Aggregate Production
Planning
(APP)
Aggregate Production Planning
(APP)





Matches market demand to company resources
Plans production 6 months to 12 months in
advance
Expresses demand, resources, and capacity in
general terms
Develops a strategy for economically meeting
demand
Establishes a companywide game plan for
allocating resources
Inputs and Outputs to Aggregate
Production Planning
Capacity
Constraints
Demand
Forecasts
Size of
Workforce
Strategic
Objectives
Aggregate
Production
Planning
Production
per month
(in units or $)
Inventory
Levels
Company
Policies
Financial
Constraints
Units or dollars
subcontracted,
backordered, or
lost
Strategies for Meeting Demand
1. Use inventory to absorb fluctuations in demand
(level production)
2. Hire and fire workers to match demand (chase demand)
3. Maintain resources for high demand levels
4. Increase or decrease working hours (over & undertime)
5. Subcontract work to other firms
6. Use part-time workers
7. Provide the service or product at a later time period
(backordering)
Strategy Details




Level production - produce at constant rate & use
inventory as needed to meet demand
Chase demand - change workforce levels so that
production matches demand
Maintaining resources for high demand levels ensures high levels of customer service
Overtime & undertime - common when demand
fluctuations are not extreme
Strategy Details
Subcontracting - useful if supplier meets
quality & time requirements
 Part-time workers - feasible for unskilled
jobs or if labor pool exists
 Backordering - only works if customer is
willing to wait for product/services

Level Production
Demand
Production
Units
Time
Chase Demand
Demand
Units
Production
Time
APP Example

The Bavarian Candy Company (BCC) makes a
variety of candies in three factories worldwide. Its
line of chocolate candies exhibits a highly
seasonal pattern with peaks in winter months and
valleys during the summer months. Given the
costs and quarterly sales forecasts, determine
whether a level production or chase demand
production strategy would be more economically
meet the demand for chocolate candies.
APP Using Pure Strategies
Quarter
Spring
Summer
Fall
Winter
Sales Forecast (kg)
80,000
50,000
120,000
150,000
Hiring cost = $100 per worker
Firing cost = $500 per worker
Inventory carrying cost = $0.50 per kilogram per quarter
Production per employee = 1,000 kilograms per quarter
Beginning work force = 100 workers
Level Production Strategy
Quarter
Spring
Summer
Fall
Winter
Sales
Forecast
80,000
50,000
120,000
150,000
400,000
Production
Plan
100,000
100,000
100,000
100,000
Inventory
20,000
70,000
50,000
0
140,000
Cost = 140,000 kilograms x $0.50 per kilogram = $70,000
Chase Demand Strategy
Quarter
Spring
Summer
Fall
Winter
Sales
Forecast
80,000
50,000
120,000
150,000
Production Workers
Plan
Needed
80,000
80
50,000
50
120,000
120
150,000
150
Workers
Hired
70
30
100
Workers
Fired
20
30
50
Cost = (100 workers hired x $100) + (50 workers fired x $500)
= $10,000 + 25,000 = $35,000
LP Formulation
Define
Ht = # hired for period t
Ft = # fired for period t
It = inventory at end of period t
Pt = Production in period t
Wt = Workforce in period t
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4) +
$0.50 (I1 + I2 + I3 + I4)
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4)+ $0.50 (I1 + I2 + I3 + I4)
Subject to
P1 - I1 = 80,000
I1 + P2 - I2 = 50,000
I2 + P3 - I3 = 120,000
I3 + P4 - I4 = 150,000
(1)
(2)
(3)
(4)
Demand
constraints
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4)+ $0.50 (I1 + I2 + I3 + I4)
Subject to
P1 - I1 = 80,000
(1)
Demand
I1 + P2 - I2 = 50,000
(2)
constraints
I2 + P3 - I3 = 120,000
(3)
I3 + P4 - I4 = 150,000
(4)
P1 - 1,000 W1 = 0
P2 - 1,000 W2 = 0
P3 - 1,000 W3 = 0
P4 - 1,000 W4 = 0
(5)
(6)
(7)
(8)
Production
constraints
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4)+ $0.50 (I1 + I2 + I3 + I4)
Subject to
P1 - I1 = 80,000
(1)
Demand
I1 + P2 - I2 = 50,000
(2)
constraints
I2 + P3 - I3 = 120,000
(3)
I3 + P4 - I4 = 150,000
(4)
P1 - 1,000 W1 = 0
(5)
Production
P2 - 1,000 W2 = 0
(6)
constraints
P3 - 1,000 W3 = 0
(7)
P4 - 1,000 W4 = 0
(8)
W1 - H1 + F1 = 100
W 2 - W 1 - H 2 + F2 = 0
W 3 - W 2 - H 3 + F3 = 0
W 4 - W 3 - H 4 + F4 = 0
(9)
(10)
(11)
(12)
Work force
constraints
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4)+ $0.50 (I1 + I2 + I3 + I4)
Subject to
P1 - I1 = 80,000
I1 + P2 - I2 = 50,000
I2 + P3 - I3 = 120,000
I3 + P4 - I4 = 150,000
P1 - 1,000 W1 = 0
P2 - 1,000 W2 = 0
P3 - 1,000 W3 = 0
P4 - 1,000 W4 = 0
W1 - H1 + F1 = 100
W2 - W1 - H2 + F2 = 0
W3 - W2 - H3 + F3 = 0
W4 - W3 - H4 + F4 = 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Demand
constraints
Production
constraints
Work force
constraints
LP Solution:
Z = $32,000
H1= 0, F1= 20, I1= 0, P1=80000;
H2= 0, F2= 0, I2= 30000, P2=80000;
H3= 10, F3= 0, I3= 0, P3=90000;
H4= 60, F4= 0, I4= 0, P4=150000;
Summary: APP By Linear
Programming
Min Z = $100 (H1 + H2 + H3 + H4) + $500 (F1 + F2 + F3 + F4)+ $0.50 (I1 + I2 + I3 + I4)
Subject to
P1 - I1 = 80,000
I1 + P2 - I2 = 50,000
I2 + P3 - I3 = 120,000
I3 + P4 - I4 = 150,000
P1 - 1,000 W1 = 0
P2 - 1,000 W2 = 0
P3 - 1,000 W3 = 0
P4 - 1,000 W4 = 0
W1 - H1 + F1 = 100
W2 - W1 - H2 + F2 = 0
W3 - W2 - H3 + F3 = 0
W4 - W3 - H4 + F4 = 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Demand
constraints
Production
constraints
where
Ht = # hired for period t
Ft = # fired for period t
It = inventory at end
of period t
LP Solution:
Z = $32,000
H1= 0, F1= 20, I1= 0, P1=80000;
H2= 0, F2= 0, I2= 30000,
Work force P =80000;
2
constraints H3= 10, F3= 0, I3= 0, P3=90000;
H4= 60, F4= 0, I4= 0, P4=150000;
APP By The Transportation
Method
Quarter
1
2
3
4
Expected
Demand
900
1500
1600
3000
Regular
Capacity
1000
1200
1300
1300
Overtime Subcontract
Capacity Capacity
100
500
150
500
200
500
200
500
Regular production cost per unit = $20
Overtime production cost per unit = $25
Subcontracting cost per unit = $28
Inventory carrying cost per unit per period = $3
Beginning inventory = 300 units
Initial Setup of the Transportation Tableau
Period of Production
Period of
1
Use
Beginning
Inventory 300
20
1 Regular
2
3
4
Un- Capa
used -city
23
26
29
1000
Overtime
25
28
31
34
100
Subcontract
28
31
34
37
500
20
23
26
1200
Overtime
25
28
31
150
Subcontract
28
31
34
500
20
23
1300
Overtime
25
28
200
Subcontract
28
31
500
20
1300
Overtime
25
200
Subcontract
28
500
2 Regular
3 Regular
4 Regular
Demand
900
1500 1600 3000
Solution: Step 1
Period of Production
Period of
1
Use
Beginning
Inventory 300
20
1 Regular
600
25
Overtime
2
3
4
Un- Capa
used -city
23
26
29
1000
28
31
34
100
31
34
37
500
20
23
26
1200
Overtime
25
28
31
150
Subcontract
28
31
34
500
20
23
1300
Overtime
25
28
200
Subcontract
28
31
500
20
1300
Overtime
25
200
Subcontract
28
500
28
Subcontract
2 Regular
3 Regular
4 Regular
Demand
900
1500 1600 3000
Solution: Step 2
Period of Production
Period of
1
2
Use
Beginning
Inventory 300
20
23
1 Regular
600 300
25
28
Overtime
28
Subcontract
2 Regular
3
4
Un- Capa
used -city
26
29
1000
31
34
100
31
34
37
500
20
23
26
1200
1200
Overtime
25
28
31
150
Subcontract
28
31
34
500
20
23
1300
Overtime
25
28
200
Subcontract
28
31
500
20
1300
Overtime
25
200
Subcontract
28
500
3 Regular
4 Regular
Demand
900
1500 1600 3000
Final Transportation Tableau
Period of Production
Period of
1
2
3
4
Un- Capa
Use
used -city
Beginning
Inventory 300
20
23
26
29
1 Regular
0 1000
600 300 100 ----25
28
31
34
Overtime
0 100
100
28
31
34
37
Subcontract
500 500
20
2 Regular
1200 -----
Overtime
25
Subcontract
28
26
-----
28
31
150
31
34
250
20
3 Regular
23
25
0
500
20
0
1300
25
0
200
28
0
500
200
Subcontract
900
250 500
31
1300
Overtime
150
200
500
4 Regular
0
0
28
28
Subcontract
1200
1300
-----
200
0
0
1300 -----
Overtime
Demand
23
500
1500 1600 3000
Production Plan
Period
1
2
3
4
Demand
900
1500
1600
3000
Total
Strategy Variable
Reg Prodn Overtime Sub End Inv
1000
100
0
500
1200
150
250
600
1300
200
500
1000
1300
200
500
0
7000 4800 650 1250
2100

Regular Production Cost = (4,800 * $20)=$96,000

Overtime Production Cost = (650 * $25)= $16,250

Subcontracting Cost = (1,250 * $28) =
$35,000

Inventory Cost = (2,100 * $3) =
$ 6,300

The Total Cost of the Plan =
$153,550
Linear Programming Formulation
Let:







Dt = units required in period t, (t = 1,…,T)
m = number of sources of product in any period
Pit = capacity, in units of product, of source i in period t, (i = 1,…,m)
Xit = planned quantity to be obtained from source i in period t
cit = variable cost per unit from source i in period t
ht = cost to store a unit from period t to period t+1
It = inventory level at the end of period t, after satisfying the requirement in period t
Minimize
z 
T

t 1
m
[  cit Xit  htIt ]
i 1
ST
Xit  Pit, (i  1,..., m; t  1,..., T )
It  It

1

m

Xit  Dt , ( t  1,..., T )
i 1
Xit  0, (i  1,..., m; t  1,..., T )
It  0, ( t  1,..., T )
Minimize
z  20( XR1  XR 2  XR 3  XR 4 )  25( XO1  XO 2  XO 3  XO 4 )
 28( XS 1  XS 2  XS 3  XS 4 )  3( I 1  I 2  I 3  I 4 )
ST
XR1  1000, XR 2  1200, XR 3  1300, XR 4  1300;
XO1  100, XO 2  150, XO 3  200, XO 4  200;
XS 1  500, XS 2  500, XS 3  500, XS 4  500;
XR1  XO1  XS 1  I 1  600,
XR 2  XO 2  XS 2  I 2  I 1  1500,
XR 3  XO 3  XS 3  I 3  I 2  1600,
XR 4  XO 4  XS 4  I 4  I 3  3000;
Optimal Value (Z) = $153,550












XR1
XR2
XR3
XR4
XO1
XO2
XO3
XO4
XS1
XS2
XS3
XS4
= 1000
= 1200
= 1300
= 1300
=0
= 150
= 200
= 200
=0
= 350
= 500
= 500
Strategies for Managing Demand

Shift demand into other periods
incentives, sales promotions, advertising
campaigns

Offer product or services with countercyclical demand patterns
create demand for idle resources
Aggregate Planning for Services
1. Most services can’t be inventoried
2. Demand for services is difficult to predict
3. Capacity is also difficult to predict
4. Service capacity must be provided at the
appropriate place and time
5. Labor is usually the most constraining
resource for services
Services Example

The central terminal at the Deutsche Cargo receives
airfreight from aircraft arriving from all over Europe and
redistributes it to aircraft for shipment to all European
destinations. The company guarantees overnight shipment
of all parcels, so enough personnel must be available to
process all cargo as it arrives. The company now has 24
employees working in the terminal. The forecasted demand
for warehouse workers for the next 7 months is 24, 26, 30,
28, 28, 24, and 24. It costs $2,000 to hire and $3,500 to lay
off each worker. If overtime is used to supply labor beyond
the present work force, it will cost the equivalent of $2,600
more for each additional worker. Should the company use
a level capacity with overtime or a matching demand plan
for the next six month?
The Level Capacity with Overtime Plan
(1)
(2)
(3)
Month
1
Number of Workers
Forecasted
24
Overtime Labor Cost
[(2)-24]*$2,600
0
2
26
$5,200
3
30
15,600
4
28
10,400
5
28
10,400
6
24
0
The Matching Demand Plan
(1)
Month
(2)
Number of
Workers
Required
0
1
2
3
4
5
6
7
24
24
26
30
28
28
24
24
(3)
Number of
Workers
Hired
(4)
Number of
Workers
Laid Off
0
2
4
0
(5)
Cost of
Hired
Workers
[(3)*$2,000]
(6)
Cost of
Laid-Off
Workers
[(4)*$3,500]
0
$4,000
8,000
2
0
4
0
$7,000
0
14,000

The cost of the Level Capacity with
Overtime = $ 41,600

The total cost of the Matching Demand plan
= $12,000 + $21,000 = $33,000

Hence, since the cost of matching demand
plan is less than the level capacity plan with
overtime and would be the preferred plan
Aggregate Planning Example 1
 A manufacturer produces a line of household products
fabricated from sheet metal.
To illustrate his
production planning problem, suppose that he makes
only four products and that his production system
consists of five production centers: stamping, drilling,
assembly, finishing (painting and printing), and
packaging. For a given month, he must decide how
much of each product to manufacture, and to aid in
this decision, he has assembled the data shown in
Tables 1 and 2. Furthermore, he knows that only 2000
square feet of the type of sheet metal used for
products 2 and 4 will be available during the month.
Product 2 requires 2.0 square feet per unit and
product 4 uses 1.2 square feet per unit.

TABLE 1 Production Data for Example 1
PRODUCTION RATES IN HOURS PER UNIT
DEPARTMENT
Production
PRODUCT 1 PRODUCT 2 PRODUCT 3 PRODUCT 4
Hours
Available

Stamping
0.03
0.15
0.05
0.10
400

Drilling
0.06
0.12
-----
0.10
400

Assembly
0.05
0.10
0.05
0.12
500

Finishing
0.04
0.20
0.03
0.12
450

Packaging
0.02
0.06
0.02
0.05
400

TABLE 2 Product Data for Example 1
NET SELLING
PRODUCT PRICE/UNIT




1
2
3
4
$10
25
16
20
VARIABLE
COST/UNIT
$6
15
11
14
SALES POTENTIAL
MINIMUM MAXIMUM
1000
----500
100
6000
500
3000
1000
A Linear Program of Example 1:
 Define xi be the number of units of Product i to be produced per month, i = 1, 2,
3, and 4.
Maximize
4 x 1  10 x 2  5 x 3  6 x 4
ST
0.03 x 1  0.15 x 2  0.05 x 3  0.10 x 4  400
0.06 x 1  0.12 x 2  0.10 x 4  400
0.05 x 1  0.10 x 2  0.05 x 3  0.12 x 4  500
0.04 x 1  0.20 x 2  0.03 x 3  0.12 x 4  450
0.02 x 1  0.06 x 2  0.02 x 3  0.05 x 4  400
2.0 x 2  1.2 x 4  2000
x 1  1000
x 1  6000
x 2  500
x 3  500
x 3  3000
x 4  100
x 4  1000

Solution of Example 1 using LINGO Software Package (get a free
copy of this package from the web site at www.lindo.com):

Objective value: 42600.00




Variable
X1
X2
X3
X4
Value
5500.000
500.0000
3000.000
100.0000
Reduced Cost
0.0000000
0.0000000
0.0000000
0.0000000














Row
PROFIT
STAMPING
DRILLING
ASSEMBLY
FINISHING
PACKAGING
SHEETMETAL
MINPROD1
MAXPROD1
MAXPROD2
MINPROD3
MAXPROD3
MINPROD4
MAXPROD4
Slack or Surplus
42600.00
0.0000000
0.0000000
13.00000
28.00000
195.0000
880.0000
4500.000
500.0000
0.0000000
2500.000
0.0000000
0.0000000
900.0000
Dual Price
1.0000000
0.0000000
66.66666
0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
2.0000000
0.0000000
5.000000
-0.6666667
0.0000000
Ranges in which the basis is unchanged:
Objective Coefficient Ranges




Variable
X1
X2
X3
X4
Current
Allowable
Coefficient Increase
4.000000
INFINITY
10.00000
INFINITY
5.000000
INFINITY
6.000000
INFINITY
Allowable
Decrease
INFINITY
INFINITY
INFINITY
INFINITY
Righthand Side Ranges:
Row













STAMPING
DRILLING
ASSEMBLY
FINISHING
PACKAGING
SHEETMETAL
MINPROD1
MAXPROD1
MAXPROD2
MINPROD3
MAXPROD3
MINPROD4
MAXPROD4
Current
RHS
400.0000
400.0000
500.0000
450.0000
400.0000
2000.000
1000.000
6000.000
500.0000
500.0000
3000.000
100.0000
1000.000
Allowable
Increase
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
INFINITY
Allowable
Decrease
100.0000
3000.000
500.0000
5500.000
0.0
13.00000
28.00000
195.0000
880.0000
4500.000
500.0000
2500.000
900.0000