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Question 8-16 (Material blending problem) Amalgamated Products has just received a contract to construct steel body frames for automobiles that are to be produced at the new Japanese factory in...

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Question 8-16 (Material blending problem)
Amalgamated Products has just received a contract to construct steel body frames for automobiles that are to be produced at the new Japanese factory in Tennessee. The Japanese auto manufacturer has strict quality control standards for all of its component subcontractors and has informed Amalgamated that each frame must have the following steel content:
MATERIAL MINIMUM PERCENTAGE (%) MAXIMUM PERCENTAGE (%)
Manganese 2.1 2.3
Silicon 4.3 4.6
Carbon 5.05 5.35

Amalgamated mixes batches of eight different available materials to produce one ton of steel used in the body frames. The table on this page details these materials.
Formulate and solve the LP model that will indicate how much each of the eight materials should be blended into a 1-ton load of steel so that Amalgamated meets its requirements while minimizing costs.

Material
Available
Manganese
(%)
Silicon
(%)
Carbon
(%)
Pounds
Available
Cost Per Pound ($)
Alloy 1 70.0 15.0 3.0 No limit 0.12
Alloy 2 55.0 30.0 1.0 300 0.13
Alloy 3 12.0 26.0 0 No limit 0.15
Iron 1 1.0 10.0 3.0 No limit 0.09
Iron 2 5.0 2.5 0 No limit 0.07
Carbide 1 0 24.0 18.0 50 0.10
Carbide 2 0 25.0 20.0 200 0.12
Carbide 3 0 23.00 25.0 100 0.09

ANSWER: (please check if this is the correct answer)
Let X1 = the number of pounds of alloy 1 in one ton of steel
X2 = the number of pounds of alloy 2 in one ton of steel
Etc., down to X8 = the number of pounds of carbide 3 in one ton of steel
Minimize .12X1 + .13X2 + .15X3 + .09X4 + .07X5 + .10X6 + .12X7 + .09X8
S.T. (Mn-min) .7X1 + .55X2 + .12X3 + .01X4 + .05X5 = 42
(Mn-max) .7X1 + .55X2 + .12X3 + .01X4 + .05X5 = 46
(Si-min) .15X1 + .30X2 + .26X3 + .10X4 + .025X5 + .24X6 + .25X7 + .23X8 = 86
(Si-max) .15X1 + .30X2 + .26X3 + .10X4 + .025X5 + .24X6 + .25X7 + .23X8 = 92
(C-min) .03X1 + .01X2 + .03X4 + .18X6 + .20X7 + .25X8 = 101
(C-max) .03X1 + .01X2 + .03X4 + .18X6 + .20X7 + .25X8 = 107
Alloy 2 lim. X2 = 300
Carbide 1 lim. X6 = 50
Carbide 2 lim. X7 = 200
Carbide 3 lim. X8 = 100
Weighs 1 ton X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2000
X1, X2, X3, X4, X5, X6, X7, X8 = 0
Question 8-17
Refer to Problem 8-16. Find the cause of the difficulty and recommend how to adjust it. Then solve the problem again.
Question 8-23 (Airline fuel problem)
Coast-to-Coast Airlines is investigating the possibility of reducing the cost of fuel purchases by taking advantage of lower fuel costs in certain cities. Since fuel purchases represent a substantial portion of operating expenses for an airline, it is important that these costs be carefully monitored. However, fuel adds weight to an airplane, and consequently, excess fuel raises the cost of getting from one city to another. In evaluating one particular flight rotation, a plane begins in Atlanta, flies from Atlanta to Los Angeles, from Los Angeles to Houston, from Houston to New Orleans, and from New Orleans to Atlanta. When the plane arrives in Atlanta, the flight rotation is said to have been completed, and then it starts again. Thus, the fuel on board when the flight arrived in Atlanta must be taken into consideration when the flight begins. Along each leg of this route, there is a minimum and a maximum amount of fuel that may be carried. This and additional information is provided in the table on this page. The regular fuel consumption is based on the plane carrying the minimum amount of fuel. If more than this is carried, the amount of fuel consumed is higher. Specifically, for each 1,000 gallons of fuel above the minimum, 5% (or 50 gallons per 1,000 gallons of extra fuel) is lost due to excess fuel consumption. For example, if 25,000 gallons of fuel were on board when the plane takes off from Atlanta, the fuel consumed on this route would be thousand gallons. If 26 thousand gallons were on board, the fuel consumed would be increased by another 0.05 thousand, for a total of 12.1 thousand gallons.
Formulate this as an LP problem to minimize the cost. How many gallons should be purchased in each city? What is the total cost of this?
Data for Problem 8-23
LEG MINIMUM FUEL REQUIRED (1,000 GAL.) MAXIMUM FUEL ALLOWED (1,000 GAL.) REGULAR FUEL CONSUMPTION (1,000 GAL.) FUEL PRICE PER GALLON

Atlanta–Los Angeles
24 36 12 $4.15

Los Angeles–Houston
15 23 7 $4.25

Houston–New Orleans
9 17 3 $4.10

New Orleans–Atlanta
11 20 5 $4.18

Question 9 (need answers to the question highlighted in yellow only) :
Semiconductor companies such as Texas Instrument produce integrated circuits in different packages such as thin-shrink small outline package(TSSOP) , small outline transistor (SOT), micro small outline package (MSOP) and quad flat no lead (QFN) package. In the testing area of theproductions, this ICs testingoperationsare separated into different areas in the productions floor. Each of them have their own operations team and handled as different entities. This is mainly due to the fact that the different packages need different type machine and handlers. All of the operations are overseen by the test operations director. For the next quarter , the testoperationsdirector got 1 million worth of budget to expand his operations. This budget will be used to purchase additionalequipment such as testers and handlers andincrease test floor space. Due to overall test operations efficiency issue and head count constraints the directors decides to spend the entire budget on one specific test operations. He plans to spend the budget on the operation with the highest expected output in the next few months. Among all the operations, the QFN area and TSSOP area has the highest and comparable outputs in the past. Based on the previous confirmed output results for both the area, we will use forecasting method in this project and determine which one of this area that he needs to choose to allocate the entire budget.
General outline of the assignment and allocation based on the outline of the question are:
1.Implementation/results- forecasting method below are performed on the raw data for both the QFN area and TSSOP area.
a) moving average
b) weighted moving average
c) exponential smoothing
d) adjusted exponential smoothing
e) Linear trend – this is already provided in the attached file .
-based on all the analysis, we pick the area with higher output forecast from the most accurate forecasting method!
2.recommendation and limitation
Answered Same Day Dec 23, 2021

Solution

Robert answered on Dec 23 2021
129 Votes
9 Ans)
A) Moving Average method:

X
QFN AREA confirmed output
volume(Y)

1 50000

2 53000

3 54000

4 57000

5 57500

6 60000

7 61000

8 62000

9 65000

10 66000

11 67500

12 70000
Total 78 723000
 F13 =


(January 13)
 F14 =


(Feb 13)
 F15 =


(March 13)
 F16 =


(April 13)
Total forecasted volume for first quarter of 2013 for QFN area = 245561


 
t
ti
it yF
11
1
12
1
X
TSSOP AREA confirmed output
volume

1 51250

2 52500

3 54500

4 56000

5 56500

6 59350

7 62000

8 63450

9 60000

10 65000

11 67250

12 70250
total 78 718050
 F13 =


(January 13)
 F14 =


(Feb 13)
 F15 =


(March 13)
 F16 =


(April 13)
Total forecasted volume for first quarter of 2013 for TSSOP area = 243399
We observe that the total forecast volume for first quarter is greater for QFN
AREA using moving average method.
B) Weighted Moving Average method:
Forecast(t+1) = ∑


Where Ci is the weights associated with the period and
Yi is the output of that period
Here we assume that the weights are equal for all hence = (1/12)
Now this case is equivalent to the moving average method and hence we will get
the same results as above.
Total forecasted volume for first quarter of 2013 for QFN area = 245561
Total forecasted volume for first quarter of 2013 for TSSOP area = 243399
We observe that the total forecast volume for first quarter is greater for QFN
AREA using moving average method
C) Exponential smoothing method:
X
QFN AREA confirmed output
volume(Y)
Forecast
( = 0.1)
1 50000 50000(assume)
2 53000 50000
3 54000 50300
4 57000 50670
5 57500 51303
6 60000 51923
7 61000 52730
8 62000 53557
9 65000 54402
10 66000 55461
11 67500 56515
12 70000 57614
13 58852 (assume) 58852
14 58852 (assume) 58852
15 58852 (assume) 58852
16 58852 (assume) 58852
Total forecasted volume for first quarter of 2013 for...
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