2Aggregate 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?
3Introduction to Aggregate Planning Goal: To plan gross work force levels and set firm-wide production plansConcept is predicated on the idea of an “aggregate unit” of production as we will see later
4Overview 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.
5Important IssuesSmoothing. Refers to the costs and disruptions that result from making changes from one period to the nextBottleneck Planning. Problem of meeting peak demand in the face of capacity restrictionsPlanning Horizon. Assumed given (T), but what is “right” value? Rolling horizons and end of horizon effect are both important issuesTreatment of Demand. Assume demand is known. Ignores uncertainty to focus on the predictable or systematic variations in demand, such as seasonality
6Relevant Costs Smoothing Costs Holding Costs Shortage Costs changing size of the work forcechanging number of units producedHolding Costsprimary component: opportunity cost of investment $’s tied up in inventoryShortage CostsCost of demand exceeding stock on hand. Why should shortages be an issue if demand is known?Other Costs: payroll, overtime, subcontracting
7Aggregate UnitsThe method is (fundamentally) based on notion of aggregate units. They may be:Actual units of productionWeight (tons of steel)Volume (gallons of gasoline)Dollars (Value of sales)Fictitious aggregated unitsthey are a composite that estimates a tangible ‘input constant’
8Developing aggregate units (Example 3.1) One plant produced 6 models of washing machines:Model # hrs Price % salesAKLLMMQuestion: How do we define an aggregate unit here?
9Example (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
10This 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) (5.8) = worker hours
11Prototype 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.
12Step 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. NetDemand Demand1(Jan)2(Feb)3(Mar)4(Apr)5(May)6(June)7(July)8(Aug)
13Step 2. Graph Cumulative Net Demand to Find Plans Graphically Cum. Net Demand
14Constant 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.
15Monthly Production = 1940/8 = 242. 2 or rounded to 243/month Monthly Production = 1940/8 = or rounded to 243/month. But notice: there are stockouts!
16How 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!
17From 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)2(Feb)3(Mar)4(Apr.)5(May)6(June)7(July)8(Aug)
18But - 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 workersSince 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!
19To 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
20Finding KSuppose 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) = average number of units produced by one worker in one day.
21Computing 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 = = 61.We find that:960/61 = units/day/.3421 = 46 workers requiredActually – 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
22Why 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?
23Tabulate Days/Production Per Worker Vs. Demand To Find Minimum Numbers Month# Work Days#Units/workerForecast Demand netMin # WorkersC. Net DemandC.Units/WorkerJan22.007.53220.0029.23Feb16.005.47280.0051.15500.0013.0038.46Mar23.007.87460.0058.46960.0020.8746.00Apr20.006.84190.0027.7727.7141.50May21.007.18310.0043.1534.8941.84Jun145.0019.2742.4237.84Jul110.0015.3149.6034.57Aug225.0029.9057.1333.96
24What Should We Look At?Cumulative Demand says March needs most workers – but will mean building inventories in Jan + Feb to fulfill the greater March demandIf we keep this number of workers we will continue to build inventory through the rest of the plan!
25Constant Work Force Production Plan: Mon # wk days Pr. Cum Cum N. EndLevel Prod Dem InvJanFebMarAprMayJunJulAug
26Lets Add some Costs Holding Cost (per unit per month): $8.50 Hiring Cost per worker: $800Firing Cost per worker: $1,250Payroll Cost: $75/worker/dayShortage Cost: $50 unit short/month
27Cost Evaluation for Constant Work Force Plan Assume that the work force at end of Dec was 40Cost to hire 6 workers: 6*800 = $4800Inventory Cost: accumulate ending inventory: ( ) = Add in 100 units netted out in Aug = Hence Inv. Cost = 2193*8.5=$18,640.50Payroll cost:($75/worker/day)(46 workers )(167days) = $576,150Cost of plan: $576,150 + $18, $4800 = $599,590.50
28An 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 demandExamining our chart (earlier) we need:Jan: 30; Feb: 51; Mar: 59; Apr: 27; May: 43 Jun: 20; Jul: 15; Aug: 30Found by: (monthly demand) (monthly pr. /worker)
29An Alternative is called the “Chase Plan” So we hire or Fire (lay-off) monthlyJan (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
30An Alternative is called the “Chase Plan” Inventory cost is essentially 165*8.5 = $Employment costs: $428325Chase Plan Total: $Betters the “Constant Workforce Plan” by:– = 40913But will this be good for your image?Can we find a better plan?
31Before We seek an Optimal … Lets try this approachIn your engineering teams, do:Problem 31, page 147
32Cost 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.
33Cost 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 $30,000But, here the holding costs are reduced to only $4,250Here then the total cost of the modified plan is (only) $467,450.
34Optimal 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.
35Exploring 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, etcHT and FT are number of workers hired and firedIT, 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
36Exploring 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
37Exploring the Optimal (L.P.) Approach Assuming we allow no idle time and will produce only on regular timeNo overtime or subcontracting valuesWe would have:9 worker variables (W0 to Waug)8 Hire Variable8 Fire Variables9 Inventory Variables (I0 to Iaug)8 Production Variables8 ‘Demands’And 1 complicated Objective function
38Exploring the Optimal (L.P.) Approach This is a toughee!Lets try Excel!Lots of variables and lots of constraints – but work is straight forward!
39Real Constraint Equation (rewritten for L.P.): Employee Constraints:Inventory Constraints:
40Real Constraint Equations (rewritten for L.P.): Production Constraints:
41Real Constraint Equations (rewritten for L.P.): Finally, we need constraints defining:Initial Workforce sizeStarting InventoryFinal Desired InventoryAnd, of course, the general constraint forcing all variables to be 0
42We will build the L.P. and use the solver in Excel! Stepping in ExcelWe will build the L.P. and use the solver in Excel!
43DisaggregationAggregate plans were built to optimal staffing levels for “families” or groups of productsDisaggregation 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 optimalLater leads to values similar to EOQ which we will explore in Chapter 4!