Discussion unit 6
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School
University of Colorado, Denver *
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Course
3304
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
docx
Pages
3
Uploaded by daveakston
The Simplex method is a manual technique for solving linear programming models, incorporating slack variables, tableaus, and pivot variables to discover the optimal solution for an optimization problem. Linear programming involves determining the most favorable result within linear constraints, where the goal is to maximize or minimize an equation (Kun, 2018).
To make it easier for me to work on this assignment I assumed:
-p for the number of pants.
-j for number of jackets.
-s for maximum sales.
Here is what I have to maximize sales by using an equation: s = 40p + 75j, where p and j ≤ 500. m2.
From here we will use the formula to calculate the number of cotton textiles to be used
:
5p + 3j ≤ 1000.
Then we will use the formula to calculate the number of polyesters used:
2p + 2j ≤ 950.
Now we will rewrite the equations to make it easier to solve them, here is what we have:
5p + 3j + x =100
2p + 2j + y = 950
p + j + z = 500
s – 40p – 75j =0
p
j
x
y
z
s
t
0.5
3
1
0
0
0
1000
2
2
0
1
0
0
950
1
1
0
0
1
0
500
-40
-75
0
0
0
1
0
We are going to use the Simplex method to calculate the outmost sales. -75 is the largest negative number value, so our pivot column is for jackets. 100 / 3 = 333.3
950 / 2 = 475
500 / 1 = 500
The smallest value we have is 333.3 so now we need to divide by 3 the whole row for “x”.
p
j
x
y
z
s
t
0.166...
1
0.333…
0
0
0
333.3
2
2
0
1
0
0
950
1
1
0
0
1
0
500
-40
-75
0
0
0
1
0
Now we need to make cells with jackets in other rows to 0.
p
j
x
y
z
s
t
0.166…
1
0.333…
0
0
0
333.333…
1.666…
0
-0.666…
1
0
0
283.333…
0.833…
0
-0.333…
0
1
0
166.666…
-27.5
0
25
0
0
1
25000
Repeating this process again, but now we need to divide by 1.666 or 5/3 the “y”row. p
j
x
y
z
s
t
0.166…
1
0.333…
0
0
0
333.333…
1
0
-0.4
0.6
0
0
170
0.833…
0
-0.333…
0
1
0
166.666…
-27.5
0
25
0
0
1
25000
Perform operations on the other rows:
p
j
x
y
z
s
t
0
1
0.4
-0.1
0
0
305
1
0
-0.4
0.6
0
0
170
0
0
0
-0.5
1
0
25
0
0
14
16.5
0
1
29675
Now we see that there are no more negative numbers in our last row. The final result to maximize profits
within the constraints, we need to produce 305 jackets and 170 pants for a total profit of $29,675.
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