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Production And Operaions Management

Content type
User Generated
Subject
Management
Type
Homework
Showing Page:
1/5
Problem 1 Solution:
Moving average = Σ demand in previous n periods
n
Week
Auto Sales
1
8
2
10
3
9
4
11
(9 + 8 + 10) ÷ 3 = 9
5
10
(9 + 10 + 11) ÷3 = 10
6
13
(11 +9 +10) ÷ 3 = 10
7
-
(10 + 11 + 13) ÷3 = 11 1/3
Problem 2 Solution:
Weighted moving average= Σ (Weight for period n) (demand in period n)
Σ Weights
Week
Auto Sales
1
8
2
10
3
9
4
11
[ (2×10) + (1×8) + (3×9)] ÷ 6 = 9. 17
5
10
[(3×11) + (1×10) + (2×9)] ÷6 = 10 .17
6
13
[(3×10) + (1×9) + (2×11)] ÷6 = 10 .17
7
-
[(3×13)+ (1×11) + (2×10)] ÷6 = 11 .7

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Problem 3 Solution:
F
t
= F
t-1
+ α (A
t-1
- F
t-1
)
500 + 0.1 (450 500)
= 495 units
Problem 4 Solution:
Based on the above analysis, smoothing constant of α = 0.8 is ideal as compared to that of α =
0.5 since its MAD is smaller.
Actual Battery
Sales
Forecast
Rounded
with α =0.8
Deviation
Absolute
with α =0.8
Forecast
Rounded
with α =0.5
Deviation
Absolute
with α =0.5
20
22
2.0
22
2.0
21
20.4
0.6
21
0.0
15
20.9
5.9
21
0.0
14
16.2
2.2
18
4.0
13
14.4
1.4
16
3.0
16
13.3
2.7
14.5
1.5
Σ = 14.8
Σ = 16.5
MAD = Σ deviations
n
2.46
2.75
MSE
8.84
11.21

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Problem 1 Solution: Moving average = Σ demand in previous n periods n Week Auto Sales 1 8 2 10 3 9 4 11 (9 + 8 + 10) ÷ 3 = 9 5 10 (9 + 10 + 11) ÷3 = 10 6 13 (11 +9 +10) ÷ 3 = 10 7 - (10 + 11 + 13) ÷3 = 11 1/3 Problem 2 Solution: Weighted moving average= Σ (Weight for period n) (demand in period n) Σ Weights Week Auto Sales 1 8 2 10 3 9 4 11 [ (2×10) + (1×8) + (3×9)] ÷ 6 = 9. 17 5 10 [(3×11) + (1×10) + (2×9)] ÷6 = 10 .17 6 13 [(3×10) + (1×9) + (2×11)] ÷6 = 10 .17 7 - [(3×13)+ (1×11) + (2×10)] ÷6 = 11 .7 Problem 3 Solution: Ft = Ft-1 + α (At-1- Ft-1) 500 + 0.1 (450 – 500) = 495 units Problem 4 Solution: Month Actual Battery Sales Forecast Rounded with α =0.8 Deviation Absolute with α =0.8 Forecast Rounded with α =0.5 Deviation Absolute with α =0.5 January 20 22 2.0 22 2.0 February 21 20.4 0.6 21 0.0 March 15 20.9 5.9 21 0.0 April 14 16.2 2.2 18 4.0 May 13 14.4 1.4 16 3.0 June 16 13.3 2.7 14.5 1.5 Σ = 14.8 Σ = 16.5 MAD = Σ deviations n 2.46 2.75 MSE 8.84 11.21 Based on the above analysis, smoothing constant of α = 0.8 is ideal as compared to that of α = 0.5 since its MAD is smaller. Problem 5 Solution Year Time period (X) Sales (Units) (Y) X2 XY 1996 1 100 1 100 1997 2 110 4 220 1998 3 122 9 336 1999 4 130 16 520 2000 5 139 25 695 2001 6 152 36 912 2002 7 164 49 1148 ΣX=28 ΣY=917 ΣX2=140 ΣXY=3961 ẍ = ...
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