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CPA
Foundation Leval
Quantitative Analysis December 2017
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1(a)

Q Outline tour applications or mathematical functions in business. (4 marks)
A

Solution


Applications of mathematics functions in business

➫ Break even analysis

➫ Linear programming
➫ Supply analysis

➫ Demand analysis

➫ Cost estimation

➫ Profit determination

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QUESTION 1(b)

Q Highlight the four components of a time series. (4 marks)
A

Solution


Components of a time series

(i) Secular trend(T)
➫ It defines the long-term movement of a time series, which may be growing, decreasing, or steady globally.

(ii) Cyclical variation(C)
➫ Cyclical variation refers to the ups and downs periodically reoccurring in a time series. These are brought on by a cycle of four phases a business go through at some point. The four stages of a business cycle are prosperity/ boom, recession, depression, and recovery.

(iii) Seasonal variation(S)
➫ Seasonal variations are changes in time series that occur as a result of repeated factors that behave in a predictable and regular way. Every year, these forces typically follow the same pattern or one that is very close.

(iv) Random/Irregular variation (R)
➫ These are alterations brought on by unanticipated and unforeseeable events. These forces don't follow any certain pattern and act in a completely random or chaotic manner.They could result from floods, famines, earthquakes etc.

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QUESTION 1(c)

Q ( i) Present the above information in a venn diagram. (4 marks)
(ii) The number of households that preferred Fex and Gcx. (l mark)
(iii) The probabilitv that a household selected at random docs not prefer any or the three goods. (l mark)
A

Solution


Let:

E = universal set.
F = Fex
G = Gex
M = Mex
Therefore;
n = 800
F = a + b + e + f = 230
G = b + c + d + e = 245
M = d + e + f + g = 325
e = 30.
d = 70
a = 110
g = 185
f = M - (d + e + g)
f = 325 - (70 + 30 + 185)
f = 40
b = F - (a + f + e)
b = 230 - (110 + 40 + 30)
b = 50
c = G - (b + e + d)
c = 245 - (50 + 30 + 70)
c = 95

Venn diagram


F
G
a = 110
e = 30
b = 50
c = 95
f = 40
d = 70
g = 185
M


(ii) The number of households that preferred Fex and Gex

n(F∩G ) = b + e = 50 + 30

80 households

(iii) The probabilitv that a household selected at random docs not prefer any or the three goods.

P(h)
=
n - (a + b + c + d + e + f + g)
n

P(h)
=
800 - (110 + 50 + 95 + 70 + 30 + 40 + 185)
800
P(h) = 220 / 800 = 0.275

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QUESTION 1(d)

Q The percentage of customers that will be in each of the three categories of accounts as at 31 December 2017. (6 marks)
A

Solution


From To
P D B Totals
P 285,000 15,000 0 300,000
D 20,000 700,000 30,000 750,000
B 0 0 450,000 450,000
305,000 715,000 480,000 1,500,000


Where;

P = Paid account
D = Delinquent account
B = Bad debt account

The proportion of customers' accounts transferred and retained between 31 december 2015 and 31 December 2017:

P D B
P 285,000 / 300,000 = 0.95 15,000 / 300,000 = 0.05 0 = 0 / 300,000 = 0.00
D 20,000 / 750,000 = 0.03 700,000 / 750,000 = 0.93 30,000 / 750,000 = 0.04
B 0 / 450,000 = 0.00 0 / 450,000 = 0.00 450,000 / 450,000 = 1.00


Matrix of transition

[
0.95
0.03
0.00
0.05
0.93
0.00
0.00
0.04
1.00
]


Initial state matrix

(300,000 / 1,500,000 750,000 / 1,500,000 450,000 / 1,500,000)

(0.2 0.5 0.3)

Hence

(0.2 0.5 0.3) x
[
0.95
0.03
0.00
0.05
0.93
0.00
0.00
0.04
1.00
]


0.205 0.475 0.32

Hence the percentage of customers that will be in each of the three categories of accounts as at 31 December, 2017:

For paid customer accounts

0.205 or 20.5%

For delinquent customer accounts

0.475 or 47.5%

For bad debt customer accounts:

0.32 or 32%

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QUESTION 2(a)

Q Enumerate three circumstances under which the Poisson distribution is most applicable. (3 marks)
A

Solution


Circumstances under which the poison distribution is most applicable

➢ It is applicable where events occur at random i.e are independent of each other.

➢ Where the probability of some event occurring [P] is very small relative to the probability of the event not occurring [q]. i.e for a rare or improbable event.

➢ Its can be used where the number of items or events is very large and perhaps not readily available or known

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QUESTION 2(b)

Q Determine whether the qualitv of the two types of electric components differ significantly. (Use a significance level of 95%). (t) ( 3 marks)
A

Solution


Z
=
(x̄1 - x̄2) - (μ1- μ2)
√
δ12
n1
+
δ22
n2


Where;

N1 size of sample 1
N2 size of sample 2.
X1 the mean of sample 1
X2 the mean of sample 2

Formulating hypotheses.

Ho:μ1 = μ2... no significant difference in quality

H1:μ1 = μ2
there's significant difference in quality of the two types of electric components

At significant level = 95%

α = 1 - 0.95 = 0.05

The alternative hypothesis (H1), formulated above shows that the test is a two- failed test

Hence;

Zα/2 = Z0.05 / 2 = Z0.025 = ±1.96

The critical Z-value i.e Zc = 1.96
X̄₁ = 1600
X̄₂ = 1528
δ₁ = 132
δ₂ = 149
n₁ = 120
n₂ = 110

Z
=
(1,600 - 1,528) - 0
√
1322
120
+
1492
110
= 3.865


The observed Z - value of 3.865 is more than the critical Z -value of 1.96 and lies in the rejection region for Ho. Hence, we reject the null hypothesis (Ho) and conclude that there is significant difference in quality of the two types of electric components manufactured.

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QUESTION 2(c)

Q (i) Is greater than Sh.500,000. (2 marks)

(ii) Is greater than Sh- I ,220,000. (2 marks)

(iii) Lies between Sh.852,OOO and Sh.1,1OO,OOO, (2 marks)
A

Solution


(i) Is greater than Sh.500,000

Z = (x - μ) / δ Where;

μ = Sh. 980,000
X = Sh. 500,000
δ = Sh. 160,000

∴ Z = ( 500,000 - 980,000) / 160,000
Z = -480,000 / 160,000 = -3

illustration



(ii) Is greater than shs. 1,220,000

Z = (x - μ) / δ Where;

μ = Sh. 980,000
X = Sh.1,220,000
δ = Sh. 160,000

∴
Z = (1,220,000 - 980,000) / 160,000
Z = 240,000 / 160,000 = 1.5

P(Z ≥ 1.5) = 0.5 - 0.4332 = 0.0668

Hence, there is 6.68% chance /probability that the annual income of the citizen selected at random is greater than Sh.1,220,000

(iii) Lies between Sh.852,OOO and Sh.1,1OO,OOO

Z = (x - μ) / δ Where;

μ = Sh. 980,000
X = Sh. 852,000
δ = Sh. 160,000

∴
Z = (852,000 - 980,000) / 160,000 = -0.8
Z = (1,100,000 - 980,000) / 160,000 = 0.75

P(-0.8 ≤ Z ≤ 0.75)

0.2881 + 0.2734 = 0.5615

Hence, there is 56.15% chance or probability that the annual income of the citizen selected at random lies between Shs. 852,000 and sh 1,100,000. .
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QUESTION 2(d)

Q Advise the management of Excellent Products Limited on the best assignment that will maximise production. (4 marks)
A

Solution


Maximum output (units 000) = 22

By subtracting every element from the maximum

Value = 22; give the opportunity cost matrix below.

Opportunity cost matrix

Output units ('000')

Machines
0
M1
M2
M3
M4
[
A
10
4
6
8
B
10
2
7
10
C
16
0
10
6
D
9
2
4
10
]


By subtracting the minimum value from each row gives

0
M1
M2
M3
M4
[
A
1
4
2
2
B
1
2
3
4
C
7
0
6
0
D
0
2
0
4
]


By subtracting the minimum value from each column gives:

A,B,C,D

0
M1
M2
M3
M4
[
A
0
3
1
1
B
0
1
2
3
C
7
0
6
0
D
0
2
0
4
]


The minimum number of lines drawn to cover the zeros ie. 3, is not equal to the order of the matrix i.e 4, hence no optimality.

The smallest uncovered element = 1. Hence, by subtracting 1 from all the uncovered elements and adding it to elements at the intersection of lines gives:
0
M1
M2
M3
M4
[
A
0
3
1
1
B
0
1
2
3
C
7
0
6
0
D
0
2
0
4
]


The number of minimum lines drawn to cover the zeros is now 4 which is equal to the order of matrix.Hence the solution is optimal.

The optimal assignment.

0
M1
M2
M3
M4
[
A
0
2
0X
0X
B
0X
0
1
2
C
8
0X
6
0
D
1
2
0
4
]


Assignment Machine Output
Product Units"000"
A M1 12
B M2 20
D M3 18
C M4 16
Maximum total production 66


There's multiple assignment, alternatively assign either B, C, D and A to M1, M2, M3 and M, respectively so as to achieve the maximum production of 66,000 units

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QUESTION 2(e)

Q (i) The profit function. ( l mark).
(ii) The output level that would maximise profit. (3 marks)
A

Solution


(i) The profit function

Total profit (TP) = total revenue (TR) Total cost (TC)
TP = TR - TC
TR = PQ
(75 - 0.180)Q

TR = 75Q - 0.180²

TC = 80Q + 5Q² + 0.03Q³

TP = (75Q - 0.18Q²) - (80Q + 5Q² - 0.030³)

75Q - 80Q - 0.18Q² - 5Q² + 0.03Q³

0.03Q³ - 5.18Q² - 5Q

(ii) The output level that would maximize profit

At maximum profit

Marginal profit (MP) = 0

MP = dTP / dQ

0.09Q² - 10.36Q - 5

Therefore;

0.09Q² - 10.36Q - 5 = 0

a = 0.09
b = -10.36
c = -5

- b
+
√
b2 - 4ac
2a

-(-10.36)
+
√
-10.362 - 4 x 0.09 x -5
2 x 0.09


∴

10.36 + 24.62

10.36 + 24.62 = 34.98 or
10.36 - 24.62 = -14.26

The output level that would maximize profit = 34.98

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QUESTION 3(a)

Q Distinguish between "regression analysis" and "con-elation analysis". (2 marks)
A

Solution


Distinguishing between "regression analysis" and "correlation analysis"

➫ Correlation analysis describes the relationship or association between two variables.
➢ Regression analysis describes how an independent variable (x) is numerically related to the dependent variable(y)

➫ The objective or purpose of regression analysis is to estimate values of random variable on the basis of the values of a fixed variable.
➢ The objective of correlation analysis is to find a numerical value expressing the relationship between variables.

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QUESTION 3(b)

Q Summarise two applications or rank correlation. (2 marks)
A

Solution


Summarising two application of rank correlation

➢ Used to measure the degree of similarity between two rankings.
➢ Used in assessing the significance of the relationship between two rankings

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QUESTION 3(c)

Q (i) The advertising cost function of the enterprise, using the normal equation. (7 marks)
(ii) The advertising cost or the enterprise in year 8. (l mark)
A

Solution


Y = abx

Log y = log a + log b

x y log y x2 xlog y
1 32 1.5051 1 1.5051
2 47 1.6721 4 1.3442
3 65 1.8129 9 5.4387
4 92 1.9638 16 7.8552
5 132 2.1206 25 10.6030
6 190 2.2788 36 13.6728
7 275 2.4393 47 17.0751
28 13.7926 140 59.5919


Log a
=
∑log y
n
-
log b∑x
n


Log b
=
n∑xlog y
n∑x2
-
∑x∑log y
(∑x)2


Log b
=
7 x 59.5919
7 x 140
-
28 x 13.7926
282


30.9505 / 196
Log b = 0.1579

∴
b = Antilog 0.1579

1.438

Log a
=
∑log y
n
-
log b∑x
n


Log a
=
13.7926
7
-
0.1579 x 28
7


1.3388

a = Antilog 1.3388 = 21.817

∴ y = 21.817 x 1.438x

(ii) The advertising cost of the enterprise in year 8.

Y(sh 000) = 21.817 x 1.438x
x = 8
Y(sh 000) = 21.817 x 1.4388 = 398.902
y = sh. 398,902

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QUESTION 3(d)

Q Simulate the average profit of product Zimsang on the basis of 10 trials.
A

Solution


selling price
Unit selling price Probability Cummulative probability Random number interval
60 0.25 0.25 00-24
80 0.45 0.70 25-69
100 0.30 1.00 70-99


Variable Cost
Variable Cost Probability Cummulative probability Random number interval
20 0.25 0.25 00-24
40 0.55 0.80 25-79
60 0.20 1.00 80-99


Sales volume
Sales volume Probability Cummulative probability Random number interval
40,000 0.30 0.30 00-29
60,000 0.35 0.65 30-64
100,000 0.35 1.00 65-99


Simulation for 10 trials
Trial
No.		 RN Unit Selling price
(1)		 RN Variable Cost(sh)
(2)		 RN Sales volume(Units)
(3)		 Fixed cost(sh)
(4)		 Profit
5 = 3(1-2) - 4
1 81 100 40 40 79 100,000 80,000 5,920,000
2 32 80 02 20 31 60,000 80,000 3,520,000
3 60 80 39 40 86 100,000 80,000 3,920,000
4 04 60 68 40 68 100,000 80,000 1,920,000
5 46 80 08 20 82 100,000 80,000 5,920,000
6 31 80 59 40 89 100,000 80,000 3,920,000
7 67 80 66 40 25 40,000 80,000 1,520,000
8 25 80 90 60 11 40,000 80,000 720,000
9 24 60 12 20 98 100,000 80,000 3,920,000
10 10 60 64 40 16 40,000 80,000 720,000
32,000,000


∴ Average profit = 32,000,000 / 10 = 3,200,000


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QUESTION 4(a)

Q (i) The average waiting cost per day. (4 marks)

(ii) (ii) Advise the management on whether thev should improve the facility.

(iii). Compare the probabilities that the total number of clients in the queue and those being served is greater than 17 in the existing and in the improved facilities.
A

Solution


Total waiting cost based on the time spent by a client, waiting on the queue is given by

αWqCw

(Total time spent waiting by all arrivals) x (cost of waiting)

Where;

α = mean number of arrivalsper day
Wq = Average time spent waiting on the queue
Cw = Waiting cost
α = 800 clients per hour

Number of working hours = 6.00 pm - 6 am
12 hours per day

Wq
=
λ
μ(μ - λ)


λ = 800
μ = 820

Wq
=
800
820(820 - 800)
= 0.0488 hours per client


Wq = shs 125 per hour

Hence;

Average waiting cost = 800 x 12 x 0.0488 x 125

Sh.58,560

(ii) Advise the management on whether thev should improve the facility.

Total cost upon improving the facility

Average waiting cost + cost of improving facility

Cost of improving facility = shs. 18,500 per day

Total average waiting cost per day

αWqCw

800 clients per day
Number of working hours = 6.00 p.m - 6.00 a.m

12 hours per day

Number of clients waiting for service on the queue

800 x 12 = 9,600 clients per day

Wq = Sh.125 per hour

Wq
=
λ
μ(μ - λ)


α = 800 clients per hour
μ = 847 clients per hour

Wq
=
800
847(847 - 800)


= 0.0201 hours per client

Total waiting cost = 9,600 x 0.0201 x 125 = 24,120 per day

Total = cost = 24,120 + 18,500 = 42,600

= Sh. 42,600

Advice
The management should consider to improve the facility so as to save on cost by (58,560 - 42,600). Sh. 15,940 per day on customer waiting time.

(iii). Compare the probabilities that the total number of clients in the queue and those being served is greater than 17 in the existing and in the improved facilities.

The probability that number of customers/clients in the system is greater than K1,Pn > k is given by

P(n > k) = [λ / μ]k + 1

The existing facility

P(n > 17) = [800 / 820]18

0.64

For the improved facility

P(n > 17) = [800 / 820]k + 1 = 0.36

There is 36% chance that the number of clients in the system is greater than 17

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QUESTION 4(b)

Q (i) The optimal strategies for each airline. (7 marks)
(ii) The value of the game. ( l mark)
A

Solution


Q= Airline Q K= Airline K
0
k1
k2
k3
[
Q1
-30
-40
80
Q2
0
-15
20
Q3
-90
-20
-50
]


Checking for saddle point by using maximin and minmax

0
k1
k2
k3
Maxmin
[
Q1
-30
-40
80
80
Q2
0
-15
20
20
Q3
-90
-20
-50
-20
]
Minmax
-90
-40
-50


0
k1
k2
k3
[
Q1
-30
-40
80
Q2
0
-15
20
Q3
-90
-20
-50
]


Optimal strategy

0
k2
k3
[
Q1
-40
80
Q3
-20
-50
]


(ii) The value of the game

Range Interchange Probability
[
-40
80
-20
-50
]
20
130
 130
20		 130 / 150 = 0.87
20 / 150 = 0.13


Range 120 30
Probability 30 / 150 120 / 150

V = -40 x 0.87 x 0.2 - 20 x 0.87 x 0.8 + 80 x 0.13 x 0.2 - 50 x 0.13 x 0.8 = -24

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QUESTION 5(a)

Q Explain the following terms as used in game theory:

(i) Saddle point.
(ii) Dominance.
A

Solution


Saddle point
Describes the point of equilibrium that happens when the maxmin value and the minmax value are equal. Typically, it is related to pure strategy games. It is also equivalent to the value of these games.

Dominance
Refers to a property that is used in mixed strategy games to convert the payoff matrix into a 2 x 2 matrix so that the best strategies for each player can be identified as well as the game's value.

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QUESTION 5(b)

Q Suggest two areas in which advanced information technology could be used to solve quantitative analysis problems.
A

Solution


Suggesting the areas in which advance information technology could be used to solve quantitative analysis problems

(i) Inventory management

(ii) Linear programming for sensitivity analysis

(iii) Resource planning

(iv) Quantitative data analysis (v) Simulation

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QUESTION 5(c)

Q (i) The normal duration and the corresponding total cost,

(ii) The minimum duration and the coresponding total cost.

(iii) The optimum duration and thc corresponding total cost.
A

Solution




Critical path is;

1-2-3-4-6

∴= 20days

Associated Total Cost

Total normal cost + total indirect cost

Total normal cost:
2,800,000
4,000,000
2,200,000
1,600,000
0
1,800,000
5,000,000
1,000,000
Total 18,400,000

Total Indirect Cost
Sh.600,000 x 20days

Sh.12,000,000

Total cost:

12,000,000
18,400,000
30,400,000


(ii) The minimum duration and the coresponding total cost.

Network diagram;using minimum durations.



Critical path is 12 days

Total Crash cost

Total normal Cost:
3,800,000
5,600,000
3,000,000
2,800,000
0
3,200,000
7,000,000
1,600,000
Total 27,000,000

Total Indirect cost
600,000x12days
7,200,000

Total cost
7,200,000
27,000,000
Total 34,200,000


(iii) The optimum duration and thc corresponding total cost.

The original critical path is

1 - 2 - 3 - 4 - 6

Critical activity Cost
Shs per day 		Duration
Crash by days		 Rank
1-2 500 2 3
2-3 400 2 2
3 - 4 0 0 1
4 - 6 500 4 3


Note;

Cost slope
=
Crash cost - Normal cost
Normal time - Crash time


Paths / Days Duration Critical
1 - 2 - 3 - 4 - 6 20 Yes
1 - 2 - 4 - 6 19 No
1 - 3 - 4 - 6 18 No
1 - 3 - 5 - 6 17 No


Hence maximum crash time is 2 days for the activity 2 - 3 so as to give the minimum cost of
18,400,000 + (2 x 400,000) + 18 x 600,000 = Sh.30,000,000

Therefore, the optimal duration is 18 days with a total cost of sh. 30,000,000

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