F1 Business Maths April

THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS & QUANTITATIVE METHODS FORMATION 1 EXAMINATION - APRIL 2006. Time Allowed: 3 hours and 10 minutes to read the paper

1.

Answer 5 questions Only the first five questions answered will be marked. All questions carry equal marks.

You are advising CPA Ltd. on two alternative proposals for implementing an IT system. The system will automate a number of existing office routines and is projected to provide an acceptable return on the investment. Proposal 1 requires an expenditure of €6,500 now and a further €4,000, at the end of 2 years. It will have a cash inflow of €18,100 in 5 years time. Proposal 2 requires 4 annual investments of €2,200 starting now, it will have a cash inflow of €16,250 in 5 years time. (i)

Calculate the Net Present Value (NPV) for each project using a cost of capital of 10%.

(8 Marks)

(ii)

Calculate the Internal Rate of Return (IRR) on both projects.

(8 Marks)

(iii)

Advise the company on the best option and the most profitable investment.

(4 Marks) [Total: 20 Marks]

2.

The battery division of DIB Ltd. manufactures a range of commercial batteries for fork-lift trucks. It claims that the average length of life for its grade 1 battery is 60 months. The company offers a guarantee on it for 36 months. If the standard deviation is known to be 10 months and the frequency distribution is known to be a standard normal distribution, assuming the company’s claim is true, determine: (i)

The percentage of the grade 1 batteries which will last more than 50 months.

(6 Marks)

(ii)

The percentage of the batteries that will last less than 40 months.

(6 Marks)

(iii)

Comment on the company’s claim if your battery lasts 37 months.

(8 Marks) [Total: 20 Marks]

Page 1

3.

NPA Remuneration Consultants have developed a data-base in which they compile labour costs for a range of categories of employees. The following data relates to the percentage change in costs of senior managers over the past 3 years. % change in costs -5 0 5 10 15 20 25 30 35 40 100

and and and and and and and and and and and

less less less less less less less less less less less

No. of Managers

0 5 10 15 20 25 30 35 40 100 170

4 32 25 10 8 3 2 5 4 3 4

You are asked to calculate: (i)

The mean and median values of the annual percentage change in labour costs.

(8 Marks)

(ii)

The standard deviation.

(8 Marks)

(iii)

Comment on the differences in values derived in (i) above.

(4 Marks) [Total: 20 Marks]

4.

At a management meeting, the marketing manager of Superior Products Ltd. contends that the level of sales is directly related to marketing expenditure. The data is provided below.

Sales (€000s)

Advertising Expenditure (€000s)

25

8

35

12

29

11

24

5

28

14

12

3

18

6

27

8

17

4

30

9

Since expenditure appears to be large relative to the volume of sales, you are asked to prepare a report on the manager’s claim. As part of your report: (i)

Derive a linear regression equation representing the data.

(8 Marks)

(ii)

Find the correlation coefficient between the data.

(8 Marks)

(iii)

Comment on the marketing manager’s claim.

(4 Marks) [Total: 20 Marks]

Page 2

5.

The sales of kitchen units (in 000s) for Sno Products Ltd. for the years 2003 – 2005 are set out in the table below:

Quarter Year

1

2

3

4

2003

3

5

8

4

2004

4

6

10

5

2005

5

8

11

7

The financial manager asks you to prepare a report containing the following information. (i)

The trend in sales using a centred moving average.

(10 Marks)

(ii)

Calculate the seasonal variations for each quarter and adjust the data for 2005.

(10 Marks) [Total: 20 Marks]

6.

As a CPA consultant, part of your brief is to assist clients to resolve a wide range of business problems. The following problems have been assigned to you for immediate delivery to a corporate client: (i)

Electricity company ELP charges customers for units of energy used on a kwhr basis. The company has a standing charge of €30.04 per two months and charges 2.39c per kwhr for electricity used. Represent this model by means of a linear equation and calculate the annual cost for 2,400 kwhrs. (6 Marks)

(ii)

According to Consumer Survey magazine, a random sample of 36 family-owned hotels taken from the Failte Ireland brochure showed that the average daily profit was €720 with a standard deviation of €110. Find the average profit of the hotels with a 95% confidence level.

(iii)

(6 Marks)

A Manufacturing Company claims that the wages of its employees have more than exceeded the rate of inflation which has increased at an average rate of 6.8% per annum. As an example it states that the wages of a typical employee 10 years ago were €12,000 and this has now risen to €25,200. Verify its claim.

(8 Marks) [Total: 20 Marks]

END OF PAPER

Page 3

SUGGESTED SOLUTIONS

BUSINESS MATHEMATICS & QUANTITATIVE METHODS FORMATION 1 EXAMINATION - APRIL 2006.

Solution 1. (i)

Net Present Value for each proposal. Cash flows for proposal 1.

Year

0 1 2 3 4 5 NPV

Investment

Inflows

Net Cash Flow

€6,500

(€6,500)

€4,000

(€4,000)

€18,100

PV Factor @ 10% 1.000 0.909 0.826 0.751 0.683 0.621

€18,100

PV @ 14% (€6,500) (€3,304)

€11,240 €1,436

PV Factor PV 1.000 0.877 0.769 0.675 0.592 0.519

PV

(€6,500) (€3,076)

€9,394 (€182)

4 Marks Cash flows for proposal 2. Year

Investment

Inflows

Net Cash Flow

PV Factor @ 10%

PV @ 14%

0 1 2 3 4 5 NPV

€2,200 €2,200 €2,200 €2,200

(€2,200) (€2,200) (€2,200) (€2,200) €16,250

€16,250

1.000 0.909 0.826 0.751 0.683 0.621

(€2,200.0) (€1,999.8) (€1,817.2) (€1,652.2) €10,091.0 €2,422

PV Factor PV

PV

1.000 0.877 0.769 0.675 0.592 0.519

(€2,200.0) (€1,929.4) (€1,691.8) (€1,485.0) €8,433.75 €1,128 4 Marks

Due to the higher net present value, proposal 2 represents the best proposal for the company.

(ii)

Internal rates of return. The IRR can be derived by calculation or graphically. Using the formula N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate I2 gives NPV N2. N1 - N2 Page 4

Proposal 1.

N1 = €1436, I1 = 10%; N2 = (€182), I2 = 14%

IRR = 1436 x 0.14 - (182) x 0.10 = 219.24 = 13.55% 1436 - (182) 1618 4 Marks

Proposal 2.

N1 = €2,422, I1 = 10%; N2 = €1,128, I2 = 14%

IRR = 2422 x 0.14 - 1128 x 0.10 = 226.28 = 17.5% 2422 - 1128 1294 4 Marks

(iii)

Initially the net present value for proposal 2 indicates that the company should accept this option. This is confirmed by the internal rate of return which provides a increased return on the investment with respect to proposal 1. Proposal 2 is therefore the most profitable investment for the company. 4 Marks [Total: 20 Marks]

Page 5

SOLUTION 2. (i)

If the distribution of length of life of the batteries is Normal with a mean of 60 months and a standard deviation of 10 months, the diagram below represents the distribution.

Relative frequency 34%

34%

13.5%

13.5%

13.5%

2.5%

2.5%

Life length (months) µ—3σ

µ—2σ

µ—σ

µ

µ+σ

µ+2σ

µ+3σ

-3

-2

-1

0

+1

+2

+3

40

50

60

70

80

This diagram represents the fact that 68% of battery lives fall within one standard deviation of the mean. i.e. x - µ, x + µ; 95% fall within two standard deviations and 99% fall within three standard deviations. The figure shows (a) that 34% of the measurements fall between the mean and one standard deviation on each side (b) 2.5% of the measurements lie beyond two standard deviations in each direction from the mean. The % of batteries lasting more than 50 months is 34% (between 50 and 60 months) plus 50% (greater than 60 months). 6 Marks (ii)

The % of batteries that lasts less than 40 months can thus be determined; 2.5% of the batteries should fail prior to 40 months. 6 Marks

(iii)

If the battery fails at 37 months a number of inferences can be made -

the battery was out of the 2.5% that fails prior to 40 months

-

something about the company’s claim is not true. The chances are so small that a battery fails before 40 months that there are substantial reasons to have doubts about the claim. A mean smaller than 60 months and/or a standard deviation greater than 10 months would both increase the likelihood of failure prior to 40 months. 8 Marks

The above probabilities can be determined by calculation using the population z-score where z = x - µ and the Normal distribution tables. σ [Total: 20 Marks]

Page 6

Solution 3. (i)

The mean and median annual values of percentage on labour costs. Assuming that the data is continuous gives the mid points in the table.

Class Interval

f

Cum freq

-5 – 0

4

4

0–5

32

5 – 10

(x – x)2

f(x – x)2

fx

x-x

- 2.5

-10.0

-19.575

383.18

1532.7

36

2.5

80.0

-14.575

212.43

6797.8

25

61

7.5

187.5

-9.575

91.68

2292.0

10 -15

10

71

12.5

125.0

-4.575

20.93

209.3

15 – 20

8

79

17.5

140.0

0.425

0.18

1.4

20 – 25

3

82

22.5

67.5

5.425

29.43

88.3

25 – 30

2

84

27.5

55.0

10.425

108.68

217.4

30 – 35

5

89

32.5

162.5

15.425

237.93

1189.6

35 – 40

4

93

37.5

150.0

20.425

417.18

1668.7

40 – 100

3

96

70.0

210.0

52.925

2801.05

8403.2

100 - 170

4

100

135.0

540.0

13906.30

55625.2

∑

Mid point x

100

Mean = x = ∑fx ∑f

117.92

1707.5

78025.6

= 1707.5 = 17.075% 100

5 Marks

Median: The median position lies in the class interval 5 - 10 Median = Lm + {N/2 - Fm-1} x Cm { fm } where Lm

= lower boundary of median class (5 - 10)

Fm-1 = cumulative frequency of class prior to median class (36) Fm

= frequency of median class (25)

Cm

= median class width (5),

Therefore, median

(ii)

N = ∑f = 100

=

5 + {100.5/2 - 36}x 5 25

=

7.85%

5 Marks

Standard Deviation. σ

= √ ∑ f (x - x)2 ∑f =

√ 780.256

=

√ 78025.6 100

=

27.9% 5 Marks

(iii)

When the data set is distributed in a symmetrical way all three types of average – the mean, median, mode – have the same value. However, when a set of data is skewed in one direction the three types of average will not be identical; the mean will always be the most heavily influenced by the direction of skew, that is, it will be most affected by a few extreme values. In this case the distribution is positively skewed and the mean is greater than the median. 5 Marks [Total: 20 Marks] Page 7

Solution 4. (i) y is the dependent variable and the data in the table can be represented by a linear regression equation. Sales (€000s)

Advertising Expenditure (€000s)

25

8

35

12

29

11

24

5

28

14

12

3

18

6

27

8

17

4

30

9

The linear regression equation is: y = a + bx where ∑y = na + b∑x ∑xy = a ∑ x + b ∑x2 ; and ∑y - b∑x n n

a =

b = n∑xy - ∑x∑y, n∑x2 - (∑x)2 Sales (€000s) y

Advert Exp (€000s) x

x2

y2

xy

25

8

64

625

200

35

12

144

1225

420

29

11

121

841

319

24

5

25

576

120

28

14

196

784

392

12

3

9

144

36

18

6

36

324

108

27

8

64

729

216

17

4

16

289

68

30

9

81

900

270

∑ 245

∑ 80

∑ 756

∑ 6437

∑ 2149

Substituting the above values b

a

=

10 x 2149 - 80 x 245 10 x 756 - 802

=

21,490 - 19,600 = 1890 = 1.63 7560 - 6400 1160

=

245 - 1.63 x 80 10 10

= 11.46

[10 Marks]

Therefore, y = 11.46 + 1.63x

Page 8

(ii)

The strength of the relationship is represented by the correlation co-efficient, r where

r

= √[n∑x2

=

n∑xy - ∑x∑y, - (∑x)2][n∑y2 - (∑y)2]

10 x 2149 – 80 x 245 √[10 x 756 – 802][10 x 6437 – 60,025]

=

1890 √5040200

=

0.842

=

1890 2245

Since the value of r is close to +1, this suggests a strong positive relationship between the variables – expenditure on advertising is directly related to sales, that is, 84.2% of the variation in sales is due to the variation in the levels of advertising expenditure. This can be confirmed prior to the calculation of r by drawing a scatter graph which will show the form and strength of the relationship.

[10 Marks] [Total: 20 Marks]

Page 9

SOLUTION 5. (i)

Derivation of the trend by means of centred moving averages.

Quarter Year

1

2

3

4

2003

3

5

8

4

2004

4

6

10

5

2005

5

8

11

7

The moving averages must coincide with the natural cycle of the series. In this case the moving average must have a period of 4. Since we are dealing with an even period, in order to calculate a satisfactory trend it is necessary to use centring where the calculated averages are averaged in successive overlapping pairs. This ensures that each trend value coincides with a time period.

Year

Quarter

Sales (000s)

Moving totals of 4

(y)

2003

Deviations

Seasonal Variation

(y – t)

s

Seasonally adjusted data

1

3

-2.047

5.047

2

5

0.079

4.921

3 4

2 3 4 1 2

-1.375

-1.671

5.671

5.75

-1.75

-2.047

6.047

6.125

-0.125

0.079

5.921

6.375

3.625

3.641

6.359

6.75

-1.75

-1.671

6.671

7.125

-2.125

-2.047

7.047

0.079

7.991

7

5 29

5.375

6.5

5 28

4.359

6.25

10 26

3.625

6

6 25

2.875

5.5

4 24

5.125 5.5

4 22

1

5

8 21

2005

Centred Moving Av

(t)

20

2004

Moving Average

7.25

8

7.5 31

0.5

7.75

3

11

3.641

7.359

4

7

-1.671

8.671

[5Marks]

Page 10

[5 Marks]

[5 Marks]

(ii)

The last three columns in the table above outline the calculations for seasonally adjusted data. Seasonal variation gives an average effect on the trend which is attributable to the ‘season’ itself. They may be expressed in terms of deviations or percentages of the trend. In this case the additative model is used, that is, y = t (trend) + s (seasonal) + r (residual variations) The s values are found by calculating the mean of (y – t) values and adjusting the averages so that their total is zero, as follows.

Q1

Q2

2003

-------

-------

2004

-1.75

2005

-2.125

0.5

------

Totals

-3.875

0.375

7.500

-3.125

Mean

-1.938

0.188

3.750

-1.562

Adjustment

-0.109

-0.109

-0.109

-0.109

Adj. average

-2.047

0.079

3.641

-1.671

-0.125

Q3

Q4

2.875

-1.375

3.625

-1.75 ----Total 0.438 Total 0

[5 Marks]

Each of the adjusted average values are placed in column 8 of the table above and modify column 3 to give the adjusted figures in column 9. [Total: 20 Marks]

Page 11

SOLUTION 6. (i)

The equation will have the general form y = a + bx (y is the dependent variable, x is the independent variable) where y = total charge, x = number of units (kwhrs) used [2 Marks]

a = fixed costs, b = cost per unit of energy used (kwhr). Therefore the electricity model can be represented by y = 30.04 + 0.0239x.

[2 Marks] The annual cost for 2400 kwhrs is y = 30.04 x 6 + 0.0239 x 2400 = 180.24 + 57.36 = €237.60 (ii)

[2 Marks]

95% confidence interval is derived from the following and limits for _

µ

x - 1.96 (s/√ n) <

< x + 1.96 (s/√n)

where n = sample size = 36; s = sample standard deviation = 110; x = average sample expenditure = €720 720 - 1.96 (110/∧36) < 720 - 35.93 €684 <

µ

<

µ

<

µ

[3 Marks]

< 720 + 1.96 (110/∧36)

< 720 + 35.93 €756

[3 Marks]

We can be 95% confident that the mean amount of profit is between €684 and €756. (iii)

The purpose is to determine the rate of interest when applied to €12.000 will achieve a level of €25,200, compounded over a 10 year period; S = P(1 + r)10 where P = €12,000, S = €25,200, r = ?

[4 Marks]

25,200 = 12,000(1+ r)10 (1+ r)10 = 25,200/12,000 = 2.1, that is, 1 + r =

10√2.1

= 1.077

Therefore r = 1.077 - 1 = 7.7%. [4 Marks] Its claim is correct – the increase in wages is greater than the rate of inflation at 6.8%. [Total: 20 Marks]

Page 12