Solutions to Linear Programming Problems Applied Statistics and Quantitative Method Assignment #1 GROUP 6 Abie Widyatmojo Billy Biondi Donny M Sitompul R Nurjaman B Vincentius Ricky T
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Problem #1 : Serendipity Introduction The three princes of Serendip went on a little trip. They could not carry too much weight. More than 300 pounds made them hesitate. They planned to the ounce. When they returned to Ceylon they discovered that their supplies were just about gone. When, what to their joy, Prince William found a pile of coconuts on the ground. “Each will bring 60 rupees,” said Prince Richard with a grin as he almost tripped over a lion skin. “Look out!” cried Prince Robert with glee as he spied some more lion skins under a tree. “These are worth even more—300 rupees each If we can just carry them all down to the beach.” Each skin weighed fifteen pounds and each coconut, five, but they carried them all and made it alive. The boat back to the island was very small. 15 cubic feet baggage capacity—that was all. Each lion skin took up one cubic foot while eight coconuts the same space took. With everything stowed they headed to sea and on the way calculated what their new wealth might be. “Eureka!” cried Prince Robert, “Our worth is so great That there’s no other way we could return in this state. Any other skins or nut that we might have brought Would now have us poorer. And now I know what I’ll write my friend Horace in England, for surely only he can appreciate our serendipity.”
Problem Formulate and solve Serendipity by graphical LP in order to calculate “what their new wealth might be.”
Solution Let: X1 X2
: :
Number of Lion Skin carried Number of Coconut carried
Since each lion skin weighed 15 pounds and each coconut 5 pounds, while they did not want to bring more than 300 pounds : 15 X1 + 5 X2 ≤ 300
…………………………………………………………………………… (1)
Also, since each lion skin took up one cubic foot while eight coconuts the same space took and the boat back to the island was very small in which only 15 cubic feet baggage capacity, then we have:
X1 + 1/8 X2 ≤ 15
…………………………………………………………………………… (2)
Every coconut will bring 60 rupees while lion skin worth 300 rupees. Then the objective function will be: MAXIMIZE 300 X1 + 60 X2
…….…………………………………….………………… (3)
Furthermore, we have the following model: Objective :
Z: f(X1,X2) = MAX {300 X1 + 60 X2 }
Constraints:
15 X1 + 5 X2 ≤ 300 X1 + 1/8 X2 ≤ 15 X1 , X2 ≥ 0
By entering the model given above to the POM software table : Serendipity Solution
Maximize
Lion Skin
Coconut
300
60
RHS
Dual
Constraint 1
15
5
<=
300
7.2
Constraint 2
1
.125
<=
15
192
12
24
Solution->
Table 1 Constraints and Maximizing Function Entry Table
The graphical LP provided by POM software is as per Figure 1.
Figure 1 Graphical Solution
5040
Conclusion By searching for maximum wealth, based on the graphical LP using POM software, the princes’ new wealth will be 5,040 rupees resulted from the value of 12 ea of lion skins and 24 ea of coconuts. In addition to that, their baggage capacity is fully utilized. The 15 cubit feet is used up by taking 12 x 1 cubit feet + 24 x 0.125 cubit feet.
Recommendation Therefore it is recommended to them to bring 12 each of lion skins and 24 each of coconuts to fully utilized the baggage capacity and maximize their “new wealth”.
Problem #2 : Mexicana Wire Works Introduction Ron Garcia felt good about his first week as a management trainee at Mexicana Wire Winding, Inc. He had not yet developed any technical knowledge about the manufacturing process, but he had toured the entire facily, located in the suburbs of Mexico City, and had met many people in various areas of the operation. Mexicana, a subsidiary of Westover Wire Works, a Texas firm, is a medium-sized producer of wire windings used in making electrical transformers. Carlos Alverez, the production control manager, described the windings to Garcia as being of standardized design. Garcia’s tour of the plant, laid out by process type, followed the manufacturing sequence for the windings : drawing, extrusion, winding, inspection and packaging. After inspection, good product is packaged and sent to finished product storage; defective product is stored separately until it can be reworked.
Figure 2 Layout of the Plant
On March 8, Vivian Espania, Mexicana’s general manager, stopped by Garcia’s office and asked him to attend a staff meeting at 1:00 P.M. “Let’s get started with the business at hand,” Vivian said, opening the meeting. “You all have met Ron Garcia, our new management trainee. Ron studied operations management in his MBA program in Southern California, so I think he is competent to help us with a problem we have been discussing for a long time without resolution. I’m sure that each of you on my staff will give Ron your full cooperation.”
Vivian turned to Jose Arroyo, production control manager. “Jose, why don’t you describe the problem we are facing?” “Well,” Jose said, “business is very good right now. We are booking more orders than we can fill. We will have some new equipment on line within the next several months, which will take care of our capacity problems, but that won’t help us in April. I have located some retired employees who used to work in the drawing department, and I am planning to bring them in as temporary employees in April to increase capacity there. Because we are planning to refinance some of our long-term debt, Vivian wants our profits to look as good as possible in April. I’m having a hard time figuring out which orders to run and which to back order so that I can make the bottom line look as good as possible. Can you help me with this?” Garcia was surprised and apprehensive to receive such an important high-profile assignment so early in his career. Recovering quickly, he said, “Give me your data and let me work with it for a day or two.” April Orders Product W0075C 1,400 units Product W0033C 250 units Product W0005X 1,510 units Product W0007X 1,116 units Note : Vivian Espania has given her word to a key customer that we will manufacture 600 units of product W0007X and 150 units of product W0075C for him during April Table 2 April Orders
Standard Cost PRODUCT MATERIAL LABOR OVERHEAD SELLING PRICE W0075C $33.00 $9.90 $23.10 $100.00 W0033C 25.00 7.50 17.5 80.00 W0005X 35.00 10.50 24.50 130.00 W0007X 75.00 11.25 63.75 175.00 Selected Operating Data Average output per month = 2,400 units Average machine utilization = 63% Average percentage of production set to rework department = 5% (mostly from Winding Department Average no. of rejected units awaiting rework = 850 (mostly from Winding Department) Table 3 Standard Cost and Selling Price
Plant Capacity (Hours) DRAWING EXTRUSION WINDING PACKAGING 4,000 4,200 2,000 2,300 Note: Inspection capacity is not a problem, we can work overtime, as necessary, to accommodate any schedule. Table 4 Plant Capacity
Bill of Labor (Hours/Unit) PRODUCT DRAWING W0075C 1.0 W0033C 2.0 W0005X 0.0 W0007X 1.0
EXTRUSION 1.0 1.0 4.0 1.0
WINDING 1.0 3.0 0.0 0.0
PACKAGING 1.0 0.0 3.0 2.0
Table 5 Bill of Labor
Problem 1. What recommendations should Ron Garcia make, with what justification? Provide a detailed analysis with charges, graphs, and computer printouts included. 2. Discuss the need for temporary workers in the drawing department. 3. Discuss the plant layout.
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
W0075C W0033C W0005X W0007X
$33.00 $25.00 $35.00 $75.00
$9.90 $7.50 $10.5 $11.25
$23.10 $17.50 $24.5 $63.75
$66.00 $50.00 $70.00 $150.00
Table 6 Cost, Selling Price and Profits per Unit
Let: X1 X2 X3 X4
: : : :
Number of Product W0075C produced Number of Product W0033C produced. Number of Product W0005C produced Number of Product W0007C produced
Then to objective function will be :
MAXIMIZE 34X1 + 30X2 + 60X3 + 25X4 ………………………………………………….(1)
SELLING PRICE $100.00 $80.00 $130.00 $175.00
PROFITS $34.00 $30.00 $60.00 $25.00
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: Maximum Plant Capacity W0075C W0033C W0005C W0007C (Hours) (X1) (X2) (X3) (X4) PRODUCTS
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
X1 + X2 +4X3 + X4 ≤ 4,200 X1 + 3X2 X1
≤ 2,000
……………………………………………………(3)
+3X3 + 2X4 ≤ 2,300
If we assume that Mexicana Wire Works works based on order then : X1 ≤ 1,400 X2 ≤ 250 X3 ≤ 1,510 X4 ≤ 1,116
…………………………………………(4)
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 X1
≤ 2,000
+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
Maximize
X1
X2
X3
X4
RHS
Dual
34
30
60
25
Constraint 1
1
0
0
0
>=
150
0
Constraint 2
0
0
0
1
>=
600
-43
Constraint 3
1
2
0
1
<=
4000
0
Constraint 4
1
1
4
1
<=
4200
0
Constraint 5
1
3
0
0
<=
2000
0
Constraint 6
1
0
3
2
<=
2300
34
Constraint 7
1
0
0
0
<=
1400
0
Constraint 8
0
1
0
0
<=
250
30
Constraint 9
0
0
1
0
<=
1510
0
Constraint 10
0
0
0
1
<=
1116
0
1100
250
0
600
Solution->
59900
Table 8 POM Software Result
Assuming that by doubling the number of worker we can reduce the labor time in drawing department by 50% .but also increase the labor cost by 32% (4,000 hours out of total 12,500 hours). The following model is the modified model after reconsidering the labor time and labor cost : Objective :
Z: f(X1,X2,X3,X4) = MAX {30.83X1 + 27.60X2 + 56.64X3 + 21.40X4 }
Constraints:
X1 ≥ 150 X4 ≥ 600 0.5X1 + X2
+ 0.5X4 ≤ 4,000
X1 + X2 +4X3 + X4 ≤ 4,200 X1 + 3X2 X1
≤ 2,000
+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
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
<=
4000
0
Constraint 4
1
1
4
1
<=
4200
0
Constraint 5
1
3
0
0
<=
2000
0
Constraint 6
1
0
3
2
<=
2300
30.83
Constraint 7
1
0
0
0
<=
1400
0
Constraint 8
0
1
0
0
<=
250
27.6
Constraint 9
0
0
1
0
<=
1510
0
Constraint 10
0
0
0
1
<=
1116
0
1100
250
0
600
Maximize
Solution->
RHS
Dual
53653
Table 9 POM Software Result After Increase the Drawing Dept Headcounts
Conclusion By searching for maximum profits, based on the LP using POM software, the maximum profits will be $59,900 resulted from the manufacturing 1,100 units of W0075C , 250 units of W0033C , and 600 units of W0007X . The result also suggests not to produce W0005C product. By adding more headcounts in the drawing department, it is going to reduce the labor time yet it increases the labor cost up to 32%. According to the result from POM software, the quantity of the
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 of W0007X. 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.
