1 Chapter 3 Aggregate Planning McGraw-Hill/Irwin Copyright © 2005 by The McGraw-Hill Companies, Inc. All rights reserved.

2 Aggregate Planning Strategies n Should inventories be used to absorb changes in demand during planning period? n Should changes be accomodated by varying the size of the workforce? n Should part-timers be used, or should overtime and idle time absorb fluctuations? n Should subcontractors be used on fluctuating orders so a stable workforce can be maintained? n Should prices or other factors be changed to influence demand?

3 Introduction to Aggregate Planning n Goal: To plan gross work force levels and set firm-wide production plans. n 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 of sales. »Can even be a fictitious quantity. (Refer to example in text and in slide below.)

4 Overview of the Problem Suppose that D 1, D 2,..., D T are the forecasts of demand for aggregate units over the planning horizon (T periods.) The problem is to determine both work force levels (W t ) and production levels (P t ) to minimize total costs over the T period planning horizon.

5 Important Issues n Smoothing. Refers to the costs and disruptions that result from making changes from one period to the next. n Bottleneck Planning. Problem of meeting peak demand because of capacity restrictions. n Planning Horizon. Assumed given (T), but what is “right” value? Rolling horizons and end of horizon effect are both important issues. n Treatment of Demand. Assume demand is known. Ignores uncertainty to focus on the predictable/systematic variations in demand, such as seasonality.

6 Relevant Costs n Smoothing Costs – changing size of the work force – changing number of units produced n Holding Costs – primary component: opportunity cost of investment n Shortage Costs – Cost of demand exceeding stock on hand. Why should shortages be an issue if demand is known? n Other Costs: payroll, overtime, subcontracting.

7 Aggregate Units The method is based on notion of aggregate units. They may be n Actual units of production n Weight (tons of steel) n Volume (gallons of gasoline) n Dollars (Value of sales) n Fictitious aggregate units

8 Example of fictitious aggregate units. (Example 3.1) One plant produced 6 models of washing machines: Model # hrs. Price % sales A K L L M M Question: How do we define an aggregate unit here?

9 Example continued n Notice: Price is not necessarily proportional to worker hours (i.e., cost): why? n One method for defining an aggregate unit: requires:.32(4.2) +.21(4.9) (5.8) = worker hours. Forecasts for demand for aggregate units can be obtained by taking a weighted average (using the same weights) of individual item forecasts.

10 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. Forecasted demands for Jan-Aug. are: 420, 280, 460, 190, 310, 145, 110, 125. Starting inventory at the end of December is 200 and the firm would like to have 100 units on hand at the end of August. Find monthly production levels.

11 Step 1: Determine “net” demand. (subtract starting inv. from per. 1 forecast and add ending inv. to per. 8 forecast.) MonthNet PredictedCum. Net Demand Demand Demand Demand 1(Jan) (Feb) (Mar) (Apr) (May) (June) (July) (Aug)

12 Step 2. Graph Cumulative Net Demand to Find Plans Graphically

13 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 planning horizon. How would we do that? One method: In previous picture, draw a straight line from origin to 1940 units in month 8: The slope of the line is the number of units to produce each month.

14 Monthly Production = 1940/8 = or rounded to 243/month. But: there are stockouts.

15 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:

16 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) (Jan) (Feb) (Mar) (Apr.) (May) (June) (July) (Aug)

17 But - may not be realistic for several reasons: n It may not be possible to achieve the production level of 320 unit/mo with an integer number of workers n Since all months do not have the same number of workdays, a constant production level may not translate to the same number of workers each month.

18 To overcome these shortcomings: n Assume number of workdays per month is given n K factor given (or computed) where K = # of aggregate units produced by one worker in one day

19 Finding K n Suppose that we are told that over a period of 40 days, the plant had 38 workers who produced 520 units. It follows that: n K= 520/(38*40) =.3421 = average number of units produced by one worker in one day. = average number of units produced by one worker in one day.

20 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 = = 61. Find 960/61 = units/day implies /.3421 = 46 workers required.

21 Why again did we pick on March? n Examining the graph we see that that was the “Trigger point” where our constant production line intersected the cumulative demand line assuring NO STOCKOUTS! n Can we “prove” this is best?

22 Tabulate Days/Production per Worker Vs. Demand to find minimum numbers Month# Work Days#Units/workerForecast Demand netMin # WorkersC. Net DemandC.Units/Worker Min # Workers Jan Feb Mar Apr May Jun Jul Aug

23 What should we look at? n Cumulative Demand says March needs most workers – but will mean building inventories in Jan + Feb n If we keep this number of workers we will continue to build inventory through the rest of the plan

24 Constant Work Force Production Plan Mo # wk days Prod. Cum Cum Nt End Inv Mo # wk days Prod. Cum Cum Nt End Inv Level Prod Dem Level Prod Dem Jan Jan Feb Feb Mar Mar Apr Apr May May Jun Jun Jul Jul Aug Aug

25 Addition of Costs n Holding Cost (per unit per month): $8.50 n Hiring Cost per worker: $800 n Firing Cost per worker: $1,250 n Payroll Cost: $75/worker/day n Shortage Cost: $50 unit short/month

26 Cost Evaluation of Constant Work Force Plan n Assume that the work force at end of Dec was 40. n Cost to hire 6 workers: 6*800 = $4800 n Inventory Cost: accumulate ending inventory: ( ) = Add in 100 units netted out in Aug = Hence Inv. Cost = 2193*8.5=$18, n Payroll cost: ($75/worker/day)(46 workers )(167days) = $576,150 n Cost of plan: $576,150 + $18, $4800 = $599,590.50

27 An Alternative is called the “Chase Plan” n Here, we hire and fire workers to keep inventory low n We would employ only the number of workers needed each month to meet demand n Examining our chart (earlier) we need: »Jan: 30; Feb: 51; Mar: 59; Apr: 27; May: 43 »Jun: 20; Jul: 15; Aug: 30

28 An Alternative is called the “Chase Plan” n 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) n Total Personnel Costs: $128950

29 An Alternative is called the “Chase Plan” n Inventory cost is essentially 165*8.5 = $ n Employment costs: $ n Chase Plan Total: $ n Betters the “Constant Workforce Plan” by: » – = n But will this be good for your image? n Can we find a better plan?

30 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.

31 Cost Evaluation of Modified Plan n I will not present all the details here. The modified plan calls for reducing the workforce to 36 at start of April and making another 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.

32 Optimal Solutions to Aggregate Planning Problems Via Linear Programming Linear Programming provides a means of solving aggregate planning problems optimally. The LP formulation is fairly complex requiring 8T variables and 3T constraints, where T is the length of the planning horizon. Clearly, this can be a formidable linear program. The LP formulation shows that the modified plan we considered with two months of layoffs is in fact optimal for the prototype problem.

33 Exploring the Optimal (L.P.) Approach n We need an Objective Function for cost of the aggregate plan (target minimization): –Here the c i ’s are cost for hiring, firing, inventory, production, etc –H T and F T are number of workers hired and fired –I T, P T, O T, S T AND U T are numbers inventoried, produced on regular time, overtime, by ‘sub-contract’ or idle worker hours respectively

34 Exploring the Optimal (L.P.) Approach n This objective Function would be subject to a series of constraints (one for each period) –Number of Worker Constraints: –Inventory Constraints: –Production Constraints:

35 Exploring the Optimal (L.P.) Approach n Assuming we allow no idle time and will produce only on regular time »No overtime or subcontracting n We would have: »9 worker variables (W 0 to W aug ) »8 Hire Variable »8 Fire Variables »9 Inventory Variables (I 0 to I aug ) »8 Production Variables »8 ‘Demands’ »And 1 complicated Objective function

36 Exploring the Optimal (L.P.) Approach n Lets try Excel! n This is a toughee! n Lots of variables and lots of constraints – but work is straight forward!

37 Disaggregation n Aggregate plans were built to optimal staffing levels for “families” or groups of products n Disaggregation is a means to build specific “Master Production Schedules” n Typically by breaking down the aggregating weights to individual parts – or working on schedules of these families as optimal n Later leads to values similar to EOQ which we will explore in Chapter 4!