Suppose that a furniture manufacturer makes chairs, sofas, and tables. Consider
ID: 3148814 • Letter: S
Question
Suppose that a furniture manufacturer makes chairs, sofas, and tables. Consider the following chart of labor hours required and available.
Chair
Sofa
Table
Daily labor available (labor-hours)
Carpentry
6
3
8
752
Finishing
1
1
2
152
Upholstery
4
4
0
336
$90,
$70,
$140.
Suppose that a furniture manufacturer makes chairs, sofas, and tables. Consider the following chart of labor hours required and available.
Chair
Sofa
Table
Daily labor available (labor-hours)
Carpentry
6
3
8
752
Finishing
1
1
2
152
Upholstery
4
4
0
336
The profit per chair is$90,
per sofa$70,
and per table$140.
How many pieces of each type of furniture should be manufactured each day to maximize the profit?Explanation / Answer
So we can write the linear programming problem by given information
X1= Number of chairs X2 = number of sofas X3 = number of tables
Maximize z = 90x1 + 70x2 + 140x3
Subject to 6x1 + 3x2 +8x3 752
x1 + x2 +2x3 152
4x1 + 4x2 336
x1,x2 ,x3>= 0
Given linear problem is a stnadard linear problem so we can add slack variables to get
Equations from given inequalities.
6x1 + 3x2 +8x3 + S1 = 752
x1 + x2 +2x3 + S2 = 152
4x1 + 4x2 + s3 = 336
-90x1 - 70x2 - 140x3 + z =0
Now we can matrix representation of initial tableau for above equations.
X1 X2 X3 S1 S2 S3 Z
6 3 8 1 0 0 0 752
1 1 2 0 1 0 0 152
4 4 0 0 0 1 0 336
-90 -70 -140 0 0 0 1 0
Here the most negative element in the bottom row will indicates the pivot column so 3rd column is the pivot column and For pivot row the least positive result when last column divided by pivot column will indicates
i.e. +min(752/8,152/2,336/0) = 152/2 so 2nd row is a pivoted row
R2 -> R2 (1/2)
X1 X2 X3 S1 S2 S3 Z
6 3 8 1 0 0 0 752
1/2 1/2 1 0 1/2 0 0 76
4 4 0 0 0 1 0 336
-90 -70 -140 0 0 0 1 0
R1-> R1 - 8R2 R4 -> R4 + 140R2
X1 X2 X3 S1 S2 S3 Z
2 -1 0 1 -4 0 0 144
1/2 1/2 1 0 1/2 0 0 76
4 4 0 0 0 1 0 336
-20 0 0 0 70 0 1 10640
Here the most negative element in the bottom row will indicates the pivot column so 1st column
And for pivot row +min(144/2,76/(1/2)) = 144/2 so 1st row is a pivot row.
R1 -> R1(1/2)
X1 X2 X3 S1 S2 S3 Z
1 -1/2 0 1/2 -2 0 0 72
1/2 1/2 1 0 1/2 0 0 76
4 4 0 0 0 1 0 336
-20 0 0 0 70 0 1 10640
R2 -> R2 - (1/2)R1 R3 -> R3 - 4R1 R4 -> R4 + 20R1
X1 X2 X3 S1 S2 S3 Z
1 -1/2 0 1/2 -2 0 0 72
0 3/4 1 -1/4 3/2 0 0 40
0 6 0 -2 8 1 0 48
0 -10 0 10 30 0 1 12080
Here the most negative element in the bottom row will indicates the pivot column so 2nd column
And for pivot row +min(40/(3/4),48/6) = 48/6 so 3rd row is a pivot row
X1 X2 X3 S1 S2 S3 Z
1 -1/2 0 1/2 -2 0 0 72
0 3/4 1 -1/4 3/2 0 0 40
0 1 0 -1/3 4/3 1/6 0 8
0 -10 0 10 30 0 1 12080
R1 -> R1 + (1/2)R3 R2 -> R2 – (3/4)R3 R4 -> R4 + 10R3
X1 X2 X3 S1 S2 S3 Z
1 0 0 1/3 -4/3 1/12 0 76
0 0 1 0 1/2 -1/8 0 34
0 1 0 -1/3 4/3 1/6 0 8
0 0 0 20/3 130/3 5/3 1 12160
So now we did not have any negative elements in bottom row so we can stop the iterations.Now the optimum solution is Maximum Z = 12160 At X1 = 76 , X2 = 8 ,X3 = 34
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