1-2. Short Questions and Multiple Choices
1. Given an actual demand of 60 for a period when forecast of 70
was anticipated, and an alpha of 0.3, what would the forecast for
the next period be using simple exponential smoothing?
F = (1-0.3)(70)+0.3(60) = 67
2. Suppose you have been asked to generate a demand forecast for
a product for year 2012 using an exponential smoothing method.
The forecast demand in 2011 was 910. The actual demand in 2011
was 850. Using this data and a smoothing constant of 0.3, which of
the following is the demand forecast for year 2012?
A) 850
B) 885
C) 892
D) 925
E) 930
F = (1-0.3)(910)+0.3(850) = 892
3. Short Questions and Multiple Choices
The president of State University wants to forecast student
enrollments for this academic year based on the following
historical data: 5 years ago ; 15,000, 4 years ago ; 16,000, 3 years
ago; 18,000, 2 years ago; 20,000, Last year; 21,000. What is the
forecast for this year using exponential smoothing with α = 0.4, if
the forecast for two years ago was 16,000?
t
1
2
3
4
5
At 15000
16000
18000
20000
21000
Ft
16000
Forecast for last year
F5 = (1-α)F4+ α(A4)
F5 = 0.6(16000)+0.4(20000)=17600
Forecast for this year
F6 = (1-α)F5+ α(A5)
F6 = 0.6(17600)+0.4(21000)=18960
17600
4.Short Questions and Multiple Choices
Given the following demand
Suppose the forecast for period 2 is equal to the actual for period
1. What is your forecast for period 4 using exponential smoothing
Period
Demand
and α=0.5?
1
300
2
500
A) 300
3
600
B) 400
C) 500
D) 550
E) none of the above
F = (1-α)F + α(A )
t+1
t
t
Ft+1 = (1-0.5)Ft+ 0.5(At)
Ft+1 = (1/2) Ft+ (1/2) (At)
Ft+1 = (Ft+At)/2
F3 = (F2+A2)/2
F3 = (300+500)/2= 400
F4 = (F3+A3)/2 = (400+600)/2 = 500
5-6. Short Questions and Multiple Choices
5. Use exponential smoothing to forecast this period’s demand if 
= 0.2, previous actual demand was 30, and previous forecast was
35.
A) 29
B) 31
C) 34
F = (1-0.2)(35)+0.2(30) = 34
D) 36
E) 37
6. Exponential smoothing is being used to forecast demand.
The previous forecast of 66 turned out to be 5 units larger
than actual demand. The next forecast is 65. Compute ?
F(t+1) = Ft +  (At-Ft)
65
66
65 = 66 +  (-5)
+5
-5
5=1
 = 0.2
7-8. Short Questions and Multiple Choices
7. For what value of α, exponential smoothing becomes naїve
method?
Ft+1 = (1-α)Ft+ α(At)
A) α =0
B) α =0.25
Ft+1 =
(At)
C) α =0.5
Ft+1 = (1-1)Ft+ 1(At)
D) α =0.75
E) α =1
8. For what value of α, exponential smoothing becomes a straight
line?
Ft+1 = (1-α)Ft+ α(At)
A) α =0
B) α =0.25
Ft+1 =
Ft
C) α =0.5
Ft+1 = (1-0)Ft+ 0(At)
D) α =0.75
E) α =1
9-10. Short Questions and Multiple Choices
9. A forecast based on the previous forecast plus a percentage of
the forecast error is:
A) a naive forecast
B) a simple moving average forecast
F = (1-α)Ft+ α(At)
C) a centered moving average forecast t+1
D) an exponentially smoothed forecast Ft+1 = Ft+ α(At-Ft )
E) an associative forecast
10. In exponential smoothing forecasting, using large values of the
smoothing coefficient  generates forecasts that are more:
A) accurate
B) responsive
C) random
Ft+1 = (1-α)Ft+ α(At)
D) stable
E) level
11-12. Short Questions and Multiple Choices
11. For what value of α the forecast for the next period is equal to
90% of the actual of this period.
A) 0.9
B) 0.1
C) 0.5
D) all of the above
E) we do not know
12. It can be mathematically proved that the Age of data in
exponential smoothing is 1/α . The larger the α, the larger the
number of periods in the moving average. True or false? Why?
False
α = 0.5  age of data is 2 periods.
α = 0.2  age of data is 5 periods.
α = 0.1  age of data is 10 periods.
Short Questions and Multiple Choices
13. Given a forecast using a 6 period moving average. What is the
average age of data?
The last piece (newest piece) of data is only 1 period old.
The first piece (oldest piece) of data in a 6 period moving average
is 6 periods old.
The average age of data is (1+6)/2 = 3.5
14. If the age of data in exponential smoothing is 1/ α, for what
value of α, exponential smoothing performs close to a six period
moving average?
The age of data in a 6 period moving average is 3.5.
The age of data in exponential smoothing is 1/ α.
1/ α = 3.5
α = 1/3.5
α = 0.29
Problem 1
Given the following demand data
Month
Feb Mar Apr May Jun
Demand
19
18
15
20
18
Jul
22
Aug
20
a) Graph the data.
b) What are your forecasts using 5-period moving average.
c) Forecast for September using Exponential smoothing. Alpha is
0.2 and forecast for march was 19.
d) Forecast for September using Naïve method
e) Compute MAD for Naïve Method and Exponential
Smoothing. Which one is preferred? Naïve Method and
Exponential Smoothing?
f) Forecast for September using Linear Regression
(a) Plot the Data
Sales (1000)
19
18
15
20
18
22
20
Monthly Sales
25
Sale (1000)
1
2
3
4
5
6
7
Month
Feb
Mar
Apr
May
Jun
Jul
Aug
20
15
Monthly Sales
10
5
0
0
2
4
Month
6
8
( b) Forecast for Sep Using 5 Period Moving Average
Feb
Mar
Apr
May
Jun
Jul
Aug
t
1
2
3
4
5
6
7
At
19
18
15
20
18
22
20
F8 =MA7= (A7+A6+A5+A4+A3)/5 = (20+22+18+20+15)/5
F8 =MA7= 19
(b) Forecast Using 5 Period Moving Average for All Periods
t
1
2
3
4
5
6
7
At
19
18
15
20
18
22
20
Moving Average
MAt
Ft
18
18.6
19
18
18.6
19
(c)Forecast for Sep Using Exponential Smoothing α =0.2 and F(Mar) = 19
t
1
2
3
4
5
6
7
At
19
18
15
20
18
22
20
F3 = (1-α)F2 + α A2
F3 = (0.8)19+ 0.2(18)
F3 = 18.8
March is period 2
(c) Forecast for Sep Using α =.2 and F(Mar) = 19
Using the same formula, we compute F4, F5, F6, F7, and finally F8
which is the demand for Sep.
1
2
3
4
5
6
7
At
19
18
15
20
18
22
20
Ft
19
18.80
18.04
18.43
18.35
19.08
19.26
(d) Forecast for Sep Using Naïve Method
F8 =A7
F(t +1) =At
F8 = 20
Forecast for all periods using Naïve Method
t
1
2
3
4
5
6
7
At
19
18
15
20
18
22
20
Ft
19
18
15
20
18
22
20
(e) Which Technique ?
When comparing several methods, we need to use the same
time horizon for all methods. We need to have actual as well as
forecasts for all methods for all periods of MAD computations
Here we have Actual for periods 1 to 7; that is 7 periods.
Regression can provide us with forecast for periods 1 to ∞
Five period moving average can only provide forecast for
periods 6 and 7; that is 2 periods
Therefore, to compare all these methods, we can compute MAD
only over 2 periods. But two period is not enough.
Naïve Method or Exponential Smoothing ?
Naïve method forecasts for periods 2 to 7; That is 6 periods
Exponential Smoothing for periods 2 to 7; That is 6 periods
We can compare NM and ES over 6 periods.
(e) Naïve Method or Exponential Smoothing ?
Period Actual Naïve Method Expo. Smoothing
2
18
19
19.00
3
15
18
18.80
4
20
15
18.04
5
18
20
18.43
6
22
18
18.35
7
20
22
19.08
NM
1
3
5
2
4
2
ES
1.00
3.80
1.96
0.43
3.65
0.92
2.83
1.96
Better
However, we need to keep all methods, because we need more
actual data. A MAD computed just 6 periods is not a reliable
measure.
It is better to have all methods for say 10-20 more periods, and
then identify the best method