Upload
eduardo-bustos-farias
View
216
Download
0
Embed Size (px)
Citation preview
7/27/2019 Respuesta.problema.3.Tipo.A
1/5
Solution
As a business entity, the maximization of profit is the primary objective. The possible secondary
objectives are commitments to customers, company reputation and quality of the products.
Profits is revenue minus costs. The revenue is taken from the sales of every single unit of products. The
costs in this case are the costs for material, labor and overhead. We can then modify Table 3 into a table
mentioning the profits for every single unit of each product.
Profits
PRODUCT MATERIAL LABOR OVERHEAD COST SELLING
PRICE
PROFITS
W0075C $33.00 $9.90 $23.10 $66.00 $100.00 $34.00
W0033C $25.00 $7.50 $17.50 $50.00 $80.00 $30.00
W0005X $35.00 $10.5 $24.5 $70.00 $130.00 $60.00
W0007X $75.00 $11.25 $63.75 $150.00 $175.00 $25.00Table 6 Cost, Selling Price and Profits per Unit
Let:
X1 : Number of Product W0075C produced
X2 : Number of Product W0033C produced.
X3 : Number of Product W0005C producedX4 : Number of Product W0007C produced
Then to objective function will be :
MAXIMIZE 34X1 + 30X2 + 60X3 + 25X4 .(1)
7/27/2019 Respuesta.problema.3.Tipo.A
2/5
Vivian has given her words to a key customer that they are going to manufacture 600 units of X4 and
150 units of X1. Then we have:
X1 150 (2)
X4 600
For each process, there are boundaries regarding the hour of plant capacity as follow:
PRODUCTS Maximum
Plant
Capacity
(Hours)W0075C W0033C W0005C W0007C
(X1) (X2) (X3) (X4)
DRAWING 1 2 - 1 4,000
EXTRUSION 1 1 4 1 4,200
WINDING 1 3 - - 2,000
PACKAGING 1 - 3 2 2,300
Table 7 Plant Capacity
X1 + 2X2 + X4 4,000
(3)X1 + X2 +4X3 + X4 4,200
X1 + 3X2 2,000
X1 +3X3 + 2X4 2,300
If we assume that Mexicana Wire Works works based on order then :
X1 1,400
X2 250 (4)
X3 1,510
X4 1,116
7/27/2019 Respuesta.problema.3.Tipo.A
3/5
Thus, we will have the following model :
Objective : Z: f(X1,X2,X3,X4) = MAX {34X1 + 30X2 + 60X3 + 25X4 }
Constraints: X1 150
X4 600
X1 + 2X2 + X4 4,000
X1 + X2 +4X3 + X4 4,200
X1 + 3X2 2,000
X1 +3X3 + 2X4 2,300
X1 1,400
X2 250
X3 1,510
X4 1,116
X1 , X2 0
By entering the model given above to the POM software table :
Mexicana Wire Works Solution
X1 X2 X3 X4 RHS Dual
Maximize 34 30 60 25
Constraint 1 1 0 0 0 >= 150 0Constraint 2 0 0 0 1 >= 600 -43
Constraint 3 1 2 0 1
7/27/2019 Respuesta.problema.3.Tipo.A
4/5
Constraints: X1 150
X4 600
0.5X1 + X2 + 0.5X4 4,000
X1 + X2 +4X3 + X4 4,200X1 + 3X2 2,000
X1 +3X3 + 2X4 2,300
X1 1,400
X2 250
X3 1,510
X4 1,116
X1 , X2 0
The result is given in the following table :
Mexicana Wire Works Solution
X1 X2 X3 X4 RHS Dual
Maximize 30.83 27.6 56.64 21.4
Constraint 1 1 0 0 0 >= 150 0
Constraint 2 0 0 0 1 >= 600 -40.26
Constraint 3 .5 1 0 .5
7/27/2019 Respuesta.problema.3.Tipo.A
5/5
product still the same, yet it is not a change for better but for less profit ($53,653) due to no increase in
the pricing but burdened by the increased cost.
Recommendation
It is for maximum profits to produce 1,100 units of W0075C , 250 units of W0033C , and 600 units ofW0007X. Company must turn down the orders to produce 1,510 units of W0005X. It is better for the
company not to increase the number of employee in the drawing department in order to keep the cost
down.