MGT 2070 Assignment

MGT 2070 Assignment #2 – Solutions

4.1 Registration numbers for an accounting seminar over the past 10 weeks are shown below:

a) Starting with week 2 and ending with week 11, forecast registrations using the naïve forecasting method.

Naïve Forecast Ft = At-1

ie F2 = A1 = 22. Carrying this down the table through to week 11 gives:

Week 1 2 3 4 5 6 7 8 9 10 11
Registrations 22 21 25 27 35 29 33 37 41 37
Forecast 22 21 25 27 35 29 33 37 41 37

(3 marks)

b) Starting with week 3 and ending with week 11, forecast registration using a two-week moving average.

Moving Average Forecast Ft = ie F3 = = = 21.5. Carrying this down the table through to week 11 gives:

Week 1 2 3 4 5 6 7 8 9 10 11
Registrations 22 21 25 27 35 29 33 37 41 37
Forecast 21.5 23 26 31 32 31 35 39 39

(3 marks)

c) Starting with week 5 and ending with week 11, forecast registrations using a four-week moving average.

Moving Average Forecast Ft = ie F5 = = = 23.75. Carrying this down the table through to week 11 gives:

Week 1 2 3 4 5 6 7 8 9 10 11
Registrations 22 21 25 27 35 29 33 37 41 37
Forecast 23.75 27 29 31 33.5 35 37

(3 marks)

d) Plot the original data and the three forecasts on the same graph. Which forecast smoothes the data the most? Which forecast responds to change the best?

We see from the graph that the four-week moving average forecast smoothes the data the most, while the naïve forecast responds to change the best. (1 mark)

4.5 Given the following data, use exponential smoothing ( = 0.2) to develop a demand forecast. Assume the forecast for the initial period is 5.

Exponential Smoothing Forecast Ft = Ft-1 + (At-1 – Ft-1)

ie F2 = F1 + (A1 – F1) = 5 + 0.2 (7 – 5) = 5.4. Carrying this through to week 7 gives:

Period Demand Exponentially Smoothed Forecast
1 7 5
2 9 5 + 0.2 (7 – 5) = 5.4
3 5 5.4 + 0.2 (9 – 5.4) = 6.12
4 9 6.12 + 0.2 (5 – 6.12) = 5.90
5 13 5.90 + 0.2 (9 – 5.90) = 6.52
6 8 6.52 + 0.2 (13 – 6.52) = 7.82
7 7.82 + 0.2 (8 – 7.82) = 7.86
(10 marks)
4.7 Calculate (a) MAD and (b) MSE for the following forecast versus actual sales figures:

a) MAD =

Actual Forecast Error | Error |
95 100 -5 5
108 110 -2 2
123 120 3 3
130 130 0 0  = 10

MAD = 10 / 4 = 2.5 (5 marks)

b) MSE =

Actual Forecast Error Error2
95 100 -5 25
108 110 -2 4
123 120 3 9
130 130 0 0  = 38

MSE = 38 / 4 = 9.5 (5 marks)

4.19 Consulting income at Dr. Thomas W. Jones Associates for the period February to July has been as follows. Use trend-adjusted exponential smoothing to forecast August’s income. Assume that the initial forecast for February is $65,000 and the initial trend adjustment is 0. The smoothing constants selected are  = .1 and  = .2.

Forecast Ft = (At-1) + (1 - )(Ft-1 + Tt-1), Trend Tt = (Ft – Ft-1) + (1 - )T t-1

Month Inc Forecast Trend FIT
Feb 70.0 65.0 0 65+0 = 65
Mar 68.5 0.1(70)+0.9(65) = 65.5 0.2(65.5–65)+(0.8)0 = 0.1 65.5+0.1=65.6
Apr 64.8 0.1(68.5)+0.9(65.6) = 65.89 0.2(65.89-65.5)+(0.8)0.1 = 0.16 65.89+0.16=66.05
May 71.7 0.1(64.8)+0.9(66.05) = 65.93 0.2(65.92-65.89)+(0.8)0.16 = 0.13 65.93+0.13=66.06
Jun 71.3 0.1(71.7)+0.9(66.06) = 66.62 0.2(66.62-65.93)+(0.8)0.13 = 0.25 66.62+0.25=66.87
Jul 72.8 0.1(71.3)+0.9(66.87) = 67.31 0.2(67.31-66.62)+(0.8)0.25 = 0.33 67.31+0.33=67.64
Aug 0.1(72.8)+0.9(67.64) = 68.16 0.2(68.16-67.31)+(0.8)0.33 = 0.43 68.16+0.43=68.60

(10 marks)

4.29 The director of the Riley County, Kansas, library system would like to forecast evening patron usage for next week. Below are the data for the past 4 weeks:

a) Calculate a seasonal index for each day of the week.

Day Week 1 Week 2 Week 3 Week 4 Avg
Monday 210 215 220 225 217.5
Tuesday 178 180 176 178 178
Wednesday 250 250 260 260 255
Thursday 215 213 220 225 218.3
Friday 160 165 175 176 169
Saturday 180 185 190 190 186.3  = 1224.1

Average Daily Demand =  Average Demand / 6 Days = 1224.1 / 6 = 204

Seasonal Index = Average Demand / Average Daily Demand

Seasonal Index for Monday = 217.5 / 204 = 1.066
Seasonal Index for Tuesday = 178 / 204 = 0.873
Seasonal Index for Wednesday = 255 / 204 = 1.25
Seasonal Index for Thursday = 218.3 / 204 = 1.07
Seasonal Index for Friday = 169 / 204 = 0.828
Seasonal Index for Saturday = 186.3 / 204 = 0.913

(5 marks)

b) If the trend equation for this problem is y = 201.74 + 0.18x, what is the forecast for each day of week 5? Round your forecast to the nearest whole number.

Note that each day is one period along the x-axis. So in week five, Monday is x = 25, Tuesday is x = 26, etc.

Day x y Seasonal Index Forecast (Rounded)
Monday 25 201.74 + 0.18(25) = 206.24 1.066 206.24 (1.066) = 220
Tuesday 26 201.74 + 0.18(26) = 206.42 0.873 206.42 (0.873) = 180
Wednesday 27 201.74 + 0.18(27) = 206.6 1.25 206.6 (1.25) = 258
Thursday 28 201.74 + 0.18(28) = 206.78 1.07 206.78 (1.07) = 221
Friday 29 201.74 + 0.18(29) = 206.96 0.828 206.96 (0.828) = 171
Saturday 30 201.74 + 0.18(30) = 207.14 0.913 207.14 (0.913) = 189

(5 marks)

4.41 Summer-month bus and subway ridership in Washington, DC, is believed to be tied heavily to the number of tourists visiting the city. During the past 12 years, the following data have been obtained:

a) Plot these data and decide if a linear model is reasonable.

As the number of tourists (x-axis) increases, the ridership appears to increase: a linear model is reasonable. (1 mark)

b) Develop a regression relationship.

Year Tourists (x) Ridership (y) x2 y2 xy
1 7 1.5 49 2.25 10.5
2 2 1.0 4 1.00 2.0
3 6 1.3 36 1.69 7.8
4 4 1.5 16 2.25 6.0
5 14 2.5 196 6.25 35.0
6 15 2.7 225 7.29 40.5
7 16 2.4 256 5.76 38.4
8 12 2.0 144 4.00 24.0
9 14 2.7 196 7.29 37.8
10 20 4.4 400 19.36 88.0
11 15 3.4 225 11.56 51.0
12 7 1.7 49 2.89 11.9 x = 132 y = 27.1 x2 = 1796 y2 = 71.59 xy = 352.9

= x / n = 132 / 12 = 11

= y / n = 27.1 / 12 = 2.26

 the relationship is y = 0.511 + 0.159x (2 marks)

c) What is expected ridership if 10 million tourists visit the city in a year?

Y = 0.511 + 0.159(10) = 2.101 or 2,101,000 persons. (2 marks)

d) Explain the predicted ridership if there are no tourists at all.

If there are no tourists at all, the model predicts a ridership of 0.511 or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists is outside the range of data used to develop the model. (1 mark)

e) What is the standard error of the estimate?

(2 marks)

f) What is the model’s correlation coefficient and coefficient of determination?

r2 = 0.9172 = 0.840

(2 marks)

A.17 Chris Suit is administrator for Lowell Hospital. She is trying to determine whether to build a large wing on the existing hospital, a small wing, or no wing at all. If the population of Lowell continues to grow, a large wing could return $150,000 to the hospital each year. If a small wing were built, it would return $60,000 to the hospital each year of the population continues to grow. If the population of Lowell remains the same, the hospital would encounter a loss of $85,000 with a large wing and a loss of $45,000 with a small wing. Unfortunately, Suit does not have any information about the future population of Lowell.

a) Construct a decision tree.

b) Construct a decision table.

Grow Stable
Build Large Wing 150 -85
Build Small Wing 60 -45
Do Not Build 0 0

c) Assuming that each state of nature has the same likelihood, determine the best alternative.

Grow P Stable P EMV
Build Large Wing 150 0.5 -85 0.5 150(0.5) – 85(0.5) = $32.5
Build Small Wing 60 0.5 -45 0.5 60(0.5) – 45(0.5) = $7.5
Do Not Build 0 0.5 0 0.5 $0

We would make the decision which results in the largest EMV – in this case, we would build the large wing.

d) If the likelihood of growth is 0.6 and that of remaining the same is 0.4 and the decision criterion is expected monetary value, which decision should Suit make?

Grow P Stable P EMV
Build Large Wing 150 0.6 -85 0.4 150(0.6) – 85(0.4) = $56
Build Small Wing 60 0.6 -45 0.4 60(0.6) – 45(0.4) = $18
Do Not Build 0 0.6 0 0.4 $0

We would make the decision which results in the largest EMV – in this case, we would build the large wing.