Spreadsheet exercise – guidelines for solution
Using the data-pivot table commands (as outlined in chapter 2) should give three tables:
(i)
Count of Customer Type
Customer Type
Employed
Student
Unemployed/Retired
Grand Total
Order Type
TV
10
20
20
50
Video
3
4
4
11
Payment Type
Direct Debit Payment Book
29
7
49
6
12
47
90
60
Grand Total
36
55
59
150
Joint
23
31
35
89
Grand Total
36
55
59
150
(ii)
Count of Customer Type
Customer Type
Employed
Student
Unemployed/Retired
Grand Total
(iii)
Count of Customer Type
Customer Type
Employed
Student
Unemployed/Retired
Grand Total
A
3
4
7
14
Delivery Area
B
25
32
37
94
C
3
6
3
12
D Grand Total
5
36
13
55
12
59
30
150
(b) Reading values from the tables:
(i) P(TV) = 50/150 = 0.33
P(video) = 11/150 = 0.07
P(joint) = 89/150 = 0.59
(ii) P(area B) = 94/150 = 0.63
(iii) P(joint package|customer is a student) = 31/55 = 0.56
(iv) (pays by direct debit|customer is employed) = 29/36 = 0.81
(c) There are several ways that a system could be set up to complete this part of the workshop: just
& one suggested solution is given here. The key aim is to get students building a simple EXCEL
(d) model so that they can perform some basic sensitivity analysis on the scenario and get some idea
of how the spreadsheet might be used to do this sort of analysis in practice.
First, the table of figures can be laid out with cells to calculate the probabilities and expected
values, then an IF statement can be used to test which of the three schemes gives the highest
returns. If this is felt to be too complicated/technical for the level of knowledge of the students
three simple IF statements, one for each scheme, could be entered instead to compare the expected
value for each scheme in turn to the maximum value
Figures 5.1 and 5.2 give screen dumps of the suggested model and the formulae used
Figure 5.1 – EXCEL model to compare schemes
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Figure 5.2 EXCEL model showing formulae used
Students can then experiment by changing the value of 0.7 for the P(high demand) to see where
each scheme gives the best return. They should find:
0.00  P(high)  0.28 gives Scheme A
0.29  P(high)  0.66 gives Scheme C
0.67  P(high)  1.00 gives Scheme B
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