answerser 3std

Operations and Production Management MGMT 405 Answer set 3

MGMT 405 Operations and Production Management

Answer set 3

(Reference chapter 3– William J. Stevenson-2007, ninth edition)

Problems and Solutions

1. A commercial bakery has recorded sales in dozen for the products as shown

below:

DayBlueberry

Muffins

Cinnamon

bunsCupcakes

1 30 18 45

2 34 17 26

3 32 19 27

4 34 19 23

5 35 22 22

6 30 23 48

12 31 28 26

13 35 29 27

© 2010/11, Sami Fethi, EMU, All Right Reserved, McGraw-Hill, 2007, 9. Ed. 1


14 34 31 24

15 33 33 22

(a) Predict orders for the following day for each of the products using an appropriate

naïve method. Also Plot each data set.

(b) What should the use of sales data instead of demand imply??

Ans:

(a) We need to use naive methods. This means that any simple appropriate method

can conducted. For example: Ploting each data set reveals that muffins orders are

almost stable, varying around an average (e.g. 33). The demand for cinnamon buns

has a trend. If we get the last three period, we realized that number increased two by

two so 33-31=2 and last one is 33+2=35. This is for the following day for cinnamon

buns. Demand for cupcakes has an apparent seasonal variation with peaks every five

days. Day 1=45, Day 6=48, Day 11=47 and the next peak would be Day 16=50.

A commercial bakery

0

10

20

30

40

50

60

1 2 3 4 5 6 12 13 14 15

Days

sale

s

Blueberry Muffins

Cinnamon buns

Cupcakes



(b) The use of sales data instead of demand implies that sales adequately reflect

demand. We are assuming that n stock-outs because demand equals sales if there are

no shortages.

2. A company records indicates that monthly sales for a seven-month period are

as follows:.

Month Sales (000, unit)

Feb 19

Mar 18

Apr 15

May 20

Jun 18

July 22

Aug 20

(a) Plot the monthly data as can be seen in the table above.

(b) Forecast the monthly sales using linear trend equation,

(c) A five-month moving average and exponential Smoothing technique? Assume

that smoothing constant and march forecast value are 0.20 and 19 respectively.

(d) The naïve approach

(e) A weight average method conducting 0.60 for August, 0.30 for July, and 0.10 for

June.

(f) Which method seems least appropriate? Why?



Ans:

(a)

Sales

0

5

10

15

20

25

Feb Mar Apr May Jun July Aug

month

000 Sales

(b)

T (or x) Square of t Y tY

1 1 19 19

2 4 18 36

3 9 15 45

4 16 20 80

5 25 18 90

6 36 22 132



7 49 20 140

Sum= 28 Sum of square= 140 Sum= 132 Sum= 542

n= 7

Or If formula b is used first, it may be used formula a in the following format:

Y = 16.86 + 0.50X

After Aug (7), For sept: Y = 16.86 + 0.50(8)=20.86

(b)

(c) Ft = Ft-1 + (At-1 - Ft-1)

Ft = forecast for period tFt-1 = forecast for the previous period = smoothing constant At-1 = actual data for the previous period

April-F=19+0.20(18-19)=18.8



May- F=18.8+0.20(15-18.8)=18.04

June- F=18.04+0.20(20-18.04)=18.43

July- F=18.43+0.20(18-18.43)=18.34

Aug- F=18.34+0.20(22-18.34) =19.07

Sep- F=19.07+0.20(20-19.07) =19.26

(d) It may be around 20 (19.26)

(e) WA=0.60 (20) + 0.30 (22) + 0.10 (18) =20.4

(f) Probably trend method because the data appear to vary around Y = 16.86 +

0.50(4) =18.86.

3. A cosmetics manufacturer’s marketing department has developed a linear trend

equation that can be used to predict annual sales of its popular hand-foot cream.

Ft = 80 + 15 t

Where Ft =Annual sales (000 bottles), and assume that t= 0 in1990

(a) Are annual sales increasing or decreasing? By how much.

(b) Predict annual sales for the year 2006 using the equation.

(c ) Predict changes in annual sales the year between 1995 and 2005 using the

equation.

Ans:

(a) Ft = 80 + 15 t



15 is the slope of the equation so this value show us whether annual sales increase or

decrease. The sign of the value is positive that’s why annual sales are increasing by

15000 bottles per year.

(b) Ft = 80 + 15 t= 80+ 15(16) = 320000

Where t=2006-1990=16

(c) Ft = 80 + 15 t= 80+ 15(5) = 155000 in 1995

Ft = 80 + 15 t= 80+ 15(15) = 305000 in 2005

Change in sales between the relevant years= 305000-155000=150000

4. New Car sales with monthly indexes (seasonal relatives) for a dealer for the past

years are shown in the following table:

Month Unit sold index Month Unit sold index

Jan 640 0.8 Jul 765 0.9

Feb. 648 0.8 Aug 805 1.15

Mar 630 0.7 Sept 840 1.20

Apr 761 0.94 Oct 828 1.20

May 735 0.89 Nov 840 1.25

Jun 850 1.00 Dec 800 1.25



(a) Plot the data. Does there seem to be a trend?

Unit Sales

0100200300400500600700800900

Jan Feb. Mar Apr May Jun Jul Aug Sept Oct Nov Dec

Month

sale

s vo

lum

e Unit sold

index

It seems that there exists a trend starting from 600 to 800.

(b) Compute the Deseasonalized car sales

Month Unit sold index Deseasonalize =actual sales/seasonal relative

Jan 640 0.8 640/0.8=800

Feb. 648 0.8 810

Mar 630 0.7 900

Apr 761 0.94 809.6

May 735 0.89 825.8



Jun 850 1.00 850

Jul 765 0.9 850

Aug 805 1.15 700

Sept 840 1.20 700

Oct 828 1.20 690

Nov 840 1.25 672

Dec 800 1.25 640

(c ) Plot both seasonalized and Deseasonalized car sales data on the same graph. Briefly

explain.

sales

0

200

400

600

800

1000

Jan Feb. Mar Apr May Jun Jul Aug Sept Oct Nov Dec

Month

volu

me seasonal

index

deseasonal

Seasonalized car sales data are upward trend in sales and Deseasonalized car sales data

are downward trend in sales.

5. Two different forecast techniques (F1 and F2) were used to forecast demand for

cases of bottled water. Actual demand the two sets of forecasts are as follows:



Period and Demand Predicted Demand

Period Unit sold F1 F2

1 68 66 66

2 75 68 68

3 70 72 70

4 74 71 72

5 69 72 74

6 72 70 76

7 80 71 78

8 78 74 80

(a) Compute MAD for each set of forecast. Which forecast is better or appears to be

more accurate? Briefly explain.

(b) Compute MES for each set of forecast. Which forecast is better or appears to be


(c) Compute MAPE for each set of forecast. Which forecast is better or appears to be


(d) Compute a tracking signal for the 8th month for each forecast using the cumulative

error for 1 and 8.

(e) Compute 2s control limits for each forecast.

Ans:



Period Demand F1 e e2 ( /actual)*100

1 68 66 2 2 4 3

2 75 68 7 7 49 9.3

3 70 72 -2 2 4 2.8

4 74 71 3 3 9 4

5 69 72 -3 3 9 4.3

6 72 70 2 2 4 2.7

7 80 71 9 9 81 11.2

8 78 74 4 4 16 5.1

586 32 176 42.4

Period Demand F2 e e2( /

actual)*100



1 68 66 2 2 4 3

2 75 68 7 7 49 10.2

3 70 70 0 0 0 0

4 74 72 2 2 4 2.7

5 69 74 -5 5 25 6.7

6 72 76 -4 4 16 5.2

7 80 78 2 2 4 2.5

8 78 80 -2 2 4 2.5

24 106 32.9

(a) Compute MAD for each set of forecast. Which forecast is better or appears to be


Ans:

MAD= ∑│Actual-Forecast│/n

MADF1= 32/8= 4

MADF2= 24/8= 3

F2 forecast results appears to be more accurate because MAD model gives less error

than F1.



(b) Compute MES for each set of forecast. Which forecast is better or appears to be


Ans:

MES= ∑│Actual-Forecast│2/ (n-1)

MESF1= 176/7= 25.14

MESF2= 106/7= 15.14

F2 forecast results appears to be more accurate because MES model gives less error

than F1.

(c) Compute MAPE for each set of forecast. Which forecast is better or appears to be


Ans:

MAPE= (∑ (│Actual-Forecast│/actual)*100)/n

MAPE F1=42.4/8=5.34

MAPE F2==32.9/8=4.11

F2 forecast results appears to be more accurate because MAPE model gives less error

than F1.

(d) Compute a tracking signal for the 8th month for each forecast using the cumulative

error as well as limits of ± 4 for 1 and 8.

Ans:

Tracking Signalt = ∑ (Actual-Forecast)/MADt

Control Limits= 0 +/- 2



Period Demand F1 eCUM

ERROR

1 68 66 2 2 2

2 75 68 7 7 9

3 70 72 -2 2 7

4 74 71 3 3 10

5 69 72 -3 3 7

6 72 70 2 2 9

7 80 71 9 9 18

8 78 74 4 4 22

© 2010/11, Sami Fethi, EMU, All Right Reserved, McGraw-Hill, 2007, 9. Ed.

Period Demand F2 eCUM

ERROR

1 68 66 2 2 2

2 75 68 7 7 9

3 70 70 0 0 9

4 74 72 2 2 11

5 69 74 -5 5 6

6 72 76 -4 4 2

7 80 78 2 2 4

8 78 80 -2 2 2

14


MADF1= 32/8= 4

MADF2= 24/8= 3

MESF1= 176/7= 25.14

MESF2= 106/7= 15.14

MAPE F1=42.4/8=5.34

MAPE F2==32.9/8=4.11

Tracking Signalt = ∑ (Actual-Forecast)/MADt =cumulative error/ MADt

Tracking Signalt for F1= 22/4=5.5

Tracking Signalt for F2= 2/3=0.66

Tracking Signal is used to monitor a forecast whether the method is biased or not.

Values can be positive or negative and zero value would be ideal. Since 5.5 >4 the forecast method 1 is biased. This means that using F1, we are underestimating (positive bias)* demand. Since 0.66 < 4, the forecast is in control.

*overestimating (negative bias).

(e) Compute 2s control limits for each forecast.

Ans:

Control Limits= 0 +/- 2

Control Limits for F1 = 0 +/- 2 =0 +/- 2 √(25.14=+10.01 and -10.01

Control Limits for F2 = 0 +/- 2 =0 +/- 2 √(15.14=+7.78 and -7.78

Since all errors for both forecasts 1 and 2 are within these limits, the forecasts are in

control (see column e for both F1 and F2 in the tables above).


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