s Value (Col. B)}
Column G (Left Over Resource) = Each Constraint {Available Resource (Col. E) – Usage (Col. F)}
Problem # 5
(2)
Col. C
(3)Items:
1 (x1)
2 (x2)
5
8
(4)Profit per Item:
Col. D
(5)Constraints:
Col. E
Col. F
Col. G
F6 = C6*B12 + D6*B13 [Resource 1]
Available
Usage
Left Over
(6)
1
3
5
50
0
50
(7)
2
2
4
40
0
40
(8)
3
1
0
8
0
8
(9)
4
0
1
10
0
10
F7 = C7*B12 + D7*B13 [Resource 2]
F8 = C8*B12 + D8*B13 [Resource 3]
F9 = C9*B12 + D9*B13 [Resource 4]
G6 = E6 – F6
Production
Col. B
(12)
1=
(13)
2=
(14)
Z=
G7 = E7 – F7
G8 = E8 – F8
G9 = E9 – F9
#6) Given the following graph of a linear programming model with a single constraint (7x1 + 6x2 = 420), and the
objective function; maximize Z = 30x1 + 50x2, determine the optimal solution point.
Slope = (Vertical Change, Δx2) ÷ (Horizontal Change, Δx1)
from point A(0, 70) to point B(60, 0)
Slope = (0 – 70) ÷ (60 - 0) = (-70) ÷ (60) = (-7/6) = (- 1.17)
Constraint (point A): Slope = (x2 – 70) ÷ (x1 – 0) = (-7/6) →
(x2 – 70) = x1*(-7/6) = 7x1 + 6x2 = 420
Constraint (point B): Slope = (x2 – 0) ÷ (x1 – 60) = (-7/6) →
(x2) = (x1 – 60)*(-7/6) = 7x1 + 6x2 = 420
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
Determine the values by which c1 and c2 must decrease or increase in order to change the current solution point to the
other extreme point (point B).
Answer: At the least at point B, the objective function (Z) is equal to the optimal value at point A (3,500) where it
would become an alternate optimal feasible solution. At point B, we can substitute the coordinates for (x1, x2) into the
objective equation while it is set equal to 3,500 to find the necessary coefficients for (Z) to give the same maximum as
at point A:
3,500 = c1*(60) + c2* (0) = (3,500) ÷ (60) = c1 = 58.33
Therefore, C1 would have to increase by 28.33 from 30, and C2 has no effect at point B since it is zero.
#7) Southern Sporting Goods Company makes basketballs and footballs. Each product is produced from two resources—
rubber and leather. The resource requirements for each product and the total resources available are as follows:
Resource Requirements per Unit
Rubber (lbs.)
Leather (ft.2)
Z = 12x1 + 16x2
Basketballs (X1) $12 ea.
3
4
3x1 + 4x2 ≤ 500 lbs.
Footballs (X2) $16 ea.
2
5
2x1 + 5x2 ≤ 800 ft.2
500 lbs.
800 ft.2
Product
Total Resources Available
Each basketball produced
results in a profit of $12, and
each football earns $16 in
profit. Z = 12x1 + 16x2
(a) Formulate a linear programming model to determine the number of basketballs and footballs to
produce in order to maximize profit.
Answer: Z = 12x1 + 16x2 where Z = profit, so maximize Z subject to the following assumptions:
let x1 = the number of basketballs to produce, x2 = the number of footballs to produce.
Rubber constraint equation: 3x1 + 4x2 ≤ 500 lbs.; Leather constraint equation: 2x1 + 5x2 ≤ 800 ft.2
(a) Transform this model into standard form (with slack/surplus variables).
Answer: Z = 12x1 + 16x2, maximize Profit (Z) subject to the following constraints: (x1, x2, s1, s2 ≥ 0)
Rubber Resource constraint equation with slack variable s1: 3x1 + 4x2 + s1 = 500 lbs.
Leather Resource constraint equation with slack variable s2: 2x1 + 5x2 + s2 = 800 ft.2
#13) Erwin Textile Mills produces two types of cotton cloth—denim and corduroy. Corduroy is a heavier grade of cotton
cloth and, as such, requires 7.5 pounds of raw cotton per yard, whereas denim requires 5 pounds of raw cotton per yard.
A yard of corduroy requires 3.2 hours of processing time; a yard of denim requires 3.0 hours. Although the demand for
denim practically unlimited, the maximum demand for corduroy is 510 yards per month. The manufacturer has 6,500
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
pounds of raw cotton and 3,000 hours of processing time available each month. The manufacturer makes a profit of
$2.25 per yard of denim and $3.10 per yard of corduroy. The manufacturer wants to know how many yards of each type
of cloth to produce to maximize profit.
(a) Formulate a linear programming model for this problem.
Types of Cotton Cloth
Resource:
Denim (X1)
Corduroy (X2)
Raw Cotton (lbs.)
5 lbs./yd.
7.5 lbs./yd.
≤ 6,500 lbs./month
Process Time (hrs.)
3 hrs./yd.
3.2 hrs./yd.
≤ 3,000 hrs./month
Maximize Profit (Z)
$2.25/yd.
$3.10/yd.
Z = 2.25x1 + 3.10x2
(b) Transform this model into standard form (with slack/surplus variables).
Answer: Z = 2.25x1 + 3.10x2, subject to the following constraints: (x1, x2, s1, s2 ≥ 0)
Raw Cotton Constraint: 5x1 + 7.5x2 + s1 = 6,500
Process Time Constraint: 3x1 + 3.2x2 + s2 = 3,000
#14) Solve the model formulated in Problem 13 for Irwin Textile Mills graphically.
Since Z = 2.25x1 + 3.10x2, at Point (456, 510) Z = 2.25*(456) + 3.10*(510) = $2,607 at the optimum point (max)
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
(a) How much extra cotton and processing time are left over at the optimal solution? Is the demand for
corduroy met?
Answer: At the optimal solution: (Z) $2,607 = $2.25*(456) + $3.10*(510); so x1 = 456 & x2 =510
5*(456) + 7.5*(510) + s1 = 6,500 → 2280 + 3825 + s1 = 6,500
→ 6,105 + s1 = 6,500, therefore, s1 = 6,500 – 6105 = 395 lbs. of (slack) Raw Cotton
3*(456) + 3.2*(510) + s2 = 3,000 → 1,368 + 1,632 + s2 = 3,000
→ 3,000 + s2 = 3,000, therefore, s2 = 0, no remaining (slack) Processing Time
The max demand for corduroy (x2) at optimum point (456 yards of denim, 510 yards of corduroy) is met fully.
What is the effect on the optimal solution if the profit per yard of denim is increased from $2.25 to $3.00 which
makes Z = 3x1 + 3.10x2? What is the effect if the profit per yard of corduroy is increased from $3.10 to $4.00
which makes Z = 2.25x1 + 4x2?
Answer: The optimum corner point of the feasible solution space moves to the point (1000, 0) and Z = 3,000
when the profit per yard of denim is increased to $3.00/yd. which requires only denim be produced since it uses
all of the available processing time of 3,000 hours but leaves 1,500 lbs. of raw cotton unused (slack).
If the profit per yard of corduroy is increased to $4.00 (with denim at $2.25),the optimum corner point of the
feasible solution space returns to the original point (456, 510) and Z = $3,066, an increase in Z of $66.
(b) What would be the effect on the optimal solution of Irwin Mills could obtain only 6,000 pounds of cotton
per month (compared to all of the original conditions)?
Answer: The optimal solution becomes corner point (507.7, 461.5), and Z = 2573.08. Profit decreases slightly.
#15) Solve the model formulated in Problem 13 for Irwin Textile Mills by using the computer.
(a) If Irwin Mills can obtain additional cotton or processing time, but not both, which should it select? How
much? Explain your answer.
Answer: Since there is unused raw cotton: s1 = 395 lbs., that resource is abundant; however, from the computer
solution the dual price of the processing time resource is worth the most at $0.75, so it is the most valuable
resource to acquire in order to raise profit beyond the current optimum value.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
(b) Identify the sensitivity ranges for the objective function coefficients and for the constraint quantity values.
Then explain the sensitivity range for the demand for corduroy.
Answer: From the computer solution summary for the original Irwin Mills linear programming model:
Z = c1*x1 + c2*x2: Objective function coefficient ranges: 0.00 ≤ c1 ≤ 2.91, 2.40 ≤ c2 ≤ +∞
Constraint available resources ranges: (6,105 ≤ q1 ≤ +∞) , (1632 ≤ q2 ≤ 3,237), (0.00 ≤ q3 ≤ 692.31)
(q1 is the available raw cotton; q2 is the available processing time; q3 is the demand for corduroy)
For q3, the demand for corduroy can be as low as 0 yards, or as large as 692.31 yards and the objective function
will still remain at the maximum profit with all other conditions unchanged.
#25) Rucklehouse Public Relations has been contracted to do a survey following an election primary in New Hampshire.
The firm must assign interviewers to conduct the survey by telephone or in person. One person can conduct 80
telephone interviews or 40 personal interviews in a single day. The following criteria have been established by the firm
to ensure a representative survey:

At least 3,000 interviews must be conducted

At least 1,000 interviews must be by telephone (total number of telephone interviews)

At least 800 interviews must be personal
(total number of interviews)
(total number of personal interviews)
An interviewer conducts only one type of interview each day. The cost is $50 per day for a telephone interviewer and
$70 per day for a personal interviewer. The firm wants to know the minimum number of interviewers to hire in order to
minimize the total cost of the survey.
Formulate a linear programming model for this problem.
Answer: Let x1 = the number of people (employees) required for telephone interviews
Let x2 = the number of people required for in-person (personal) interviews
Let the objective function Z represent the total cost of the survey: minimize Z (cost)
The linear programming model becomes: (Note that c1 = $50 and c2 = $70 for later reference)
Z = $50x1 + $70x2 Subject to the following constraints:
80x1 + 40x2 ≥ 3,000 (number of interviews) becomes:
80x1 + 40x2 + s1 = 3,000 (1)
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
80x1 + 0x2 ≥ 1,000 (number of telephone interviews) becomes:
80x1 + 0x2 + s2 = 1,000 (2)
0x1 + 40x2 ≥ 800 (number of personal interviews) becomes:
0x1 + 40x2 + s3 = 800 (3)
and the non-negativity (positive or zero) constraints: (x1,x2) ≥ 0, (s1, s2, s3) ≥ 0
#26) Solve the linear programming model formulated in Problem 25 for Rucklehouse Public Relations graphically.
The minimum Z-value (cost) occurs at the point (x1=27.5, x2=20) where Z = $50(27.5) + $70(20) = $2,775. At the other
potential corner point at the intersection of the constraint #2 line (light green) and the constraint #1 line (red) at the
point (x1=12.5, x2=50), Z = $50(12.5) + $70(50) = $4,125 which is not the minimum between the two available corner
point values of the feasible solution space as shown in the graph.
(a) Determine the sensitivity ranges for the daily cost of a
X2
80
0
60
0
telephone interviewer (c1) and the number of personal
80X1 + 0X2 ≥ 1,000
interviews required (q3).
Z @ (12.5, 50) = $4,125
Answer: Let c1 = the daily cost of a telephone interviewer
and q3 = the number of personal interviews. We start by
80X1 + 40X2 ≥ 3,000
40
0
finding the slopes of the three constraint equations and
Min Z @ (27.5, 20) = $2,775
20
0
0X1 + 40 X2 ≥ 800
use only those that that sensibly bound the objective
function: Z = 50x1 + 70x2, the slope of the objective line
is therefore determined algebraically from the standard
20
0
40
0
of constraint 2: (-80/0) =(
60
0
80
0
X1
ratio (-c1/c2) = (-c1/70). Next, we find the slope of
constraint 1 similarly: (-80/40) = (-2). Then for the slope
∞ ) however, the objective function line pivots around the current optimal corner
point and is therefore bound between the constraint #1 line 80x1 + 40x2 = 3,000 and the constraint #3 line 0x1 +
40x2 = 800, which has a slope of: -(0/40) = ( 0 ). Then we have the range: (-2) ≤ (-c1/70) ≤ (0), then we multiply
by (-70) and reverse the inequality to get: (0) ≤ (c1) ≤ (140).
(a) Does the firm conduct any more telephone and personal interviews than are required, and if so, how many
more?
Answer: Ordinarily, it is desired to conduct a minimum of interviews to minimize the associated costs. Since the
optimal value of x1 is 27.5 telephone interviewers, it would be necessary to round up to 28 because it is not
possible to have a fractional part of an interviewer and the optimum x1 is given as more than 27 in order to
satisfy all of the specified constraints.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
The resulting objective function value is now Z = (28)$50 + (20)$70 = $2,800 for the minimum cost instead of
$2,775. Only integer values for the objective function make sense in this case since they represent whole people
conducting two types of interviews.
NOTE: The answers to #26 could not be confirmed (no answer in textbook) so these are my best attempts at
making sense of the questions related to the model in problem #25.
(b) What would be the effect on the optimal solution if the firm was required by the client to increase the
number of personal interviews conducted
from 800 to a total of 1,200?
X2
80
0
Answer: In general, requiring more interviews of either
80X1 + 0X2 ≥ 1,000
type to be conducted will increase the total cost of
conducting interviews. In the case of personal interviews,
60
0
80X1 + 40X2 ≥ 3,000
40
0
this will change the constraint #3 equation to become:
Min Z @ (22.5, 30) = $3,225
0X1 + 40 X2 ≥ 1200
20
0
0x1 + 40x2 = 1,200 and this will change the minimum Z
corner point values to (22.5, 30) which changes the
objective function value to: Z = (22.5)$50 + (30)$70 =
$3,225; slope = -22.5/30 = -0.75. Again, as before, x1
20
0
40
0
60
0
80
0
X1
should be rounded up to 23 which changes the solution to
Z = $50(23) + $70(30) = $3,250.
#27) Solve the linear programming model formulated in Problem 25 for the Rucklehouse Public Relations by
using the computer.
(a) If the firm could reduce the minimum interview requirement for either the telephone (q2) or personal
interviews (q3), which should the firm select? How much would a reduction of one interview in the
requirement you selected reduce the total cost? Solve the model again, using the computer, with the
reduction of this one interview in the constraint requirement to verify your answer.
Answer: The best choice for reducing the total cost is to reduce the number of interviews that has the largest
shadow price. According to the computer solution, constraint #2 (telephone interviews) has a shadow price of
$0.00 and constraint #3 (personal interviews) has a shadow price of $1.13; a decrease in the minimum number
of personal interviews, within sensitivity limits, lowers the total minimum cost accordingly, $1.13 for each single
decrease in the number of personal interviews.
(Note that the textbook answer says “personal interviews” at $0.625 ea. but Excel Solver says that it’s $1.13 ea.)
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 3
Darrell Hicks
Spring 2013
(b) Identify the sensitivity ranges for the cost of a personal interview (c2) and the number of total interviews
required (q1).
Answer: Let c2 = the daily cost of a telephone interviewer in the objective function (cost).
Let q1 = the number of total interviews in the constraint #1 equation.
The sensitivity ranges for c2 is determined by the comparing the ranges of the slopes of the constraint equations
to the slope of the objective function: Slope = -c1/c2 where c1 is for x1 and c2 is for x2.
Z = 50x1 + 70x2, the slope of the objective line = -50/c2 (solve for c2)
(Total # of interviews) 80x1 + 40x2 ≥ 3,000 (q1), the slope of constraint 1 = -80/40 = -2
(# of Telephone interviews) 80x1 + 0x2 ≥ 1,000 (q2), the slope of constraint 2 = -80/0 =
∞
(# of Personal interviews) 0x1 + 40x2 ≥ 800 (q3), the slope of constraint 3 = 0/-40 = 0
The range for c2 in the objective function Z is between the
X2
Constraint #1 line and the Constraint #3 line as shown in
80
0
60
0
40
0
Constraint #2
the graph below and is based on the optimal corner point
of the solution region. Then we arrange the two extremes
as: (-2) ≤ (-50/c2) ≤ 0 now invert the relation, which then
Constraint #1 Slope=-2
Z
Constraint #3 Slope=0
20
0
becomes: (-1/2) ≥ (c2/-50) ≥ ∞ then finally, multiplying
through by -50 gives the sensitivity range for c2:
(Note that infinity times anything equals infinity)
(25) ≤ (c2) ≤ (∞). As shown in the graph, the objective
20
40
80 X1
60
function line (dashed line) can pivot around the optimal
0
0
0
0
solution corner point and is bound between the slope of the Constraint #1 line and the slope of the Constraint
#3 line which is confirmed by Excel Solver results.
The range for q1 is given in the computer solution as having a minimum of 1800 and maximum of infinity.
Therefore, the range for the minimum number of total interviews is: (1800) ≤ (c1) ≤ (∞). So at the current
optimal point of the feasible solution space, the cost of a telephone interviewer can be as low as $25 each,
which is preferred over the original $50 each, and maintain the current optimal solution while the minimum
number of total interviews can be as low as 1800.