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
Registrations
Forecast
1
22
2
21
22
3
25
21
4
27
25
5
35
27
6
29
35
7
33
29
8
37
33
9
41
37
10
37
41
11
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 =
 previous n registrations
n
A1  A2 22  21
=
= 21.5. Carrying this down the table through to week 11 gives:
2
2
Week
Registrations
Forecast
1
22
2
21
3
25
21.5
4
27
23
5
35
26
6
29
31
7
33
32
8
37
31
9
41
35
10
37
39
11
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 =
 previous n registrations
n
A  A2  A3  A4 22  21  25  27
ie F5 = 1
=
= 23.75. Carrying this down the table through to
4
4
week 11 gives:
Week
Registrations
Forecast
1
22
2
21
3
25
4
27
5
35
23.75
6
29
27
7
33
29
8
37
31
9
41
33.5
10
37
35
11
37
(3 marks)
Registrations
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?
45
40
35
30
25
20
15
10
5
0
Registration
Naïve
2-Week
Moving
4-Week
Moving
1
3
5
7
9
11
Week
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
1
7
2
9
3
5
4
9
5
13
6
8
7
(10 marks)
Exponentially Smoothed Forecast
5
5 + 0.2 (7 – 5) = 5.4
5.4 + 0.2 (9 – 5.4) = 6.12
6.12 + 0.2 (5 – 6.12) = 5.90
5.90 + 0.2 (9 – 5.90) = 6.52
6.52 + 0.2 (13 – 6.52) = 7.82
7.82 + 0.2 (8 – 7.82) = 7.86
4.7 Calculate (a) MAD and (b) MSE for the following forecast versus actual sales figures:
a) MAD =
 forecast errors
n
Actual
95
108
123
130
Forecast
100
110
120
130
Error
-5
-2
3
0
| Error |
5
2
3
0
 = 10
Error
-5
-2
3
0
Error2
25
4
9
0
 = 38
MAD = 10 / 4 = 2.5 (5 marks)
 forecast errors 
2
b) MSE =
n
Actual
95
108
123
130
Forecast
100
110
120
130
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
Feb
Mar
Apr
May
Jun
Jul
Aug
Inc
70.0
68.5
64.8
71.7
71.3
72.8
(10 marks)
Forecast
65.0
0.1(70)+0.9(65) = 65.5
0.1(68.5)+0.9(65.6) = 65.89
0.1(64.8)+0.9(66.05) = 65.93
0.1(71.7)+0.9(66.06) = 66.62
0.1(71.3)+0.9(66.87) = 67.31
0.1(72.8)+0.9(67.64) = 68.16
Trend
0
0.2(65.5–65)+(0.8)0 = 0.1
0.2(65.89-65.5)+(0.8)0.1 = 0.16
0.2(65.92-65.89)+(0.8)0.16 = 0.13
0.2(66.62-65.93)+(0.8)0.13 = 0.25
0.2(67.31-66.62)+(0.8)0.25 = 0.33
0.2(68.16-67.31)+(0.8)0.33 = 0.43
FIT
65+0 = 65
65.5+0.1=65.6
65.89+0.16=66.05
65.93+0.13=66.06
66.62+0.25=66.87
67.31+0.33=67.64
68.16+0.43=68.60
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
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Week 1
210
178
250
215
160
180
Week 2
215
180
250
213
165
185
Week 3
220
176
260
220
175
190
Week 4
225
178
260
225
176
190
Avg
217.5
178
255
218.3
169
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
Seasonal Index for Tuesday
Seasonal Index for Wednesday
Seasonal Index for Thursday
Seasonal Index for Friday
Seasonal Index for Saturday
= 217.5 / 204
= 178 / 204
= 255 / 204
= 218.3 / 204
= 169 / 204
= 186.3 / 204
= 1.066
= 0.873
= 1.25
= 1.07
= 0.828
= 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
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
(5 marks)
x
25
26
27
28
29
30
y
201.74 + 0.18(25) = 206.24
201.74 + 0.18(26) = 206.42
201.74 + 0.18(27) = 206.6
201.74 + 0.18(28) = 206.78
201.74 + 0.18(29) = 206.96
201.74 + 0.18(30) = 207.14
Seasonal Index
1.066
0.873
1.25
1.07
0.828
0.913
Forecast (Rounded)
206.24 (1.066) = 220
206.42 (0.873) = 180
206.6 (1.25) = 258
206.78 (1.07) = 221
206.96 (0.828) = 171
207.14 (0.913) = 189
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.
5
Ridership
4
3
2
1
0
0
5
10
15
20
25
Tourists
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
1
2
3
4
5
6
7
8
9
10
11
12
Tourists (x)
7
2
6
4
14
15
16
12
14
20
15
7
x = 132
x = x / n = 132 / 12 = 11
y = y / n = 27.1 / 12 = 2.26
Ridership (y)
x2
1.5
49
1.0
4
1.3
36
1.5
16
2.5
196
2.7
225
2.4
256
2.0
144
2.7
196
4.4
400
3.4
225
1.7
49
2
y = 27.1
x = 1796
y2
2.25
1.00
1.69
2.25
6.25
7.29
5.76
4.00
7.29
19.36
11.56
2.89
2
y = 71.59
xy
10.5
2.0
7.8
6.0
35.0
40.5
38.4
24.0
37.8
88.0
51.0
11.9
xy = 352.9
b
 xy - nxy  352.9  12(11)(2.26)  54.58  0.159
1796  12(11)
344
 x  nx
2
2
2
a  y - bx  2.26 - 0.159(11)  0.511
 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?
Sy,x 
y
2
 a  y - b xy
n2

71.59  0.511(27.1)  0.159(352.9)
1.63

 0.163  0.404
12  2
10
(2 marks)
f) What is the model’s correlation coefficient and coefficient of determination?
r

n x
n xy   x y
2

  x  n y 2   y 
2
2

12(352.9)  (132)( 27.1)
 12(1796)  (132) 12(71.59)  (27.1) 
2
2
657.6
657.6

 0.917
514637.76 717.38
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.
Grow 150
Build Large
Stable -85
Grow 60
Build Small
Stable -45
Grow 0
Do Not Build
Stable 0
b) Construct a decision table.
Grow
150
60
0
Build Large Wing
Build Small Wing
Do Not Build
Stable
-85
-45
0
c) Assuming that each state of nature has the same likelihood, determine the best alternative.
Build Large Wing
Build Small Wing
Do Not Build
Grow
150
60
0
P
0.5
0.5
0.5
Stable
-85
-45
0
P
0.5
0.5
0.5
EMV
150(0.5) – 85(0.5) = $32.5
60(0.5) – 45(0.5) = $7.5
$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?
Build Large Wing
Build Small Wing
Do Not Build
Grow
150
60
0
P
0.6
0.6
0.6
Stable
-85
-45
0
P
0.4
0.4
0.4
EMV
150(0.6) – 85(0.4) = $56
60(0.6) – 45(0.4) = $18
$0
We would make the decision which results in the largest EMV – in this case, we would build the
large wing.
5.13 The product planning group of Hawkes Electrical Supplies, Inc., has determined that it
needs to design a new series of switches. It must decide upon one of three design strategies. The
market forecast is for 200,000 units. The better and more sophisticated the design strategy, and
the more time spent on value engineering, the less will be the variable cost. The chief of
engineering design, Dr. Gerry Johnson, has decided that the following costs are a good estimate
of the initial and variable costs connected with each approach. These are:
a) Low-tech: a low-technology, low-cost process consisting of hiring several new junior
engineers. This option has a cost of $45,000, and variable cost probabilities of .2 for $55 each,
.5 for $.50, and .1 for $.45.
b) Subcontract: a medium-cost approach using a good outside design staff. This approach would
have an initial cost of $65,000 and variable cost probabilities of .7 of $.45, .2 of $.40 and .1 of
$.35.
c) High-tech: a high-technology approach using the very best of the inside staff and the latest
computer-aided design technology. This approach has a fixed cost of $75,000 and variable cost
probabilities of .9 of $.40 and .1 of $.35.
What is the best decision based on an expected monetary value (EMV) criterion? (Note: we want
the lowest EMV as we are dealing with costs in this problem.)
Construct a decision tree:
Probability
Low Tech
Subcontract
High Tech
(0.2)
(0.5)
(0.3)
(0.7)
(0.2)
(0.1)
(0.9)
(0.1)
Fixed
Costs
$45,000 +
$45,000 +
$45,000 +
$65,000 +
$65,000 +
$65,000 +
$75,000 +
$75,000 +
Variable Costs
(200,000 x $0.55) =
(200,000 x $0.50) =
(200,000 x $0.45) =
(200,000 x $0.45) =
(200,000 x $0.40) =
(200,000 x $0.35) =
(200,000 x $0.40) =
(200,000 x $0.35) =
Total
Costs
$155,000
$145,000
$135,000
$155,000
$145,000
$135,000
$155,000
$145,000
Multiplying the total cost by the probability for each branch, we find the following EMV’s:
- Low-tech: 0.2 x $155,000 + 0.5 x $145,000 + 0.3 x $135,000 = $144,000
- Subcontract: 0.7 x $155,000 + 0.2 x $145,000 + 0.1 x $135,000 = $151,000
- High-tech: 0.9 x $155,000 + 0.1 x $145,000
= $154,000
Since we want the choice with the lowest EMV (ie the lowest potential costs to us) we would
choose the low-tech option.