Better Products, Inc., manufactures three products on two machines. In a typical
ID: 470429 • Letter: B
Question
Better Products, Inc., manufactures three products on two machines. In a typical week, 40 hours are available on each machine. The profit contribution and production time in hours per unit are as follows:
Category Product 1, Product 2, Product 3
Profit/unit $31, $60, $16
Machine 1 time/unit 0.5, 1.5, 1.5
Machine 2 time/unit 1, 1.5, 1
Two operators are required for machine 1; thus, 2 hours of labor must be scheduled for each hour of machine 1 time. Only one operator is required for machine 2. A maximum of 120 labor-hours is available for assignment to the machines during the coming week. Other production requirements are that product 1 cannot account for more than 55% of the units produced and that product 3 must account for at least 25% of the units produced.
(a) How many units of each product should be produced to maximize the total profit contribution? If required, round your answers to the nearest integer.Explanation / Answer
Let x1, x2 and x3 be the number of units of Product 1, Product 2 and Product 3 respectively.
Thus, the objective function is
Maximize p = 31x1 + 60x2 + 16x3
Subject to
0.5x1 + 1.5x2 + 1.5x3 <= 40 (Machine 1 constraint)
x1 + 1.5x2 + x3 <= 40 (Machine 2 constraint)
3x1 + 4.5x2 + 4x3 <= 120 (Labour hours constraint)
x1 <= 0.55 (x1 + x2 + x3) or 0.45x1 - 0.55x2 - 0.55x3 <= 0 (Marketing constraint for Product 1)
x3 >= 0.25 (x1 + x2 + x3) or 0.25x1 + 0.25x2 - 0.75x3 <=0
x1, x2, x3 >= 0
We proceed with the simplex algorithm for solving:
Tableau #1
x1 x2 x3 s1 s2 s3 s4 s5 p
0.5 1.5 1.5 1 0 0 0 0 0 40
1 1.5 1 0 1 0 0 0 0 40
3 4.5 4 0 0 1 0 0 0 120
0.45 -0.55 -0.55 0 0 0 1 0 0 0
0.25 0.25 -0.75 0 0 0 0 1 0 0
-31 -60 -16 0 0 0 0 0 1 0
Tableau #2
x1 x2 x3 s1 s2 s3 s4 s5 p
-1 0 6 1 0 0 0 -6 0 40
-0.5 0 5.5 0 1 0 0 -6 0 40
-1.5 0 18 0 0 1 0 -18 0 120
1 0 -2.2 0 0 0 1 2.2 0 0
1 1 -3 0 0 0 0 4 0 0
29 0 -200 0 0 0 0 240 1 0
Tableau #3
x1 x2 x3 s1 s2 s3 s4 s5 p
-0.17 0 1 0.17 0 0 0 -1 0 6.7
0.42 0 0 -0.92 1 0 0 -0.5 0 3.3
1.4 0 0 -2.9 0 1 0 -0.5 0 3.3
0.63 0 0 0.37 0 0 1 0 0 15
0.5 1 0 0.5 0 0 0 1 0 20
-3.7 0 0 33 0 0 0 44 1 1300
Tableau #4
x1 x2 x3 s1 s2 s3 s4 s5 p
0 0 1 -0.18 0 0.12 0 -1.1 0 7.1
0 0 0 -0.059 1 -0.29 0 -0.35 0 2.4
1 0 0 -2.1 0 0.71 0 -0.35 0 2.4
0 0 0 1.7 0 -0.45 1 0.22 0 13
0 1 0 1.5 0 -0.35 0 1.2 0 19
0 0 0 25 0 2.6 0 43 1 1300
Optimal Solution: p = 1300; x1 = 2.4, x2 = 19, x3 = 7.1
Rounding off to the nearest integers:
Units of Product 1 = x1 = 2
Units of Product 2 = x2 = 19
Units of Product 3 = x3 = 7
Projected weekly profit = 31* 2 + 60*19 + 16*7 = $1314
Time on Machine 1 = 0.5x1 + 1.5x2 + 1.5x3 = 40 hours
Time on Machine 2 = x1 + 1.5x2 + x3 = 37.5 hours
Pj - Cj = -2.6 for s3 (Labour constraint).
Thus, value of each hour of labour = $ 2.6
On adding labour, the new labour constraint becomes:
3x1 + 4.5x2 + 4x3 <= 130
We again solve by the simplex method.
Tableau #1
x1 x2 x3 s1 s2 s3 s4 s5 p
0.5 1.5 1.5 1 0 0 0 0 0 40
1 1.5 1 0 1 0 0 0 0 40
3 4.5 4 0 0 1 0 0 0 130
0.45 -0.55 -0.55 0 0 0 1 0 0 0
0.25 0.25 -0.75 0 0 0 0 1 0 0
-31 -60 -16 0 0 0 0 0 1 0
Tableau #2
x1 x2 x3 s1 s2 s3 s4 s5 p
-1 0 6 1 0 0 0 -6 0 40
-0.5 0 5.5 0 1 0 0 -6 0 40
-1.5 0 18 0 0 1 0 -18 0 130
1 0 -2.2 0 0 0 1 2.2 0 0
1 1 -3 0 0 0 0 4 0 0
29 0 -200 0 0 0 0 240 1 0
Tableau #3
x1 x2 x3 s1 s2 s3 s4 s5 p
-0.17 0 1 0.17 0 0 0 -1 0 6.7
0.42 0 0 -0.92 1 0 0 -0.5 0 3.3
1.4 0 0 -2.9 0 1 0 -0.5 0 13
0.63 0 0 0.37 0 0 1 0 0 15
0.5 1 0 0.5 0 0 0 1 0 20
-3.7 0 0 33 0 0 0 44 1 1300
Tableau #4
x1 x2 x3 s1 s2 s3 s4 s5 p
0 0 1 -0.2 0.4 0 0 -1.2 0 8
1 0 0 -2.2 2.4 0 0 -1.2 0 8
0 0 0 0.2 -3.4 1 0 1.2 0 2
0 0 0 1.8 -1.5 0 1 0.76 0 9.6
0 1 0 1.6 -1.2 0 0 1.6 0 16
0 0 0 25 8.8 0 0 40 1 1300
Optimal Solution: p = 1300; x1 = 8, x2 = 16, x3 = 8
Thus,
Units of Product 1 = 8
Units of Product 2 = 16
Units of Product 3 = 8
Profit = 1300
As we do not see any increase in the profit, we will not use the additional labour.
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