Ain Shams University

Faculty of Engineering

Credit Hours Engineering Program

Senior (MANF)

Industrial Engineering

Spring 2014

Assignment (1)

FORECASTING

1) Auto sales at Carmen’s Chevrolet are shown below. Develop a 3-week moving average to

forecast the auto sales at the 7th week

Week 1 2 3 4 5 6 7

Auto Sales 8 10 9 11 10 13 ?

forecast the auto sales at the 7th week by weighting the three weeks as follows:

Weights Applied Period

3 Last week

2 Two weeks ago

1 Three weeks ago

6 Total

Answer:

Using Weighted Moving Average:

𝐹𝑖+1 =∑(𝑊𝑒𝑖𝑔ℎ𝑡 𝑖𝑛 𝑝𝑒𝑟𝑖𝑜𝑑 𝑖) × (𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑝𝑒𝑟𝑖𝑜𝑑)

∑ 𝑊𝑒𝑖𝑔ℎ𝑡

week Auto Sales

3-Month Weighted Moving Average

1 08 2 10

3 09

4 11 =(3×09+2×10+1×08)/6 9.166667 09 1/6

5 10 =(3×11+2×09+1×10)/6 10.16667 10 1/6

6 13 =(3×10+2×11+1×09)/6 10.16667 10 1/6

7 ?? =(3×13+2×10+1×11)/6 11.66667 11 2/3

Forecasted the auto sales of the 7th week by weighting the three weeks is 11 2/3

y = 0.7714x + 7.4667R² = 0.70217

8

9

10

11

12

13

14

0 1 2 3 4 5 6 7

Au

to S

ales

Week

2) A firm uses simple exponential smoothing with 01. to forecast demand. The forecast for

the week of January 1 was 500 units whereas the actual demand turned out to be 450 units.

Calculate the demand forecast for the week of January 8.

Answer: Using Simple Exponential Smoothing Forecast: 𝐹𝑡+1 = 𝐹𝑡 + 𝛼(𝐴𝑡 – 𝐹𝑡)

New forecast = last period’s forecast + α (last period’s actual demand – last period’s forecast) 𝛼 = 0.1 𝐹𝑡 = 500 𝐴𝑡 = 450

𝐹𝑡+1 = 500 + 0.1(450 – 500) = 500 − 0.1 × 50 = 495 𝑢𝑛𝑖𝑡𝑠 3) Exponential smoothing is used to forecast automobile battery sales. Two value of are

examined, 0 8. and 0 5. . Evaluate the accuracy of each smoothing constant. Which is

preferable? (Assume the forecast for January was 22 batteries.) Actual sales are given

below:

Month Actual

Battery

Sales

Forecast

January 20 22

February 21

March 15

April 14

May 13

June 16

Answer:

Forecasting Performance Measures:

Mean Forecast Error

(MFE or Bias)

Mean Absolute Deviation

(MAD)

Standard Squared Error (MSE)

Mean Absolute Percentage Error

(MAPE)

𝑀𝐹𝐸 =1

𝑛∑(𝐴𝑡 – 𝐹𝑡)

𝑛

𝑡=1

𝑀𝐴𝐷 =1

𝑛∑|𝐴𝑡 – 𝐹𝑡|

𝑛

𝑡=1

𝑀𝑆𝐸 =1

𝑛∑(𝐴𝑡 – 𝐹𝑡)2

𝑛

𝑡=1

𝑀𝐴𝑃𝐸 =100

𝑛∑ |

𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

𝑛

𝑡=1

y = -1.2857x + 21R² = 0.5407

10

12

14

16

18

20

22

0 1 2 3 4 5 6 7

Act

ual

Bat

tery

Sal

es

Mounth

495 units

𝜶 = 𝟎. 𝟖

Month Actual Battery

Sales

Rounded Forecast

MFE MAD MSE MAPE

Error

(𝐴𝑡 – 𝐹𝑡)

Absolute Error

|𝐴𝑡 – 𝐹𝑡|

Error ^2

(𝐴𝑡 – 𝐹𝑡)2

Error percentage

|𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

January 20 22 -2 02 04 0.10

February 21 20 1 01 01 0.05

March 15 21 -6 06 36 0.40 April 14 16 -2 02 04 0.14

May 13 14 -1 01 01 0.08 June 16 13 3 03 09 0.19

Summation -7 15 55 0.96 Average -1.17 2.50 9.17 0.16

MFE MAD MSE MAPE

-1.17 2.50 9.17 16 % 𝜶 = 𝟎. 𝟓

Month Actual Battery

Sales

Rounded Forecast

MFE MAD MSE MAPE

Error

(𝐴𝑡 – 𝐹𝑡)

Absolute Error

|𝐴𝑡 – 𝐹𝑡|

Error ^2

(𝐴𝑡 – 𝐹𝑡)2

Error percentage

|𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

January 20 22 -2 02 04 0.10

February 21 21 0 00 00 0.00 March 15 21 -6 06 36 0.40

April 14 18 -4 04 16 0.29

May 13 16 -3 03 09 0.23 June 16 15 2 02 04 0.09

Summation -13 17 69 1.11 Average -2.167 2.833 11.500 0.185

MFE MAD MSE MAPE

-2.167 2.833 11.500 18.5 %

MFE MAD MSE MAPE

𝛼 = 0.8 -1.17 2.50 9.17 16 %

𝛼 = 0.5 -2.167 2.833 11.500 18.6 %

Smoothing constant of 𝜶 = 0.8 is better than of 𝛼 = 0.5 because it has a smaller error.

0

5

10

15

20

25

0 1 2 3 4 5 6 7

BA

TTER

Y SA

LES

MOUNTH

actual

0.8

0.5

4) The annual sales of the Engineering Co. are shown in the following table:

Year 1 2 3 4 5 6 7 8 9 10

Sales (1000

L.E.) 45 45.5 50.1 50.6 62 58 64.3 70.2 72.4 75.6

a. Construct a 2 year moving average.

b. Construct a 5 year moving average.

c. Find the regression line.

d. Forecast the sales for year 11.

Show your results graphically.

Answer:

Sales (1000 L.E.)

2 years moving average

5 years average

1 45

2 45.5

3 50.1 =(45.0+45.5)/2 45.25

4 50.6 =(45.5+50.1)/2 47.80

5 62 =(50.1+50.6)/2 50.35

6 58 =(50.6+62.0)/2 56.30 =(45.0+45.5+50.1+50.6+62.0)/5 50.64

7 64.3 =(62.0+58.0)/2 60.00 =(45.5+50.1+50.6+62.0+58.0)/5 53.24

8 70.2 =(58.0+64.3)/2 61.15 =(50.1+50.6+62.0+58.0+64.3)/5 57.00

9 72.4 =(64.3+70.2)/2 67.25 =(50.6+62.0+58.0+64.3+70.2)/5 61.02

10 75.6 =(70.2+72.4)/2 71.30 =(62.0+58.0+64.3+70.2+72.4)/5 65.38

11 - =(72.4+75.6)/2 74.00 =(58.0+64.3+70.2+72.4+75.6)/5 68.10

y = 3.6442x + 39.327R² = 0.9566

30

35

40

45

50

55

60

65

70

75

80

0 2 4 6 8 10 12

Sale

s (1

00

0 L

.E.)

year

Linear Trend Equation:

𝑌𝑡 = 𝑎 + 𝑏𝑡

𝑏 = 𝑛 ∑ 𝑡𝑦−∑ 𝑡 ∑ 𝑦

𝑛 ∑ 𝑡2−(∑ 𝑡)2

𝑎 =∑ 𝑦−𝑏 ∑ 𝑥

𝑛

year

Sales (1000 L.E.)

t2 ty

1 45.0 001 045.0 2 45.5 004 091.0

3 50.1 009 150.3

4 50.6 016 202.4

5 62.0 025 310.0

6 58.0 036 348.0 7 64.3 049 450.1

8 70.2 064 561.6 9 72.4 081 651.6

10 75.6 100 756.0

∑ 55 593.7 385 3566

30

35

40

45

50

55

60

65

70

75

80

0 2 4 6 8 10 12

actual

2

5

𝑏 = 𝑛 ∑ 𝑡𝑦−∑ 𝑡 ∑ 𝑦

𝑛 ∑ 𝑡2−(∑ 𝑡)2 𝑎 =

∑ 𝑦−𝑏 ∑ 𝑥

𝑛

𝑏 =10(3566) − (55 × 593.7)

(10 × 385) − (55)2 = 3.6442

𝑎 =593.7 − (3.6442 × 55)

10= 39.3269

𝑌𝑡 = 𝑎 + 𝑏𝑡

𝒀𝒕 = 𝟑𝟗. 𝟑𝟐𝟔𝟗 + 𝟑. 𝟔𝟒𝟒𝟐 𝒕

y = 3.6442x + 39.32730

40

50

60

70

80

90

0 2 4 6 8 10 12

Sale

s (1

00

0 L

.E.)

Year

5) The following table shows the demand for the products of the General Batteries Co. and

the number of automobiles in Egypt in the last 10 years.

Year 78 79 80 81 82 83 84 85 86 87

Demand (1000 L.E.) 100 180 240 350 500 470 610 800 900 1000

No of Autos(10,000) 20 55 81 125 190 175 250 320 360 390

a. Find the coefficient of correlation.

b. Forecast the demand for year 88, if the expected number of automobiles is 4,300,000

Answer:

Correlation (r):

𝑟 =𝑛(∑ 𝑥𝑦) − (∑ 𝑥)(∑ 𝑦)

√𝑛(∑ 𝑥2) − (∑ 𝑥)2 ∙ √𝑛(∑ 𝑦2) − (∑ 𝑦)2

Year time

period (x)

Demand (1000 L.E.)

(y) x^2 y^2 x*y

78 1 100 1 10000 100 79 2 180 4 32400 360

80 3 240 9 57600 720 81 4 350 16 122500 1400

82 5 500 25 250000 2500

83 6 470 36 220900 2820

84 7 610 49 372100 4270

85 8 800 64 640000 6400 86 9 900 81 810000 8100

87 10 1000 100 1000000 10000

sum 55 5150 385 3515500 36670

n 10

r 0.988851 r^2 0.977827

y = 101.15x - 7830R² = 0.9778

y = 42.545x - 3313.4R² = 0.9798

0

100

200

300

400

500

600

700

800

900

1000

1100

7 6 7 8 8 0 8 2 8 4 8 6 8 8 YEAR

Demand (1000 L.E.)

No of Autos(10,000)

Year time

period (x)

No of Autos(10,000)

(y) X^2 y^2 x*y

78 1 20 1 400 20

79 2 55 4 3025 110 80 3 81 9 6561 243

81 4 125 16 15625 500 82 5 190 25 36100 950

83 6 175 36 30625 1050

84 7 250 49 62500 1750 85 8 320 64 102400 2560

86 9 360 81 129600 3240 87 10 390 100 152100 3900

sum 55 1966 385 538936 14323

n 10 r 0.989825

r^2 0.979754

6) Given:

Year 1 2 3 4 5 6 7 8

Demand 90 100 107 113 123 136 144 155

a)Plot the data and establish a forecast for year 9 using the least square method.

b)Compare forecast using exponential smoothing with α = 0.15 and F1 = 85 with the

forecasted values obtained by the regression equation established in part (a).

Answer:

a)

year Demand t2 ty

1 90 01 0090

2 100 04 0200 3 107 09 0321

4 113 16 0452

5 123 25 0615 6 136 36 0816

7 144 49 1008 8 155 64 1240

∑ 36 968 204 4742

𝑏 = 𝑛 ∑ 𝑡𝑦−∑ 𝑡 ∑ 𝑦

𝑛 ∑ 𝑡2−(∑ 𝑡)2 𝑎 =

∑ 𝑦−𝑏 ∑ 𝑥

𝑛 𝑌𝑡 = 𝑎 + 𝑏𝑡

𝑏 =8(4742) − (36 × 968)

(8 × 204) − (36)2 = 9.19

𝑎 =968 − (9.19 × 36)

8= 79.65

𝑌𝑡 = 79.65 + 9.19 𝑡

y = 9.1905x + 79.643R² = 0.992

60

80

100

120

140

160

180

0 1 2 3 4 5 6 7 8 9

dem

and

year

At t=9

𝑌𝑡 = 79.65 + 9.19 (9) = 𝟏𝟔𝟐. 𝟑𝟔

b)

𝐹𝑡+1 = 𝐹𝑡 + 𝛼(𝐴𝑡 – 𝐹𝑡) New forecast = last period’s forecast + α (last period’s actual demand – last period’s forecast)

𝛼 = 0.15 𝐹𝑡 = 85 𝐴𝑡 = 155

𝐹𝑡+1 = 85 + 0.15(155 – 85) = 85 + (0.1 × 70) = 95.5

a) using the least square method Demand =162.36 b) using exponential smoothing with α = 0.15 and F1 = 85 Demand =155

y = 9.1905x + 79.64360

80

100

120

140

160

180

0 2 4 6 8 10

Dem

and

year

7) Use exponential smoothing and the data in the following table to determine the monthly

forecasts (marked by X) using a smoothing factor of 0.1, 0.3 and 0.6

Year Month Demand α= 0.1 α= 0.3 α= 0.6

2002

April 120 120 120 120

May 140 x x x

June 160 x x x

July 110 x x x

August 120 x x x

September 110 x x x

October X x x x

Which is the best smoothing factor of these three?

Answer:

𝐹𝑡+1 = 𝐹𝑡 + 𝛼(𝐴𝑡 – 𝐹𝑡) New forecast = last period’s forecast + α (last period’s actual demand – last period’s forecast)

Mean Forecast Error (MFE or Bias)

Mean Absolute Deviation (MAD)

𝑀𝐹𝐸 =1

𝑛∑(𝐴𝑡 – 𝐹𝑡)

𝑛

𝑡=1

𝑀𝐴𝐷 =1

𝑛∑|𝐴𝑡 – 𝐹𝑡|

𝑛

𝑡=1

Standard Squared Error (MSE)

Mean Absolute Percentage Error (MAPE)

𝑀𝑆𝐸 =1

𝑛∑(𝐴𝑡 – 𝐹𝑡)2

𝑛

𝑡=1

𝑀𝐴𝑃𝐸 =100

𝑛∑ |

𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

𝑛

𝑡=1

y = -4.5714x + 156.38R² = 0.1892

100

110

120

130

140

150

160

170

3 4 5 6 7 8 9 10

Dem

and

month

Month

Demand

α= 0.1 error error ^2 α= 0.3 error error^2 α= 0.6 error error^2

4 120 120 0.00 0000.00 120.00 0.00 0000.00 120 0.00 0.00

5 140 120 20.00 0400.00 120.00 20.00 0400.00 120.00 20.00 400.00

6 160 122 38.00 1444.00 126.00 34.00 1156.00 132.00 28.00 784.00

7 110 125.8 -15.80 0249.64 136.20 -26.20 0686.44 148.80 -38.80 1505.44

8 120 124.22 -4.22 0017.81 128.34 -8.34 0069.56 125.52 -5.52 30.47

9 110 123.798 -13.80 0190.38 125.84 -15.84 0250.84 122.21 -12.21 149.04

10 X 122.42 121.09 114.88

Summation 24.18 2301.83 3.62 2562.84 -8.53 2868.95

Error

MAD 15.303 MAD 17.396 MAD 17.421

MSE 383.639 MSE 427.140 MSE 478.158

MAPE 11.41 % MAPE 13.45 % MAPE 13.45 %

Smoothing constant of α = 0.1 is preferred to that of α = 0.3 and α = 0.5 because it has a smaller error.

90

100

110

120

130

140

150

160

170

3 4 5 6 7 8 9 10 11

actual

0.1

0.3

0.6

8) The table below shows the demand for a particular brand of fax machine in a department

store in each of the last twelve months.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Demand 12 15 19 23 27 30 32 33 37 41 49 58

Calculate the four month moving average for months 4 to 12. What would be your forecast

for the demand in month 13?

Apply exponential smoothing with a smoothing constant of 0.2 to derive a forecast for the

demand in month 13.

Which of the two forecasts for month 13 do you prefer and why?

What other factors, not considered in the above calculations, might influence demand for the

fax machine in month 13?

For =0.2, compute 2S control limits (confidence level 95%). Plot the data and check that

all errors are within the limits.

Answer:

a) The four month moving average:

Month Demand four month

moving average

error abs error ^2 |𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

1 12

2 15

3 19

4 23

5 27 =(12+15+19+23)/4 17.250 09.750 9.75 095.0625 0.361111

6 30 =(15+19+23+27)/4 21.000 09.000 9.00 081.0000 0.300000

7 32 =(19+23+27+30)/4 24.750 07.250 7.25 052.5625 0.226563

8 33 =(23+27+30+32)/4 28.000 05.000 5.00 025.0000 0.151515

9 37 =(27+30+32+33)/4 30.500 06.500 6.50 042.2500 0.175676

10 41 =(30+32+33+37)/4 33.000 08.000 8.00 064.0000 0.195122

11 49 =(32+33+37+41)/4 35.750 13.250 13.25 175.5625 0.270408

12 58 =(33+37+41+49)/4 40.000 18.000 18.00 324.0000 0.310345

13 - =(37+41+49+58)/4 46.250

Error Calculations

Sum 76.750 76.750 859.4375 1.990739

Average n=8

9.594

MAD

9.594

MSE

107.430

MAPE 0.249

=24.9 %

y = 3.6923x + 7.3333R² = 0.9582

0

10

20

30

40

50

60

70

0 2 4 6 8 10 12 14

Dem

and

Month

The forecast for month 13 = 46.25

b) Exponential smoothing:

𝐹𝑡+1 = 𝐹𝑡 + 𝛼(𝐴𝑡 – 𝐹𝑡) New forecast = last period’s forecast + α (last period’s actual demand – last period’s forecast)

𝛼 = 0.2 𝐹𝑡 = 12

Month Demand Exponential smoothing

error abs error ^2 |𝐴𝑡 – 𝐹𝑡

𝐴𝑡|

1 12 2 15 12.000 03.000 03.000 009.000 0.200

3 19 12.600 06.400 06.400 040.960 0.337 4 23 13.880 09.120 09.120 083.174 0.397

5 27 15.704 11.296 11.296 127.600 0.418 6 30 17.963 12.037 12.037 144.885 0.401

7 32 20.371 11.629 11.629 135.244 0.363

8 33 22.696 10.304 10.304 106.163 0.312 9 37 24.757 12.243 12.243 149.887 0.331

10 41 27.206 13.794 13.794 190.282 0.336 11 49 29.965 19.035 19.035 362.347 0.388

12 58 33.772 24.228 24.228 587.012 0.418

13 - 38.617

Error Calculations

Sum 133.087 133.0867 1936.554 3.902151

Average n=11

12.099

MAD

12.099

MSE

176.050

MAPE 0.355 35.5 %

To compare the two forecasts we calculate the mean squared deviation (MSD) and Mean Absolute Percentage Error (MAPE).

(MSD) (MAPE)

Four month moving average 107.430 24.9 %

Exponential smoothing 176.050 35.5 %

Four month moving average is preferred, because it has a smaller error.

Other factors:

Seasonal Demand Price Changes, Both This Brand And Other Brands General Economic Situation New Technology Advertising

0

10

20

30

40

50

60

70

0 2 4 6 8 10 12 14

DEM

AN

D

MONTH

Series1

4 month

Exponential

For four month moving average:

Month 1 2 3 4 5 6 7 8 9 10 11 12

Sum Demand 12 15 19 23 27 30 32 33 37 41 49 58

four month moving average 17.25 21.00 24.75 28.00 30.50 33.00 35.75 40.00

Error 9.75 9.00 7.25 5.00 6.50 8.00 13.25 18.00 76.75

Error -Average 0.16 -0.59 -2.34 -4.59 -3.09 -1.59 3.66 8.41 0

(Error –Average)2 0.02 0.35 5.49 21.10 9.57 2.54 13.37 70.67 123.12

Average error = ∑ 𝑒𝑟𝑟𝑜𝑟

𝑛=

76.75

8= 9.594

𝑠 = √𝑀𝑆𝐸 = √123.12

8 − 1= 4.19

0.00 ± 2𝑠 = 0 ± (2 × 4.19) = 0 ± 8.39 𝐿𝐶𝐿 = −8.39 𝑈𝐶𝐿 = +8.39

0.16-0.59

-2.34

-4.59

-3.09

-1.59

3.66

8.41

y = - 8.39

y = 8.39

-10.00

-8.00

-6.00

-4.00

-2.00

0.00

2.00

4.00

6.00

8.00

10.00

3 4 5 6 7 8 9 10 11 12 13

9) Quarterly unit demand for a product are given below:

Year Winter Spring Summer Autumn

2001 190 350 250 200

2002 250 400 300 150

2003 300 340 260 180

2004 280 400 220 200

2005 350 420 340 240

Determine a seasonal index, and establish a forecast for each quarter of the next year.

Answer:

10) A gift shop in a tourist center is open on weekends (Friday, Saturday, and Sunday). The owner

manager hopes to improve scheduling of part time employees by determining seasonal

relatives for each of these days. Data on recent activity at the store have been tabulated and

are shown in the table below.

-Develop seasonal relatives for the shop.

-Forecast the sales for week 7.

Week

1 2 3 4 5 6

Friday 149 154 152 150 159 163

Saturday 250 255 260 268 273 276

Sunday 166 162 171 173 176 183

Answer: