Advanced P.O.M.
Chap. 3
Aggregate Planning
By: Prof. Y. Peter Chiu
9 / 1 / 2010
1
§. A1: Introduction
~ Aggregate Planning ~
 Macro production planning
 The problem of deciding how many
employees the firm should retain
— for a manufacturing firm to decide the
quantity and mix of products to be produced
 e.g. In Service organizations：
— Airlines plans staffing levels for flight
attendants & pilots
— Hospitals plans staffing levels for nurses
2
§. A1: Introduction
( page 2 )
 Macro planning begins with the forecast of
demand
 The aggregate planning methodology we
discussed later, requires the assumption that
〝demand is deterministic, or known in
advance〞.
 Ways to satisfy demand
— in house production
— out-sourced / subcontracting
3
§. A1: Introduction
■
( page 3 )
Competing objectives of Aggregate planning：
— To react quickly to anticipated changes in
demand（i.e. making frequent and potentially
large changes in the size of the labor force
– A chase strategy, may be cost effective
short term.）
— To retain a stable workforce.
— To develop a production plan for the firm that
maximizes profit over the planning horizon
subject to constraints on capacity.
4
§. A1: Introduction
( page 4 )
 Aggregate planning methodology is
designed to translate demand forecasts into
a blueprint for planning staffing and
production levels for the firm over a
predetermined planning horizon
 Production planning may be viewed as a
hierarchical process in
— purchasing
— production
— staffing
decisions must be made at several levels in
the firm
5
Fig. 3-1
p.128
Fig.1 The hierarchy of prodution planning decision
Forcast of aggregate demand
for t period planning horizon
Aggregate production plan
Determination of aggregate production
& Workforce levels for t period planning
horizon
Master Production Schedule
Production levels by item by time
period
Materials Req. Planning System
Detailed timetable for production &
assembly of components and
subassemblies
6
§. A1: Introduction
( page 6 )
 Aggregate Units of Production
— amount of work required（worker-years）
— weight（tons of steel）
— volume（gallons of gasoline）
— dollar value（value of inventory in dollars）
~ Not Always Obvious
7
§. A2: Example ~ Aggregate Planning ~
 Example 3-1
( p.127 )
A plant manager working for a large national
appliance firm is considering implementing an
aggregate planning system to determine the
workforce and production levels in his plant. This
particular plant produces 6 models of TVs. The
characteristics of the TVs are :
Model #
1
2
3
4
5
6
Number of Worker - Hours Required
to produce
4.2
4.9
5.1
5.2
5.4
5.8
Selling Price
$285
$345
$395
$425
$525
$725
8
§. A2: Example
( page 2 )
~ Aggregate Planning ~
 Example 3-1 （continued）
The manger notices that the percentages of the
total number of sales for these six models have been
fairly constant：
Model
1
2
3
4
5
6
#
% of the total numbers of sales
32%
21%
17%
14%
10%
6%
9
§. A2: Example
( page 3 )
~ Aggregate Planning ~
Eg. 3-1 / Solution:
 To find the particular aggregation scheme
(1) Selling price / Number of worker-hours required
= $ per Input-Hour
Model #
1
2
3
4
5
6
$/hr
$285/4.2 = $67.86
$345/4.9 = $70.41
$395/5.1 = $77.45
$425/5.2 = $81.73
$525/5.4 = $97.22
$725/5.8 = $125.00
10
§. A2: Example
~ Aggregate Planning ~
( page 4 )
Eg. 3-1 / Solution:
then
Model #
1
2
3
4
5
6
$/hr * % of Sales
$67.86* 0.32 = $21.72
$70.41* 0.21 = $14.79
$77.45* 0.17 = $13.17
$81.73* 0.14 = $11.44
$97.22* 0.10 = $9.72
$125.00* 0.06 = $7.50
 $78.34
what is $78.34?
“Average dollars of output / worker-hour input”
in this particular production plant
11
§. A2: Example
( page 5 )
~ Aggregate Planning ~
 Eg. 3-1 / Solution: (continued)
(2) The manager decides to define an aggregate unit of
production as a fictitious TV
Model #
1
2
3
4
5
6
Number of worker-Hours Required
4.2*0.32 = 1.34
4.9*0.21 = 1.03
5.1*0.17 = 0.87
5.2*0.14 = 0.73
5.4*0.10 = 0.54
5.8*0.06 = 0.35
  $=4.864.86
Sum
12
§. A2: Example
( page 6 )
~ Aggregate Planning ~
◆ what is 4.86 hours?
“ Average worker-hours required to produce
a fictitious TV”
What if we like to know
“ how many fictitious TV can one worker – one
day (8hrs) produce ? ”
[ 1 / 4.86 ] x 8 = 1.646
— Applications of this Avg. worker-hours/ TV ：
～ If the manager can obtain sales forecast of overall models, then
he can use this to plan workforce ～
13
§. A3:
Hierarchical production planning (HPP)
 Hierarchy for Aggregate planning ( by Hax ＆ Meal 1975 )
(1) Items：Final products to be delivered to the customer.
(SKU-stock keeping unit)
(2) Families：A group of items that share a common
manufacturing setup cost
(3) Types：Groups of families with production quantities
that are determined by a single aggregate
production plan
■
■
In our previous example, different models of TVs are families,
while type might be large appliances.
Hax ＆ Meal aggregation scheme will not necessarily work
in every situation
14
§. A4: Overview of the
Aggregate planning problem
 The goal of aggregate planning is to determine
aggregate production quantities and the levels of
resources required to achieve these production goals
 The primary issues related to the aggregate planning
problem include：
◆ Smoothing
— 2 key components of smoothing costs are the
costs that result from hiring and firing workers
◆ Bottleneck problems
— System unable to respond to sudden changes
in demand as a result of capacity restrictions
15
§. A4: Overview of the
Aggregate planning problem
( page 2 )
 The primary issues related to the aggregate planning
problem (continued) :
◆ Planning horizon — Not too small T, Not too large T
— End-of-horizon effect
◆ Treatment of demand
— Assumption of deterministic or known, ignores
the possibility of forecast errors
— Needs a buffer for forecast errors
— incorporates the effects of seasonal
fluctuations & business cycles
16
§. A5: Costs in Aggregate planning
◆ Smoothing Costs — laid off (firing) workers
— hiring workers
◆ Holding Costs — capital tied up in inventory
◆ Shortage Costs — excess demand normally
assumes backlogged
◆ Regular time Costs
— the cost of producing one unit of output
during regular working hours
◆ Overtime and subcontracting costs
17
§. A6: A Prototype problem
Example 3.2：
( p.133)
Densepack company is to plan workforce and
production levels for the six-month period January to
June. The firm produces a line of disk drives for
mainframe computers that are plug compatible with
several computers produced by major manufacturers.
Forecast demands for the next 6 months for a
particular line of drives produced in their plant 1, are
1280, 640, 900,1200, 2000, and 1400. There are
currently (end of December) 300 workers employed in
plant 1. Ending inventory in December is expected to
be 500 units, and the firm would like to have 600 units
on hand at the end of June. And It is estimated that：
18
§. A6: A Prototype problem
( page 2 )
Example 3.2：(continued)
Cost of hiring 1 worker, C H  $500
Cost of firing 1 worker, CF  $1000
Cost of holding 1 unit of Inventory for 1month, C I  $80
The plant manager observed that in the past, over 22
working days, with the workforce level constant at 76
workers, the firm produced 245 disk drives.
What is ‘’Number of aggregate units produced by 1
worker in 1 day ?’’
Evaluate : (1) the chase strategy (zero inventory plan)
and (2) the constant workforce plan
19
§. A6: A Prototype problem
( page 3 )
Example 3.2 : Solution
Starting：500 units
300 workers
300 workers
1
1280
1
780(-)
2
640
2
640
1420
3
900
3
900
2320
4
1200
4 1200
3520
5
2000
5 2000
5520
6
1400
6 2000(+) 7520
Ending：600 units
x worker？
780
x worker？
20
§. A6: A Prototype problem
( page 4 )
Example 3.2 : Solution
K= # of aggregate units produced by one worker in one
day
76 workers work 22 days producing 245 disk drives
245
K
 0.14653
76 (22)
21
§. A6: A Prototype problem
( page 5 )
Example 3.2 : Solution
(1) Evaluation of chase strategy
Table 1: Initial Calculation for Chase Strategy Zero Inventory Plan for Densepack
A
month
January
February
March
April
May
June
B
Number of
Working Days
20
24
18
26
22
15
C
D
E
Number of
Units Produced
per Worker
(B ×.14653)
Forecasted
Net Demand
Minimum
Number of
Workers Required
(D/C Rounded Up)
2.931
3.517
2.638
3.810
3.224
2.198
780
640
900
1200
2000
2000
267
182
342
315
621
910
22
§. A6: A Prototype problem
( page 6 )
Example 3.2 : Solution
(1) Evaluation of chase strategy
Table 2: Chase Strategy Zero Inventory Plan for Densepack
A
F
G
H
I
Number of
Number of
Units
Ending
Number of Number Number
unit
Produced Cumulative Cumulative Inventory
month
Worker
Hired Fired per Worker (B ×E) Production Demand
(G-H)
January 267
33
2.931
783
783
780
3
February 182
85
3.517
640
1423
1420
3
March
342
160
2.638
902
2325
2320
5
April
315
27
3.810
1200
3525
3520
5
May
621
306
3.224
2002
5527
5520
7
June
910
289
2.198
2000
7527
7520
7
Total
B
C
755
D
145
E
30
23
§. A6: A Prototype problem
( page 7 )
Example 3.2 : Solution
(1) Evaluation of chase strategy
Table 2: Chase Strategy Zero Inventory Plan for Densepack
A
B
C
F
G
H
I
Number of
Number of
Units
Ending
Number of Number Number
unit
Produced Cumulative Cumulative Inventory
month
Worker
Hired Fired per Worker (B ×E) Production Demand
(G-H)
January 267
33
2.931
782
782
780
2
February 182
85
3.517
640
1422
1420
2
March
342
160
2.638
902
2324
2320
4
April
315
27
3.810
1200
3525
3520
5
May
621
306
3.224
2002
5527
5520
7
June
910
289
2.198
2000
7527
7520
7
Total
755
753
CH  $500
C F  $1000
C I  $ 80
D
E
145
144
30
13
755*(500)+145*(1000)+30*(80)=$524,900
600*(80)=4800
+ 48,000
753 & 144 & 13 
572,900
$569,540/910 workers
24
§. A6: A Prototype problem
( page 8 )
Example 3.2 : Solution
(2) Evaluation of Constant workforce plan
Table 3: Computation of the Minimum Work Force Required by Densepack
A
month
January
February
March
April
May
June
B
Cumulative
Net Demand
780
1420
2320
3520
5520
7520
C
D
Cumulative
Number of units
Produced per Worker
2.931
6.448
9.086
12.896
16.120
18.318
Ratio
B/C
(Rounded Up)
Hire to max. 411
workers initially 
267
221
256
273
343
411
25
§. A6: A Prototype problem
( page 9 )
Example 3.2 : Solution
(2) Evaluation of Constant workforce plan
Table 4: Inventory Levels for Constant Work Force Schedule
A
B
Number of
unit Produced
month
per Worker
January
2.931
February
3.517
March
2.638
April
3.810
May
3.224
June
2.198
Total
C
Monthly
Production
(B ×411)
1,205
1,445
1,084
1,566
1,325
903
D
E
Cumulative
Production
1,205
2,650
3,734
5,300
6,625
7,528
Cumulative
Net Demand
780
1,420
2,320
3,520
5,520
7,520
F
Ending
Inventory
(D - E)
425
1,230
1,414
1,780
1,105
8
5,962
Constant Work Force Plan hires 111 workers in January
111*(500)+5962*(80)+600*(80)=$580,460
26
§. A6: A Prototype problem
( page 10 )
Example 3.2 : Solution
(3) Comparison of 2 plans
Chase Strategy ( Zero Inventory Plan )
■ Hiring & Firing Constantly Appropriate?
H：755(-2) $522,500
F：145(-1)
■ Minimum Inventory Level, total I = 30 or 13 (-17)
$48,000+$2,400=$50,400
■ Ending at desirable work force level ? 910
Total costs = $569,540 & ending 910 workers
27
§. A6: A Prototype problem
( page 11 )
Example 3.2 : Solution
(3) Comparison of 2 plans (continued)
Constant Work Force Plan
■ Minimum hiring & firing (one time) $55,500
 H : 111 
F :0
 55,500


 one time 
■ More carryovers units, total I = 5962
$476,960
■ Ending at better work force level . 411
Total costs = 580,460 & ending 411 workers
28
§. A6: A Prototype problem
( page 12 )
Example 3.2 : Solution
(4) Other Suggestions：(A) CHIU’s – Suggestion Ⅰ
A
January
February
March
April
May
June
B
 Di
C
D
 Ki
'
780
2.931
1420
6.448
2320
9.086
3520
12.896
5520 2000 16.120 3.224
7520 2000 18.318 2.198
E
B/C
267
221
256
273
343
411
300
Inv
300
300
300
300
513
910
99
514
405
348
1
1
H：610
610*(500)+1368*(80)=$414,440
$ 414,440
F
I：1368
910
910 workers
+600*($80)=$462,440
(19% reductions)
29
§. A6: A Prototype problem
( page 13 )
Example 3.2 : Solution
(4) Other Suggestions： (B) CHIU’s – Suggestion II
A
January
February
March
April
May
June
B
 Di
C
D
 Ki
'
780
2.931
1420
6.448
2320
9.086
3520
12.896
5520 2000 16.120 3.224
7520 2000 18.318 2.198
E
B/C
267
221
256
273
343
411
F
300
Inv
300
300
300
300
674
674
99
514
405
348
520
1
H：374
374*(500)+1887*(80)=$337,960
$ 337,960
I：1887
674
674 workers
+600*($80)=$385,960
(33% reductions)
30
§. A6.1: Class
Problems Discussion
Chapter 3 :
( # 12, 14 )
p.139-140
Preparation Time : 10 ~ 15 minutes
Discussion
: 10 minutes
31
The End
By: Prof. Y.P. Chiu
32