Question

Problem 2: T & L Manufacturing produces 3 unique products: X, Y, and Z. The office manager has recently been approached by upper management and asked to create an optimal production schedule for the coming month. The office manager knows the following about products X, Y, and Z. There are 2 steps to the production of each product: Plating and Heating. The plating process coats the product in a substance that will eventually harden and protect the outside of the product. Once the product has been plated it is heated until the outer shell has reached a temperature high enough to harden the material used in plating. Due to production stage, the manufacturer needs to produce at least 1500 units for each product. If made in house they cost $2.20, $1.50, and $1.20. Products X, Y, and Z need 12 minutes, 20 minutes, and 6 minutes each of plating. Under the current staffing situation there is currently 1500 hours of plating available for the coming month. Products X, Y, and Z need 6, 12, and 3 minutes of heating. There are currently 1000 hours of heating available. Due to limited resources all three of the products may either be built on site or bought from another manufacturer. It costs $3.00, $2.80, and $2.20 to buy these products from another manufacturer. The demand for products X, Y, and Z in the coming month is 3000, 5000, and 4000. Develop a plan to minimize the total costs for the coming month.

Answer 1

Let’s say the number of products made in house is denoted by X1, Y1, Z1 and the number of products purchased from outside is X2, Y2, Z2

Then we can write our total cost function as

Minimize Z where

Z = 2.2X1 + 1.1Y1 + 1.2Z1 + 3X2 + 2.8Y2 +2.2Z2

Subject to constraints

12X1 + 20Y1 + 6Z1 <= 1500*60

6X1 + 12Y1 + 3Z1 <= 1000*60

X1 >= 1500

Y1 >= 1500

Z1 >= 1500

X1 + X2 >= 3000

Y1 + Y2 >= 5000

Z1 + Z2 >= 4000

X1, Y1, Z1, X2, Y2, Z2 >= 0

The solution is shown below from Lindo. The optimal solution is to

Produce (in house)

X = 1500, Y = 2400, Z = 4000

Purchase (outside)

X = 1500, Y = 2600, Z = 0

This will cause a cost of 22520 and this is the optimal cost.

LINDO File Edit Solve Reports Window Help Reports Window | <untitled> Hninize 2.2%1 + 1.1Y1 1.221 + 3X2 + 2 8Y2 +2 222 ST LP OPTINUN FOUND AT STEP 12X1 + 20Y1 + 621 90000 6x1 121 321-60000 11500 Y1 >= 15001 Z1 - 1500 1 +X23000 Y1Y2-5000 +22 >= 4000 OBJECTIVE FUNCTION VALUE 22520.00 ARIABLE X1 ALUE 1500.000000 2400.000000 4000, 000000 1500.000000 2600.000000 0.000000 REDUCED COST 0.000000 0.000000 Z1 12 Y2 0.000000 0.000000 0.000000 0.490000 Z2 ROW SLACK OR SURPLUS 0.000000 10200,000000 0.000000 DUAL PRICES 0.085000 0,000000 -0.220000 900.000000 2500.000000 0,000000 0.000000 0.000000 0.000000 0.000000 -3,000000 -2.800000 -1.710000 NO, ITERATIONS- RANGES IN WHICH THE BASIS IS UNCHANGED OBJ COEFFICIENT RaNGES VARIABLE CURRENT COEF 2.200000 1.100000 1.200000 3.000000 2,800000 2.200000 ALLOTABLE DECREASE INCREASE INEINIT 0.366667 0.490000 0.220000 1,633333 INEINITS 0.220000 1.633333 1.710000 3.000000 0.36666 0.490000 21 X2 12 Z2 RIGHTHAND SIDE RANGES ALLOUABLE ROW CURRENT RHS ALLOUABLE DECREASE 18000.000000 10200.000000 INCREASE 17000.000000 INFINITS 2 90000.000000 360000.000000