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 difference between short and long run assignment help-homework help by online classification of production function tutors
tutorsglobe.com chloroplast assignment help-homework help by online cell organelles tutors
Theory and lecture notes of Normal Forms all along with the key concepts of normal forms, Chomsky normal form, reibach normal form, Context Free Grammars & Languages. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Normal Forms.
tutorsglobe.com cost-of-living adjustments assignment help-homework help by online intermediate microeconomics tutors
Want to secure top-notch marks? Avail American Foreign Policy Assignment Help service with skilled tutors for notable grades!
tutorsglobe.com artificial technique of vegetative propagation assignment help-homework help by online reproduction in angiosperms tutors
malvaceae family involves about 82 genera and more than 1,500 species. the plants are cosmopolitan in distribution, more abundant in tropical and subtropical regions.
Theory and lecture notes of Money Stock, Money Market and LM Curve all along with the key concepts of Money demand, nominal interest rate, demand for money. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Money Stock, Money Market and LM Curve.
Systematic Classification of Algae tutorial all along with the key concepts of Rhodophycophyta, Xanthophycophyta, Chrysophycophyta, Phaeoophycophyta, Bacillariophycophyta, Euglenophycophyta, Chlorophycophyta, Cryptophycophyta and Pyrrophycophyta
Refrigerators tutorial all along with the key concepts of Heat Pumps, Air conditioners, Coefficient of Performance of Refrigerators, Coefficient of Performance of Carnot-Cycle Refrigerator, Coefficient of Performance an Ideal Stirling-Cycle Engine
tutorsglobe.com population explosion assignment help-homework help by online human population and explosion-issues tutors
www.tutorsglobe.com offers area and volume of plane figures homework help, assignment help, online tutoring assistance, geometry mathematics solutions by online qualified tutor's help.
Free AP Biology Study Guide, AP Biology Test Papers, AP Biology Practice papers, AP Biology Test pattern and general information, Find AP Biology exam information and resource, material free at Tutorsglobe.com
The dissimilar areas to be covered are relies on requirement of uniformity in the reporting and accuracy of the comparison needed through the various units participated in the uniform costing system.
b) If battery eliminator is employed, test its output dc voltage. If output voltage is acquired, a) Clean the battery contacts and test the battery.
1954389
Questions Asked
3689
Tutors
1465721
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!