Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
Chapter 4 – Linear Programming: Modeling Examples
(#3)(a) In the investment example in this chapter, how would the solution become affected if the requirement that the
entire $70,000 be invested was relaxed such that it is the maximum amount available for investment?
Answer: Original solution: x3 = $38,181.82 invested in treasury bonds, x4 = $31,818.18 invested in growth stock fund
(max return on investments)
Z = $6,618.18 (as shown in Exhibit 4.7).
(a) Changes to constraints #1 and #3 must be made into algebraic expressions (pgs. 122, 123). Changing
constraint #1 to: 0.8x1 - 0.2x2 - 0.2x3 - 0.2x4 ≤ 0, constraint #3 to: 0.3x1+ 0.7x2+ 0.7x3 - 0.3x4≥ 0, and constraint #5
to: x1 + x2 + x3 + x4 ≤ 70,000produces the following results: x1 = $0.00, x2 = $0.00,
x3 = $38,181.82 invested in treasury bills, x4 = $31,818.18 invested in growth stock fund
Z = $6,618.18 (no change from the original problem)
(b) If the entire amount available for investment does not have to be invested and the amount available is
increased by $10,000 (to $80,000), how much will the total optimal return increase?
Answer: Changing only constraint #5 to: x1 + x2 + x3 + x4 ≤ 80,000produces the following results:
x3 = $43,636.36 invested in treasury bills, x4 = $36,363.64 invested in growth stock fund
(max return on investments)
Z = $7,563.64; the increase is $7,563.64 - $6,618.18 = $945.46 increase
(c) Will the entire $10,000 increase be invested in one alternative?
Answer: (c) No, both x3 and x4 increase in the amounts invested respectively: x3 invests $354.54 more than
before and x4 invests $590.91 more (than with $70,000 total). The sum of these two investments (treasury bills +
stock funds) is $7,563.63. The optimal investments are now in the same arrangement as in the original problem.
Treasury bills and growth stock funds should be the only investments:x3 = $43,636.36 invested in treasury
bonds, x4 = $36,363.64 invested in growth stock fund (with $80,000 total).
The sum is $43,636.36 + $36,363.64 = $80,000 total investments; no municipal bonds & no certificates of deposit.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
(#5) For the transportation example in this chapter, suppose that television sets not shipped were to incur storage costs
of $9.00 at Cincinnati, $6.00 at Atlanta, and $7.00 at Pittsburgh. How would these storage costs be reflected in the linear
programming model for this example problem, and what would the new solution be if any?
Original Transportation Example(pg.127)
Warehouse
1 Cincinnati
2
Atlanta
3
Pittsburgh
Actual TV Sets Shipped
Constraints
City Demand
Cost = $
New York A
0
0
150
150
=
150
7300
Store
Dallas B
0
200
50
250
=
250
Detroit C
200
0
0
200
=
200
TV Sets
Shipped
200
200
200
Constraint
<=
<=
<=
Supply
300
200
200
Answer: The new solution would simply add the additional cost of the storage fee according to the unshipped TVs
remaining in each of the three warehouses. Since the capacity of the warehouses in Atlanta and Pittsburgh are used up
by the results from Excel Solver, there are only 200 TVs shipped from Cincinnati to Detroit (more supply than demand)
with 100 unshipped TVs left in Cincinnati’s warehouse costing $9 for each TV to store. The total storage cost is $9
multiplied by100 TVs = $900 which would raise the total cost of shipping and storing the TVs but not change the optimal
solution for this linear programming model. Only the total shipping & storing costs, Z = $8200 would change with an
additional 900 dollars of storage cost for 100 TVs not shipped from Cincinnati.
◊ The Zephyr Television Company is considering leasing a new warehouse in Memphis. The new
warehouse would have a supply of 200 television sets, with shipping costs of $18 to New York, $9 to Dallas,
and $12 to Detroit. If the total transportation cost for the company (ignoring the cost of leasing the warehouse)
is less than with the current warehouses, the company will lease the new warehouse. Should the warehouse
be leased?
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
Answer: Yes, the new Memphis warehouse should be leased because using it to ship 200 TVs to Dallas as in the
original linear programming model reduced the total shipping cost to $6,550 from $7,300. The cost of shipping
from Memphis to Dallas for $9 is clearly the lowest cost of shipping to Dallas among all other warehouses.
Consequently, it is cheaper to ship all 200 TVs from the Memphis warehouse to Dallas with the remaining 50
(demand = 250) coming from the next cheapest shipping to Dallas: Atlanta for $12.
If supply could be increased at any one warehouse, which should it be? What restrictions would there
be on the amount of the increase?
Answer: Since the Memphis warehouse has the lowest shipping cost to Dallas, its capacity should be increased by
50 TVs (to 250) to accommodate the entire demand for TVs in Dallas and ship all 250 from Memphis to Dallas for $9.
(#6) For the blend example in this chapter, if the requirement that “at least 3,000 barrels of each grade of motor oil” was
changed to exactly 3,000 barrels of each grade of oil, how would this affect the optimal solution?
Answer: The optimal solution remains unchanged except for the shadow prices, which change for some of the
nine decision variables. Listed below are the results for the original problem and also with the change to equalities: Note
that there are alternative optimal solutions.
Variable
Value Reduced Cost Original Val Lower Bound
Upper Bound
X1s
1500.0 0
11
8
Infinity
X2s
450
13
13
13
X3s
1050.0 0
9
9
9
X1p
1200
0
8
-Infinity
11
X2p
1050
0
10
-8
10
X3p
750
0
6
6
10.8
X1e
1800.0 0
6
-Infinity
17.3333
X2e
1200
0
8
8
25
X3e
0
0
4
-Infinity
14
0
Constraint
Dual Value
Slack/Surplus Original Val Lower Bound
Upper Bound
Constraint 1
20
0
4500
4500
6200
Constraint 2
4
0
2700
2250
3150
The Dual value for all nine constraints are lowered by the change to equalities but the optimal solution remains
unchanged at Z = $76,800.
If the company could acquire more of one of the three components, which should it be? What would be the
effect on the total profit of acquiring more of this component?
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
Answer: In general, we want the item with the largest shadow price which contributes the most to the increases
in total profit. Here, it is Component Number one with a shadow price of $20. Acquiring one more unit of
component 1 increases profit by $20 (for each additional unit of component #1).
(#7) On their farm, the Friendly family grows apples that they harvest each fall and make into three products—apple
butter, applesauce, and apple jelly. They sell these three items at several local grocery stores, at craft fairs in the region,
and at their own Friendly Farm Pumpkin Festival for 2 weeks in October. Their three primary resources are cooking time
in their kitchen, their own labor time, and the apples. They have a total of 500 cooking hours available, and it requires
3.5 hours to cook a 10-gallon batch of apple butter, 5.2 hours to cook 10 gallons of applesauce, and 2.8 hours to cook 10
gallons of jelly. A 10-gallon batch of apple butter requires 1.2 hours of labor, a batch of sauce takes 0.8 hour, and a batch
of jelly requires 1.5 hours. The Friendly family has 240 hours of labor available during the fall. They produce about 6,500
apples each fall. A batch of apple butter requires 40 apples, a 10-gallon batch of applesauce requires 55 apples, and a
batch of jelly requires 20 apples. After the products are canned, a batch of apple butter will generate $190 in sales
revenue, a batch of applesauce will generate a sales revenue of $170, and a batch of jelly will generate sales revenue of
$155. The Friendlys want to know how many batches of apple butter, applesauce, and apple jelly to produce in order to
maximize their revenues.
(a) Formulate a linear programming model for this problem.
Let x1 = the # of batches (10 gal) of Apple Butter, x2 = the # of batches (10 gal) of Apple Sauce,
x3 = the # of batches (10 gal) of Apple Jelly
Now we construct the linear program from the given information for price, resources, materials, etc.:
maximize profit Z = 190x1 + 170x2 + 155x3 subject to the following constraints:
3.5x1 + 5.2x2 + 2.8x3 ≤ 500
(cooking time each product ≤ max available hours)
1.2x1 + 0.8x2 +1.5x3 ≤ 240
(labor time each product ≤ max available labor hrs)
40x1 + 55x2 + 20x3 ≤ 6,500
(required apples to use ≤ max available apples)
(b) Solve the model by using the computer.
x1 = 41.27 (≈ 42), x2 = 0, x3 = 126.98 (≈ 127), Profit Z = $190(42) + $170(0) + $155(127) = $27,665
The computer solution suggests that the maximum profit occurs with 42 (10 gal) batches of Apple Butter
(rounding up), no Apple Sauce should be made and 127 (10 gal) batches of Apple Jelly. With the largest Shadow
Price, cooking time is the most valuable resource; for each additional hour of cooking time, profit will increase by
$52.38 up to a max of 500 hrs.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
(#13) The Kalo Fertilizer Company produces two brands of lawn fertilizer—Super Two and Green Grow—at plants in
Fresno, California, and Dearborn, Michigan. The plant at Fresno has resources available to produce 5,000 pounds of
fertilizer daily; the plant at Dearborn has enough resources to produce 6,000 pounds daily. The cost per pound of
producing each brand at each plant is as follows:
(cost of Production) Plant
Product
Fresno
Max Demand
Dearborn
Available/day
Price per Item
Super Two
$2
$4
6,000 lbs/day
$9/lb.
Green Grow
$2
$3
7,000 lbs/day
$7/lb.
The company has a daily budget of $45,000 for both plants combined. Based on past sales, the company knows
the maximum demand (converted to a daily basis) is 6,000 pounds for Super Two and 7,000 pounds for Green Grow. The
selling price is $9 per pound for Super Two and $7 per pound for Green Grow. The company wants to know the number
of pounds of each brand of fertilizer to produce at each plant in order to maximize profit.
(a) Formulate a linear programming model for this problem.
Answer:
Let x1 = the # of lbs. of Super Two at Fresno,
Let x2 = the # of lbs. of Super Two at Dearborn
Letx3 = the # of lbs. of Green Grow at Fresno,
Let x4 = the # of lbs. of Green Grow at Dearborn
(cost of production)
$2x1 + $4x2 + $2x3 + $3x4≤ $45,000
(Fresno capacity)
(Super Two capacity)
1x1+1x2≤ 6,000 (Dearborn capacity)
1x2+
(Green Grow capacity) 1x3+ 1x4 ≤ 7,000
1x1+ 1x3≤ 5,000 lbs./day
1x4 ≤ 6,000 lbs./day
(non-negativity)
(x1, x2, x3, x4) ≥ 0
Let Z (profit) = [Revenue] – (costs) = [$9(x1 + x2) + $7(x3 + x4)] – ($2x1 +$4x2 + $2x3 + $3x4); Z = $7x1 +$5x2 + $5x3 + $4x4
(b) Solve the model by using the computer.
Answer:According to the results for this linear
programming model from QM for Windows™,the max profitZ =
$7(5,000) + $5(1,000) + $5(0) + $4(5,000) = $60,000
x1 = 5,000 lbs. x2 = 1,000 lbs. x3 = 0 lbs. x4 = 5,000 lbs.
This result indicates the optimal solution is to make 5,000
pounds of Super Two at Fresno but not any Green Grow there
and to make 1,000 pounds of Super Two at Dearborn and 5,000
pounds of Green Grow also at Dearborn.
X1
X2
X3
X4
RHS
Dual
Maximize
7
5
5
4
Constraint
2
4
2
3
1
Constraint
<=
45000
0
1
1
0
0
<=
6000
1
2
Constraint
0
0
3
Constraint
1
1
<=
7000
0
1
0
1
0
<=
5000
6
4
Constraint
5
Solution->
0
1
0
1
<=
6000
4
5000
1000
0
5000
Optimal
60000
Z->
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
(#14) Grafton Metalworks Company produces metal alloys from six different ores it mines. The company has an order
from a customer to produce an alloy that contains four metals according to the following specifications: at least 21% of
metal A, no more than 12% of metal B, no more than 7% of metal C, and between 30% and 65% of metal D. The
proportions of the four metals in each of the six ores and the level of impurities in each ore are provided in the following
table:
 When the metals are processed and refined, the impurities are removed.
Metal (%)
Ore
1
2
3
4
5
6
A
19
43
17
20
0
12
B
15
10
0
12
24
18
C
12
25
0
0
10
16
D
14
7
53
18
31
25
Impurities
40
(%)
15
30
50
35
29
Cost/Ton
$27
25
32
22
20
24
The company wants to know the amount of each ore
to use per ton of the alloy that will minimize the cost per ton
of the alloy.
(a) Formulate a linear programming model for this
problem.Answer: Let x1 = # of tons of ore #1, x2 = # of tons of
ore #2, x3 = # of tons of ore #3, … , x6 = # of tons of ore #6
The total cost per ton of alloy (Z) is then ($Cost/Ton)(# of Tons
of ore1) + ($Cost/Ton)(# of Tons of ore2) + ($Cost/Ton)(# of Tons of ore 3) + … + ($Cost/Ton)(# of Tons of ore 6) = Z
Minimize cost: Z = 27x1 + 25x2 + 32x3 + 22x4 + 20x5 + 24x6
Constraint 1: (Metal A %)
19x1 + 43x2 + 17x3 + 20x4 + 0x5 + 12x6≥ 21
[at least 21%]
Constraint 2: (Metal B %)
15x1 + 10x2 + 0x3 + 12x4 + 24x5+ 18x6≤ 12
[no more than 12%]
Constraint 3: (Metal C %)
12x1 + 25x2 + 0x3 + 0x4 + 10x5 + 16x6≤ 7
[no more than 7%]
Constraint 4: (Metal D % ≥)
14x1 + 7x2 + 53x3 + 18x4 + 31x5 + 25x6≥ 30
[more than 30%]
Constraint 5: (Metal D % ≤)
14x1 + 7x2 + 53x3 + 18x4 + 31x5 + 25x6≤ 65
[less than 65%]
(x1, x2, x3, x4, x5, x6) ≥ 0
(b) Solve the model using the computer.
[non-negativity]
Answer:
Minimize cost: Z = 27x1 + 25x2 + 32x3 + 22x4 + 20x5 + 24x6 = $23.91 per Ton of ore for x2 = 0.28 ton, x3 = 0.53 ton
Z = $25(0.28) + $32(0.53) = $23.96 per ton of ore
Shadow price of (Metal A) = $0.51 per ton, (Metal D ≥) = $0.44 per ton, Metal B, Metal C, Metal D ≤ = $0.00.
(#15) The Roadnet Transport Company expanded its shipping capacity by purchasing 90 trailer trucks from a bankrupt
competitor. The company subsequently located 30 of the purchased trucks at each of its shipping warehouses in
Charlotte, Memphis, and Louisville. The company makes shipments from each of these warehouses to terminals in St.
Louis, Atlanta, and New York. Each truck is capable of making one shipment per week. The terminal managers have
indicated their capacity for extra shipments. The manager at St. Louis can accommodate 40 additional trucks per week,
the manager at Atlanta can accommodate 60 additional trucks, and the manager at New York can accommodate 50
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
additional trucks. The company makes the following profit per truckload shipment from each warehouse to each
terminal. The profits differ as a result of differences in products shipped, shipping costs, and transport rates:
The company wants to know how many trucks to
Terminal
assign to each route (i.e. warehouse to terminal)
to maximize profit.
Warehouse
(a) Formulate a linear
programming model for this problem. Answer:
St. Louis
Atlanta
New York
Charlotte
$1,800 (route 1) $2,100 (route 2) $1,600 (route 3)
Memphis
$1,000 (route 4)
$700 (route 5)
$900 (route 6)
Louisville
$1,400 (route 7)
$800 (route 8)
$2,200 (route 9)
Let x1 = # of trucks on route 1, Let x2 = # of trucks
on route 2, Let x3 = # of trucks on route 3, … , Let
x9 = # of trucks on route 9. Maximize profit Z.
Z = 1800x1 + 2100x2 + 1600x3 +1000x4 + 700x5 + 900x6 + 1400x7 + 800x8 + 2200x9
Constraint 1: (# of trucks from Charlotte)
x1 + x2 + x3 = 30
Constraint 2: (# of trucks from Memphis)
x4 + x5 + x6 = 30
Constraint 3: (# of trucks from Louisville)
x7 + x8 + x9 = 30
Constraint 4: (# of trucks to St. Lous)
x1 + x4 + x7≤ 40
Constraint 5: (# of trucks to Atlanta)
x2 + x5 + x8 ≤ 60
Constraint 6: (# of trucks to New York)
x3 + x6 + x9 ≤ 50
(b) Solve the model by using the computer.
Only the most profitable route from each
warehouse to each terminal is needed,using all of its trucks to maximize profit.
Answer: Z = $159,000 (10 iterations); x2 = 30, x4 = 30, x9 = 30; (x1, x3, x5, x6, x7, x8)= 0
(#24) Brooks City has three consolidated high schools, each with a capacity of 1,200 students. The school board has
partitioned the city into five busing districts—north, south, east, west, and central—each with different high school
student populations. The three schools are located in the central, west, and south districts. Some students must be
bused outside their districts, and the school board wants to minimize the total bus distance traveled by these students.
The average distances from each district to the three schools and the total student population in each district are:
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
Distance (Miles)
Central
West
South
District
H.S.
H.S.
H.S.
North
8 (x1)
11 (x2)
14 (x3)
700
South
12 (x4)
9 (x5)
0 (x6)
300
East
9 (x7)
16 (x8)
10 (x9)
900
West
8 (x10)
0 (x11)
9 (x12)
600
Student
Population
The school board wants to determine the number of
students to bus from each district to each school to minimize the
total busing miles traveled.
(a) Formulate a linear programming model for this
problem.
Answer: Let x1 = the # of bused students on route 1, x2 =
the # of bused students on route 2, x3 = the # of bused students
on route 3, … , x15 = the # of students bused on route 15.
Minimize the total # of miles travelled but also determine the
number of students being bused on each route and to which High
Central 0 (x13) 8 (x14) 12 (x15)
500
School. We ultimately want only one bus route from each district
to its closest high school, maybe also to the next closest High
School if necessary. We will have unusual units of student-miles.
Therefore, we calculate (Route length)*(# of students on Route)
The student-miles can be used to count the number of students and the average number of miles the students are
bused but the Z-value will need to be divided by the appropriate unit (students or miles) or reconciled with the given
information.
Z = (Route #1 miles)*(x1) + (Route #2 miles)*(x2) + … + (Route #14 miles)*(Z=x14) + (Route #15 miles)*(x15)
Z = 8x1 + 11x2 + 14x3 + 12x4 + 9x5 + 0x6 + 9x7 + 16x8 + 10x9 + 8x10 + 0x11 + 9x12 + 0x13 + 8x14 + 12x15
(Constraint 1)
x1 + x4 + x7 + x10 + x13 ≤ 1,200
[# of students on Routes to Central H.S.] ≤ Capacity
(Constraint 2)
x2 + x5 + x8 + x11 + x14 ≤ 1,200
[# of students on Routes to West H.S.] ≤ Capacity
(Constraint 3)
x3 + x6 + x9 + x12 + x15 ≤ 1,200
[# of students on Routes to South H.S.] ≤ Capacity
(Constraint 4)
x1 + x2 + x3 = 700
[# of students & Routes departing North District]
(Constraint 5)
x4 + x5+ x6= 300
[# of students & Routes departing South District]
(Constraint 6)
x7 + x8 + x9 = 900
[# of students & Routes departing East District]
(Constraint 7)
x10 + x11 + x12 = 600
[# of students & Routes departing West District]
(Constraint 8)
x13 + x14 + x15 = 500
[# of students & Routes departing Central District]
(Constraint 9)
(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, x13, x14, x15) ≥ 0 [non-negativity]
(b) Solve the model by using the computer.
Answer: Minimum Student-Miles Z = 14,600; the sum of all districts = 3,000 students. See Figure 1
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
North
(700)
South
(300)
Darrell Hicks
Spring 2013
700
X1 (8 mi.)
X2 (11 mi.)
X3 (14 mi.)
Central H.S.
X4 (12 mi.)
(700+500)
X5 (9 mi.)
X6 (0 mi.)
East
300
X7 (9 mi.)
West H.S.
X8 (16 mi.)
(900)
West
(600)
X9 (10 mi.)
900
(600)
X10 (8 mi.)
600
X11 (0 mi.)
X12 (9 mi.)
Central
X13 (0 mi.)
500
X14 (8 mi.)
(500)
South H.S.
(300+900)
X15 (12 mi.)
Figure 1 Problem 24 Total Number of Students=3,000 Z=14,600
x1 = 700, x6 = 300, x9 = 900, x11 = 600, x13 = 500, and (x2, x3, x4, x5, x7, x8, x10, x12, x14, x15) = 0 (students per route)
(#25) (a) In Problem #24 the school board decided that because all students in the north and east districts must be
bused, then at least 50% of the students who live in the south, west, and central districts must also be bused to another
district. Reformulate the linear programming model to reflect this new set of constraints and solve by using a computer.
Additional constraints are for the South, West, and Central districts to include the within-district students as
available to be bused to other districts’ schools which were not considered before (because they were not to be
bused):
South to South:
x6 ≤ 150
(at most 50% of the South District’s students can be bused)
West to West:
x11≤ 300
(at most 50% of the West District’s students can be bused)
Central to Central:
x13≤ 250
(at most 50% of the Central District’s students can be bused)
All of the previous constraints remain unchanged and in effect in addition to the new constraints that
include students to be bused outside their own districts that previously went to their local district high school.
The solution is illustrated in Figure 2 (a) to include the 3 additional constraints. Since the total number of
students being bused increases, the cost increases as well.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
North
(700)
South
(50/50)
700
X1 (8 mi.)
X2 (11 mi.)
X3 (14 mi.)
Central H.S.
X4 (12 mi.)
(700+250+250)
150
X5 (9 mi.)
X6 (0 mi.)
East
Darrell Hicks
Spring 2013
X7 (9 mi.)
150
250
West H.S.
X8 (16 mi.)
(900)
West
(50/50)
X9 (10 mi.)
X11 (0 mi.)
X13 (0 mi.)
X14 (8 mi.)
(50/50)
(150+300+250)
X10 (8 mi.)
X12 (9 mi.)
Central
650
300
300
250
South H.S.
250
(150+650+300)
X15 (12 mi.)
Figure 2 Problem 25 (a) Total Number of Students=3,000 Z=20,400
x1 = 700, x5 = 150, x6 =150, x7 = 250, x9 = 650, x11 = 300, x12 = 300, x13 = 250, x14 = 250, and x2, x3, x4, x8, x10, x15 = 0
(b) The school board has further decided that the enrollment in all three high schools should be equal.
Formulate this additional restriction in the linear programming model.
Answer: The constraints for each of the three high schools now changes to a maximum student capacity of 1000.
(Constraint 1)
x1 + x4 + x7 + x10 + x13= 1,000
[# of students on Routes to Central H.S.] = Capacity
(Constraint 2)
x2 + x5 + x8 + x11 + x14= 1,000
[# of students on Routes to West H.S.] = Capacity
(Constraint 3)
x3 + x6 + x9 + x12+ x15 = 1,000
[# of students on Routes to South H.S.] = Capacity
 Note: Due to the fact that originally, there was more student capacity (3,600) than there are students available,
the original constraints were written to not exceed the maximum capacity while now all of each high school’s
student capacity must be used to accommodate all 3,000 students. See Figure 2 (b) for the results:
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
North
(700)
South
(300)
400
X1 (8 mi.)
X2 (11 mi.)
300
X3 (14 mi.)
Central H.S.
X4 (12 mi.)
(1,000
X5 (9 mi.)
X6 (0 mi.)
East
Darrell Hicks
Spring 2013
X7 (9 mi.)
150
150
350
West H.S.
X8 (16 mi.)
(900)
West
(600)
X9 (10 mi.)
(1,000)
X10 (8 mi.)
X11 (0 mi.)
X12 (9 mi.)
X13 (0 mi.)
Central
X14 (8 mi.)
(500)
550
300
300
250
250
South H.S.
(1,000)
X15 (12 mi.)
Figure 2 Problem 25 (b) Total Number of Students=3,000 Z=21,200
x1 = 400, x2 = 300, x5 = 150, x6 = 150, x7 = 350, x9 = 550, x11 = 300, x12 = 300, x13 = 250, x14 = 250, and x3, x4, x8, x10, x15 = 0
(#26)The Southfork Feed Company makes a feed mix from four ingredients—oats, corn, soybeans, and a vitamin
supplement. The company has 300 pounds of oats, 400 pounds of corn, 200 pounds of soybeans, and 100 pounds of
vitamin supplements for the mix. The company has the following requirements for the mix:





At least 30% of the mix must be soybeans.
At least 20% of the mix must be the vitamin supplement.
The ratio of corn to oats cannot exceed 2 to 1.
The amount of oats cannot exceed the amount of soybeans.
The mix must weigh at least 500 pounds.
(constraint 1)
(constraint 2)
(constraint 3)
(constraint 4)
(constraint 5)
A pound of oats costs $0.50; a pound of corn, $1.20; a pound of soybeans, $0.60; a pound of vitamin
supplement, $2.00. The feed company wants to know the number of pounds of each ingredient to put in the mix
in order to minimize the total costs.
(a) Formulate a linear programming model for this problem.
Let x1 = # of pounds of oats, x2 = # of pounds of corn, x3 = # of pounds of soybeans, x4 = # of pounds of vitamins
The “mix” = (x1 + x2 + x3 + x4); the sum of the number of pounds of all ingredients combined.
Minimize cost (Z) = 0.50x1 + 1.20x2 + 0.60x3 + 2.00x4; (cost per pound of ingredient * # of pounds of ingredient)
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
(constraint #1) x3 ≥ 0.30(x1 + x2 + x3 + x4) → x3 ≥ 0.30x1 + 0.30x2 + 0.30x3 + 0.30x4 →
-0.3x1 -0.3x2 + 0.7x3 – 0.3x4 ≥ 0 : at least 30% of the mix must be soybeans.
(constraint #2) x4 ≥ 0.20(x1 + x2 + x3 + x4) → x4 ≥ 0.20x1 + 0.20x2 + 0.20x3 + 0.20x4 →
-0.2x1 -0.2x2- 0.2x3+ 0.8x4 ≥ 0
: at least 20% of the mix must be soybeans.
(constraint #3) 2 pound of corn ≤ 1 pound of oats → 2x2 ≤ 1x1 →
-1x1 + 2x2 ≤ 0
: the ratio of corn to oats cannot exceed (≤) 2 to 1.
(constraint #4) x1 ≤ x3 → x1 – x3 ≤ 0
: the amount of oats cannot exceed the amount of soybeans.
(constraint #5) x1 + x2 + x3 + x4 ≥ 500
: the mix must weigh at least 500 pounds.
(b) Solve the model by using the computer. (7 iterations) x1 = 200, x2 = 0, x3 = 200, x4 = 100
Minimum cost (Z) = 0.50x1 + 1.20x2 + 0.60x3 + 2.00x4 = $0.50(200) + $1.20(0) + $0.60(200) + $2.00(100) = $420
(#27)The United Charities annual fund-raising drive is scheduled to take place next week. Donations are collected during
the day and night, by telephone, and through personal contact. The average donation resulting from each type of
contact is as follows:
Phone Interview
Personal Interview
Day
Avg. Donation: $2.00 (x1)
Avg. Donation: $4.00 (x3)
Night
Avg. Donation:$3.00 (x2)
Avg. Donation $7.00 (x4)
The charity group has enough donated gasoline and cars to make at most 300 personal contacts during one day
and night combined. The volunteer minutes required to conduct each type interview are as follows:
Phone Interview Time (min.)
Personal Interview Time (min.)
Day
Minutes: 6 (x1)
Minutes: 15 (x3)
Night
Minutes: 5 (x2)
Minutes: 12 (x4)
The charity has 20 volunteer hours available each day and 40 volunteer hours each night. The chairperson of the
fund-raising drive wants to know how many different types of contacts to schedule in a 24-hour period (i.e., 1 day and 1
night) to maximize the total donations.
Introduction to Management Science (ISOM 3123)
Practice Homework Problems
Chapter 4
Darrell Hicks
Spring 2013
(a) Formulate a linear programming model for this problem. Answer: Maximize total donations (Z).
(Z) = 2x1 + 3x2 + 4x3 + 7x4 (sum of all types of donations)
1 hour = 60 minutes, 20 hours = 1200 minutes, 40 hours = 2400 hours;
Let x1 = number of Phone Interviews (Day), x2 =number of Phone Interviews (Night),
x3 = number of Personal interviews (Day), x4 = Personal Interviews (Night)
Resources: funding for 20 hours of labor per day, 40 hours of labor per night.
(x1 + x2) = total # of Phone Interviews; (x3 + x4) = total # of Personal Interviews
(constraint 1)
(x3 + x4) ≤ 300
[gas and car supply at most 300 personal interviews]
(constraint 2)
6x1 + 15x3 ≤ 1,200 minutes
[available Day labor time = 20 hours], interview time in mins.
(constraint 3)
5x2 + 12x4 ≤ 2400 minutes[
[available Night labor time = 40 hours], interview time in mins.
(constraint 4)
(x1, x2, x3, x4) ≥ 0
[non-negativity requirement for linear programs]
(b) Solve the model by using the computer.
Answer: Maximum donations: Z = $1,840; x1 = 200, x2 = 480, x3= 0, x4 = 0
This solution suggests that only Phone Interviews should be conducted because they take less time and
are therefore more profitable. Notice the labor hours available are used up by the Phone Interviews:
Phone Interviews (Day) x1(6 mins) + Phone Interviews (Night) x2(5 mins) = (200)(6) + (480)(5) =
1,200 minutes (Day maximum labor hours) + 2,400 minutes (Night maximum labor hours)
Z = 2x1 + 3x2 + 4x3 + 7x4 = $2(200) + $3(480) + $4(0) + $7(0) = $1,840