Forecasting

IE 214: Operations ManagementKAMAL

Lecture3

EXERCISE 4.3

Solution:

a) There is no pattern of any kind

0 1 2 3 4 5 6 7 8 9 10 110

2

4

6

8

10

12

14

EXERCISE 4.3

Solution:

Year 1 2 3 4 5 6 7 8 9 10 11 Frcast

Demand 7 9 5 9 13  8 12 13  9 11  7

(b) 3-year moving 7  7.7  9 10 11 11 11.3 11 9

(c) 3-year weighted 6.4  7.8 11  9.6 10.9 12.2 10.5 10.6 8.4

EXERCISE 4.3

Solution:

d) The 3-year moving average method gives better results

EXERCISE 4.5

Solution: Week of Pints UsedAug 31 360Sept 7 389Sept 14 410Sept 21 381Sept 28 368Oct 5 374

a) (381+368+374)/3 = 374.33

b) (381*0.1)+(368*0.3)+(374*0.6) = 372.9

EXERCISE 4.5

Solution:

Week of Actual ForecastAug 31 360 360Sept 7 389 360Sept 14 410 365.8Sept 21 381 374.64Sept 28 368 375.91Oct 5 374 374.33Oct 12 374.264

c) Ft = F t-1 + α(At-1 – Ft-1) ; α = 0.2

EXERCISE 4.8

Given:

Days 1 2 3 4 5 6 7 8At 93 94 93 95 96 88 90 ?

EXERCISE 4.8

Solution:

Days 1 2 3 4 5 6 7 8At 93 94 93 95 96 88 90 ?Ft

a) Ft = (96+88+90)/3 = 91.3333

91.3333

EXERCISE 4.8

Days 1 2 3 4 5 6 7 8At 93 94 93 95 96 88 90 ?Ft

b) Ft = (88+90)/2 = 89

89

EXERCISE 4.8

Days 1 2 3 4 5 6 7At 93 94 93 95 96 88 90Ft 93.5 93.5 94 95.5 92|At-Ft| 0.5 1.5 2 7.5 2

c) MAD = ∑|At-Ft|/n = 13.5/5 = 2.7

EXERCISE 4.8

Days 1 2 3 4 5 6 7At 93 94 93 95 96 88 90Ft 93.5 93.5 94 95.5 92|At-Ft| 0.5 1.5 2 7.5 2(At-Ft)^2 0.25 2.25 4 56.25 4

d) MSE = ∑(At-Ft)^2/n = 66.75/5 = 13.35

EXERCISE 4.8

Days 1 2 3 4 5 6 7

At 93 94 93 95 96 88 90

Ft 93.5 93.5 94 95.5 92

|At-Ft| 0.5 1.5 2 7.5 2(At-Ft)^2 0.25 2.25 4 56.25 4|At-Ft|/At 0.00538 0.01579 0.02083 0.08523 0.02222

e) MAPE = (∑|At-Ft|/At)*100/n

= 0.149449*100/5 = 2.98897%

EXERCISE 4.19

Solution:Ft = F t-1 + α(At-1 – Ft-1)

F3 = 48 = 50 + α(42 - 50) → -8α = -2 → α = 0.25

F5 = F4 + α(A4 – F4) = 50 + 0.25*(46 - 50) = 49

HW

4.1

4.4

4.14

4.45

EXERCISE 4.20

Given:Month feb mar apr may jun julAt 70 68.5 64.8 71.7 71.3 72.8

α = 0.1 , ß = 0.2

EXERCISE 4.20

Solution:

Ft= α(At-1) + (1- α)(Ft-1 + Tt-1)Tt= ß(Ft - Ft-1) + (1- ß)*Tt-1

FIT = Ft + Tt

EXERCISE 4.20

feb mar apr may jun julAt 70 68.5 64.8 71.7 71.3 72.8Ft 65Tt 0

FITt 65

Ft= 0.1*(70) + (0.9)(65 + 0) = 65.5Tt= 0.2(65.5 - 65) + (0.8)*0 = 0.1FIT = 65.5 + 0.1 = 65.6

65.50.1

65.6

EXERCISE 4.20

feb mar apr may jun jul AugAt 70 68.5 64.8 71.7 71.3 72.8Ft 65 65.5 65.89 65.9232 66.62062 67.30988 68.16Tt 0 0.1 0.158 0.13304 0.245915 0.334585 0.4377

FITt 65 65.6 66.048 66.05624 66.86653 67.64446 68.5977

EXERCISE 4.21

feb mar apr may jun jul AugAt 70 68.5 64.8 71.7 71.3 72.8Ft 65 65.5 65.89 65.9232 66.62062 67.30988 68.16Tt 0 0.1 0.158 0.13304 0.245915 0.334585 0.4377

FITt 65 65.6 66.048 66.05624 66.86653 67.64446 68.5977Error^2 25 6.76 3.87 21.89 8.91 9.76

MSE= ∑(Error^2)/n = 12.7

EXERCISE 4.26

Given:Month x No. of accidents(y)Jan 1 30Feb 2 40Mar 3 60Apr 4 90

a= y – b x

Y = a + b x = 5 + (20)(5) = 105^

EXERCISE 4.26

Month x y xy x^2Jan 1 30 30 1

Feb 2 40 80 4

Mar 3 60 180 9Apr 4 90 360 16

sum 5 650 30

avg 2.5 55 7.5

Solution:

EXERCISE 4.26

Month x y xy x^2Jan 1 30 30 1

Feb 2 40 80 4

Mar 3 60 180 9Apr 4 90 360 16

sum 5 650 30

avg 2.5 55 7.5

Solution:

b= (650)-(4)(2.5)(55)/[(30)-(4*(2.5^2))] = 100/5 = 20

a= 55 – (20*2.5) = 5

Y = a + b x = 5 + (20)(5) = 105^

HW

4.38

4.48

EXERCISE 4.30

EXERCISE 4.30

Season 2009 2010Winter 350 300Spring 150 165Summer 300 285Fall 200 250

Given:

• 1000 radials produced each year.

• Production planned to be 1200 radials next year.

EXERCISE 4.30

Season 2009 2010 Avg soldAvg seasonaly sold

Seasonal Index

Winter 350 300 325 250 1.3Spring 150 165 157.5 250 0.63Summer 300 285 292.5 250 1.17Fall 200 250 225 250 0.9

Solution:

1200 / 4 = 300 radials planned to be sold each season next year

300 * Is = Forecast for next year’s season

EXERCISE 4.30

Season 2009 2010Avg. sold

Avg. seas. sold

Seasonal Index 2011

Winter 350 300 325 250 1.3 390Spring 150 165 157.5 250 0.63 189Summer 300 285 292.5 250 1.17 351Fall 200 250 225 250 0.9 270

Solution:

HW PROBLEMS

1. 4.27

2. 4.29