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Chapter 3
Aggregate Planning
McGraw-Hill/Irwin Copyright © 2005 by The McGraw-Hil l Companies, Inc. All r ights reserved.
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Aggregate Planning Strategies
Should inventories be used to absorb changes indemand during planning period?
Should changes be accomodated by varying thesize of the workforce?
Should part-timers be used, or should overtimeand idle time absorb fluctuations?
Should subcontractors be used on fluctuating
orders so a stable workforce can be maintained? Should prices or other factors be changed to
influence demand?
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Introduction to Aggregate Planning
Goal: To plan gross work force levels and set
firm-wide production plans.
Concept is predicated on the idea of an“aggregate unit ” of production:
» May be actual units,
» May be measured in weight (tons of steel), volume
(gallons of gasoline), time (worker-hours), or dollars ofsales.
» Can even be a fictitious quantity. (Refer to example intext and in slide below.)
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Important Issues
Smoothing. Refers to the costs and disruptions that
result from making changes from one period to the
next.
Bottleneck Planning . Problem of meeting peak demand because of capacity restrictions.
Planning Horizon. Assumed given (T), but what is
“right” value? Rolling horizons and end of horizon
effect are both important issues.
Treatment of Demand . Assume demand is known.
Ignores uncertainty to focus on the
predictable/systematic variations in demand, such as
seasonality.
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Relevant Costs
Smoothing Costs
– changing size of the work force
– changing number of units produced
Holding Costs
– primary component: opportunity cost of investment
Shortage Costs
– Cost of demand exceeding stock on hand. Whyshould shortages be an issue if demand is known?
Other Costs: payroll, overtime, subcontracting.
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Aggregate Units
The method is based on notion of aggregate
units. They may be
Actual units of production
Weight (tons of steel)
Volume (gallons of gasoline)
Dollars (Value of sales)
Fictitious aggregate units
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Example of fictitious aggregate units.
(Example 3.1)
One plant produced 6 models of washing machines:
Model # hrs. Price % sales
A 5532 4.2 285 32
K 4242 4.9 345 21
L 9898 5.1 395 17
L 3800 5.2 425 14
M 2624 5.4 525 10
M 3880 5.8 725 06
Question: How do we define an aggregate unit here?
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Example continued
Notice: Price is not necessarily proportional
to worker hours (i.e., cost): why?
One method for defining an aggregate unit:
requires: .32(4.2) + .21(4.9) + . . . + .06(5.8)
= 4.8644 worker hours. Forecasts fordemand for aggregate units can be obtained
by taking a weighted average (using the
same weights) of individual item forecasts.
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Prototype Aggregate Planning Example
(this example is not in the text)
The washing machine plant is interested in
determining work force and production
levels for the next 8 months. Forecasteddemands for Jan-Aug. are: 420, 280, 460,
190, 310, 145, 110, 125. Starting inventory
at the end of December is 200 and the firmwould like to have 100 units on hand at the
end of August. Find monthly production
levels.
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Step 1: Determine “net” demand.
(subtract starting inv. from per. 1 forecast and
add ending inv. to per. 8 forecast.)Month Net Predicted Cum. Net
Demand Demand
1(Jan) 220 2202(Feb) 280 500
3(Mar) 460 960
4(Apr) 190 1150
5(May) 310 1460
6(June) 145 1605
7(July) 110 1715
8(Aug) 225 1940
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Step 2. Graph Cumulative Net Demand
to Find Plans Graphically
0
200
400
600
800
1000
1200
1400
1600
1800
2000
1 2 3 4 5 6 7 8
Cum Net Dem
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Constant Work Force Plan
Suppose that we are interested in determining
a production plan that doesn‟t change the
size of the workforce over the planninghorizon. How would we do that?
One method: In previous picture, draw astraight line from origin to 1940 units in
month 8: The slope of the line is the number
of units to produce each month.
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Constant Workforce Plan (zero ending inv)
0
500
1000
1500
2000
1 2 3 4 5 6 7 8
Monthly Production = 1940/8 = 242.2 or rounded to
243/month.
But: there are stockouts.
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How can we have a constant work force plan
with no stockouts?
Answer: using the graph, find the straight line that goes
through the origin and lies completely above the
cumulative net demand curve:
Constant Work Force Plan With No Stockouts
0
500
1000
1500
2000
2500
3000
1 2 3 4 5 6 7 8
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From the previous graph, we see that cum. net demand curve
is crossed at period 3, so that monthly production is 960/3 =
320. Ending inventory each month is found from:
Month Cum. Net. Dem. Cum. Prod. Invent.
1(Jan) 220 320 100
2(Feb) 500 640 1403(Mar) 960 960 0
4(Apr.) 1150 1280 130
5(May) 1460 1600 1406(June) 1605 1920 315
7(July) 1715 2240 525
8(Aug) 1940 2560 620
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But - may not be realistic for several
reasons:
It may not be possible to achieve the
production level of 320 unit/mo with an
integer number of workers
Since all months do not have the same
number of workdays, a constant productionlevel may not translate to the same number
of workers each month.
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To overcome these shortcomings:
Assume number of workdays per month is
given
K factor given (or computed) where
K = # of aggregate units produced by one
worker in one day
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Finding K
Suppose that we are told that over a period
of 40 days, the plant had 38 workers who
produced 520 units. It follows that:
K= 520/(38*40) = .3421
= average number of units produced by
one worker in one day.
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Computing Constant Work Force
Assume we are given the following # working
days per month: 22, 16, 23, 20, 21, 22, 21,
22. March is still the critical month. Cum.net demand thru March = 960. Cum #
working days = 22+16+23 = 61. Find
960/61 = 15.7377 units/day implies15.7377/.3421 = 46 workers required.
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Why again did we pick on March?
Examining the graph we see that that was
the “Trigger point” where our constant
production line intersected the cumulativedemand line assuring NO STOCKOUTS!
Can we “prove” this is best?
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Tabulate Days/Production per Worker Vs.
Demand to find minimum numbers
Month # Work Days #Units/worker Forecast Demand net Min # Workers C. Net Demand C.Units/Worker
Min #
Workers
Jan 22.00 7.53 220.00 29.23 220.00 7.53 29.23
Feb 16.00 5.47 280.00 51.15 500.00 13.00 38.46
Mar 23.00 7.87 460.00 58.46 960.00 20.87 46.00
Apr 20.00 6.84 190.00 27.77 1150.00 27.71 41.50
May 21.00 7.18 310.00 43.15 1460.00 34.89 41.84
Jun 22.00 7.53 145.00 19.27 1605.00 42.42 37.84
Jul 21.00 7.18 110.00 15.31 1715.00 49.60 34.57
Aug 22.00 7.53 225.00 29.90 1940.00 57.13 33.96
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What should we look at?
Cumulative Demand says March needs
most workers – but will mean building
inventories in Jan + Feb If we keep this number of workers we will
continue to build inventory through the rest
of the plan
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Constant Work Force Production Plan
Mo # wk days Prod. Cum Cum Nt End Inv
Level Prod Dem
Jan 22 346 346 220 126
Feb 16 252 598 500 98
Mar 23 362 960 960 0
Apr 20 315 1275 1150 125
May 21 330 1605 1460 145Jun 22 346 1951 1605 346
Jul 21 330 2281 1715 566
Aug 22 346 2627 1940 687
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Addition of Costs
Holding Cost (per unit per month): $8.50
Hiring Cost per worker: $800
Firing Cost per worker: $1,250
Payroll Cost: $75/worker/day
Shortage Cost: $50 unit short/month
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Cost Evaluation of Constant Work Force Plan
Assume that the work force at end of Dec was 40.
Cost to hire 6 workers: 6*800 = $4800
Inventory Cost: accumulate ending inventory:(126+98+0+. . .+687) = 2093. Add in 100 units netted
out in Aug = 2193. Hence Inv. Cost =
2193*8.5=$18,640.50
Payroll cost:($75/worker/day)(46 workers )(167days) = $576,150
Cost of plan: $576,150 + $18,640.50 + $4800 =
$599,590.50
A Al i i ll d h “Ch
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An Alternative is called the “Chase
Plan”
Here, we hire and fire workers to keep
inventory low
We would employ only the number ofworkers needed each month to meet
demand
Examining our chart (earlier) we need:» Jan: 30; Feb: 51; Mar: 59; Apr: 27; May: 43
» Jun: 20; Jul: 15; Aug: 30
A Al i i ll d h “Ch
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An Alternative is called the “Chase
Plan”
So we hire or Fire (lay-off) monthly» Jan (starts with 40 workers): Fire 10 (cost $8000)
» Feb: Hire 21 (cost $16800)
» Mar: Hire 8 (cost $6400)
» Apr: Fire 31 (cost $38750)
» May: Hire 15 (cost $12000)
» Jun: Fire 23 (cost $28750)
» Jul: Fire 5 (cost $6250)» Aug: Hire 15 (cost $12000)
Total Personnel Costs: $128950
A Alt ti i ll d th “Ch
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An Alternative is called the “Chase
Plan”
Inventory cost is essentially 165*8.5 =$1402.50
Employment costs: $428325 Chase Plan Total: $558677.50
Betters the “Constant Workforce Plan” by:
» 599590.50 – 558677.50 = 40913 But will this be good for your image?
Can we find a better plan?
Cost Reduction in Constant Work Force Plan
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Cost Reduction in Constant Work Force Plan
& Chase Plan
In the original cum net demand curve, consider making
reductions in the work force one or more times over the
planning horizon to decrease inventory investment.
Plan Modified With Lay Offs in March and May
0
500
1000
1500
2000
1 2 3 4 5 6 7 8
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Cost Evaluation of Modified Plan
I will not present all the details here. The
modified plan calls for reducing the
workforce to 36 at start of April and makinganother reduction to 22 at start of June. The
additional cost of layoffs is $30,000, but
holding costs are reduced to only $4,250.The total cost of the modified plan is
$467,450.
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Optimal Solutions to Aggregate Planning
Problems Via Linear Programming
Linear Programming provides a means of solvingaggregate planning problems optimally. The LPformulation is fairly complex requiring 8Tvariables and 3T constraints, where T is the lengthof the planning horizon. Clearly, this can be aformidable linear program. The LP formulationshows that the modified plan we considered withtwo months of layoffs is in fact optimal for the
prototype problem.
E l i th O ti l (L P )
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Exploring the Optimal (L.P.)
Approach
We need an Objective Function for cost of the aggregate
plan (target minimization):
– Here the ci‟s are cost for hiring, firing, inventory, production,
etc – HT and FT are number of workers hired and fired
– IT, PT, OT, ST AND UT are numbers inventoried, produced on
regular time, overtime, by „sub-contract‟ or idle worker hours
respectively
1
T
H H F F I T R R o T u T S T
t
c N c N c I c P c O c U c S
E l i th O ti l (L P )
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Exploring the Optimal (L.P.)
Approach
This objective Function would be subject to
a series of constraints (one for each period)
– Number of Worker Constraints:
– Inventory Constraints:
– Production Constraints:
1t t t tW W H F
1t t t t t I I P S D
Where: k*n is the number of units produced by a workert
on regular time during a period
t t t t t P k n W O U
E l i th O ti l (L P )
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Exploring the Optimal (L.P.)
Approach
Assuming we allow no idle time and will produceonly on regular time
» No overtime or subcontracting
We would have:» 9 worker variables (W0 to Waug)
» 8 Hire Variable
» 8 Fire Variables
» 9 Inventory Variables (I0 to Iaug)» 8 Production Variables
» 8 „Demands‟
» And 1 complicated Objective function
E l i th O ti l (L P )
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Exploring the Optimal (L.P.)
Approach
Lets try Excel!
This is a toughee!
Lots of variables and lots of constraints –
but work is straight forward!
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Disaggregation
Aggregate plans were built to optimal staffinglevels for “families” or groups of products
Disaggregation is a means to build specific“Master Production Schedules”
Typically by breaking down the aggregatingweights to individual parts – or working onschedules of these families as optimal
Later leads to values similar to EOQ which wewill explore in Chapter 4!