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 reproduction assignment help-homework help by online human physiology tutors
tutorsglobe offers geography homework help, geography assignment help, online geography tutoring assistance, geography writing solutions.
tutorsglobe.com atopy assignment help-homework help by online hypersensitivity-anaphylaxsis tutors
Don’t hesitate to avail the impeccable Gender in Health and Body Assignment Help service from the qualified tutors at low prices to score high.
tutorsglobe.com shapes of orbitals assignment help-homework help by online wave nature of electrons tutors
Hire Environmental Management Assignment Help tutors and obtain non-plagiarized papers from industry best experts at viable prices!
Radiation tutorial all along with the key concepts of Properties of Radiation, Detecting Heat Radiation, Black Body Radiation, Provost's Theory of Heat Exchange, Stefan-Boltzmann Law of Radiation, Thermos Flask
tutorsglobe.com vivipary assignment help-homework help by online process of seed germination tutors
Semi-Automatic Washing Machine is comparatively simple in operation and construction. It contains two separate tubs that are wash tub and spin tub.
Theory and lecture notes of Introduction and Bio-amplifier requirements all along with the key concepts of Ideal Amplifier, Non-Ideal Amplifier, Negative Feedback and Operational Amplifier. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Introduction and Bio-amplifier requirements.
Fibre Optics tutorial all along with the key concepts of Optical Fibres, principles of light propagation, Propagation of Meridional Rays, Dispersion, types of fibers and Applications of Optical Fibres
Introduction to the Systematic of Insects and Mites tutorial all along with the key concepts of Objectives of Classification, Elements of Classification, General Classification of Insects, Class Insecta, Subclass Apterygota, Subclass Pterygota and General Classification of Mites
a software development life cycle is an abstract representation of gradual development and evolution of the software that undergoes a series of sequential or concurrent steps of the software development process.
tutorsglobe.com fleshy xerophytes assignment help-homework help by online xeric habitats characterization tutors
satellites can be categorized through their functions. satellites are launched into space to perform a particular task.
1950959
Questions Asked
3689
Tutors
1475373
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!