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MEPM 531 Operation Research - American University of Ras Al Khaimah
Solve the problems using LP Modeling and the Simplex Algorithm.
Problem 1: Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit.
Solution:
Reddy Mikks produces both interior and exterior paints from two raw materials, ??1 and ??2. The following table provides the basic data of the problem:
|
|
Exterior Paint
|
Interior Paint
|
Maximum daily availability(tons)
|
|
Raw Material,M1
|
6
|
4
|
24
|
|
Raw Material,M2
|
1
|
2
|
6
|
|
Profit per ton(1000$)
|
5
|
4
|
|
A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton. Also, the maximum daily demand of interior paint is 2 tons. Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit.
Constraints
6M1+4M2<=24
M1+2M2<=6
M2-M1<=1
M2<=2
Objective function(Maximize Profit)
Z=5M1+4M2
6M1+4M2=24
|
M1
|
0
|
4
|
|
M2
|
6
|
0
|
|
Points
|
(0,6)
|
(4,0)
|
M1+2M2=6
|
M1
|
0
|
6
|
|
M2
|
3
|
0
|
|
Points
|
(0,3)
|
(6,0)
|
M2-M1<=1
|
M1
|
0
|
-1
|
|
M2
|
1
|
0
|
|
Points
|
(0,1)
|
(-1,0)
|
Constraints
6M1+4M2+ S1=24
M1+2M2+ S2=6
M2-M1+ S3=1
M2+ S4=2
Objective function(Maximize Profit)
Z=5M1+4M2+0 S1+ 0S2+0 S3+0 S4
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
6
|
4
|
1
|
0
|
0
|
0
|
24
|
|
0
|
S2
|
1
|
2
|
0
|
1
|
0
|
0
|
6
|
|
0
|
S3
|
-1
|
1
|
0
|
0
|
1
|
0
|
1
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
Cj-- zj
|
5
|
4
|
0
|
0
|
0
|
0
|
|
For optimal solution, Cj-- zj<=0
Maximum positive value of Cj-- zj=5
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
6
|
4
|
1
|
0
|
0
|
0
|
24
|
|
0
|
S2
|
1
|
2
|
0
|
1
|
0
|
0
|
6
|
|
0
|
S3
|
-1
|
1
|
0
|
0
|
1
|
0
|
1
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
Cj-- zj
|
5
|
4
|
0
|
0
|
0
|
0
|
|
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
6
|
4
|
1
|
0
|
0
|
0
|
24
|
24/6=4(MIN)
|
|
0
|
S2
|
1
|
2
|
0
|
1
|
0
|
0
|
6
|
6/1=6
|
|
0
|
S3
|
-1
|
1
|
0
|
0
|
1
|
0
|
1
|
1/-1=-1(REJECT)
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
0/2=0(REJECT)
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
Cj-- zj
|
5
|
4
|
0
|
0
|
0
|
0
|
0
|
|
New first row= old first row/pivot key
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
|
|
0
|
S2
|
|
|
|
|
|
|
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New second row =old second row -(pivot key for second row * new line)
=(1 2 0 1 0 0 6) -(1) * ( 1 2/3 1/6 0 0 0 4)
= (0 4/3 -1/6 1 0 0 2)
|
c
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
|
|
0
|
S2
|
0
|
4/3
|
-1/6
|
1
|
0
|
0
|
2
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New third row =old third row -(pivot key for third row * new line)
=(-1 1 0 0 1 0 1) -(-1) * ( 1 2/3 1/6 0 0 0 4)
= (0 5/3 1/6 0 1 0 5)
|
c
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
|
|
0
|
S2
|
0
|
4/3
|
-1/6
|
1
|
0
|
0
|
2
|
|
|
0
|
S3
|
0
|
5/3
|
1/6
|
0
|
1
|
0
|
5
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New third row =old third row -(pivot key for third row * new line)
=(0 1 0 0 0 1 2) -(0) * ( 1 2/3 1/6 0 0 0 4)
= (0 1 0 0 0 1 2)
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
|
|
0
|
S2
|
0
|
4/3
|
-1/6
|
1
|
0
|
0
|
2
|
|
|
0
|
S3
|
0
|
5/3
|
1/6
|
0
|
1
|
0
|
5
|
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
|
|
Zj
|
5
|
10/3
|
5/6
|
0
|
0
|
0
|
20
|
|
|
Cj-- Zj
|
0
|
2/3
|
-5/6
|
0
|
0
|
0
|
|
|
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
6
|
|
0
|
S2
|
0
|
4/3
|
-1/6
|
1
|
0
|
0
|
2
|
3/2(MIN)
|
|
0
|
S3
|
0
|
5/3
|
1/6
|
0
|
1
|
0
|
5
|
3
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
2
|
|
Zj
|
5
|
10/3
|
5/6
|
0
|
0
|
0
|
20
|
|
|
Cj-- Zj
|
0
|
2/3
|
-5/6
|
0
|
0
|
0
|
|
|
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
6/6=1
|
4/6=2/3
|
1/6
|
0/6=0
|
0/6=0
|
0/6=0
|
4
|
|
|
0
|
M2
|
0
|
4/3
|
-1/6
|
1
|
0
|
0
|
2
|
|
|
0
|
S3
|
0
|
5/3
|
1/6
|
0
|
1
|
0
|
5
|
|
|
0
|
S4
|
0
|
1
|
0
|
0
|
0
|
1
|
2
|
|
|
Zj
|
5
|
10/3
|
5/6
|
0
|
0
|
0
|
20
|
|
|
Cj-- Zj
|
0
|
2/3
|
-5/6
|
0
|
0
|
0
|
|
|
NEW SECOND ROW
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
|
|
|
|
|
|
|
|
|
4
|
M2
|
0
|
1
|
-1/8
|
3/4
|
0
|
0
|
3/2
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
Zj
|
|
|
|
|
|
|
|
|
|
Cj-- Zj
|
|
|
|
|
|
|
|
|
NEW FIRST ROW
New FIRST row =old FIRST row -(pivot key for FIRST row * new line)
=(1 2/3 1/6 0 0 0 4) -(2/3) * ( 0 1 - 1/8 3/4 0 0 3/2)
= (1 0 1/4 -1/2 0 0 3)
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
1
|
0
|
1/4
|
-1/2
|
0
|
0
|
3
|
|
|
4
|
M2
|
0
|
1
|
-1/8
|
3/4
|
0
|
0
|
3/2
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
Zj
|
|
|
|
|
|
|
|
|
|
Cj-- Zj
|
|
|
|
|
|
|
|
|
NEW THIRD ROW
New THIRD row =old THIRD row -(pivot key for THIRD row * new line)
=(0 5/3 1/6 0 1 0 5) -(5/3) * ( 0 1 - 1/8 3/4 0 0 3/2)
= (0 0 9/24 -5/4 1 0 5/2)
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
1
|
0
|
1/4
|
-1/2
|
0
|
0
|
3
|
|
|
4
|
M2
|
0
|
1
|
-1/8
|
3/4
|
0
|
0
|
3/2
|
|
|
0
|
S3
|
0
|
0
|
9/24
|
-5/4
|
1
|
0
|
5/2
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
Zj
|
|
|
|
|
|
|
|
|
|
Cj-- Zj
|
|
|
|
|
|
|
|
|
NEW FOURTH ROW
New FOURTH row =old FOURTH row -(pivot key for THIRD row * new line)
=(0 1 0 0 0 1 2) -(1) * ( 0 1 - 1/8 3/4 0 0 3/2)
= (0 0 1/8 -3/4 0 1 1/2)
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
1
|
0
|
1/4
|
-1/2
|
0
|
0
|
3
|
|
|
4
|
M2
|
0
|
1
|
-1/8
|
3/4
|
0
|
0
|
3/2
|
|
|
0
|
S3
|
0
|
0
|
9/24
|
-5/4
|
1
|
0
|
5/2
|
|
|
0
|
S4
|
0
|
0
|
1/8
|
-3/4
|
0
|
1
|
1/2
|
|
|
Zj
|
|
|
|
|
|
|
|
|
|
Cj-- Zj
|
|
|
|
|
|
|
|
|
|
CB
|
|
5
|
4
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
M1
|
M2
|
S1
|
S2
|
S3
|
S4
|
|
5
|
M1
|
1
|
0
|
1/4
|
-1/2
|
0
|
0
|
3
|
|
|
4
|
M2
|
0
|
1
|
-1/8
|
3/4
|
0
|
0
|
3/2
|
|
|
0
|
S3
|
0
|
0
|
9/24
|
-5/4
|
1
|
0
|
5/2
|
|
|
0
|
S4
|
0
|
0
|
1/8
|
-3/4
|
0
|
1
|
1/2
|
|
|
Zj
|
5
|
4
|
3/4
|
1/2
|
0
|
0
|
21
|
|
|
Cj-- Zj
|
0
|
0
|
-3/4
|
-1/2
|
0
|
0
|
|
|
This is optimal solution as Cj-- Zj <=0
M1=3, M2=3/2
Z=5(3)+4(3/2)=21
ORDER NEW COPY OF MEPM 531 OPERATION RESEARCH ASSIGNMENT & GET HIGH QUALITY SOLUTIONS FROM SUBJECT'S TUTORS!
Problem 2
A firm manufactures two types of products A and B and sells them at a profit of $2 on type A and $3 on type B. Each product is processed on two machines G and H. Type A requires one minute of processing time on G and two minutes on H; type B requires one minute on G and one minute on H. The machine G is available for not more than 6 hour 40 minutes while machine H is available for not more than 10 hours during any working day. Formulate the problem as a linear programming problem and determine the optimum product mix of types A and B products.
Solution:
|
|
Machine G(min)
|
Machine H(min)
|
Profit
|
|
Type A,X1
|
1
|
2
|
$2
|
|
Type B,X2
|
1
|
1
|
$3
|
|
Availability of each machine
|
400
|
600
|
|
Constraints
X1 +X2 <=400
2X1+X2<=600
Objective function
Z=2X1+3X2
Converting inequalities into equations
X1+X2+S1=400
2X1+X2+S2=600
Z=2X1+3X2+0S1+0S2
|
CB
|
|
2
|
3
|
0
|
0
|
Solution
|
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
400
|
|
|
0
|
S2
|
2
|
1
|
0
|
1
|
600
|
|
|
zj
Cj-- zj
|
0
|
0
|
0
|
0
|
0+
|
|
|
2
|
3
|
0
|
0
|
|
|
|
CB
|
|
2
|
3
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
400
|
400(min)
|
|
0
|
S2
|
2
|
1
|
0
|
1
|
600
|
600
|
|
zj
Cj-- zj
|
0
|
0
|
0
|
0
|
0
|
0
|
|
2
|
3
|
0
|
0
|
|
|
New table
First row
|
CB
|
|
2
|
3
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
|
2
|
X2
|
1
|
1
|
1
|
0
|
400
|
|
|
0
|
S2
|
|
|
|
|
|
|
|
zj
Cj-- zj
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Second row
New second row =(2 1 0 1 600) -(1)*(1 1 1 0 400)
=(1 0 -1 1 200)
|
CB
|
|
2
|
3
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
|
3
|
X2
|
1
|
1
|
1
|
0
|
400
|
|
|
0
|
S2
|
1
|
0
|
-1
|
1
|
200
|
|
|
zj
Cj-- zj
|
3
|
3
|
3
|
0
|
1200
|
|
|
0
|
0
|
-2
|
0
|
|
|
This is optimal solution as Cj-- zj <=0
X1=0 ,X2=400 , z=1200
EXPERTSMINDS.COM ACCEPTS INSTANT AND SHORT DEADLINES ORDER FOR MEPM 531 OPERATION RESEARCH ASSIGNMENT - ORDER TODAY FOR EXCELLENCE!
Problem 3
A company produces two types of hats. Each hat of the first type requires twice as much labor time as the second type. The company can produce a total of 500 hats a day. The market limits daily sales of the first and second type to 150 and 250 hats respectively. Assuming that the profits per hat are $8 for type A and $5 for type B, formulate the problem as a linear programming model in order to determine the number of hats to be produced of each type so as to maximize the profit.
Solution:
Let say first type hats are X1 and second type hats are X2
2X1+X2<=500
X1<=150
X2<=250
X1<=2X2
Objective Function (Maximize)
Z=8X1+5X2
Converting inequalities into equations
2X1+X2+S1=500
X1+S2=150
X2+S3=250
Z=8X1+5X2+0S1+0S2+0S3
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
2
|
1
|
1
|
0
|
0
|
500
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
0
|
150
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
250
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
|
|
|
Cj-- zj
|
8
|
5
|
0
|
0
|
0
|
|
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
2
|
1
|
1
|
0
|
0
|
500
|
250
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
0
|
150
|
150(Min)
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
250
|
infinite
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
|
|
|
|
Cj-- zj
|
8
|
5
|
0
|
0
|
0
|
|
|
New second row
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
|
|
|
|
|
|
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
New first row=(2 1 1 0 0 250)-(2)(1 0 0 1 0 150)= (0 1 1-2 0 200)
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-2
|
0
|
-50
|
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
New third row =(0 1 0 0 1 250)-(0)(1 0 0 1 0 150)= (0 1 0 0 1 250)
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-2
|
0
|
-50
|
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
100
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-2
|
0
|
-50
|
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
100
|
|
|
|
zj
|
8
|
0
|
0
|
8
|
0
|
1200
|
|
|
|
Cj-- zj
|
0
|
5
|
0
|
-8
|
0
|
|
|
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-2
|
0
|
-50
|
-50
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
infinite
|
|
0
|
X2
|
0
|
1
|
0
|
0
|
1
|
100
|
100(min)
|
|
|
zj
|
8
|
0
|
0
|
8
|
0
|
1200
|
|
|
|
Cj-- zj
|
0
|
5
|
0
|
-8
|
0
|
|
|
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-2
|
0
|
-50
|
-50
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
infinite
|
|
5
|
X2
|
0
|
1
|
0
|
0
|
1
|
100
|
100(min)
|
|
|
zj
|
8
|
0
|
0
|
8
|
0
|
1200
|
|
|
|
Cj-- zj
|
0
|
5
|
0
|
-8
|
0
|
|
|
|
CB
|
|
8
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
X1
|
X2
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
0
|
1
|
-2
|
-1
|
-150
|
|
|
8
|
X1
|
1
|
0
|
0
|
1
|
0
|
150
|
|
|
5
|
X2
|
0
|
1
|
0
|
0
|
1
|
100
|
|
|
|
zj
|
8
|
5
|
0
|
8
|
5
|
1700
|
|
|
|
Cj-- zj
|
0
|
0
|
0
|
-8
|
-5
|
|
|
This is optimal solution
X1=150
X2=100
Z=1700
SAVE DISTINCTION MARKS IN EACH MEPM 531 OPERATION RESEARCH ASSIGNMENT WHICH IS WRITTEN BY OUR PROFESSIONAL WRITER!
Problem 4
The manufacturer of patent medicines is proposed to prepare a production plan for medicines A and B. There are sufficient ingredients available to make 20,000 bottles of medicine A and 40,000 bottles of medicine B, but there are only 45,000 bottles into which either of the medicines can be filled.
Solution:
1. Formulate this problem as L.P.P.
2. How the manufacturer schedule his production in order to maximize profit
Objective function
Z=8x+7y
Constraints
x+y<=.45000
x<=20000
y<=40000
3x+y<=66000
Converting into equations
x+y+S1=45000
x+S2=20000
y+S3=40000
3x+y+S4=66000
Z=8x+7y+0S1+0S2+0S3+0S4
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
0
|
0
|
45000
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
|
0
|
S4
|
3
|
1
|
0
|
0
|
0
|
1
|
66000
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
Cj-- zj
|
8
|
7
|
0
|
0
|
0
|
0
|
|
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
0
|
0
|
45000
|
45000
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
20000(Min)
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
Infinity
|
|
0
|
S4
|
3
|
1
|
0
|
0
|
0
|
1
|
66000
|
22000
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
Cj-- zj
|
8
|
7
|
0
|
0
|
0
|
0
|
|
|
New second row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
|
|
|
|
|
|
|
|
|
8
|
x
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New third row and first row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
|
0
|
S3
|
-1
|
1
|
0
|
-1
|
1
|
0
|
20000
|
|
|
0
|
S4
|
|
|
|
|
|
|
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New fourth row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
|
|
0
|
S4
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
|
|
0
|
S4
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
|
|
|
zj
|
8
|
0
|
0
|
8
|
0
|
0
|
160000
|
|
|
|
Cj-- zj
|
0
|
7
|
0
|
-8
|
0
|
0
|
|
|
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
25000
|
|
8
|
X
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
infinite
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
40000
|
|
0
|
S4
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
6000(min)
|
|
|
zj
|
8
|
0
|
0
|
8
|
0
|
0
|
160000
|
|
|
|
Cj-- zj
|
0
|
7
|
0
|
-8
|
0
|
0
|
|
|
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
|
|
|
|
|
|
|
|
|
8
|
X
|
|
|
|
|
|
|
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
7
|
Y
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
6000(min)
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New first row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
0
|
1
|
2
|
0
|
-1
|
19000
|
|
|
8
|
x
|
|
|
|
|
|
|
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
7
|
y
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
6000(min)
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New second row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
0
|
0
|
1
|
0
|
0
|
20000
|
|
|
0
|
S3
|
|
|
|
|
|
|
|
|
|
7
|
y
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
6000(min)
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
New third row
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
-1
|
0
|
4
|
0
|
-1
|
14000
|
|
|
0
|
S3
|
0
|
0
|
0
|
3
|
1
|
-1
|
34000
|
|
|
7
|
y
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
6000(min)
|
|
|
zj
|
|
|
|
|
|
|
|
|
|
|
Cj-- zj
|
|
|
|
|
|
|
|
|
|
CB
|
|
8
|
7
|
0
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
Y
|
S1
|
S2
|
S3
|
S4
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
0
|
0
|
25000
|
|
|
8
|
x
|
1
|
-1
|
0
|
4
|
0
|
-1
|
14000
|
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
0
|
40000
|
|
|
7
|
y
|
0
|
1
|
0
|
-3
|
0
|
1
|
6000
|
|
|
|
zj
|
8
|
-1
|
0
|
11
|
0
|
-1
|
140000
|
|
|
|
Cj-- zj
|
0
|
8
|
0
|
-53
|
0
|
1
|
|
|
This has multiple optimal solutions.
24/7 AVAILABILITY OF TRUSTED MEPM 531 OPERATION RESEARCH ASSIGNMENT WRITERS! ORDER ASSIGNMENTS FOR BETTER RESULTS!
Problem 5
A toy company manufactures two types of doll, a basic version - doll A and a deluxe version - doll B. Each doll of type B takes twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day. The supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). The deluxe version requires a fancy dress of which there are only 600 per day available. If the company makes a profit of $3 and $5 per doll, respectively on doll A and B, then how many of each doll should be produced per day in order to maximize the total profit. Formulate this problem in order to determine the number of dolls to be produced of each type so as to maximize the profit.
Solution:
Maximize
Z=3x +5y
x+2y<=2000
x+y<=1500
Y<=600
Adding slack variables
Z=3x+5y+0S1+0S2+0S3
X+2y+S1=2000
X+y+S2=1500
y+S3=600
|
CB
|
|
3
|
5
|
0
|
0
|
0
|
Solution
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
|
0
|
S1
|
1
|
2
|
1
|
0
|
0
|
2000
|
|
0
|
S2
|
1
|
1
|
0
|
1
|
0
|
1500
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
600
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
Cj-- zj
|
3
|
5
|
0
|
0
|
0
|
|
|
CB
|
|
3
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
|
0
|
S1
|
1
|
2
|
1
|
0
|
0
|
2000
|
1000
|
|
0
|
S2
|
1
|
1
|
0
|
1
|
0
|
1500
|
1500
|
|
0
|
S3
|
0
|
1
|
0
|
0
|
1
|
600
|
600(min)
|
|
|
zj
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
Cj-- zj
|
3
|
5
|
0
|
0
|
0
|
|
|
NEW table
|
CB
|
|
3
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
0
|
1400
|
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
-1
|
900
|
|
|
5
|
y
|
0
|
1
|
0
|
0
|
1
|
600
|
|
|
|
zj
|
0
|
5
|
0
|
0
|
5
|
3000
|
|
|
|
Cj-- zj
|
3
|
0
|
0
|
0
|
-5
|
|
|
|
CB
|
|
3
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
|
0
|
S1
|
1
|
1
|
1
|
0
|
0
|
1400
|
1400
|
|
0
|
S2
|
1
|
0
|
0
|
1
|
-1
|
900
|
900(min)
|
|
5
|
y
|
0
|
1
|
0
|
0
|
1
|
600
|
infinite
|
|
|
zj
|
0
|
5
|
0
|
0
|
5
|
3000
|
|
|
|
Cj-- zj
|
3
|
0
|
0
|
0
|
-5
|
|
|
New table
|
CB
|
|
3
|
5
|
0
|
0
|
0
|
Solution
|
Ratio
|
|
B.v
|
x
|
y
|
S1
|
S2
|
S3
|
|
0
|
S1
|
0
|
1
|
1
|
-1
|
1
|
500
|
|
|
3
|
x
|
1
|
0
|
0
|
1
|
-1
|
900
|
|
|
5
|
y
|
0
|
1
|
0
|
0
|
1
|
600
|
|
|
|
zj
|
3
|
5
|
0
|
3
|
2
|
5700
|
|
|
|
Cj-- zj
|
0
|
0
|
0
|
-3
|
-2
|
|
|
This is optimal solution as Cj-- zj<=0
X=900
Y=600
Z=5700
MOST RELIABLE AND TRUSTWORTHY MEPM 531 OPERATION RESEARCH ASSIGNMENT HELP & HOMEWORK WRITING SERVICES AT YOUR DOORSTEPS!
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