Worked Examples
Solve with the help of simplex method
Example 1
Maximize Z = 80x1 + 55x2
Subject to
4x1 + 2x2 ≤ 40
2x1 + 4x2 ≤ 32
& x1 ≥ 0, x2 ≥ 0
Answer
SLPP
Maximize Z = 80x1 + 55x2 + 0s1 + 0s2
4x1 + 2x2+ s1= 40
2x1 + 4x2 + s2= 32
x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0
Cj → 80 55 0 0
Basic Variables
CB XB
X1 X2 S1 S2
Min ratio
XB /Xk
s1
s2
0 40
0 32
4 2 1 0
2 4 0 1
40 / 4 = 10→ outgoing
32 / 2 = 16
Z= CB XB = 0
↑incoming
Δ1= -80 Δ2= -55 Δ3=0 Δ4=0
x1
80 10
0 12
(R1=R1 / 4)
1 1/2 1/4 0
(R2=R2- 2R1)
0 3 -1/2 1
10/1/2 = 20
12/3 = 4→ outgoing
Z = 800
Δ1=0 Δ2= -15 Δ3=40 Δ4=0
x2
80 8
55 4
(R1=R1- 1/2R2)
1 0 1/3 -1/6
(R2=R2 / 3)
0 1 -1/6 1/3
Z = 860
Δ1=0 Δ2=0 Δ3=35/2 Δ4=5
As all Δj ≥ 0, optimal basic feasible solution is achieved. Hence the solution is Max Z = 860, x1 = 8 and x2 = 4
Example 2
Maximize Z = 5x1 + 3x2
3x1 + 5x2 ≤ 15
5x1 + 2x2 ≤ 10
Maximize Z = 5x1 + 3x2 + 0s1 + 0s2
3x1 + 5x2+ s1= 15
5x1 + 2x2 + s2= 10
Cj → 5 3 0 0
0 15
0 10
3 5 1 0
5 2 0 1
15 / 3 = 5
10 / 5 = 2 → outgoing
Δ1= -5 Δ2= -3 Δ3=0 Δ4=0
0 9
5 2
(R1=R1- 3R2)
0 19/5 1 -3/5
(R2=R2 /5)
1 2/5 0 1/5
9/19/5 = 45/19 →
2/2/5 = 5
Z = 10
↑
Δ1=0 Δ2= -1 Δ3=0 Δ4=1
3 45/19
5 20/19
(R1=R1 / 19/5)
0 1 5/19 -3/19
(R2=R2 -2/5 R1)
1 0 -2/19 5/19
Z = 235/19
Δ1=0 Δ2=0 Δ3=5/19 Δ4=16/19
As all Δj ≥ 0, optimal basic feasible solution is attained. So the solution is Max Z = 235/19, x1 = 20/19 and x2 = 45/19
Example Sample Assignments 3
Maximize Z = 5x1 + 7x2
x1 + x2 ≤ 4
3x1 - 8x2 ≤ 24
10x1 + 7x2 ≤ 35
Maximize Z = 5x1 + 7x2 + 0s1 + 0s2 + 0s3
x1 + x2 + s1= 4
3x1 - 8x2 + s2= 24
10x1 + 7x2 + s3= 35
x1 ≥ 0, x2 ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0
Cj → 5 7 0 0 0
X1 X2 S1 S2 S3
s3
0 4
0 24
0 35
1 1 1 0 0
3 -8 0 1 0
10 7 0 0 1
4 /1 = 4→outgoing
-
35 / 7 = 5
-5 -7 0 0 0
←Δj
7 4
0 56
0 7
(R2 = R2 + 8R1)
11 0 8 1 0
(R3 = R3 - 7R1)
3 0 -7 0 1
Z = 28
2 0 7 0 0
Because all Δj ≥ 0, optimal basic feasible solution is achieved
Thus the solution is Max Z = 28, x1 = 0 and x2 = 4
Sample Assignment 4
Maximize Z = 2x - 3y + z
3x + 6y + z ≤ 6
4x + 2y + z ≤ 4
x - y + z ≤ 3
& x ≥ 0, y ≥ 0, z ≥ 0
Solution
Maximize Z = 2x - 3y + z + 0s1 + 0s2 + 0s3
3x + 6y + z + s1= 6
4x + 2y + z + s2= 4
x - y + z + s3= 3
x ≥ 0, y ≥ 0, z ≥ 0 s1 ≥ 0, s2 ≥ 0, s3 ≥ 0
Cj → 2 -3 1 0 0 0
X Y Z S1 S2 S3
0 6
0 3
3 6 1 1 0 0
4 2 1 0 1 0
1 -1 1 0 0 1
6 / 3 = 2
4 / 4 =1→ outgoing
3 / 1 = 3
Z = 0
-2 3 -1 0 0 0
x
2 1
0 2
0 9/2 1/4 1 -3/4 0
1 1/2 1/4 0 1/4 0
0 -3/2 3/4 0 -1/4 1
3/1/4=12
1/1/4=4
8/3 = 2.6→
Z = 2
0 4 1/2 0 1/2 0
z
0 7/3
2 1/3
1 8/3
0 5 0 1 -2/3 -1/3
1 1 0 0 1/3 -1/3
0 -2 1 0 -1/3 4/3
Z = 10/3
0 3 0 0 1/3 2/3
As all Δj ≥ 0, optimal basic feasible solution is achieved. Consequently the solution is Max Z = 10/3, x = 1/3, y = 0 and z = 8/3
Example 5
Maximize Z = 3x1 + 5x2
3x1 + 2x2 ≤ 18
x1 ≤ 4
x2 ≤ 6
Maximize Z = 3x1 + 5x2 + 0s1 + 0s2 + 0s3
3x1 + 2x2 + s1= 18
x1 + s2= 4
x2 + s3= 6
Cj → 3 5 0 0 0
CB
XB
X1
X2
S1
S2
S3
0
18
3
2
1
18 / 2 = 9
4
4 / 0 = ∞ (neglect)
6
6 / 1 = 6→
-3
-5
(R1=R1-2R3)
-2
6 / 3 = 2 →
4 / 1 = 4
5
--
Z = 30
(R1=R1 / 3)
1/3
-2/3
(R2=R2 - R1)
-1/3
2/3
Z = 36
As find that, all Δj ≥ 0, optimal basic feasible solution is achieved. As a result the solution is Max Z = 36, x1 = 2, x2 = 6
Example 6
Minimize Z = x1 - 3x2 + 2x3
3x1 - x2 + 3x3 ≤ 7
-2x1 + 4x2 ≤ 12
-4x1 + 3x2 + 8x3 ≤ 10
& x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
Min (-Z) = Max Z? = -x1 + 3x2 - 2x3 + 0s1 + 0s2 + 0s3
3x1 - x2 + 3x3 + s1 = 7
-2x1 + 4x2 + s2 = 12
-4x1 + 3x2 + 8x3 + s3 = 10
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0 s1 ≥ 0, s2 ≥ 0, s3 ≥ 0
Cj → -1 3 -2 0 0 0
X3
7
-1
12
3→
10
-4
8
10/3
Z' = 0
(R1 = R1 + R2)
5/2
1/4
4→
(R2 = R2 / 4)
-1/2
(R3 = R3 - 3R2)
-5/2
-3/4
Z' = 9
3/4
(R1 = R1 / 5/2)
6/5
2/5
1/10
(R2 = R2 + 1/2 R1)
3/5
1/5
3/10
(R3 = R3 + 5/2R1)
11
Z' = 11
As all Δj ≥ 0, optimal basic feasible solution is achieved
Thus the solution is Z' =11 which means Z = -11, x1 = 4, x2 = 5, x3 = 0
Example 7
Max Z = 2x + 5y
x + y ≤ 600
0 ≤ x ≤ 400
0 ≤ y ≤ 300
Max Z = 2x + 5y + 0s1 + 0s2 + 0s3
x + y + s1 = 600
x + s2 = 400
y + s3 = 300
x1 ≥ 0, y ≥ 0, s1 ≥ 0, s2 ≥ 0, s3 ≥ 0
Cj → 2 5 0 0 0
X
Y
600
600 / 1 = 600
400
300
300 /1 = 300→
(R1 = R1 - R3)
400 / 1 = 400
y
Z = 1500
(R2 = R2 - R1)
100
Z = 2100
As given that, all Δj ≥ 0, optimal basic feasible solution is achieved. Thus the solution is Z = 2100, x = 300, y = 300
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