Tools and Techniques of Linear Programming

Assignment Problem:

An automobile tyre company has the ability to produce both nylon and fiberglass tyres.

During the next three months they have agreed to deliver tyres as follows:

The company has two presses, a Wheeling machine and aRegal machine, and appropriate moulds that can be used to produce these tyres, with the following production hours available in the upcoming months:

The production rates for each machine-and-tyre combination, in terms of hours per tyre, are as follows:

The variable costs of producing tyres are $5.00 per operating hour, regardless of which machine is being used or which tyre is being produced. There is also an inventory-carrying charge of $0.10 per tyre per month. Material costs for the nylon and fiberglass tyres are $3.10 and $3.90 per tyre, respectively. Finishing, packaging and shipping costs are $0.23 per tyre. Prices have been set at $7.00 per nylon tyre and $9.00 per fiberglass tyre.

Formulate as a Linear Programme which could be used to provide a production schedule in order to meet the delivery requirements at minimum costs?

Note: Book to refer Managerial Decision Modeling with Spreadsheets 3rd Edition by Balakrishnan | Render | Stair

Model Formulation

Selection of a Time Horizon

In this particular situation the time horizon covers three months, divided into three time periods of a months duration each. More realistic production-planning models normally have a full-year time horizon.

Selection of Decision Variables and Parameters

We can determine what decision variables are necessary by defining exactly what information the plant foreman must have in order to schedule the production. Essentially, he must know the number of each type of tire to be produced on each machine in each month and the number of each type of tire to place in inventory at the end of each month. Hence, we have the following decision variables:

Wn,t = Number of nylon tires to be produced on the Wheeling machine during month t;

Rn,t = Number of nylon tires to be produced on the Regal machine during month t;

Wg,t = Number of fiberglass tires to be produced on the Wheeling machine in month t;

Rg,t = Number of fiberglass tires to be produced on the Regal machine in month t;

In,t = Number of nylon tires put into inventory at the end of month t;

Ig,t = Number of fiberglass tires put into inventory at the end of month t.

In general there are six variables per time period and since there are three months under consideration, we have a total of eighteen variables. However, it should be clear that it would never be optimal to put tires into inventory at the end of August since all tires must be delivered by then. Hence, we can ignore the inventory variables for August.

The parameters of the problem are represented by the demand requirements, the machine availabilities, the machine productivity rates, and the cost and revenue information. All these parameters are assumed to be known deterministically.

Definition of the Constraints

There are two types of constraint in this problem representing production-capacity available and demand requirements at each month.

Let us develop the constraints for the month of June. The production-capacity constraints can be written in terms of production hours on each machine. For the Wheeling machine in June we have:

0.15Wn,1 + 0.12Wg,1 700,

while, for the Regal machine in June, we have:

0.16Rn,1 + 0.14Rg,1 1500.

The production constraints for future months differ only in the available number of hours of capacity for the righthand side.

Now consider the demand constraints for June. For each type of tire produced in June we must meet the demand requirement and then put any excess production into inventory. The demand constraint for nylon tires in June is then:

Wn,1 + Rn,1 In,1 = 4000,

while for fiberglass tires in June it is:

Wg,1 + Rg,1 Ig,1 = 1000.

In July, however, the tires put into inventory in June are available to meet demand. Hence, the demand constraint for nylon tires in July is:

In,1 + Wn,2 + Rn,2 In,2 = 8000,

while for fiberglass tires in July it is:

Ig,1 + Wg,2 + Rg,2 Ig,2 = 5000.

In August it is clear that tires will not be put into inventory at the end of the month, so the demand constraint for nylon tires in August is:

In,2 + Wn,3 + Rn,3 = 3000,

while for fiberglass tires in August it is:

Ig,2 + Wg,3 + Rg,3 = 5000.

Finally, we have the nonnegativity constraints on all of the decision variables:

Wn,t 0, Wg,t 0, (t = 1, 2, 3);

Rn,t 0, Rg,t 0, (t = 1, 2, 3);

In,t 0, Ig,t 0, (t = 1, 2).

Selection of the Objective Function

The total revenues to be obtained in this problem are fixed, because we are meeting all the demand requirements, and maximization of profit becomes equivalent to minimization of cost. Also, the material-cost component is fixed, since we know the total amount of each product to be produced during the model time horizon. Thus, a proper objective function to select is the minimization of the variable relevant cost components: variable production costs plus inventory-carrying costs.

Now, since each kind of tire on each machine has a different production rate, the cost of producing a tire on a particular machine will vary, even though the variable cost per hour is constant for each tire-and-machine combination. The variable production cost per tire for the fiberglass tires made on the Regal machine can be determined by multiplying the production rate (in hours/tire) by the variable production cost (in $/hour) resulting in (0.14)*(5) = $0.70/tire. The remaining costs for producing each tire on each machine can be computed similarly, yielding

Wheeling machine Regal machine

Nylon 0.75 0.80

Fiberglass 0.60 0.70

Table 1. Multistage Planning Model

June July August

Nylon Glass Inventory Nylon Glass Inventory Nylon Glass Wn,1 Rn,1 Wg,1 Rg,1 In,1 Ig,1 Wn,2 Rn,2 Wg,2 Rg,2 In,2 Ig,2 Wn,3 Rn,3 Wg,3 Rg,3 RHS HoursavailableMachinetimeconstraintsJuneWheeling0.15 0.12 700 Regal 0.16 0.14 1500 JulyWheeling 0.15 0.12 300 Regal 0.16 0.14 400 AugWheeling 0.15 0.12 1000 Regal 0.16 0.14 300 Demand constrantsJune Nylon 1 1 -1 = 4000 NumberoftiresdemandGlass 1 1 -1 = 1000 JulyNylon 1 1 1 -1 = 8000 Glass 1 1 1 -1 = 5000 AugNylon 1 1 1 = 3000 Glass 1 1 1 = 5000 Objective0.75 0.8 0.6 0.7 0.1 0.1 0.75 0.8 0.6 0.7 0.1 0.1 0.75 0.8 0.6 0.7 Minimum

Given the inventory-carrying cost of $0.10 per tire per month, we have the following objective functionfor minimizing costs:

min: F=t=13(0.75Wn,t+0.80Rn,t+0.60Wg,t+0.70Rg,t+0.10In,t+0.10Ig,t);and we understand that In, 3 = 0 and Ig, 3 = 0.

The formulation of this problem is summarized in Table 1. This problem is what we call a multistagemodel, because it contains more than one time period. Note that the constraints of one time period are linkedto the constraints of another only by the inventory variables. Note that there are very few elements different from 0, 1, and 1 in thetableau given in Table 1. This problem structure can be exploited easily in the design of a matrix generator,to provide the input for the linear-programming computation.

Computer Results

We will now present the computer output obtained by solving the linear-programming model by SOLVER Ms Excel.

Production Scheduling

Examination of the optimal values of the decision variables in thelinear-programming solution reveals thatthe appropriate production schedule should be:

Wheeling machine Regal machine

June No. of nylon tires 1867 7633

No. of fiberglass tires 3500 0

July No. of nylon tires 0 2500

No. of fiberglass tires 2500 0

August No. of nylon tires 2667 333

No. of fiberglass tires 5000 0

The resulting inventory at the end of each month for the two types of products is as follows:

June July August

Inventory of nylon tires 5500 0 0

Inventory of fiberglass tires 2500 0 0

Figure 1.Computer printout of solution of the problem.