Mgt 2070 Assignment 2 – Solutions
4.13 As you can see in the following table, demand for heart transplant surgery at Washington
General Hospital has increased steadily in the past few years:
Year
Heart Transplants
1
45
2
50
3
52
4
56
5
58
6
?
The director of medical services predicted 6 years ago that demand in year 1 would be 41
surgeries.
a)
Use exponential smoothing, first with a smoothing constant of .6 and then with one of .9,
to develop forecasts for years 2 through 6.
Exponential Smoothing α = 0.6
Year
1
2
3
4
5
6
A
45
50
52
56
58
?
Ft = Ft-1 + (At-1 – Ft-1)
41 (given in question)
41.0 + 0.6(45 - 41)
43.4 + 0.6(50 – 43.4)
47.4 + 0.6(52 – 47.4)
50.2 + 0.6(56 – 50.2)
53.7 + 0.6(58 – 53.7)
Ft
41
43.4
47.4
50.2
53.7
56.3
| Error |
4.0
6.6
4.6
5.8
4.3
Σ = 25.3
MAD = 25.3 / 5 = 5.06
(5 marks)
Exponential Smoothing α = 0.9
Year
1
2
3
4
5
6
A
45
50
52
56
58
?
Ft = Ft-1 + (At-1 – Ft-1)
41 (given in question)
41.0 + 0.9(45 - 41)
44.6 + 0.9(50 – 44.6)
49.5 + 0.9(52 – 49.5)
51.8 + 0.9(56 – 51.8)
55.6 + 0.9(58 – 55.6)
Ft
41
44.6
49.5
51.8
55.6
57.8
| Error |
4.0
5.4
2.5
4.2
2.4
Σ = 18.5
MAD = 18.5 / 5 = 3.7
(5 marks)
b)
Use a 3-year moving average to forecast demand in years 4, 5, and 6.
3-Year Moving Average
Year
1
2
3
4
5
6
A
45
50
52
56
58
?
3-Year Moving Average
| Error |
(45 + 50 + 52) / 3 = 49
(50 + 52 + 56) / 3 = 52.7
(52 + 56 + 58) / 3 = 55.3
7
5.3
Σ = 12.3
MAD = 12.3 / 2 = 6.2
(5 marks)
c)
Use the trend-projection method to forecast demand in years 1 through 6.
Year (X)
1
2
3
4
5
Σ = 15
x
b
Demand (Y)
45
50
52
56
58
Σ = 261
 x  15 / 5 = 3
n
 xy - nxy =
 x  nx
2
2
y
 y  261 / 5 = 52.2
n
815  5(3)(52.2)
= 3.2
55  5(9)
a  y - bx = 52.2 – 3.2(3) = 42.6
Our equation is then Y = a + bX, ie Y = 42.6 + 3.2X
X2
1
4
9
16
25
Σ = 55
XY
45
100
156
224
290
Σ = 815
Year
(X)
1
2
3
4
5
6
Demand
Trend Projection
Forecast(Y)
| Error |
45
50
52
56
58
?
Y = 42.6 + 3.2(1)
Y = 42.6 + 3.2(2)
Y = 42.6 + 3.2(3)
Y = 42.6 + 3.2(4)
Y = 42.6 + 3.2(5)
Y = 42.6 + 3.2(6)
45.8
49.0
52.2
55.4
58.6
61.8
0.8
1.0
0.2
0.6
0.6
Σ = 3.2
MAD = 3.2 / 5 = 0.64
(5 marks)
4.14 Refer to problem 4.13. With MAD as the criterion, which of the four forecasting methods
is best?
Forecast Methodology
Exponential Smoothing, α = 0.6
Exponential Smoothing, α = 0.9
3-year Moving Average
Trend Projection
MAD
5.06
3.7
6.2
0.6
Based on the Mean Absolute Deviation criterion, trend projection is the best fit.
(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
Mo
Inc
Forecast
Trend
Feb
Mar
Apr
May
Jun
Jul
Aug
70.0
68.5
64.8
71.7
71.3
72.8
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
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
MSE = 113.2 / 6 = 18.87
(5 marks)
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
Σ=
Err2
25.0
8.4
1.6
31.9
19.7
26.6
113.2
4.20 Resolve problem 4.19 with  = .1 and  = .8. Using MSE, which smoothing constants
provide a better forecast?
Mo
Inc
Forecast
Trend
Feb
Mar
Apr
May
Jun
Jul
70.0
68.5
64.8
71.7
71.3
72.8
65.0
0.1(70)+0.9(65) = 65.5
0.1(68.5)+0.9(65.9) = 66.16
0.1(64.8)+0.9(66.77) = 66.57
0.1(71.7)+0.9(67.02) = 67.49
0.1(71.3)+0.9(68.31) = 68.61
0
0.8(65.5–65)+(0.2)0 = 0.4
0.8(66.16-65.5)+(0.2)0.4 = 0.61
0.8(66.57-66.16)+(0.2)0.61 = 0.45
0.8(67.49-66.57)+(0.2)0.45 = 0.82
0.8(68.61-67.49)+(0.2)0.82 = 1.06
Err2
FIT
65+0 = 65
65.5+0.4=65.9
66.16+0.61=66.77
66.57+0.45=67.02
67.49+0.82=68.31
68.61+1.06=69.68
Σ=
25.0
6.76
3.87
21.89
8.91
9.76
76.19
MSE = 76.19 / 6 = 12.70
Based upon the MSE criterion, exponential smoothing with  = .1 and  = .8 provides a better
forecast.
(5 marks)
4.37 The accountant at Tick Wing Coal Distributors, Inc., in San Francisco notes that the
demand for coal seems to be tied to an index of weather severity developed by the U.S. Weather
Bureau. When weather was extremely cold in the U.S. over the past five years (and the index was
thus high), coal sales were high. The accountant proposes that one good forecast of next year’s
coal demand could be made by developing a regression equation and then consulting the
Farmer’s Almanac to see how severe next year’s winter would be.
For the data in the following table, derive a least squares regression and compute the coefficient
of correlation of the data. Also compute the standard error of the estimate.
Coal Sales, y
(in millions of tons)
Weather Index, x
X
2
1
4
5
3
Σ = 15
1
4
6
5
2
1
4
5
3
X2
4
1
16
25
9
Σ = 55
Y
4
1
4
6
5
Σ = 20
x
 x  15 / 5 = 3
b
 xy - nxy =
 x  nx
n
2
4
2
y
 y  20 / 5 = 4
n
70  5(3)( 4)
= 1.0
55  5(9)
Y2
16
1
16
36
25
Σ = 94
XY
8
1
16
30
15
Σ = 70
a  y - bx = 4 – 1(3) = 1.0
Our equation is then Y = a + bX, ie Y = 1.0 + 1.0X
Correlation Coefficient:
n  xy   x  y
=
r
2
2
2
2
n  x   x  n  y   y 

r=

5(70)  (15)( 20)
 5(55)  15 5(94)  20 
350  300
=
275  225470  400
2
50
50
=
= 0.845
50  70 59.16
Standard Error of the Estimate:
Sy,x 
y
(10 marks)
2
 a  y - b xy
n2
=
94 1(20)  1(70)
= 1.333 = 1.15
52
2