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
tutorsglobe.com factors governing ionization energy assignment help-homework help by online ionisation potential tutors
Relativity of Mass tutorial all along with the key concepts of Variation of Mass with Velocity, Momentum and Force in Relativistic Mechanics, glancing collision
Theory and lecture notes of Changes in Macro-Economy all along with the key concepts of changes in macro-economy, Occupational Distribution of the Labor Force. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Changes in Macro-Economy.
accounting is regarded along with communicating, analysing and collecting financial information. the reason is to assist people who use this information to make more right decisions.
Expansion of Gases tutorial all along with the key concepts of Kinetic Molecular Theory of Gas, Boyle's Law, Charles' Law, Cubic Expansivity of a Gas, Gay - Lussac's Law, Gas Law or Equation, Ideal Gas Equation, Intermolecular Energy and Forces
tutorsglobe.com contraception assignment help-homework help by online birth control tutors
kinetics and equilibrium tutorial all along with the key concepts of factors influencing reaction rate, experimental methods, kinetics and the equilibrium constant and free energy
tutorsglobe.com characteristics of an isoquant assignment help-homework help by online isoquants tutors
tutorsglobe.com protein-lecithin as carrier assignment help-homework help by online carrier concept tutors
tutorsglobe.com edgeworth box assignment help-homework help by online pure exchange and pareto optimality tutors
Classification of Multicellular Animals tutorial all along with the key concepts of Parazoa-Phylum Porifera, Classification of Phylum porifera, Phylum Cnidaria, Coral Reefs, Phylum Ctenophora, Kinds of Coral Reefs
www.tutorsglobe.com offers ambident enolate anions homework help, assignment help, online tutoring assistance, organic chemistry solutions by online qualified tutor's help.
tutorsglobe.com determination of equilibrium price assignment help-homework help by online monopolistic competition tutors
Electric Potential tutorial all along with the key concepts of Equipotential Surfaces, Potential due to a point charge, Potential due to a system of charges, Potential Difference, Electric Field and Electric Potential and Electric Dipole
tutorsglobe.com heart transplantation assignment help-homework help by online circulation tutors
1933266
Questions Asked
3689
Tutors
1482647
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!