Bachelor of Science in Information Systems
Department of Information Systems
College of Computer & Information Sciences
Imam Mohammad Bin Saud University
Course Title: Quantitative methods
Instructor: Mjdah Al Shehri
ASSIGNEMENT # 4
Problem#1
“ THE GALAXY INDUSTRY PRODUCTION”
•
Decisions variables:
–
X1 = Production level of Space Rays (in dozens per week).
–
X2 = Production level of Zappers (in dozens per week).

Mathematical model :
Max 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = 1200
3X1 + 4X2 < = 2400
X1 + X2 < = 800
X1 - X2 < = 450
Xj> = 0, j = 1,2
(Plastic)
(Production Time)
(Total production)
(Mix)
(Nonnegativity)
a. Write the problem in standard form.
b. solve the problem using the simplex algorithms
Solution:
a.
It should be define the following:
S1= the amount of unused plastic
S2= the amount of unused production time
S3= the amount by which total production falls below 800 dozen
S4= the amount by which the difference in production of space rays and zappers
falls below 450 dozen
The standard form for the Galaxy industries model is then:
MAX 8X1+ 5X2 (profit)
ST
2X1+ X2+S1=1200 (plastic)
3X1+ 4X2+S2=2400 (production time)
X1+X2+S3=800 (total production)
X1-X2+S4=450(Mix)
All Xj>=0, and All Sj>=0
b.
the initial basic feasible solution, found by setting the nonbasic variables X1 and
X2 to 0; is X1=0, X2=0, S1=1200, S2=2400, S3=800 and S4= 450
The initial tableau is:
Basis
S1
S2
S3
S4
X1
2
3
1
1
-8
X2
1
4
1
-1
-5
S1
1
0
0
0
0
S2
0
1
0
0
0
S3
0
0
1
0
0
S4
0
0
0
1
0
RHS
1200
2400
800
450
0
Iteration # 1:

Step1:
Select the variables that objective function becomes large number
number)
Basis
X1
X2
S1
S2
S3
S4
S1
2
1
1
0
0
0
S2
3
4
0
1
0
0
S3
1
1
0
0
1
0
S4
1
-1
0
0
0
1
-8
-5
0
0
0
0
(negative
RHS
1200
2400
800
450
0
X1 is the entry variables
 Step 2
Find the minimizing ratio between the right hand side and positive entering in the
entering column
Basis
S1
S2
S3
S4
X1
2
3
1
1
-8
X2
1
4
1
-1
-5
S1
1
0
0
0
0
S2
0
1
0
0
0
S3
0
0
1
0
0
S4
0
0
0
1
0
RHS
1200
2400
800
450
0
Ratio
600
800
800
450
600 is meaning the maximum value of X1 before S1 reaches 0
800 is meaning the maximum value of X1 before S2 reaches 0
800 is meaning the maximum value of X1 before S3 reaches 0
450 is meaning the maximum value of X1 before S4 reaches 0
The S4 is the leaving variables

Step 3:
For pivot row: Find the pivot elements
(1,-1,0,0,0,1,450)/1= (1,-1,0,0,0,1,450)
Basis
X1
X2
S1
S2
S3
S4
RHS
Ratio
S4
1
-8
-1
-5
0
0
0
0
0
0
1
0
450
0
450
For other rows:
New row= current row- { ( pivot column co efficient) * (new pivot row)
2
2
0
1
-2
3
1
0
1
0
0
0
0
0
0
0
1
-2
1200
900
300
Basis
X1
0
X2
3
S1
1
S2
0
S3
0
S4
-2
RHS
300
S4
1
-8
-1
-5
0
0
0
0
0
0
1
0
450
0
3
3
0
4
-3
7
0
0
0
1
0
1
0
0
0
0
3
-3
2400
1350
1050
1
1
0
1
-1
2
0
0
0
0
0
0
1
0
0
0
1
-1
800
1350
350
Basis
S1
S2
S3
X1
X1
0
0
0
1
0
X2
3
7
2
-1
-13
S1
1
0
0
0
0
S2
0
1
0
0
0
S3
0
0
1
0
0
S4
-2
-3
-1
1
8
RHS
300
1050
350
450
3600
S1
1/3
-7/3
-2/3
1/3
13/3
S2
0
1
0
0
0
S3
0
0
1
0
0
S4
-2/3
5/3
1/3
1/3
-2/3
RHS
100
350
150
550
4900
X1= 450, X2=0.
objective function= 3600
Iteration # 2:
Basis
X2
S2
S3
X1
X1
0
0
0
1
0
X2
1
0
0
0
0
X2=100, X1=550
Objective function is 4900
Iteration# 3:
Basis
X2
S4
S3
X1
X1
0
0
0
1
0
X2
1
0
0
0
0
S1
-3/5
-7/5
-1/5
4/5
17/5
S2
2/5
3/5
-1/5
-1/5
2/5
S3
0
0
1
0
0
X2= 240, X1= 480
Objective function is 5040. [Optimal solution]
S4
0
1
0
0
0
RHS
240
210
80
480
5040