F1 - Business Maths August 08

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - AUGUST 2008 NOTES

You are required to answer any 5 questions. (If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to be marked. Otherwise, only the first 5 answers to hand will be marked). All questions carry equal marks. STATISTICAL FORMULAE TABLES ARE PROVIDED DEPARTMENT OF EDUCATION MATHEMATICS TABLES ARE AVAILABLE UPON REQUEST

TIME ALLOWED:

3 hours, plus 10 minutes to read the paper.

INSTRUCTIONS:

During the reading time you may write notes on the examination paper but you may not commence writing in your answer book. Marks for each question are shown. The pass mark required is 50% in total over the whole paper. Start your answer to each question on a new page. You are reminded that candidates are expected to pay particular attention to their communication skills and care must be taken regarding the format and literacy of the solutions. The marking system will take into account the content of the candidates' answers and the extent to which answers are supported with relevant legislation, case law or examples where appropriate. List on the cover of each answer booklet, in the space provided, the number of each question(s) attempted.

The Institute of Certified Public Accountants in Ireland, 17 Harcourt Street, Dublin 2.

THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS & QUANTITATIVE METHODS FORMATION 1 EXAMINATION - AUGUST 2008

Time Allowed: 3 hours, plus 10 minutes to read the paper. You are required to answer any 5 questions. (If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to be marked. Otherwise, only the first 5 answers to hand will be marked). All questions carry equal marks.

1.

The accountant in DIY Products Ltd. estimates that the company will have to replace some of the automatic production machines at the year end. There are three machines available, each costing €20,000 but the projected annual cash flows for each machine are different. The machines have no residual value at the end of year 5. The bank has agreed to fund the cost of the machine at an overdraft rate of 10% subject to the company carrying out an appropriate analysis. The projected cash flows are:

You are asked to:

(i)

(ii)

(iii)

Year 1 2 3 4 5

Machine A €3,000 €4,000 €7,000 €8,000 €8,500

Machine B €5,000 €6,000 €6,000 €4,000 €4,000

Machine C €6,000 €6,000 €6,500 €4,000 €1,000

Compare the alternatives using Net Present Value.

Derive the Internal Rate of Return on the most viable proposal. Describe the critical factors affecting the investment decision.

1

(8 Marks)

(8 Marks)

(4 Marks)

[Total: 20 Marks]

2.

The Impart Union represents junior and middle management administrative staff. It compiles earnings data to ensure that the terms and conditions of employment of its members over a range of companies are similar. The data below has been provided by two companies relating to 100 employees. In negotiations the Union claims that there is a major variation in the weekly wages paid to employees in both companies. Weekly Earnings €

300 360 420 480 540 600 650 700 800 900 1100

Analyse the data by: (a)

(b)

3.

(i)

(ii)

– 360 – 420 – 480 – 540 – 600 – 650 – 700 – 800 – 900 – 1100 – 1400

No. Employees In Company A 2 6 14 18 16 10 7 10 10 6 1

No Employees In Company B 2 3 7 3 7 9 3 16 10 20 20

Calculating the mean and standard deviation for both companies.

Deriving the co-efficient of variation.

Comment on the Unionʼs claim.

(12 Marks) (4 Marks)

(4 Marks)

[Total : 20 Marks]

The table below gives the number of disposable digital thermometers manufactured by Superior Products Ltd. in thousands over the past 4 years. Year 2004 2005 2006 2007

1

14 16 18

Quarters 2

16 17 20

By using the method of moving averages you are asked to: (i)

(ii)

3

9 12 13

4 13 14 17

Smooth this data by means of a centred four quarterly moving average.

Calculate the average seasonal variations.

(10 Marks)

(10 Marks)

[Total: 20 Marks]

2

4.

DIY Ltd. produces a range of components for manufacturing printed circuit boards. The quantities produced over the past number of years are set out in the table below: Component 1 2 3 4

2004 Average Price € 3.00 4.50 1.00 7.00

2004 Quantity (000) 90 180 1000 20

2005 Quantity (000) 95 200 1080 20

2006 Quantity (000) 100 150 1160 20

2007 Quantity (000) 135 90 1600 20

The Management Accountant wishes to analyse this data. As a first step, you are required to:

(i)

(ii)

5.

Explain the purpose of preparing this index and comment on your results.

(12 Marks) (8 Marks)

[Total: 20 Marks]

As the Managing Partner of CPA consultants, you are asked to advise on a large range of business problems. (i)

(ii)

(iii)

6.

Construct a quantity index for the components listed for the period 2004 – 2007 using 2004 prices as weights.

The Cost Accountant of DIB Ltd. wishes to purchase office equipment for €37,500 on 1st October 2008. The equipment will last for 5 years and will be replaced on 1st October 2013. The bank has provided a 5 year loan compounded annually at 12%. You are required to determine the size of the equal annual loan repayments. (6 Marks)

DIB Ltd. produces a new product where its sales are expected to be 70 units per week when the price is €50 per unit. However, the price was advertised at €70 per unit and only 30 units were sold. The fixed costs are €1,500 per week and the variable costs per unit are €10. You are asked to develop a demand relationship between the price and quantity demanded and to find the cost of producing 100 units. (6 Marks)

DIB Ltd. also manufactures batteries. Its states that each battery will last for approximately 200 hours with a standard deviation of 4 hours. You are asked to determine the probability that a random sample will last not less than 198 hours and to support your calculation graphically, assuming that the data is normally distributed. (8 Marks) [Total: 20 Marks]

You have attended a management conference on the subject “Methods of Describing Sets of Data”. The presentation covered: ● ● ●

●

The data needed for management decision-making. Methods of describing qualitative data such as charts and graphs. Methods of describing quantitative data such as histograms, scattergrams, measures of central tendency and dispersion and time series analysis. How presentations may be used to distort the data.

Discuss the content of the presentation with appropriate descriptions of the material outlined.

[Total: 20 Marks]

END OF PAPER 3

SUGGESTED SOLUTIONS

THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

SOLUTION 1 (i) Net Present Value. Year

0 1 2 3 4 5 NPV

Cash Flow A

(€20,000) €3,000 €4,000 €7,000 € 8,000 € 8,500

FORMATION 1 EXAMINATION - AUGUST 2008

Cash Flow B

(€20,000) €5,000 €6,000 €6,000 €4,000 €4,500

Cash Flow C

Discount Factor @ 10%

(€20,000) €6,000 €6,000 €6,500 €4,000 €1,000

1.000 0.909 0.826 0.751 0.683 0.621

PV of A

(€20,000) €2,727 €3,304 €5,257 €5,464 €5,279 €2,031

(2 Marks)

PV of B

(€20,000) €4,545 €4,956 €4,506 €2,732 €2,484 (€777)

(2 Marks)

PV of C

(€20,000) €5,454 €4,956 €4,881 €2,732 € 621 (€1,356)

(2 Marks)

The only viable option for the company is purchase of machine A which provides a positive net present value. (2 Marks) (ii)

[8 Marks]

Since A has the highest NPV, this is the proposal for which the IRR is calculated. To derive the IRR by interpolation or otherwise a discount factor of 14% is used. Year

Cash Flow A

Discount Factor @14%

2

€4,000

0.769

0

1

3

4

5

NPV

(€20,000) €3,000

1.000

(€20,000)

0.675

€4,725

0.877

€7,000

€ 8,000

0.592

€ 8,500

PV of A

0.519

Using the following formula for IRR

N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate I2 gives NPV N2. N1 - N2

€2,631 €3,076

€4,736

€4,412 (€420)

(2 Marks)

Where N1 = €2,031, I1 = 10%; N2 = (€420), I2 = 14%

(2 Marks)

The IRR for the project (Machine A) is 13.3%.

(4 Marks) [8 Marks]

IRR = 2031 x 0.14 - 420 x 0.10 2031 - 420

= 326.34 = 13.3% 2451

5

(iii)

The process of investment for capital expenditure decisions is a vital part of policy making and top management usually assumes direct responsibility for the authorisation of all large capital investment decisions because

- the sums involved are usually very large - the companyʼs resources may be tied up for a considerable period of time - investment decisions are long-term and have a major impact on the viability of the company. The critical factors affecting the capital investment decision are

- the projected future increases in income or savings in costs - the net amount of the investment. The investment should not be so large that the failure of the investment could have an impact on the survival of the company - a satisfactory rate of return. This return should be sufficient to finance the cost of borrowings - the level of risk. It is necessary to consider the risk factor since, in all business areas, the future is uncertain. (4 Marks)

[Total: 20 Marks] SOLUTION 2 (i)

The mean and standard deviation for both companies. Weekly

Freq (f)

Mid point x

360-420

6

390

Earnings € 300-360

420-480

480-540 540-600

600-650

650-700 700-800

2

14

18 16

10 7

10

800-900

10

∑

100

900-1100

1100-1400 Mean

6

1

= x = ∑fx ∑f

330

450

Company A fx

(x – x)

(x – x)2

f(x – x)2

2340

-228

51984

311904

-48

2304

660

6300

510 570

-108

4725

57

6250

675 750

7500

850

8500

1000

6000

1250

1250

61825

= 61825 = €618.25 100

Standard deviation = σ = √ ∑ f (x - x)2 = √ 3130425 ∑f 100 =

√ 31304.25

=

-168

9180 9120

625

-288

€176.93

6

7

132

232

382

632

(000)

82944

165888

28224

395136

49

490

11664 3249

17424

53824

145924 399424

209952 36864

22743

174240

538240

875544 399424

3130425 (2 Marks)

(4 Marks)

Weekly

Freq (f)

Mid point x

360-420

3

390

Earnings € 300-360

2

420-480

7

480-540

3

540-600

650-700

3

700-800

16

1100-1400

20

800-900

900-1100 ∑

= x = ∑fx ∑f

625

5625

1250

(ii)

Co-efficient of variation. Company A:

Company B:

σ/x

€176.93/€618.09

€273.93/€837.05

€273.93

=

-327

106929

-387

-267

-212

28.62%

32.72%

-87 13

413

514098

149769

1048383

44944

449440

169

1690

71289

26569

83650

=

257049

163

25000

=

599427

-507

20000

= 83650 = €836.50 100

√ 75041.55

199809

(000)

26244

Standard deviation = σ = √ ∑ f (x - x)2 = √ 7504155 ∑f 100 =

f(x – x)2

-162

8500

1000

(x – x)2

2025

12000

850

100

-447

3990

750

20

1170

660

1530

675

10

(x – x)

3150

570

9

fx

450

510

7

600-650

Mean

330

Company B

7569

170569

320787

427734 78732

121104

531380

3411380

7504155 (2 Marks)

(4 Marks)

(2 Marks)

(2 Marks)

When considering different distributions with significantly different means, it is necessary to use other means than the variance or standard dev deviation to make a realistic comparison. The co-efficient of variation provides a method of comparison of spread relative to the magnitude of data under consideration. In the present case the employees in company B appear to show substantial variability in both mean and standard deviation. This is confirmed by calculating the co-efficient of variation which shows that employees in company B show a greater variation in earnings. However, the co-efficient of variation is not of such a level as to indicate that may be difficulties with the data or its interpretation. There may be other factors involved. It is obvious that in company B a large number of the sample are at the higher end of the earnings scale. This may be due to age and seniority factors in this company while the sample from company A may contain a substantial number of younger and less senior employees. Therefore there may be distortions in the sample data which require further investigation. Using the mean and standard deviations in isolation may indicate that a problem exists with the earning data which may not be correct. It is obvious that the data in both companies is positively skewed, that is, where the mean is greater than the median, therfore the Union claim is not necessarily correct. (4 Marks) [Total: 20 Marks]

7

SOLUTION 3 (i)

The original data is presented below. Year 1984 1985 1986 1987

Quarters 2

1

14 16 18

16 17 20

3

4 13 14 17

9 12 13

Using a four quartered centred moving average gives the trend in the table below in Col. 6.

Year

Quarter

Col 1

Col 2

1984 1985

Col 3

Moving annual total Col 4

Moving pair total

2

16

1

16

4

13

1

14

3

9

4

1986

Trend

Deviation

Col 5

Col 6

Col 7

52

105

13.125

2.875

56

115

14.375

53

14

108

55

111

1

18

67

135

16.875

17

2

64

20

3

13

2 Marks

131

68

2 Marks

2 Marks

2.375

2.375

15.125

4

1.375

1.875

13.875

121

126

0.375

- 4.125

59

62

Col 8

- 4.500

17

12

Seasonal variation

13.500

2

3

1987

(ii)

Data 000s

0.125

0.375

1.625

15.750

1.375

- 3.750

16.375

- 4.125

0.625

0.375

1.125

1.375

2.375

- 4.125

(10 Marks)

4 Marks

The deviation can be derived by using either the additive or multiplicative models. Using the additive model, the deviation is found in Col. 7 by deducting the trend from the original data. This deviation is then adjusted to provide the seasonal variation as below. The total average deviation is zero. If there is an adjustment the original data can be adjusted by the required amount to provide data incorporating the seasonal adjustments. Quarter 1985 1986 1987 Total Average Adjustment Seas Var

1 -----1.625 1.125 2.750 1.375 0 1.375

2 2.875 1.875 ------4.750 2.375 0 2.375

3 - 4.500 - 3.750 -------- 8.250 - 4.125 0 - 4.125

4 0.125 0.625 ------0.750 0.375 0 0.375

0 (10 Marks)

[Total: 20 Marks]

8

SOLUTION 4

(i) Using base prices as weights gives the quantity index as the base weighted index. Component

2004 Average Price €

2

4.50

1

3

7.00 P0 3.00 4.50 1.00 7.00

Q0 90 180 1000 20

2005 Quantity (000)

2006 Quantity (000)

2007 Quantity (000)

1000

1080

1160

1600

90

95

180

1.00

4 Com. 1 2 3 4 Total

3.00

2004 Quantity (000)

200

20

Q1 95 200 1080 20

20

Q2 100 150 1160 20

Q3 135 90 1600 20

Deriving the Index of Production where 2004 is the base:

2005 = 2405/2220 = 108.3

135

150

90

20

P0Q0 270 810 1000 140 2220

P0Q1 285 900 1080 140 2405

20

P0Q2 300 675 1160 140 2275

P0Q3 405 405 1600 140 2550

(2 Marks) (2 Marks) (2 Marks) (2 Marks)

(2 Marks)

2006 = 2275/2220 = 102.4

(2 Marks)

2007 = 2550/2220 = 114.8 (ii)

100

The index for the three periods indicates that the index reduced for 2006. The change in production can be due to changes in quantities produced and these changes can be measured by a volume or quantity index. Using base prices as weights gives the Laspeyres quantity index. This index uses base year statistics as weights as in the present case. This implies that the quantities do not vary over time and in many cases this is not correct, as in the present case. If the prices are kept the same, any variation in production must be due to changes in quantities produced so a quantity or volume index can be calculated. This index tends to overstate increases in P. To attempt to overcome this current year quantities may be used such as in the Paasche index. However, this tends to understate the affects of P. The Laspeyres index generally exceeds the Paasche index. (8 Marks) [Total: 20 Marks]

9

SOLUTION 5 (i)

The amortization method of debt repayment is required. If the amount of money is borrowed over a period, the money may be repaid by means of an amortization annuity. This consists of a regular payment in which each payment consists of both capital and interest. The debt is said to be amortized if this method is used. Many of the mortgages for home purchases are of this type. The amount borrowed, P, is equal to the net present value of the annuity payments, A, over the period of the debt, that is, P = A/(1 + i) + ……….. A/(1 n + i) , where i = interest rate and n = period of loan. (3 Marks) Since P

= €37,500, i = 12% = 0.12, n = 5

€37,500

= A/1.12 + A/1.122 + A/1.123 + A/1.124 + A/1.125

= A(0.893 + 0.797 + 0.712 + 0.636 + 0.567)

Therefore, A (ii)

= A(3.605)

(3 Marks)

= 37,500/3.605 = €10,402.

Since the relationship is linear, it can be represented as y = a + bx where y = price, x = quantity.

Therefore, when x = 70, y = 50; 50 = a + 70b.

when x = 30, y = 70; 70 = a + 30b.

Solving the equations for values of a and b gives a = 85 and b = - 0.5.

(3 Marks)

The demand equation is y = 85 - 0.5x.

The cost of producing 100 units: C = Fixed costs + Variable costs per unit

(3 Marks)

C = 1500 + 10 x 100 = €2500

(iii)

Any random variable which follows a normal distribution can be transformed to the standard normal variable z by using the transformation z = (x - µ)/σ. Therefore when x = 198, z = (x - µ)/σ (198 - 200)/4 = - 2/4 = 0.5.

Hence, P(x < 198) = P(z < - 0.5) = 0.5 - 0.1915 (from Normal Tables)

(4 Marks)

= 0.3085, that is 30.85%.

The appropriate Normal Distribution is

x1 198 -0.5

x

µ= 200 0

y

10

(4 Marks) [Total : 20 Marks]

SOLUTION 6

Methods for Describing Sets of Data In all accountancy, management and consultancy areas, a large volume of data is subject to analysis, interpretation and description. The methods and techniques used can distort the description of the data and ultimately the decision made. Characteristics of a data set may contain the most frequent score, the variability in the score, the ʻshapeʼ of the data, the highest and lowest scores, and whether or not the data set contains any unusual data. Interpreting or extracting data visually is, at a minimum, difficult since it may not be possible to comprehend large volumes of information. Some formal methods for summarising and characterising the information in a data set are essential. Most populations are large data sets. Therefore, methods for describing such data sets are also essential for statistical inference. There are two key methods used for describing data – one graphical and the other numerical. Both play an important role in statistics and both methods can be used for describing both qualitative and quantitative data. (5 Marks)

Describing qualitative data. When data is grouped into non-numerical categories, the resulting table is a categorical or qualitative distribution. Therefore the value of a qualitative variable can be classified into categories called classes. Such data can be summarized numerically in two ways – by computing the class frequency, that is, the number of observations in the data set that fall into each class or by computing the class relative frequency, that is, the proportion of the total number of observations falling into each class. Two of the most widely used graphical methods for describing qualitative data are bar graphs and pie charts. The bar graph plots the class frequency against the class where the height of the ʻbarʼ is equal to the class frequency. A pie chart shows the relative frequencies of the classes where the size of the ʻsliceʼ apportioned to each class is proportional to the class relative frequency. (5 Marks) Describing quantitative data. When data are grouped according to numerical size, the resulting table is a categorical or quantitative distribution. For describing, summarising and detecting patterns in such data the most common graphical methods for describing frequency distributions are histograms, frequency curves (such as polygons or ogives) and scattergrams.

Histograms. Histograms can be used to display either the frequency or relative frequencies of the measurements falling into specified intervals. By looking at a histogram two important facts are apparent. The proportion of the total area above the interval is equal to the relative frequency of measurements falling in the interval. As the number of elements in the data set increase, a better description of the data set can be obtained by decreasing the width of the class intervals. When the class intervals become small enough a relative frequency histogram will appear as a smooth curve. While histograms provide good visual descriptions of data sets they do not allow the identification of individual measurements.

Another form of presentation is the frequency polygon. The frequencies are plotted at the class marks and the successive points are connected. Applying a similar technique to a cumulative distribution (usually a ʻless thanʼ distribution) an ogive can be obtained. This is where the cumulative frequencies are plotted at the class boundaries.

Scattergram. A common method to describe the relationship between two quantitative variables is to plot the data on a scattergram. Both a relationship, and the strength of that relationship, can be determined between the variables by linear regression form of analysis. This can be visually and quantitatively presented and is an effective display.

A large number of numerical methods are available to describe quantitative data sets. Most of these methods measure one of two data characteristics 1) the central tendency of the set, that is, the tendency of the data to cluster and 2) the variability of the set, that is, the spread of the data. The most popular measure of central tendency is the arithmetic mean. This is the sum of the measurements divided by the number of measurements contained in the data set. Another important measure of central tendency is the median. This is the middle number when the quantitative data set is arranged in ascending or descending order. This is of most value in describing large data sets and is the point where 50% of the data lies above the mid point and 50% below it. In certain situations the median may be a better measure of central tendency than the mean. It is less sensitive to extremely large or small values. In general, extreme values (large or small) affect the mean more than the median since these values are used explicitly to calculate the mean. The median is not affected directly by extreme values since only the middle value is explicitly used to calculate the median. 11

The mode is particularly useful for describing qualitative data. The modal category is the class that occurs most frequently. Because it emphasises data concentrations, the mode is also used with quantitative data sets to locate the region in which much of the data is concentrated. However, for some quantitative data sets, the mode may not be very meaningful since there may be more than one mode in the sample. A more meaningful measure is the modal class. This can be obtained from a relative frequency histogram.

These measures of central tendency provide only a partial description of a quantitative data set. The description is incomplete without a measure of the variability of the data set. Knowledge of the dataʼs variability along with its centre can help us visualise the shape of the data as well as its extreme values. The simplest measure of the variability of a quantitative set is its range. This is equal to the largest less the smallest measurement. The range is easy to compute and to understand but it is an insensitive measure of data variation when the data sets are large – two data sets can have the same range but be vastly different with respect to data variation. The variation of such data can be obtained by measuring the distance between each measurement and the mean. To cater for the + and – signs of the deviations, the deviations are squared to provide the sample variance, s2 = ∑(xi – x)2/(n – 1). This is the preliminary step in calculating the standard deviation of the data set, √s2. Sample statistics like s2 are primarily used to estimate population parameters like _2; (n -1) is preferred to n when defining the sample variance.

However, data of interest to managers are often produced over a time period. When data is produced over time it is important to record both the measurements and the time period associated with each measurement. With this information a time series plot can be constructed to describe the time series data and to learn about the process that generated the data and to monitor the movement (trend) and changes (variations) in the variable being examined. (5 Marks)

Many of the presentations outlined can misrepresent or distort, or allow the data presented to be misinterpreted. One common way to change the impression created by a pictorial or graphical presentation is to change the scale on either one or both axes. By stretching the vertical axis or by increasing the distance between vertical units can give a misleading visual impression of the data. In one case a histogram may appear to be vertically elongated and horizontally compressed or vice versa and may lead to incorrect conclusions. A visual distortion can be achieved with bar graphs by making the width of the bars proportional to the height. A similar effect can be achieved by using a scale break for the vertical axis. Further distortions can also occur with numerical descriptive measures. If a measure of central tendency only is reported in a sample, this can lead to a distortion of the information. Both a measure of central tendency and a measure of variability are needed to obtain an accurate mental image of a data set. (5 Marks) [Total : 20 Marks]

12