Chapter 8
Strategic Allocation of Resources
(Linear Programming)
A company makes 3 products: A, B and C.
A
B
C
Available
Profit
35
45
25
Labor Hrs
5
7
3
2000 hrs
Fiberglass 18
25
12
7000 lbs
At least 100 units each must be made of A, B, C
How many A’s, B’s, and C’s should be produced in
order to maximize total profits?
Incorrect Strategy: make as much as possible of the most
profitable product (B), so make as little as possible of the
other products (100 A’s and 100 C’s)
available: 2000
7000
Labor Fiberglass
Profit
500
1800
3500
make 100 A’s
300
1200
2500
make 100 C’s
4000
remaining: 1200
We run out of
4000/25
How many B’s? 1200/7
fiberglass 1st
= 171
= 160
make 160 B’s 1120
remaining:
80
4000
0
7200
$13,200 Total Profit
Optimal solution is $13,625 using LP (100 A, 100 B, 225 C)
 Difference of $425
Linear Programming Formulation
A = # of units of product A to produce
B = # of units of product B to produce
C = # of units of product C to produce
Max Z = 35A + 45B + 25C
ST
5A + 7B + 3C ≤ 2000
labor hours
18A + 25B + 12C ≤ 7000 fiberglass
A ≥ 100
minimum A
B ≥ 100
minimum B
C ≥ 100
minimum C
Linear Programming using Lindo software
Max
35 A + 45 B + 25 C
Subject to
2)
3)
4)
5)
6)
5 A + 7 B + 3 C <= 2000
18 A + 25 B + 12 C <= 7000
A >= 100
B >= 100
C >= 100
End
LP Optimum found at step 4
Objective Function Value
1)
13625.000
Variable
A
B
C
Row
2)
3)
4)
5)
6)
No. Iterations =
Value
100.000000
100.000000
225.000000
Slack or Surplus
125.000000
.000000
.000000
.000000
125.000000
4
Reduced Cost
.000000
.000000
.000000
Dual Prices
.000000
2.083333
-2.500000
-7.083333
.000000
Ranges in which the basis is unchanged:
Variable
A
B
C
Row
2
3
4
5
6
Obj Coefficient Ranges
Current
Allowable
Coef
Increase
35.000000
2.500000
45.000000
7.083333
25.000000
Infinity
Righthand Side Ranges
Current
Allowable
RHS
Increase
2000.000000
Infinity
7000.000000
500.000000
100.000000
83.333340
100.000000
60.000000
100.000000
125.000000
Allowable
Decrease
Infinity
Infinity
1.666667
Allowable
Decrease
125.000000
1500.000000
100.000000
100.000000
Inifinity
Example Using Excel Solver
10. A local brewery produces three types of beer: premium, regular,
and light. The brewery has enough vat capacity to produce 27,000
gallons of beer per month. A gallon of premium beer requires 3.5
pounds of barley and 1.1 pounds of hops, a gallon of regular
requires 2.9 pounds of barley and .8 pounds of hops, and a gallon of
light requires 2.6 pounds of barley and .6 pounds of hops. The
brewery is able to acquire only 55,000 pounds of barley and 20,000
pounds of hops next month. The brewery’s largest seller is regular
beer, so it wants to produce at least twice as much regular beer as it
does light beer. It also wants to have a competitive market mix of
beer. Thus, the brewery wishes to produce at least 4000 gallons
each of light beer and premium beer, but not more than 12,000
gallons of these two beers combined. The brewery makes a profit of
$3.00 per gallon on premium beer, $2.70 per gallon on regular beer,
and $2.80 per gallon on light beer. The brewery manager wants to
know how much of each type of beer to produce next month in order
to maximize profit.
Example Using Excel Solver
LP Formulation:
Max Z = 3P + 2.7R + 2.8L
ST
P + R + L < 27000
3.5P + 2.9R + 2.6L < 55000
1.1P + .8R + .6L < 20000
R – 2L > 0
P > 4000
L > 4000
P + L < 12000
capacity
R
2
barley

L
1
hops
R  2L
R  2L 
2:1 ratio
minimum P requirement
minimum L requirement
maximum requirement
0
Instructions for Using Excel to Solve LP Models
1.
2.
3.
4.
5.
6.
7.
8.
Set up spreadsheet like example in packet.
(Z-value and LHS column should be formulas)
Select “Tools” on menu bar. Then select “Solver…”.
“Set Target Cell:” should be the cell of your Z-value
formula.
Select “Min” or “Max”.
“By Changing Cells:” should be the range of cells for
your decision variables values.
Select “Options…”
Check 2 boxes: “Assume Linear Model” and “Assume
Non-Negative”. Then click “OK”.
Select “Add” to add constraints.
9. In “Cell Reference:” box point to LHS formula of first
constraint. Select <, =, or >. Click on “Constraint:” box
and point to RHS value of first constraint. Click “Add”
for next constraint or “OK” if finished.
10. Repeat Step 9 for each other constraint.
11. Select “Solve”.
12. If it worked okay you should get the message “Solver
found a solution. All constraints and optimality
conditions are satisfied.” If you do not get this
message you should modify your formulation or check
for mistakes.
13. In the Solver Results window under “Reports” click on
“Answer”. Then hold down the ‘Ctrl’ button while you
click on “Sensitivity”. Then click “OK”.
14. Print your final worksheet showing the new values,
print the Answer Report and print the Sensitivity
Report.
A
1
B
C
D
E
P
R
L Objective
0 Value (Z)
2
Dec Vars
0
0
3
Obj Coef
3
2.7
F
G
<,=,>
RHS
2.8 =sumproduct(B3:D3,B$2:D$2)
4
5
Constraints
6
capacity
7
LHS
1
1
1 =sumproduct(B6:D6,B$2:D$2)
<
27000
barley
3.5
2.9
2.6 =sumproduct(B7:D7,B$2:D$2)
<
55000
8
hops
1.1
0.8
0.6 =sumproduct(B8:D8,B$2:D$2)
<
20000
9
2:1 ratio
1
-2 =sumproduct(B9:D9,B$2:D$2)
>
0
10
min req. P
=sumproduct(B10:D10,B$2:D$2)
>
4000
11
min req. L
1 =sumproduct(B11:D11,B$2:D$2)
>
4000
12
max req.
1 =sumproduct(B12:D12,B$2:D$2)
<
12000
1
1
=sumproduct(B3:D3,B2:D2) is equivalent to =B3*B2 + C3*C2 + D3*D2
A
1
B
C
D
E
P
R
L
Objective
Value (Z)
2
Dec Vars
4000
9761.905
4880.952
3
Obj Coef
3
2.7
2.8
F
G
<,=,>
RHS
52023.81
4
5
Constraints
6
capacity
7
LHS
1
1
1
18642.86
<
27000
barley
3.5
2.9
2.6
55000
<
55000
8
hops
1.1
0.8
0.6
15138.1
<
20000
9
2:1 ratio
1
-2
0
>
0
10
min req. P
4000
>
4000
11
min req. L
1
4880.952
>
4000
12
max req.
1
8880.952
<
12000
1
1
Microsoft Excel 10.0 Answer Report
Worksheet: [Book1]Sheet1
Report Created: 1/15/2003 9:35:20 AM
Target Cell (Max)
Cell
$E$3
Name
Original Value
Obj Coef Value (Z)
Final Value
0
52023.80952
Adjustable Cells
Cell
Name
Original Value
Final Value
$B$2
Dec Vars P
0
4000
$C$2
Dec Vars R
0
9761.904762
$D$2
Dec Vars L
0
4880.952381
Constraints
Cell
Name
$E$6
capacity LHS
$E$7
barley LHS
$E$8
hops LHS
$E$9
2:1 ratio LHS
$E$10
min req. P LHS
$E$11
$E$12
Cell Value
Formula
Status
Slack
18642.85714
$E$6<=$G$6
Not Binding
8357.142857
55000
$E$7<=$G$7
Binding
15138.09524
$E$8<=$G$8
Not Binding
0
$E$9>=$G$9
Binding
0
4000
$E$10>=$G$10
Binding
0
min req. L LHS
4880.952381
$E$11>=$G$11
Not Binding
880.952381
max req. LHS
8880.952381
$E$12<=$G$12
Not Binding
3119.047619
0
4861.904762
Microsoft Excel 10.0 Sensitivity Report
Worksheet: [Book1]Sheet1
Report Created: 1/15/2003 9:35:20 AM
Adjustable Cells
Cell
Name
Final
Reduced
Objective
Allowable
Allowable
Value
Cost
Coefficient
Increase
Decrease
$B$2
Dec Vars P
4000
0
3
0.416666667
1E+30
$C$2
Dec Vars R
9761.904762
0
2.7
0.423076923
0.5
$D$2
Dec Vars L
4880.952381
0
2.8
1E+30
0.379310345
Constraints
Cell
Name
$E$6
capacity LHS
$E$7
barley LHS
$E$8
hops LHS
$E$9
2:1 ratio LHS
$E$10
Final
Shadow
Constraint
Allowable
Allowable
Value
Price
R.H. Side
Increase
Decrease
18642.85714
0
27000
1E+30
8357.142857
55000
0.976190476
55000
18563.63636
7400
15138.09524
0
20000
1E+30
4861.904762
0
-0.130952381
0
2551.724138
9034.482759
min req. P LHS
4000
-0.416666667
4000
2114.285714
4000
$E$11
min req. L LHS
4880.952381
0
4000
880.952381
1E+30
$E$12
max req. LHS
8880.952381
0
12000
1E+30
3119.047619
1. The Ohio Creek Ice Cream Company is planning
production for next week. Demand for Ohio Creek
premium and light ice cream continue to outpace the
company’s production capacities. Ohio Creek earns a
profit of $100 per hundred gallons of premium and $100
per hundred gallons of light ice cream. Two resources
used in ice cream production are in short supply for next
week: the capacity of the mixing machine and the
amount of high-grade milk. After accounting for required
maintenance time, the mixing machine will be available
140 hours next week. A hundred gallons of premium ice
cream requires .3 hours of mixing and a hundred gallons
of light ice cream requires .5 hours of mixing. Only
28,000 gallons of high-grade milk will be available for
next week. A hundred gallons of premium ice cream
requires 90 gallons of milk and a hundred gallons of light
ice cream requires 70 gallons of milk.
P = # of gallons of Premium ice cream to make
L = # of gallons of Light ice cream to make
Max Z = 100P + 100L
ST
.3P + .5L ≤ 140
90P + 70L ≤ 28000
capacity of mixing machine
max milk available
Solution: P = 175; L = 175; Z = 35,000
2. The Sureset Concrete Company produces
concrete in a continuous process. Two
ingredients in the concrete are sand, which
Sureset purchases for $6 per ton, and gravel,
which costs $8 per ton. Sand and gravel
together must make up exactly 75% of the
weight of the concrete. Furthermore, no more
than 40% of the concrete can be sand, and at
least 30% of the concrete must be gravel. Each
day 2,000 tons of concrete are produced.
S = # tons of sand to add to mixture
G = # tons of gravel to add to mixture
Min Z = 6S + 8G
ST
S + G = 1500
S ≤ 800
G ≥ 600
sand & gravel are 75% of 2000
sand no more than 40% of 2000
gravel at least 30% of 2000
Solution: S = 800; G = 700; Z = 10,400
3. A ship has two cargo holds, one fore and one aft. The
fore cargo hold has a weight capacity of 70,000 pounds
and a volume capacity of 30,000 cubic feet. The aft hold
has a weight capacity of 90,000 pounds and a volume
capacity of 40,000 cubic feet. The shipowner has
contracted to carry loads of packaged beef and grain.
The total weight of the available beef is 85,000 pounds;
the total weight of the available grain is 100,000 pounds.
The volume per mass of the beef is 0.2 cubic foot per
pound, and the volume per mass of the grain is 0.4 cubic
foot per pound. The profit for shipping beef is $0.35 per
pound, and the profit for shipping grain is $0.12 per
pound. The shipowner is free to accept all or part of the
available cargo; he wants to know how much meat and
grain to accept in order to maximize profit.
BF = # lbs beef to load in fore cargo hold
BA = # lbs beef to load in aft cargo hold
GF = # lbs grain to load in fore cargo hold
GA = # lbs grain to load in aft cargo hold
Max Z = .35 BF + .35BA + .12GF + .12 GA
ST
BF + GF ≤ 70000
fore weight capacity – lbs
BA + GA ≤ 90000
aft weight capacity – lbs
.2BF + .4GF ≤ 30000
for volume capacity – cubic feet
.2BA + .4GA ≤ 40000
for volume capacity – cubic feet
BF + BA ≤ 85000
max beef available
GF + GA ≤ 100000
max grain available
4. The White Horse Apple Products Company purchases
apples from local growers and makes applesauce and
apple juice. It costs $0.60 to produce a jar of
applesauce and $0.85 to produce a bottle of apple juice.
The company has a policy that at least 30% but not more
than 60% of its output must be applesauce.
The company wants to meet but not exceed the
demand for each product. The marketing manager
estimates that the demand for applesauce is a maximum
of 5,000 jars, plus an additional 3 jars for each $1 spent
on advertising. The maximum demand for apple juice is
estimated to be 4,000 bottles, plus an additional 5 bottles
for every $1 spent to promote apple juice. The company
has $16,000 to spend on producing and advertising
applesauce and apple juice. Applesauce sells for $1.45
per jar; apple juice sells for $1.75 per bottle. The
company wants to know how many units of each to
produce and how much advertising to spend on each in
order to maximize profit.
S = # jars apple Sauce to make
J = # bottles apple Juice to make
SA = $ for apple Sauce Advertising
JA = $ for apple Juice Advertising
Max Z = 1.45S + 1.75J - .6S - .85J – SA – JA
ST
S ≥ .3(S + J)
at least 30% apple sauce
S ≤ .6(S + J)
no more than 60% apple sauce
S ≤ 5000 + 3SA
don’t exceed demand for apple sauce
J ≤ 4000 + 5JA
don’t exceed demand for apple juice
.6S + .85J + SA + JA ≤ 16000
budget
5. Dr. Maureen Becker, the head administrator at Jefferson County
Regional Hospital, must determine a schedule for nurses to make
sure there are enough nurses on duty throughout the day. During
the day, the demand for nurses varies. Maureen has broken the day
into 12 two-hour periods. The slowest time of the day encompasses
the three periods from 12:00 A.M. to 6:00 A.M., which, beginning at
midnight, require a minimum of 30, 20, and 40 nurses, respectively.
The demand for nurses steadily increases during the next four
daytime periods. Beginning with the 6:00 A.M. – 8:00 A.M. period, a
minimum of 50, 60, 80, and 80 nurses are required for these four
periods, respectively. After 2:00 P.M. the demand for nurses
decreases during the afternoon and evening hours. For the five twohour periods beginning at 2:00 P.M. and ending at midnight, 70, 70,
60, 50, and 50 nurses are required, respectively. A nurse reports for
duty at the beginning of one of the two-hour periods and works eight
consecutive hours (which is required in the nurses’ contract). Dr.
Becker wants to determine a nursing schedule that will meet the
hospital’s minimum requirements throughout the day while using the
minimum number of nurses.
12 variables
(one for each time block)
X1 = # of nurses starting at Midnight & working 8 hours
X2 =
“
2am
“
X3 =
“
4am
“
X4 =
“
6am
“
X5 =
“
8am
“
X6 =
“
10am
“
X7 =
“
Noon
“
X8 =
“
2pm
“
X9 =
“
4pm
“
X10 =
“
6pm
“
X11 =
“
8pm
“
X12 =
“
10pm
“
Min Z = X1 + X2 + X3 + X4 + X5 + X6 + ……. + X11 + X12
ST
X1
+ X10 + X11 + X12 ≥ 30 midn – 2am
X1 + X2
+ X11 + X12 ≥ 20 2am – 4am
X1 + X2 + X3
+ X12 ≥ 40 4am – 6am
X1 + X2 + X3 + X4
≥ 50 6am – 8am
X2 + X3 + X4 + X5
≥ 60 8am – 10am
X3 + X4 + X5 + X6
≥ 80 10am–Noon
X4 + X5 + X6 + X7
≥ 80 Noon – 2pm
X5 + X6 + X7 + X8
≥ 70 2pm – 4pm
X6 + X7 + X8 + X9
≥ 70 4pm – 6pm
X7 + X8 + X9 + X10
≥ 60 6pm – 8pm
X8 + X9 + X10 + X11
≥ 50 8pm – 10pm
X9 + X10 + X11 + X12 ≥ 50 10pm – midn
6. The Donnor meat processing firm produces wieners from four ingredients: chicken,
beef, pork, and a cereal additive. The firm produces three types of wieners: regular,
beef, and all-meat. The company has the following amounts of each ingredient
available on a daily basis.
_____________________________________________
lb/Day
Cost/lb($)
Chicken
200
.20
Beef
300
.30
Pork
150
.50
Cereal Additive
400
.05
Each type of wiener has certain ingredient specifications, as follows.
________________________________________________________________________________
Regular
Beef
All-Meat
Specifications
Not more than 10% beef and pork combined
Not less than 20% chicken
Not less than 75% beef
No cereal additive
Not more than
50% beef and pork combined
The firm wants to know the amount of wieners of each type to produce.
Selling Price/lb($)
$0.90
1.25
1.75
14 variables
(you could also formulate it with 11 variables)
CR = # lbs Chicken ingredient in Regular wiener per day
CB = # lbs Chicken ingredient in Beef wiener per day
CM = # lbs Chicken ingredient in all-Meat wiener per day
BR = # lbs Beef ingredient in Regular wiener per day
BB = # lbs Beef ingredient in Beef wiener per day
BM = # lbs Beef ingredient in all-Meat wiener per day
PR = # lbs Pork ingredient in Regular wiener per day
PB = # lbs Pork ingredient in Beef wiener per day
PM = # lbs Pork ingredient in all-Meat wiener per day
AR = # lbs Additive ingredient in Regular wiener per day
AB = # lbs Additive ingredient in Beef wiener per day
R = total lbs of Regular wiener
B = total lbs of Beef wiener
M = total lbs of all-Meat wiener
Max Z = 0.90R + 1.25 B + 1.75 M - .2CR - .2CB - .2CM - .3BR - .3BB
- .3BM - .5PR - .5PB - .5PM - .05AR - .05AB
ST
CR + BR + PR + AR = R
R is sum of all ingredients in Regular
CB + BB + PB + AB = B
B is sum of all ingredients in Beef
CM + BM + PM = M
M is sum of all ingredients in Meat
CR + CB + CM ≤ 200
max Chicken ingredient available
BR + BB + BM ≤ 300
max Beef ingredient available
PR + PB + PM ≤ 150
max Pork ingredient available
AR + AB ≤ 400
max Additive ingredient available
BR + PR ≤ .1R
not more than 10% BR+PR combined
CR ≥ .2R
not less than 20% CR in Regular
BB ≥ .75B
not less than 75% BB in Beef
BM + PM ≤ .5M
not more than 50% BM+PM combined
7. The Jane Deere Company manufactures tractors in Provo, Utah.
Jeremiah Goldstein, the production planner, is scheduling tractor
production for the next three months. Factors that Mr. Goldstein
must consider include sales forecasts, straight-time and overtime
labor hours available, labor cost, storage capacity, and carrying cost.
The marketing department has forecasted that the number of
tractors shipped during the next three months will be 250, 305, and
350. Each tractor requires 100 labor hours to produce. In each
month 29,000 straight-time labor hours will be available, and
company policy prohibits overtime hours from exceeding 10% of
straight-time hours. Straight-time labor cost rate is $20 per hour,
including benefits. The overtime labor cost rate is 150% (time-anda-half) of the straight-time rate. Excess production capacity during a
month may be used to produce tractors that will be stored and sold
during a later month. However, the amount of storage space can
accommodate only 40 tractors. A carrying cost of $600 is charged
for each month a tractor is stored (if not shipped during the month it
was produced). Currently, no tractors are in storage.
How many tractors should be produced in each month using
straight-time and using overtime in order to minimize total labor cost
and carrying cost? Sales forecasts, straight-time and overtime labor
capacities, and storage capacity must be adhered to. (Tip: During
each month, all “sources” of tractors must exactly equal “uses” of
tractors.)
9 variables
S1 = # tractors produced in month 1 using straight-time
S2 = # tractors produced in month 2 using straight-time
S3 = # tractors produced in month 3 using straight-time
V1 = # tractors produced in month 1 using overtime
V2 = # tractors produced in month 2 using overtime
V3 = # tractors produced in month 3 using overtime
C1 = # tractors carried in warehouse at end of month 1
C2 = # tractors carried in warehouse at end of month 2
C3 = # tractors carried in warehouse at end of month 3
sources of tractors = uses of tractors
(for each month)
production + beg.inv. = sales + end.inv.
Min Z = 2000S1 + 2000S2 + 2000S3 + 3000V1 + 3000V2
+ 3000V3 + 600C1 + 600C2 + 600C3
ST
S1 + V1 + 0 = 250 + C1
month 1: sources = uses
S2 + V2 + C1 = 305 + C2
month 2: sources = uses
S3 + V3 + C2 = 350 + C3
month 3: sources = uses
100S1 ≤ 29000
straight-time capacity month 1
100S2 ≤ 29000
straight-time capacity month 2
100S3 ≤ 29000
straight-time capacity month 3
100V1 ≤ 2900
overtime capacity month 1
100V2 ≤ 2900
overtime capacity month 2
100V3 ≤ 2900
overtime capacity month 3
C1 ≤ 40
storage capacity month 1
C2 ≤ 40
storage capacity month 2
C3 ≤ 40
storage capacity month 3
8. MadeRite, a manufacturer of paper stock for copiers and printers,
produces cases of finished paper stock at Mills 1, 2, and 3. The
paper is shipped to Warehouses A, B, C, and D. The shipping cost
per case, the monthly warehouse requirements, and the monthly mill
production levels are:
Monthly Mill
Destination
Production
A
B
C
D
(cases)
Mill 1
$5.40
$6.20
$4.10
$4.90
15,000
Mill 2
4.00
7.10
5.60
3.90
10,000
Mill 3
4.50
5.20
5.50
6.10
15,000
Monthly Warehouse
Requirement (cases) 9,000
9,000
12,000
10,000
How many cases of paper should be shipped per month from each mill
to each warehouse to minimize monthly shipping costs?
A1 = # of units shipped from Mill 1 to Destination A
C3 = # of units shipped from Mill 3 to Destination C
(12 variables)
Min Z = 5.4A1 + 6.2B1 + 4.1C1 + 4.9D1 + 4.0A2 + 7.1B2
+ 5.6C2 + 3.9D2 + 4.5A3 + 5.2B3 + 5.5C3 + 6.1D3
ST
A1 + B1 + C1 + D1 ≤ 15000
Mill 1 capacity
A2 + B2 + C2 + D2 ≤ 10000
Mill 2 capacity
A3 + B3 + C3 + D3 ≤ 15000
Mill 3 capacity
A1 + A2 + A3 = 9000
Destination A demand
B1 + B2 + B3 = 9000
Destination B demand
C1 + C2 + C3 = 12000
Destination C demand
D1 + D2 + D3 = 10000
Destination D demand
9. A company has three research projects that it wants to
do, and has three research teams that can do the
projects. Any team could do any project but can only do
one project. Some teams are better skilled at certain
projects and could do them at lower costs. The
estimated cost of each team doing each project (in
$,000s) is shown below. Which team should do which
project?
Team
A
B
C
Project
1
2
87
62
81
76
77
54
3
76
64
70
A1 = 1 if team A does project 1
= 0 if not
(9 variables)
Min Z = 87A1 + 62A2 + 76A3 + 81B1 + 76B2 + 64B3
+ 77C1 + 54C2 + 70C3
ST
A1 + A2 + A3 = 1
B1 + B2 + B3 = 1 Each team does exactly one project
C1 + C2 + C3 = 1
A1 + B1 + C1 = 1
A2 + B2 + C2 = 1 Each project is done exactly once
A3 + B3 + C3 = 1