Linear Programming ⚫ SUPPLEMENT D ⚫
D-21
The objective function and constraints are
Maximize:
$29.$17.$17.40 20 05A B C+ +
Subject to:
( )
( )
( )
( )
( )
0.25 0.20 0.10 1,600 machine 1
0.10 0.15 0.05 1,400 machine 2
0.05 0.10 0.15 1,500 machine 3
12 16 9 85, 000 material 2
1, 200 minimum for product B
, , 0
A B C
A B C
A B C
A B C
B
A B C
+ +
+ +
+ +
+ + 
b. Using POM for Windows, we get:
Thus the optimal solution is:
A
C
=
=
5275 units
0
,
units
Total Profit = +
=
( ) ( )
$29. , $17. ,
$175,
40 5275 20 1200
725
18. Maxine’s Hat Company.
a. False. The reduced cost of variable
x3
is $7, which means the price of hat 3 must increase
from $6 to $13 in order to warrant the production of hat 3 and maximize profits. A selling
price of $11.50 would not be high enough.
⚫ PART 2 ⚫ Managing Customer Demand
D-22
19. Washington Chemical Company
Let
annual production quantity of product 1
X2=
annual production quantity of product 2
annual production quantity of product 3
X4=
annual production quantity of product 4
X5=
annual production quantity of product 5
Then the formulation becomes:
1 2 3 4 5
Subject to:
1 2 3 4 5
12345
1 2 3 4 5
24
1 2 3
3
4
1 2 3 4 5
0.05 0.10 0.80 0.57 0.15 7,500
0.20 0.02 0.20 0.09 0.30 7,500
0.20 0.50 0.10 0.40 0.18 10,000
0.70 0.50 6, 000
0.10 0.20 0.40 7,000
3, 000
3, 000
, , , ,
X X X X X
X X X X X
X X X X X
XX
X X X
X
X
X X X X X
+ + + +
+ + + + 
+ + + + 
+
+ + 
0
Using POM for Windows, we get:
a. The annual production quantities are:
Product
Annual Production Quantity
1
17,311
2
6,429
3
3,000
4
3,000
5
10,131
Linear Programming ⚫ SUPPLEMENT D ⚫
D-23
b. With 10 changeovers per year, the lot sizes are:
Product
Lot sizes
1
1,731
2
643
3
300
4
300
5
1,013
The Transportation Method
20. Warwick Manufacturing Company
Most students need help formulating the model for Problems 20, 21, and 22, even if they have
reviewed the section on Sales and Operations Plans in Chapter 10. Here are some thought
starters on the variables and constraints to use. For Problem 20, the model would be:
Let
It=
number of shovels to be left over as inventory in period t
W
t=
number of workers in period t (equivalent to 4000 shovels/quarter and $3500/1000
shovels)
Ht = number of workers hired in period t
Use three types of constraints:
1. A constraint for each period t, which forces the demand Di to be met with the regular
time production of the Workers (Wt), Overtime Production (Ot), Subcontracting (St), and
2. A constraint of each period that link the number of workers (Wt) with the number of
hires (Ht) and fires (Ft). For example, the constraint for the first period would be:
If the number of workers this period exceeds the number last period, then the number
hired must equal the difference. If the number of workers this period is less than the
number last period, then the number fired must equal the difference. For the first
period’s constraint, W0 is known and can be handled with the shortcut used for I0, adding
it as a positive number to the right of the equality sign.
3. Capacity constraints on regular time, overtine, subcontracting, and inventory as
appropriate.
⚫ PART 2 ⚫ Managing Customer Demand
D-24
Objective Function and Constraints
The following model formulation seeks to meet demand at minimum cost by using the best mix
of regular-time production, overtime, subcontracting, and anticipation inventory. Constraints
1−4 force demand to be met and the inventory variables to behave properly. Constraints 5−8
assure for quarter t that an increase in Ht (for hiring) accompanies any increase in Wt and that an
increase Ft (for layoffs) accompanies any decrease in Wt. One worker produces the equivalent
4,000 shovels, at a cost of $3,500 per 1,000, so the cost for Wt is $14,000 (or $3,500 4).
Minimize:
tttttt
tFHSOWI 600000,1200,4700,3000,14280$
4
1+++++
=
Subject to:
1 1 1 1 0
2 2 2 1 2
3 3 3 2 3
4 4 4 3 4
1 1 1 0
2 1 2 2
3 2 3 3
4 3 4
4 40 ( ( ) 30.)
4 150
4 320
4 100
0 30 ( ( ) 30 .)
0
0
W O S I Thecurrent inventory level I is
W O S I I
W O S I I
W O S I I
W H F Thecurrrent workforce W consists of workers
W W H F
W W H F
W W H
+ + − =
+ + + − =
+ + + − =
+ + + − =
− + = +
− − + =
− − + =
−− 40F+=
30
30
30
30
4,3,2,15
4,3,2,115
4
0
4
0
=
=
=
=
=
=
I
I
W
W
tS
tOt
t
The optimal solution found using POM for Windows is shown below.
⚫ PART 3 ⚫ Managing Supply Chains
D-26
Optimal Solution
I0 = 30
W0 = 30
W1 = 30
W2= 46.875
W3= 46.875
W4=30
O1=0
O2=0
O3=15
O4=10
S1=0
S2=0
S3=0
S4=0
H1=0
H2=16.875
H3=0
H4=0
F1=0
F2=0
F3=0
F4=16.875
I1=80
I2=117.5
I3=25
I4=30
Production Plan
Anticipated
Quarter
Regular Time
Overtime
Subcontracting
Inventory
1
30,000
0
0
80,000
2
46,875
0
0
117,500
3
46,875
15,000
0
0
4
30,000
10,000
0
30,000
Total cost = $2,335,700
21. Warwick Manufacturing Company, continued
The model is the same as for Problem 20, except for the demand constraints:
4W1 + O1 + S1 − I1 = 90 (or 120 − 30)
4W4 + O4 + S4 −I4 = 160
The optimal solution found using POM for Windows is shown on the next page.
Linear Programming ⚫ SUPPLEMENT D ⚫
D-27
⚫ PART 3 ⚫ Managing Supply Chains
D-28
Optimal Solution
I0 = 30
W0 = 30
W1 = 30
W2= 46.25
W3= 46.25
W4=30
O1=0
O2=0
O3=15
O4=15
S1=0
S2=0
S3=0
S4=0
H1=0
H2=16.25
H3=0
H4=0
F1=0
F2=0
F3=0
F4=16.25
I1=30
I2=35
I3=55
I4=30
Production Plan
Anticipated
Quarter
Regular Time
Overtime
Subcontracting
Inventory
1
30,000
0
0
30,000
2
46,250
0
0
35,000
3
46,250
15,000
0
55,000
4
30,000
15,000
0
30,000
Total cost = $2,314,000
The cost of this plan is $143,200 lower than for the original demand structure plan in Problem
20. Demand management can reduce the costs of production, primarily through reduced
inventory costs in this problem.
22. Bull Grin Company
a. There are three sets of decision variables required for this problem
Pi = Regular time production in pounds in quarter i
Oi = Overtime production in pounds in quarter i
Ii = Beginning inventory in pounds in quarter i
No mention is made of hiring and layoff costs, so we assume that they are negligible and can
be ignored. The LP formulation is:
Minimize Z = .81P1 + .81P2 + .81P3 + .81P4 + .9O1 + .9O2 + .9O3 + .9O4 + .11I1 + .11I2 + .11I3 + .11I4
Subject to: Production in Q1: P1 + O1 + I0 – I1 = 100,000
Production in Q2: P2 + O2 + I1 – I2 = 410,000
Production in Q3: P3 + O3 + I2 – I3 = 770,000
Linear Programming ⚫ SUPPLEMENT D ⚫
D-29
b. The minimized cost to satisfy all demand is $1,490,900 and is achieved by executing the
following plan:
Beginning
Inventory
Regular Time
Production
Overtime
Production
Ending
Inventory
Demand
Satisfied
Quarter 1
40,000
400,000
0
340,000
100,000
Quarter 2
340,000
400,000
40,000
370,000
410,000
Quarter 3
370,000
400,000
40,000
40,000
770,000
Quarter 4
40,000
400,000
40,000
40,000
440,000
The formulation and solution are provided in the POM for Windows output below:
⚫ PART 3 ⚫ Managing Supply Chains
D-30
c. A new formulation is required to include subcontracting.
New decision variable Si = Subcontracted production in pounds for quarter i
Minimize Z = .81P1 + .81P2 + .81P3 + + .81P4 + 1.1S1 + 1.1S2 + 1.1S3 + 1.1S4
+ .9O1 + .9O2 + .9O3 + .9O4 + .11I1 + .11I2 + .11I3 + .11I4
Subject to:
Production in Q1: P1 + S1 + O1 + I1 – I2 = 100,000
Production in Q2: P2 + S2 + O2 + I2 – I3 = 410,000
Production in Q3: P3 + S3 + O3 + I3 – I4 = 770,000
The minimized cost to satisfy all demand is now $1,489,300 and is achieved by executing
the following plan:
Beginning
Inventory
Regular
Time
Production
Overtime
Production
Subcontracted
Production
Ending
Inventory
Demand
Satisfied
Quarter 1
40,000
360,000
0
0
300,000
100,000
Quarter 2
300,000
400,000
40,000
0
330,000
410,000
Quarter 3
330,000
400,000
40,000
0
0
770,000
Quarter 4
0
400,000
40,000
40,000
40,000
440,000
Linear Programming ⚫ SUPPLEMENT D ⚫
D-31
The formulation and solution are provided in the POM for Windows output below: