Case Scenario Boulevard Industries Bulevard Industries has d

Case Scenario: Boulevard Industries

Bulevard Industries has decided to provide safety training to its 1,500+ employees who work in the Boulevard main office. Part of its training is a 60-minute presentation by a member of the Boulevard human resources (HR) department and is held in a conference room at the main office building. Boulevard plans to complete this process by holding five sessions a day on Thursday (Th) and Fridays (F) each week for the next 4 months. During the first week of sessions, the HR department recorded the following information concerning attendance.

Questions

1. During this first week of sessions, what was the mean error of the HR department\'s forecasts (ME)? What was the mean absolute deviation (MAD)? What was the mean absolute percent error (MAPE)?

2. Assume that the HR department\'s forecast for the Thursday 9 a.m. session was 10, as shown in the table. If HR had used simple exponential smoothing with an alha (a) of 0.2 to forecast the remainder of the sessions that week, what would the forecast for the Friday 9 a.m. session have been?

3. Someone in HR noticed that overall attendance on Friday was somewhat less than on Thursday and believes this pattern will continue. Calculate the two seasonal relatives that express this fluctuation, based on this data.

4. Using your answer from question 3, how many people would you forecast to attend next Friday\'s sessions, if a totall of 100 people were forecast to attend all 10 sessions next week?

Session	Day	Time	Attendance Forecast	Actual Attendance
1	Th	9 a.m.	10	13
2	Th	11 a.m.	10	10
3	Th	2 p.m.	10	12
4	Th	3 p.m.	8	12
5	Th	4 p.m.	7	8
6	F	9 a.m.	10	11
7	F	11 a.m.	10	9
8	F	2 p.m.	10	9
9	F	3 p.m.	8	4
10	F	4 p.m.	7	7

Solution

To Find out the solution,

1- Let\'s take it one by one To find out the mean error, firstly we calculate the mean by adding the actual attendance / total sessions.

Therefore 95/10=9.5

Now find out deviation from the actual 13-9.5= 3.5

10-9.5=0.5 and so on

Then we will do the square of deviation and add on like( 3.5)²=12.25 and so on when we add all the digit we get 66.5

Now to find out the standard error we first calculate the standard deviation Which is underroot of variance =66.5/10

Underroot of 6.65= 2.578

Standard error =2.578/3.162

(UNDERROOT of 10=3.162)

S. E= 0.81530

Now, MEAN ABSOLUTE DEVIATION

firstly, we find the error= Actual attendance - FOrecast attendance

13-10=3,

10-10=0 and so on

Then we calculate IerrorI to make negative value into positive one.

Now adding up these values we get 17

MAD=1/n sum of error

Therefore, 17/10= 1.7

To find out Mean Absolute Percent Error= IerrorI/ actual* 100%

3/13*100%= 23.07%

0/10*100%=0

And so on.. To add all these values we get 216.88

Now MAPE= 216.88/10=21.688

2- Exponential smoothing where alpha=0.2

Forecasted value= Last Forecasted value + alpha ( last actual value - last forecasted value)

As per given question,

Forecasted value for friday= 10+0.2(13-10)

=10+0.6

=10.6 will be the forecasted value for Friday at 9 am.

3- To find the seasonal relatives,

We have 5 hour and 2 days in a week.

Thereby we have to make one new table showing hours 9,11,2,3,4 respect to Thursday Friday session with the actual values as per the data.

When we sum up the Actual values of Thursday = 55

Sum up of the actual values of Friday=40

Now to calculate average of session / hour

For Thursday =55/5

=11/hour

For Friday =40/5

=8/hour

Now the seasonal indices will be calculated by dividing actual values with the average

For Thursday 13/11=1.181

10/11=0.90

1.09,

1.09,

0.72

For Friday 11/8= 1.375

9/8= 1.125

1.125,

0.5,

0.875

Now to find seasonal indices we find the average by add up these two values and divided by 2 ,

We get the values 1.181+1.375/2= 1.278

1.0125

1.1075

1.34

0.7975

By adding up these values it will be equal to 5

As we have 5 slot of hour.

4- To do the forecast for Friday session

Overall 100 people were about to attend now divide this by 5 As we have 5 slot of hour

100/5= 20

For friday FOrecast multiply this 20 with the seasonal indices that we have calculated above In 3 part

20*1.278= 25.56

20.25

22.15

26.8

15.95

By this we can sum up our forecast that about 110 people will attend the Friday session.