Operations Research Models and Methods / Models / Linear Programming

Aggregate Planning Problem

### Problem

 A company wants a high level, aggregate production plan for the next 6 months. Projected orders for the company's product are listed in the table. Over the 6-month period, units may be produced in one month and stored in inventory to meet some later month's demand. Because of seasonal factors, the cost of production is not constant, as shown in the table. The cost of holding an item in inventory for 1 month is \$4/unit/mo. Items produced and sold in the same month are not put in inventory. The maximum number of units that can be held in inventory is 250. The initial inventory level at the beginning of the planning horizon is 200 units; the final inventory level at the end of the planning horizon is to be 100. The problem is to determine the optimal amount to produce in each month so that demand is met while minimizing the total cost of production and inventory. Shortages are not permitted. The aggregate planning problem is interesting not only because it represents an important application of linear programming, but because it also illustrates how multiperiod planning problems are approached. For this model we use subscripted variables and summations to simplify the model presentation. This is often useful for multiperiod problems and for other problems that have repetitive aspects.

Aggregate planning data

 Demand Production Month (units) cost (\$/unit) 1 1300 100 2 1400 105 3 1000 110 4 800 115 5 1700 110 6 1900 110

Model

 VARIABLE DEFINITIONS The dimensions of all variables are units of product. The index t ranges from 1 to 6. P(t) : production level in month t I(t): inventory level at the end of month t PARAMETERS The problem statement identifies the following parameters: D(t) : demand in month t I(0) : initial inventory level I(6) : final inventory level CONSTRAINTS Conservation of flow: A basic requirement in production planning problems is that product or material must be conserved. In our case, this leads to the following production constraint I(t-1) + P(t) — I(t) = D(t), t = 1,...,6 that states that the demand in month t must be met by the production in month t plus the net change in inventory. Maximum inventory: This is simply an upper bound constraint on the inventory levels. I(t) < 250, t = 1,...,6 Initial and final conditions: I(0) = 200, I(6) = 100 Although I(0) and I(6) have constant values because of these constraints, we leave them as variables in the model. Aggregate planning models, as well as many others, are meant to be solved over and over again as time advances and as parameters change. It is easier to treat the initial and final values as constraints rather than replace the two variables by their equivalent values.   NONNEGATIVITY I(t) > 0, P(t) > 0 for all t   OBJECTIVE FUNCTION Minimize Z = 100P1 + 105P2 + 110P3 + 115P4 + 110P5 + 110P6 + 4I1 + 4I2 + 4I3 + 4I4 + 4I5 + 4I6

### Solution

 To see the model constructed with the Math Programming add-in click on the image below. The model has been solved with the Jensen LP add-in. Note that each equality constraint is placed on the worksheet by using the same number for the upper and lower bound. For the example, the demands are the constraint limits. The optimum solution for the problem produces an extra 50 units in the first period and holds the resulting inventory for three periods until it used in period 4.

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Updated 2/2/00
Operations Research Models and Methods
by Paul A. Jensen and Jon Bard, University of Texas, Copyright by the Authors