(l5) Decision Making Techniques

8/16/2019 (l5) Decision Making Techniques

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CHARTERED ACCOUNTANTS EXAMINATIONS

LICENTIATE LEVEL ________________________

L5: DECISION MAKING TECHNIQUES _______________________

WEDNESDAY 16 DECEMBER 2015

_______________________

TOTAL MARKS – 100: TIME ALLOWED: THREE (3) HOURS ________________________

INSTRUCTIONS TO CANDIDATES

1. You have fifteen (15) minutes reading time. Use it to study the examination papercarefully so that you understand what to do in each question. You will be told when tostart writing.

2. There are SEVEN questions in this paper. You are required to attempt any FIVEquestions. ALL questions carry equal marks.

3. Enter your student number and your National Registration Card number on the front ofthe answer booklet. Your name must NOT appear anywhere on your answer booklet.

4. Do NOT write in pencil (except for graphs and diagrams).

5. Cell Phones are NOT allowed in the Examination Room.

6. The marks shown against the requirement(s) for each question should be takenas an indication of the expected length and depth of the answer.

7. All workings must be done in the answer booklet.

8. Present legible and tidy work.

9. Graph paper (if required) is provided at the end of the answer booklet.


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Attempt any five (5) questions in this paper.

QUESTION ONE

(a) A husband and a wife appear in an interview for two vacancies in the same post. The

probability of husband’s selection is1

7 and that of wife’s selection is

1

5. Find the

probability that:

(i) Both of them will be selected. (2 marks)

(ii) Only one of them will be selected. (2 marks)

(iii) None of them will be selected. (2 marks)

(b) In a razor blade factory, machines A, B and C manufacture 25%, 35% and 40% of the

total output respectively. Of their outputs, 5%, 4% and 2% are defective razor blades. A

razor blade is chosen at random and found to be defective.

(i) Find the probability that the razor blade is defective. (3 marks)

(ii) A razor blade is selected at random and it is found to be defective.

Find the probability that it was manufactured by machine B. (2 marks)

(c) The XYZ Furniture Company produces chairs and tables from two resources, labour and

wood. The company has 80 hours of labour and 36 pounds of wood available each day.

Demand for chairs is limited to 6 per day. Each chair requires 8 hours of labour and 2

pounds of wood, whereas a table requires 10 hours of labour and 6 pounds of wood.

The profit derived from each chair is K400 and from each table is K100. The company

wants to determine the number of chairs and tables to produce each day in order to

maximize profit.

(i) Formulate a linear programming model for this problem (3 marks)

(ii) Solve the model by using graphical method. (6 marks)

[TOTAL: 20 Marks]


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QUESTION THREE

An investigator obtains scores for a person’s age in years and their score on a measure ofmemory function. Higher scores on the memory task reflect better memories. The investigatorhas predicted that the two variables will be significantly correlated. Using the data presentedbelow, please address the following questions:

Age Memory

23

45

18

61

56

79

41

33

67

30

16

18

21

9

12

9

15

10

8

14

Required:

(a) Calculate the least squares regression equation. (9 marks)(b) Calculate the product moment correlation coefficient and interpret your result.

(5 marks)(c) Compute the coefficient of determination and interpret your result. (4 marks)(d) Explain one limitation of linear regression analysis. (2 marks)

[Total: 20 Marks]


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QUESTION FOUR

(a) KASTA Mutual Fund has K600,000 for one of the three investment options in the stock

market – a growth fund, government bonds or a blue chip company. KASTA feels that

the stock market can assume the states as listed below. The company’s estimate of the

associated payoffs is also given in the table

Stock market trend

Type ofinvestment

Boom Moderate growth Moderate decline Collapse

Growth fund 475 000 250 000 -60 000 -500 000

Governmentbonds

600 000 200 000 -350 000 -600 000

Blue ChipCompany

350 000 200 000 0 -400 000

Using the minimax regret criterion, decide which option KASTA should adopt. (5 marks)

(b) A wholesaler stocks an item for which demand is uncertain. He wishes to assess two

re – ordering policies: to order 10 units at reorder level of 10 orders, 15 orders at a reorder

level of 15 orders, to see which is most economical over a 10 day period.

The following information is available

Demand per day (units) Probability

4 0.15

5 0.256 0.30

7 0.20

8 0.10

Carrying costs K10 per unit per day. Ordering cost K50 per order. Loss of goodwill for each unit

out of stock is K30. Lead time 3 days. Opening stock is 17 units. The probability distribution is

to be based on the following random numbers

63 88 55 46 55 69 13 17 36 81

84 63 70 06 20 41 72 37 53 90

Note: the reorder level is physical stock plus any replenishment outstanding.

Required:

Determine the most economic order policy (15 marks)

[Total: 20 Marks]


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QUESTION FIVE

(a) A university administrator wishes to estimate within ±0.05 and with 95% confidence theproportion of students enrolled in MBA programs and have undergraduate degrees in

business.

(i) Find the sample size to be collected, as minimum, if there is no basis for

estimating the approximate value of the proportion before the sample is taken.

(5 marks)

(ii) Find the minimum sample size required if prior data and information indicate that

the proportion will be no larger than 0.20. (5 marks)

(b) Suppose that 30% of the employees in a large firm are in favour of union

representation, and a random sample of 8 employees are contacted and asked for an

anonymous response.

Find the probability that:

(i) A majority of the respondents will be in favour of union representations.

(5 marks)

(ii) Fewer than half of the respondents will be in favour of union representation.

(5 Marks)

[Total: 20 Marks]

QUESTION SIX

(a) Four types of advertising display were set up in 12 retail outlets, with three outlets

randomly assigned to each of the displays, for the purpose of studying the point – of –

sale impact of the displays. With reference to the data in the table below test the null

hypothesis that there are no differences among the mean sales values for the four types

of displays, using 5% level of significance.

Type display Sales Total sales A 45 49 48 142

B 58 59 64 181

C 53 43 51 147

D 53 66 52 171

(14 marks)


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(b) The amount of time required for car transmission service is normally distributed with the

mean 46 and the standard deviation 9.0 . The service managerplans to have work begin on the transmission of a customer’s car 12 minute after the

car is dropped off, and the customer is told that the car will be ready within 1 hour total

time.

Find the probability that the service manager will be wrong. (6 marks)

[Total: 20 Marks]

QUESTION SEVEN

(a) The monthly incomes of a large group of middle managers are normally distributed with

a mean of K800 and a standard deviation of K45.

Find the:

(i)

probability of finding a middle manager with a weekly income of between K750and K850. (3 marks)

(ii) percentage of middle managers that earn less than K905. (3 marks)

(iii) limit of income above which the top 10% of the managers earn. (4 marks)

(b) A project consists of a series of tasks labeled A, B, C, D, E, F, G, H, I with the following

relationships ( ,W X Y means X and Y cannot start until W is completed; , X Y W

means W cannot start until both X and Y are completed). With this notation construct

the network diagram having the following constraints:

, ; A D E , ; B D F ;C G ; B H , F G I (6 marks)

(c) In a certain factory producing cycle tyres, there is a small chance of 1 in 500 tyres to be

defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the

approximate number of lots containing two defective tyres. (4 marks)

[Total: 20 Marks]

END OF PAPER


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L5: DECISION MAKING TECHNIQUES

SUGGESTED SOLUTIONS

DECEMBER 2015 EXAMINATIONS

SOLUTION ONE

(a) P (husband’s selection)1

7

P (wife’s selection)1

5

(i) P (both selected)1 1 1

7 5 35

(ii) P (only one selected) P (only husband’s selection) P (only wife’s

selection)

1 4 1 6

7 5 5 7

10

35

2

7

(iii) P (none of them will be selected6 4

7 5

24

35

(b)

(i) Use total probability for A, B and C from a partition because they are

disjointLet A = event that a razor blade was manufactured by machine A

B = event that a razor blade was manufactured by machine B

C = event that a razor blade was manufactured by machine C

D = event that the selected razor blade is defective

0.25, P A 0.35 P B and 0.40 P C


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/ 0.05, P D A / 0.04 P D B and / 0.02 P D C

/ / / P D P A P D A P B P D B P C P D C

0.25 0.05 0.35 0.04 0.4 0.02

0.0345

(ii)

//

P B P D B P B D

P D

0.35 0.04

0.0345

0.4058

(c)

(i) Let1

x number of chairs the company can produce each day

2 x number of tables the company can produce each day

Maximize1 2

400 100 Z x x

Subject to1 2

8 10 80 x x (labour constraint)

1 22 6 36 x x (would constraint)

1 6 x (demand for chairs)

Where1 2, 0 x x

(ii)

Tables of values

1 28 10 80 x x ….. 1 mark 1 22 6 36 x x ………. 1 mark

1 x 0 10

2 x 8 0

Test point 0,0

0 80 (True) 0 36 (True)

1 x 0 18

2 x 6 0


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1

2

max

6

3.2

2720

x

x

Z

Therefore, the company should produce 6 chairs and 3 tables in order to make

the profit of K2,720.


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SOLUTION TWO

(a)

Capital cost 7,000C K . Let it be profitable to replace the machine after n

years. Then n should be determined by the minimum value of T. Values of T for

various years are computed in the table below.

(1)

Year

of

service

n

(2)

Resale

value

S

k

(3)

Purchase

price –

resale

value

C S

k

(4)

Annual

maintenance

cost f t

k

(5)

Summation

of

maintenance

cost 0

n

t

f t

k

(6)

Total cost

0

n

t

C S f t

k

(7)

Average annual

cost

0

1 n

t

C S f n

k

1

2

3

4

5

6

7

8

4,000

2,000

1,200

600

500

400

400

400

3,000

5,000

5,800

6,400

6,500

6,600

6,600

6,600

900

1,200

1,600

2,100

2,800

3,700

4,700

5,900

900

2,100

3,700

5,800

8,600

12,300

17,000

22,900

3,900

7,100

9,500

12,200

15,100

18,900

23,600

29,500

3,900

3,550

3,166.67

3,050

3,020

3,150

3,371.43

3,687.50

We observe from the table that average annual cost is minimum (K3, 020) in the

5th year. Hence the machine should be replaced at the end of 5 years of service

(b)

(i) 0.4 p and 1 0.6q p

Let X be the random variable ‘the number of fine days in a week’.

The , X Bin n p where 7n and 0.4 p

Now 7 0.4 2.8 E X np


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(ii) 7 0.4 0.6 1.68Var X npq

Standard deviation 1.68 1.30 days

(c)

(i)

Let X be the random variable ‘the number of bacteria in 1ml of liquid’

Then 0 4 X P so that

4

4, 0,1,2,...

!

xe

P X x x x

4 0

440 0.0183

0!

e P X e

(ii) 4 4

44 0.195

4!

e P X


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SOLUTION THREE

(a)

X Y XY X2 Y2

23 16 368 529 256

45 18 810 2025 324

18 21 378 324 441

61 9 549 3721 81

56 12 672 3136 144

79 9 711 6241 81

41 15 615 1681 225

33 10 330 1089 100

67 8 536 4489 64

30 14 420 900 196

453 132 5389 24135 1912

1634.0

1.3614

6.590

10

45324135

10

132*4535389

2

b

602.20

10

4531634.0

10

132

a


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x y 1634.0602.20

(b)

7544.0

121023.7829

5906

1321912104532413510132453538910

22

r

(c) (i) Coefficient of determination = (-0.7544)² x 100% = 56.9%.

(ii) Therefore 56.8% of the memory function can be attributed to a person’s age in years.

There is an assumption that there is always one independent variable when in fact more than

one independent variable can affect the dependent variable.

SOLUTION FOUR

(a)

Table of Regret (Amounts are in thousands of kwacha)

Stock market trend

Type ofinvestment

Boom Moderategrowth

Moderatedecline

Collapse Maximum Min

Growth fund 125 0 60 100 125 125

Governmentbonds

0 50 350 200 350

Blue ChipCompany

250 50 0 0 250

Adopt the Growth fund


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(b)

The random numbers are allocated to the demand as follows

Demand Probability Cumulative

Probability

Random number

intervals4 0.15 0.15 00 – 14

5 0.25 0.40 15 – 39

6 0.30 0.70 40 – 69

7 0.20 0.90 70 – 89

8 0.10 1.00 90 – 99

Random numbers for order quantity of 15 units

Randomnumbers

63 88 55 46 55 69 13 17 36 81

demand 6 7 6 6 6 6 4 5 5 7

Results of the simulation using order quantity and reorder level of 15 units.

Day Openingstock

Demand Closingstock

Ordercosts@K50

Carryingcosts@K15

Stock outcosts@K30

Totalcost

1 17 6 11 K50 K165 2152 11 7 4 K60 60

3 4 6 K60 60

4 +15 6 9 K50 K135 185

5 9 6 3 K45 45

6 3 6 K90 90

7 +15 4 11 K50 K165 215

8 11 5 6 K90 90

9 6 5 1 K15 15

10 +15+1 7 9 K135 135

K150 K810 K150 K1110


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Random numbers for order quantity of 15 units

Randomnumbers

84 63 70 06 20 41 72 37 53 90

demand 7 6 7 4 5 6 7 5 6 8

Results of the simulation using order quantity and reorder level of 10 units.

Day Openingstock

Demand Closingstock

Ordercosts@K50

Carryingcosts@K15

Stock outcosts@K30

Totalcost

1 17 7 10 K50 K150 200

2 10 6 4 K60 60

3 4 7 K90 90

4 +10 4 6 K50 50

5 6 5 1 K15 156 1 6 K150 150

7 +10 7 3 K50 K45 95

8 3 5 K60 60

9 - 6 K180 180

10 +10 8 2 K50 K30 80

K200 K300 K480 K980

The simulation shows that the ‘order 10’ policy is more economical over the ten days

simulated.

SOLUTION FIVE

(a)(i) The sample size occurs at a point of maximum sample size where

0.5; and ±0.05; where 0.05 .

...

384.16 ≈ 385


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(ii) 0.20; 0.80 ±0.05 0.05

...

. 245.8624

≈ 246

(b)(i) 0.30; ; 0.70 ; 8

≥ 5 ∑ (8) 0.30

=0.70−

0.04668 + 0.0100 + 0.00122472 + 0.00006561 0.05797

(ii)

≤ 3 ∑ (3)

=0.300.70−

0.0576+0.1977+0.2965+0.2541 0.8059


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SOLUTION SIX

(a)

34799

641

12 558.917

142 + 181 + 147 + 171

3 641

12 351.584

55.917351.584 207.334 Source ofvariation

SS Df Ms

SS(Typedisplay)

351.584 3 117.19 4.52

SSE 207.333 8 25.92SSTO 558.917 11

Reject the null hypothesis 4.52 is greater .,, 4.07 and conclude that there aredifferences among the mean sales values for the four types of display.

(b)

> 48 , since actual work is to begin in 12 minutes

48 46

9 29 ≈ 0.2222

> 48 > 0.222 0.5 0.0871 0.4129


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SOLUTION SEVEN

(a)

Let X be the random variable ‘the monthly income of middle managers’.

Then 800, 45 X N

We standardize X so that800

45

X Z

(i) 750 800 800 850 800

750 85045 45 45

X P X P

1.11 1.11 P Z

1 1.11 1.11Q Q

1 2 1.11Q

1 2 0.1335

0.7330

(ii) 800 905 800

90545 45

X P X P

2.333 P Z

1 2.333Q

1 0.0099

0.9901 100%

99.01%

(iii) 0.1 P X a

800 8000.1

45 45

X a P

8000.1

45

a P Z

8000.1

45

aQ

From tables

1.28 0.1Q

So800

1.2845

a

800 45 1.28a

800 57.6a


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57.6 800a

857.60a K

(b) For the given precedence relationships, the project network shown below is

obtained since H is preceded by B while F is preceded by B and D, a dummyactivity must be incorporated to draw the network.

(c)

1

,500 P 10n 1 1

10 0.02500 50

np

!

xe

P X x x

20.020.02

22!

e P X

0.9802 0.0002

0.00019604

Number of lots containing 2 defectives

10, 000 0.00019604

2 lots

END OF SOLUTIONS

(l5) Decision Making Techniques

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