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
General Embryology tutorial all along with the key concepts of Gametogenesis, Kinds of Egg Membranes, Fertilization, Mechanism of Fertilization, Cleavage, Kinds of Cleavage, Planes of Cleavage, Patterns of Cleavage, Blastulation, Gastrulation and Organogenesis
tutorsglobe.com keynesian theory of income determination assignment help-homework help by online simple theory of income determination tutors
Theory and lecture notes of Location Game all along with the key concepts of location game, Hotelling’s Model, Economic Implication. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Location Game.
tutorsglobe.com respiration in bacteria assignment help-homework help by online bacteria tutors
www.tutorsglobe.com offers cycloalkanes homework help, cycloalkanes assignment help, online tutoring assistance, organic chemistry solutions by online qualified tutor's help.
tutorsglobe.com imperfections in solids assignment help-homework help by online solid state chemistry tutors
Quantum Mechanics tutorial all along with the key concepts of Introduction to quantum mechanics, Three revolutionary principles, A brief history
Fats and Oil tutorial all along with the key concepts of Occurrence and Composition of Fats, Formation of Fats and Oils, Structure of Fats and Oils, Properties of Fats and Oils, Analysis of Fats and Oils, Uses of Fats and Oil and Soaps
tutorsglobe.com choice under uncertainty assignment help-homework help by online intermediate microeconomics tutors
tutorsglobe.com formation of the cell wall assignment help-homework help by online cell wall tutors
Theory and lecture notes of Markov algorithm simulates Post machine all along with the key concepts of markov algorithm simulates post machine, homework help, assignment help, equivalence of tms, pms and markov algorithms tutors. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Markov algorithm simulates Post machine.
tutorsglobe.com timber yielding plant–teak assignment help-homework help by online economic importance tutors
www.tutorsglobe.com offers answering questions to shut-down situations in short-run, perfect competition assignment help - homework help by online tutor;s help.
Introductory development cell biology tutorial all along with the key concepts of principles of development, history of Developmental Biology, progress in the field of Developmental Biology, Descriptive embryology, Experimental Embryology
tutorsglobe.com carbon family assignment help-homework help by online p block elements tutors
1932419
Questions Asked
3689
Tutors
1446579
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!