IE 3265 Production & Operations Planning Ch. 3 – Aggregate Planning R. Lindeke UMD

Aggregate Planning Strategies Should inventories be used to absorb changes in demand during planning period? Should demand changes be accommodated by varying the size of the workforce? Should part-timers be used, or should overtime and/or machine idle time be used to 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?

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 as we will see later

Overview of the Aggregation Problem Suppose that D1, D2, . . . , DT are the forecasts of demand for aggregate units over the planning horizon (T periods.) The problem is to determine both work force levels (Wt) and production levels (Pt ) to minimize total costs over the T period planning horizon.

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 in the face 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 or systematic variations in demand, such as seasonality

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

Aggregate Units The method is (fundamentally) 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 aggregated units they are a composite that estimates a tangible ‘input constant’

Developing 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?

Example (continued) Notice: Price is not necessarily proportional to worker hours (i.e., cost): why? For aggregating, we can use individual product demand forecasts modified with a weighted average (sales weights) of individual item forecasts to develop a aggregate production forecast

This Aggregate Units needs a measurable Labor Input: Thus, Agg. Demand = .32*(DA5532) + .21(DK4242) + … + .06(DM3880) This method for defining an aggregate unit points to an aggregate labor requirement (/Agg. Unit) of: .32(4.2) + .21(4.9) + . . . + .06(5.8) = 4.8644 worker hours

Prototype Aggregate Planning Example The washing machine plant is interested in determining work force and production levels for the next 8 months. Forecasted aggregate 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.

Step 1: Determine “net” demand. (subtract starting inv. from period Step 1: Determine “net” demand. (subtract starting inv. from period. 1 forecast and add ending inv. to per. 8 forecast.) Month Net Predicted Cum. Net Demand Demand 1(Jan) 220 220 2(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

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

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.

Monthly Production = 1940/8 = 242. 2 or rounded to 243/month Monthly Production = 1940/8 = 242.2 or rounded to 243/month. But notice: there are stockouts!

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!

From The Previous Graph, We See That Cum From The Previous Graph, We See That Cum. Net Demand Curve Is Crossed At Period 3: Use monthly production of 960/3 = 320. Ending inventory each month is found from: C. Prod – C.N. Dem. Month Cum. Net. Dem. Cum. Prod. Invent. 1(Jan) 220 320 100 2(Feb) 500 640 140 3(Mar) 960 960 0 4(Apr.) 1150 1280 130 5(May) 1460 1600 140 6(June) 1605 1920 315 7(July) 1715 2240 525 8(Aug) 1940 2560 620

But - This Solution 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 production level may not translate to the same number of workers each month!

To Overcome These Shortcomings: Assume number of workdays per month is given (reasonable!) Compute a “K factor” given by: K = number of aggregate units produced by one worker in one day

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 then that: K= 520/(38*40) = .3421 average number of units produced by one worker in one day.

Computing Constant Work Force -- Realistically 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. We find that: 960/61 = 15.7377 units/day 15.7377/.3421 = 46 workers required Actually 46.003 – here we truncate because we are set to build inventory so the low number should work (check for March stock out) – however we must use care and typically ‘round up’ any fractional worker calculations thus building more inventory

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 cumulative demand line assuring NO STOCKOUTS! Can we “prove” this is best?

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 Jan 22.00 7.53 220.00 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 145.00 19.27 1605.00 42.42 37.84 Jul 110.00 15.31 1715.00 49.60 34.57 Aug 225.00 29.90 1940.00 57.13 33.96

What Should We Look At? Cumulative Demand says March needs most workers – but will mean building inventories in Jan + Feb to fulfill the greater March demand If we keep this number of workers we will continue to build inventory through the rest of the plan!

Constant Work Force Production Plan: Mon # wk days Pr. Cum Cum N. End Level Prod Dem Inv 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 145 Jun 22 346 1951 1605 346 Jul 21 330 2281 1715 566 Aug 22 346 2627 1940 687

Lets Add some 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

Cost Evaluation for 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

An Alternative is called the “Chase Plan” Here, we hire and fire (layoff) workers to keep inventory low! We would employ only the number of workers 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 Found by: (monthly demand)  (monthly pr. /worker)

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

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?

Before We seek an Optimal … Lets try this approach In your engineering teams, do: Problem 31, page 147

Cost Reduction in Constant Work Force Plan & Chase Plan In the original C. N. demand curve, consider making reductions in the work force one (or more times) over the planning horizon to decrease inventory investment.

Cost Evaluation of Modified Plan The modified plan calls for reducing the workforce to 36 at the start of April and making another reduction to 22 at the start of June. The additional cost of layoffs is $30,000 But, here the holding costs are reduced to only $4,250 Here then the total cost of the modified plan is (only) $467,450.

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 8*T variables and (at least) 3*T 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 using layoffs at two points is in fact optimal for the prototype problem.

Exploring the Optimal (L.P.) Approach We need an Objective Function for cost of the aggregate plan (target is to minimize costs): 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 units inventoried, produced on regular time, on overtime, by ‘sub-contract’ or the number of units that could be produced on idled worker hours respectively

Exploring the Optimal (L.P.) Approach This objective Function would be subject to a series of constraints (one of each type for each period) ‘Number of Workers’ Constraints: Inventory Constraints: Production Constraints: Where: nt * k is the number of units produced by a worker in a given period of nt days

Exploring the Optimal (L.P.) Approach Assuming we allow no idle time and will produce only on regular time No overtime or subcontracting values 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

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

Real Constraint Equation (rewritten for L.P.): Employee Constraints: Inventory Constraints:

Real Constraint Equations (rewritten for L.P.): Production Constraints:

Real Constraint Equations (rewritten for L.P.): Finally, we need constraints defining: Initial Workforce size Starting Inventory Final Desired Inventory And, of course, the general constraint forcing all variables to be  0

We will build the L.P. and use the solver in Excel! Stepping in Excel We will build the L.P. and use the solver in Excel!

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