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Chapter 3
Aggregate Planning
By: Prof. Y. Peter Chiu
9 / 1 / 2011
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 Fig.1 The hierarchy of prodution planning decision
p.128
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 # Number of Worker - Hours Required Selling Price
to produce
1 4.2 $285
2 4.9 $345
3 5.1 $395
4 5.2 $425
5 5.4 $525
6 5.8 $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 # % of the total numbers of sales
1 32%
2 21%
3 17%
4 14%
5 10%
6 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 # $/hr
1 $285/4.2 = $67.86
2 $345/4.9 = $70.41
3 $395/5.1 = $77.45
4 $425/5.2 = $81.73
5 $525/5.4 = $97.22
6 $725/5.8 = $125.00
10
§. A2: Example ~ Aggregate Planning ~ ( page 4 )
Eg. 3-1 / Solution:
then
Model # $/hr * % of Sales
1 $67.86* 0.32 = $21.72
2 $70.41* 0.21 = $14.79
3 $77.45* 0.17 = $13.17
4 $81.73* 0.14 = $11.44
5 $97.22* 0.10 = $9.72
6 $125.00* 0.06 = $7.50
what is $78.34?
$78.34
―Average dollars of output / worker-hour input‖
in this particular production plant
11
§. A2: Example ( page 5 )
~ Aggregate Planning ~
Eg. 3-1 / Solution: (continued)
(2) The manager decides to define an aggregate unit of
production as a fictitious TV
Model # Number of worker-Hours Required
1 4.2*0.32 = 1.34
2 4.9*0.21 = 1.03
3 5.1*0.17 = 0.87
4 5.2*0.14 = 0.73
5 5.4*0.10 = 0.54
6 5.8*0.06 = 0.35
4 4.86
Sum$=.86
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, C F $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(-) 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？ 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 B C D E
Number of Minimum
Units Produced Number of
Number of per Worker Forecasted Workers Required
month Working Days (B ×.14653) Net Demand (D/C Rounded Up)
January 20 2.931 780 267
February 24 3.517 640 182
March 18 2.638 900 342
April 26 3.810 1200 315
May 22 3.224 2000 621
June 15 2.198 2000 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 B C D E 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 755 145 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 D E 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 755 145 30
753 144 13
CH $500 755*(500)+145*(1000)+30*(80)=$524,900
600*(80)=4800 + 48,000
C F $1000
C I $ 80 572,900
$569,540/910 workers 24
753 & 144 & 13
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 D E F G H I
Number of
Number of Units
Number 0f Number Number unit per Produced Cumulative Cumulativ Ending
month Worker Hired Fired Worker (B × E) Production e Demand Inventory
January 267 33 2.931 783 783 780 3
February 182 85 3.517 640 1423 1420 3
March 341 159 2.683 900 2323 2320 3
April 315 26 3.810 1200 3523 3520 3
May 620 305 3.224 1999 5522 5520 2
June 910 290 2.198 2000 7522 7520 2
Total 754 144 16
CH=$500 745*(500)+(144)*(1000)+16*(80)= $522,280
CF=$1000 600*(80)=48,000 + 48,000
CI=$80 $570,280
25
§. 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 B C D
Cumulative Ratio
Cumulative Number of units B/C
month Net Demand Produced per Worker (Rounded Up)
January 780 2.931 267
February 1420 6.448 221
March 2320 9.086 256
April 3520 12.896 273
May 5520 16.120 Hire to max. 411
343
June 7520 18.318 workers initially 411
26
§. 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 C D E F
Number of Monthly Ending
unit Produced Production Cumulative Cumulative Inventory
month per Worker (B ×411) Production Net Demand (D - E)
January 2.931 1,205 1,205 780 425
February 3.517 1,445 2,650 1,420 1,230
March 2.638 1,084 3,734 2,320 1,414
April 3.810 1,566 5,300 3,520 1,780
May 3.224 1,325 6,625 5,520 1,105
June 2.198 903 7,528 7,520 8
Total 5,962
Constant Work Force Plan hires 111 workers in January
111*(500)+5962*(80)+600*(80)=$580,460
27
§. 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) $524,900
F：145(-1)
■ Minimum Inventory Level, total I = 30 or 13 (-17)
$48,000
■ Ending at desirable work force level ? 910
Total costs = $569,540 & ending 910 workers
28
§. 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
29
§. A6: A Prototype problem ( page 12 )
Example 3.2 : Solution
(4) Other Suggestions：(A) CHIU’s – Suggestion Ⅰ
A B C D E F
Di Ki
'
B/C 300 Inv
January 780 2.931 267 300 99
February 1420 6.448 221 300 514
March 2320 9.086 256 300 405
April 3520 12.896 273 300 348
May 5520 2000 16.120 3.224 343 513 1
June 7520 2000 18.318 2.198 411 910 1
H：610 I：1368
610*(500)+1368*(80)=$414,440 910
$ 414,440 910 workers
+600*($80)=$462,440 30
§. A6: A Prototype problem ( page 13 )
Example 3.2 : Solution
(4) Other Suggestions： (B) CHIU’s – Suggestion II
A B C D E F
Di Ki
'
B/C 300 Inv
January 780 2.931 267 300 99
February 1420 6.448 221 300 514
March 2320 9.086 256 300 405
April 3520 12.896 273 300 348
May 5520 2000 16.120 3.224 343 674 520
June 7520 2000 18.318 2.198 411 674 1
H：374 I：1887
374*(500)+1887*(80)=$337,960 674
$ 337,960 674 workers
+600*($80)=$385,960 31
(4) Other Suggestions： (C) CHIU’s – Suggestion III
A B C D E F
D K
'
i i B/C 300 Inv
J anuar y 780 2. 931 267 300
273 99
20
Febr uar y 1420 6. 448 221 300
273 514
340
ar
M ch 2320 9. 086 256 300
273 405
160
Apr i l 3520 12. 896 273 300
273 348
1
M ay 5520 2000 16. 120 3.224 343 674
738 520
380
J une 7520 2000 18. 318 2.198 411 674
738 1
2
F:27 H: 465
H：374 Inv.: 903
I ：1887
374*( 500) +1887*( 80) =$337, 960
465*($500)+903*($80)= $304,740 674
$ 304,740 +$27000 738 workers
+600*($80)=$379,740
32
§. A6.1: Class
Problems Discussion
Chapter 3 :
( # 12, 14 ) pp.139-140
Preparation Time : 10 ~ 15 minutes
Discussion : 10 minutes
33
The End
34
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