Simplex Method 1
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Transcript of Simplex Method 1
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LINEAR PROGRAMMING Example 1
0
160
32 0
48 0
64 0
80 0
96 0
0 16 0 32 0 48 0 640 800 96 0
x
y Maximise I = x+ 0.8ysubject to x+ y 1000
2x+ y 1500
3x+ 2y 2400
Initial solution:
I= 0
at (0, 0)
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LINEAR PROGRAMMING Example 1
Maximise I = x+ 0.8ysubject to x+ y 1000
2x+ y 1500
3x+ 2y
2400
Maximise I
where I- x- 0.8y = 0
subject to x+ y+ s1 = 1000
2x+ y + s2 = 1500
3x+ 2y + s3 = 2400
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I x y s1 s2 s3 RHS
1 -1 -0.8 0 0 0 0
0 1 1 1 0 0 1000
0 2 1 0 1 0 1500
0 3 2 0 0 1 2400
SIMPLEX TABLEAU
I = 0, x= 0, y= 0, s1 = 1000, s2 = 1500, s3 = 2400
Initial solution
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I x y s1 s2 s3 RHS
1 -1 -0.8 0 0 0 0
0 1 1 1 0 0 1000
0 2 1 0 1 0 1500
0 3 2 0 0 1 2400
PIVOT 1 Choosing the pivot column
Most negative number in objective row
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I x y s1 s2 s3 RHS
1 -1 -0.8 0 0 0 0
0 1 1 1 0 0 1000 1000/1
0 2 1 0 1 0 1500 1500/2
0 3 2 0 0 1 2400 2400/3
PIVOT 1 Choosing the pivot element
Ratio test: Min. of 3 ratios gives 2 as pivot element
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I x y s1 s2 s3 RHS
1 -1 -0.8 0 0 0 0
0 1 1 1 0 0 1000
0 1 0.5 0 0.5 0 750
0 3 2 0 0 1 2400
PIVOT 1 Making the pivot
Divide through the pivot row by the pivot element
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 1 1 1 0 0 1000
0 1 0.5 0 0.5 0 750
0 3 2 0 0 1 2400
PIVOT 1 Making the pivot
Objective row + pivot row
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 3 2 0 0 1 2400
PIVOT 1 Making the pivot
First constraint row - pivot row
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 0 0.5 0 -1.5 1 150
PIVOT 1 Making the pivot
Third constraint row
3 x pivot row
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 0 0.5 0 -1.5 1 150
PIVOT 1 New solution
I = 750, x= 750, y= 0, s1
= 250, s2
= 0, s3
= 150
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0
160
32 0
48 0
64 0
80 0
96 0
0 160 320 480 640 800 960
x
y
LINEAR PROGRAMMING Example
Maximise I = x+ 0.8ysubject to x+ y 1000
2x+ y 1500
3x+ 2y 2400
Solution afterpivot 1:
I= 750
at (750, 0)
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 0 0.5 0 -1.5 1 150
PIVOT 2
Most negative number in objective row
Choosing the pivot column
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250 250/0.5
0 1 0.5 0 0.5 0 750 750/0.5
0 0 0.5 0 -1.5 1 150 150/0.5
PIVOT 2 Choosing the pivot element
Ratio test: Min. of 3 ratios gives 0.5 as pivot element
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I x y s1 s2 s3 RHS
1 0 -0.3 0 0.5 0 750
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 0 1 0 -3 2 300
PIVOT 2 Making the pivot
Divide through the pivot row by the pivot element
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0.5 1 -0.5 0 250
0 1 0.5 0 0.5 0 750
0 0 1 0 -3 2 300
PIVOT 2 Making the pivot
Objective row + 0.3 x pivot row
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100
0 1 0.5 0 0.5 0 750
0 0 1 0 -3 2 300
PIVOT 2 Making the pivot
First constraint row 0.5 x pivot row
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100
0 1 0 0 2 -1 600
0 0 1 0 -3 2 300
PIVOT 2 Making the pivot
Second constraint row 0.5 x pivot row
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100
0 1 0 0 2 -1 600
0 0 1 0 -3 2 300
PIVOT 2 New solution
I = 840, x= 600, y= 300, s1
= 100, s2
= 0, s3
= 0
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0
160
32 0
48 0
64 0
80 0
96 0
0 160 32 0 4 8 0 64 0 8 0 0 96 0
x
y
LINEAR PROGRAMMING Example
Maximise I = x+ 0.8ysubject to x+ y 1000
2x+ y 1500
3x+ 2y 2400
Solution afterpivot 2:
I= 840
at (600, 300)
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100
0 1 0 0 2 -1 600
0 0 1 0 -3 2 300
PIVOT 3 Choosing the pivot column
Most negative number in objective row
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100 100/1
0 1 0 0 2 -1 600 600/2
0 0 1 0 -3 2 300
PIVOT 3 Choosing the pivot element
Ratio test: Min. of 2 ratios gives 1 as pivot element
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I x y s1 s2 s3 RHS
1 0 0 0 -0.4 0.6 840
0 0 0 1 1 -1 100
0 1 0 0 2 -1 600
0 0 1 0 -3 2 300
PIVOT 3 Making the pivot
Divide through the pivot row by the pivot element
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I x y s1 s2 s3 RHS
1 0 0 0.4 0 0.2 880
0 0 0 1 1 -1 100
0 1 0 0 2 -1 600
0 0 1 0 -3 2 300
PIVOT 3 Making the pivot
Objective row + 0.4 x pivot row
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I x y s1 s2 s3 RHS
1 0 0 0.4 0 0.2 880
0 0 0 1 1 -1 100
0 1 0 -2 0 1 400
0 0 1 0 -3 2 300
PIVOT 3 Making the pivot
Second constraint row 2 x pivot row
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I x y s1 s2 s3 RHS
1 0 0 0.4 0 0.2 880
0 0 0 1 1 -1 100
0 1 0 -2 0 1 400
0 0 1 3 0 -1 600
PIVOT 3 Making the pivot
Third constraint row + 3 x pivot row
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0
160
32 0
48 0
64 0
80 0
96 0
0 160 3 20 480 640 800 960
x
y
LINEAR PROGRAMMING Example
Maximise I = x+ 0.8y
subject to x+ y 1000
2x+ y 1500
3x+ 2y 2400
Optimal solutionafter pivot 3:
I= 880
at (400, 600)
