1 OPRE 3333 Practice Problems for Exam 3 Chapter 5, 8 and Chapter 12 Chapter 5: 1. Which of the following data patterns best describes the scenario shown in the given time series plot? a. Linear trend pattern b. Logarithmic trend c. Exponential trend d. Seasonal pattern Answer: D
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OPRE 3333
Practice Problems for Exam 3
Chapter 5, 8 and Chapter 12
Chapter 5:
1. Which of the following data patterns best describes the scenario shown in the given time series plot?
a. Linear trend pattern b. Logarithmic trend c. Exponential trend d. Seasonal pattern Answer: D
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2. all forecasts to two decimal places.). Compute MSE and a forecast for month 11.
3. The below time series gives the indices of Industrial Production in U.S for 10 consecutive years.
Year IP
1 79.62
2 86.54
3 88.14
4 89.23
5 93.45
6 97.4
7 99.34
8 96.98
9 100.22
10 103.56
a. What type of pattern exists in the data based on the time series plot.? b. Use the excel output and find the forecast for t = 11?
0
20
40
60
80
100
120
0 2 4 6 8 10 12
IP
Year (t)
4
Excel output:
Answer:
a. The time series plot shows a linear trend. From the above output, the regression estimates for the y-intercept and slope that minimize MSE for this time series are b0 = 80.458 and b1 = 2.36, which result in the following forecasts, errors: b. �̂�11= b0 + b1t = 80.458 + 2.36(11) = 106.438.
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4. The following times series shows the demand for a particular product over the past 10 months.
Month Value
1 324
2 311
3 303
4 314
5 323
6 313
7 302
8 315
9 312
10 326
a. The exponential smoothing values for the time series has been computed using α = 0.2. Compute MSE
and a forecast for month 11. b. The three-month moving average forecast for the time series has been. Compute MSE and a forecast for
month 11. b. Compare the three-month moving average forecast with the exponential smoothing forecast using α = 0.2. Which appears to provide the better forecast based on MSE?
a. Smoothing constant α = 0.2
α 0.2
Month Value Forecast Forecast Error Squared Forecast Error
1 324
2 311 324.00 -13.00 169.00
3 303 321.40 -18.40 338.56
4 314 317.72 -3.72 13.84
5 323 316.98 6.02 36.29
6 313 318.18 -5.18 26.84
7 302 317.14 -15.14 229.36
8 315 314.12 0.88 0.78
9 312 314.29 -2.29 5.26
10 326 313.83 12.17 148.01
Total = 967.94
MSE = 967.94/9 = 107.55 The forecast for month 11 is �̂�11 = αy10 + (1- α) �̂�10 = 0.2(326) + (1 - 0.2)313.83 = 316.27.
b.
Month Demand Forecast Forecast
Error Squared Forecast
Error
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1 324
2 311
3 303
4 314 312.67 1.33 1.78
5 323 309.33 13.67 186.78
6 313 313.33 -0.33 0.11
7 302 316.67 -14.67 215.11
8 315 312.67 2.33 5.44
9 312 310.00 2.00 4.00
10 326 309.67 16.33 266.78
Total 680.00
MSE= 680/7= 97.14 The forecast for month 11 is �̂�11 = (315 + 312 + 326)/3= 317.66
c. Comparing the MSE for three-month moving average (calculated in the previous problem) and the MSE for exponential smoothing, the three-month moving average provides a better forecast as it has a smaller MSE.
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Chapter 8:
5. A garden shop wishes to prepare a supply of special fertilizer at a minimal cost by mixing two fertilizers,
A and B.
The mixture is to contain:
At least 45 units of phosphate,
At least 36 units of nitrate,
At least 40 units of ammonium
Fertilizer A costs the shop $0.97 per pound and contains 5 units of phosphate, 2 units of nitrate and 2
units of ammonium
Fertilizer B costs the shop $1.89 per pound and contains 3 units of phosphate, 3 units of nitrate and 5
units of ammonium.
How many pounds of each fertilizer should the shop use in order to minimize their cost? Use LP to solve
6. Hire-a-Car System rents three types of cars at two different locations. The profit made per day
for each car type and company at the two locations is listed below:
Car Type
Company Type 1 Type 2 Type 3
A $25 $40 $10
B $30 $35 $45
The management forecasts the demand per day by car type. The demand forecast for a particular day is
125 rentals for Type 1 cars, 55 rentals for Type 2 cars, and 40 rentals for Type 3 cars. The company has
100 cars in location A and 120 cars in location B. Formulate a linear programming model to maximize
profit to determine how many reservations to accept for each type of car.
Answer:
Let A1 = number of reservations made for Type 1 car of Company A
A2 = number of reservations made for Type 2 car of Company A
A3 = number of reservations made for Type 3 car of Company A
B1 = number of reservations made for Type 1 car of Company B
B2 = number of reservations made for Type 2 car of Company B
B3 = number of reservations made for Type 3 car of Company B
Max 25A1 + 40A2 + 10A3 + 30B1 + 35B2 + 45B3
s.t.
A1 + A2 + A3 ≤ 100
B1 + B2 + B3 ≤ 120
A1 + B1 ≤ 125
A2 + B2 ≤ 55
A3 + B3 ≤ 40
A1, A2, …, B3 ≥ 0
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7 .A manager of a quality testing team wanted to test different lots of products using three resources,
R1, R2, and R3. Each lot can be tested for quality using any one of the three procedures, P1, P2, or P3.
The product once tested will be sent for packaging. The profit contribution per lot for each of these
procedures varies and they are $4, $5, and $8, respectively. Also, resource R1 requires 2 hours, 3 hours,
and 4 hours to test a lot using the procedure P1, P2, and P3, respectively. Resource R2 requires 3 hours,
2 hours, and 3 hours using the procedure P1, P2, and P3, respectively. Lastly, resource R3 requires 2
hours, 3 hours, and 4 hours using the procedure P1, P2, and P3, respectively. The available times for
these three resources are 80 hours, 90 hours, and 65 hours, respectively. Formulate and solve a linear
program and solve for the optimal solution for the above scenario by maximizing the profit.
a. What will be the change in total profit, if machine M3 is given an extra hour of production time? Answer: Let x11: number of lots tested by R1 using procedure P1
x12: number of lots tested by R1 using procedure P2
.
.
.
x33: number of lots tested by R3 using procedure P3