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CPA
Foundation Leval
Quantitative Analysis May 2019
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1a

Q (i) The profit function

(ii) The maximum profit.
A

Solution


(i) The Profit function


∫MR = 40q - 3q2

TR = 1 / 2 x 40q² - 1 / 3 x 3q³

= 20q² - q³

= Tc = AC x Q

Q(2q - 10 + 25/q)

2Q² - 10q + 25

Profit function = TR - TC

= 20q² - q³ - (2q² - 10q + 25)

20q² - q³ - 2q² + 10q - 25

18q² - q³ + 10q - 25

(ii) Maximum profit


Δπ / ΔQ = 36q - 3q² + 10 = 0

q
=
- b+- √ b 2- 4ac
2x - 3


q
=
-36 +- √ 36 2- (4 x -3 x 10)
2x - 3


q
=
-36 +- √ 1,416
-6


q
=
-36 +- 37.63
-6


q
=
-73.63
-6
or
1.6
-6


q = 12.27 or -0.27

Since output cannot be negative then profit maximizing output is 12.27

Maximum profit = (36 x 12.27) - 3 x 12.272 + 10

= 441.72 - 150.5529 + 10

= 301.1671 x 1,000,000 = Sh.301,167,100

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QUESTION 1b

Q (i) The fixed monthly basic salary of the salesman.

(ii) Commission earned per unit sold

(iii) Total earnings of the salesman in the month of April 2019.
A

Solution


(i) The fixed monthly salary of the salesman


Assume Total earnings equals Y
Fixed earnings equals a
Commission per unit b
Units sold x

Therefore

a + 100b = 60,000
a + 250b = 70,000

[
1100
1250
]
[
a
b
]
=
[
60,000
70,000
]

a
=
[
60,000
70,000
]
[
1100
1250
]


=
(60,000 x 250) - (70,000 x 100)
(1 x 250) - (1 x 100)


53,333.33

Fixed salary per month Sh.53,333.33

(ii) Using Cramer's rule


b
=
(
160,000
170,000
)
(
1100
1250
)


=
(1 x 70,000) - (60,000 x 1)
(1 x 250) - (1 x 100)


10,000 / 150 = 66.67

(iii) The total earnings of the sales man in the month of April 2019


Now earnings equation

Y = a + bx

Y = 53,333.333 + 66.67x

April 2019 earnings

53,333.333 + 66.67 x 400

53,333.33 + 26,668

Sh. 80,001.33

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QUESTION 1c

Q (i) Present the above information in a venn diagram.

(ii) The probability that a customer selected at random operates all the three types of accounts.

(iii) The probability that a customer selected at random operates only two types of accounts.
A

Solution


(i) Present the above information in a venn diagram.


Let;

A = Customer operating savings account

B = Customers operating current accounts

C = denote customers operating fixed deposit accounts

A + B + C = A + B + C - AB - BC - AC + ABC

200 = 84 + 109 + 106 - 45 - 36 - 43 + ABC

25 = ABC

Customers operating all accounts = 25

Number operating savings and current accounts = 45 - 25 = 20

Savings and fixed deposits accounts but not current accounts = 36 - 25 = 11

Fixed and current accounts but not saving accounts = 43 - 25 = 18

Customers operating savings accounts but not current accounts = 84 - 45 = 39

Customers operating current accounts but not fixed accounts = 109 - 36 = 73

Customers operating fixed accounts but not current accounts = 106 - 43 = 63

Venn diagram


A(Savings)
B(Current)
A28
25
20
B46
18
11
C52
C(Fixed deposit)


(ii) The probability that a customer selected at random operates all the three types of accounts.


25 / 100 = 0.125

(iii) The probability that a customer selected at random operates only two types of accounts.


(11 + 20 + 18) / 200 = 0.245

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QUESTION 2a

Q (a) Enumerate four assumptions of:

(i) A normal distribution.

(ii) A binomial distribution.
A

Solution


Assumption of normal distribution

1. It is a perfectly bell shaped curve

2. It is a continuous distribution

3. The tails of the curve continually approach but never touch the horizontal axis.

4. Mean, Mode and median pass through the peak of the curve and precisely bisect the area under the curve into two equal halves.

5. The first and the third quartiles are equidistant from the median

6. Height of normal curve is a maximum at the mean value

Binomial Distribution assumptions.

1. It is a discrete distribution of the occurrences of events with two outcomes - Success or failure

2. The number of trials are finite

3. The probability remain the same from one trial to another

4. The trials must be independent of one another .

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QUESTION 2b

Q The probability that there are no customers in the queuing system at a given point in time.
A

Solution


Probability that there are no customers in the queuing system at a given point in time

This is a Multiserver waiting line mode:

Arrival Rate λ = 12

Service Rate μ = 5

Number of Servers s = 3

Po
=
[
s-1
∑
n=0
1
n!
(
λ
μ
)n
+
1
s!
(
λ
μ
)S
sμ
sμ-λ
]
-1


Po
=
[
s-1
∑
n=0
1
n!
+
1
s!
(
λ
μ
)S
sμ
sμ-λ
]
-1


Po
=
[
(12 / 5)0
0!
+
(12 / 5)1
1!
+
(12/5)2
2!
+
(
(12/5)3
30!
x
3 x 5
15 - 12
)
]
-1


Po
=
[
(2.4)0
0!
+
(2.4)1
1!
+
(2.4)2
2!
+
(
(2.4)3
30!
x
3 x 5
15 - 12
)
]
-1


Po
=
[
1
1
+
(2.4)1
1!
+
5.76
2
+
(
13.824
6
x
3 x 5
15 - 12
)
]
-1


Po = [6.28 + 11.52]-1

Po = [17.8]-1

Po = 0.0562


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QUESTION 2c

Q (i) A decision tree showing the payoff and expected monetary value of each alternative action.

(ii) Advise the management of the company on the best product to introduce into the market.
A

Solution


(i) A decision tree showing the payoff and expected monetary value of each alternative action.




Product P1


(0.35 x 15,000,000) + (0.4 x 25,000,000) + (-5,000,000 x 0.25)

5,250,000 + 10,000,000 - 1,250,000 = 14,000,000

Product P2


(0.35 x 34,000,000) + (0.4 x 30,000,000) + (0.25 x -3,000,000)

11,900,000 + 12,000,000 - 750,000 = 23,150,000

Product P3


(0.35 x 22,000,000) + (0.4 x 15,000,000) + (0.25 x 8,000,000)

7,700,000 + 6,000,000 + 2,000,000 = 15,700,000

(ii) Advising the management of the company on the best product to introduce into the market


Best product to introduce to the market is product P2 because it has the highest expected payoff


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QUESTION 3a

Q (i) Free float.

(ii) Total float

(iii) Project crashing.
A

Solution


Explaining the following terms as used in network planning and analysis


(i) Free float


Free float is the amount of time an activity can be delayed without affecting the earliest start time of subsequent activities.

Free Float = Earliest Finish Time - Earliest Start Time - Activity Duration

(ii) Total Float


Total Float is the amount of time a path of activities could be delayed without affecting the overall project duration.

Total Float = Latest Finish Time - Earliest Start Time - Activity Duration

(iii) Project clashing


Project clashing refers to reducing activity duration by increasing resource utilization (e.g. hiring more workers) .

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QUESTION 3b

Q (i) In relation to hypothesis testing and estimations distinguish between "null hypothesis" and "alternative hypothesis"

(ii) Determine whether the advertising campaign increased the usage of "safi" amongst the residents. (Use a significance level of 5%).
A

Solution


(i) Distinguish between "null hypothesis" and "alternative hypothesis".


The null hypothesis of the test always predicts no effect or relationship between the variables, while the alternative hypothesis indicates what your study predicts about the effect or relationship.

(ii) Determine whether the advertising campaign increased the usage of "safi" amongst the residents. (Use a significance level of 5%).


p
=
0.1 x 500
500
+
0.51 x 600
600
= 0.13

Q = 1 - p

= 0.87

n1=500, n2 = 600, p1=0.1, p2 = 0.15

p
=
p1n1 + p2n2
n1 + n2


S (p1 - p2)
=
√
pq
n1
+
pq
n2


=
√
0.87 x 0.13
500
+
0.87 x 0.13
600


0.02036

Zcalc
=
p1 - p2
S(p1 - p2)
=
0.1 - 0.15
0.02036


Z value (Critical) from table ± 1.96

Since Z calculated 2.45587 > 1.96, then the advertising campaign increased usage of "safi"

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QUESTION 3c

Q (i) The optimal strategies for companies A and B.

(ii) The value of the game.
A

Solution


(i) The optimal strategies for companies A and B


Using dominance rule to reduce 3 by 3 matrix to a two by two matrix

Company A
 Company B
Newspaper Radio Television Minimum
Newspaper 40 50 -17 -17
Radio 10 25 -10 -10
Television 100 30 60 30
Maximum 100 50 60


No saddle point therefore it is a mixed strategy game.

TV dominates radio for company A therefore a new matrix

Company A
 Company B
Newspaper Radio Television
Newspaper 40 50 -17
Television 100 30 60


Newspaper dominate Tv for Company B therefore eliminate Newspaper

Company A
 Company B
Radio Television Probability
Newspaper 50 -17 30/97
Television 30 60 67/97
Probability 77/97 20/97


Hence company A should play strategies Newspaper: Radio: Television in the ratio

30/97 : 0 : 67/97

While company B should play strategies Newspaper: Radio: Television in the ratio

0:77/97:20/97

(ii) The value of the game


(30 / 97 x 50) + (67 / 97 x 30)

15.46 + 20.72 = 36.18

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QUESTION 4a

Q Highlight two differences between "transportation" and ''assignment" models of linear programming.
A

Solution


Differences between "transportation" and "assignment" models of linear programming


1. In the transportation model, a single source can be used to supply several destinations, whereas in the assignment model, one person can only be assigned one job at a time.

2. The assignment model's aim is to either maximize profits or minimize costs. A transportation problem only aims at minimizing transportation costs.

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QUESTION 4b

Q Summarise three applications of shadow prices in decision making.
A

Solution


Analysts and economists utilize "shadow pricing" to place a monetary value on non-marketed items such production costs and intangible assets.

Applications of shadow prices in decision making.


1.Shadow pricing can help companies better understand the costs and benefits associated with a project.

2. Determination of limiting factors

3. Sensitivity analysis

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QUESTION 4c(i)

Q The coefficient of correlation. Interpret your result.
A

Solution


The coefficient of correlation and Interpretation of the result


Let years of experience be x

Let mean monthly sales be y

X Y Y2
"000"		 XY X2
6 180,000 32,400,000 1,080,000 36
4 150,000 22,500,000 600,000 16
2 80,000 6,400,000 160,000 4
10 500,000 250,000,000 5,000,000 100
7 190,000 36,100,000 1,330,000 49
4 100,000 10,000,000 400,000 16
6 200,000 40,000,000 1,200,000 36
7 220,000 48,400,000 1,540,000 49
12 600,000 360,000,000 7,200,000 144
8 200,000 40,000,000 1,600,000 64
∑x=66 ∑y=2,420,000 ∑y2=845,800,000 ∑xy=20,110,000 ∑x2=514


r
=
n∑xy - ∑x∑y
√n∑x2 - (∑x)2 x √n∑y2 - (∑y)2


r
=
(10 x 20,110,000) + (66 x 2,420,000)
√(10 x 514) - 662 x √(10 x 845,800,000,000)- 2,420,0002


41,380,000 / 45,162,533.1444

0.9162

There is a strong positive correlation between years of experience and the mean monthly sales.

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QUESTION 4c(ii)

Q Using ordinary least squares method, predict the mean monthly sales that would be realised by a salesperson having 15 years of experience.
A

Solution


Using ordinary least squares method to predict the mean monthly sales that would be realised by a salesperson having 15 years of experience

∴
b
=
nΣxy - ExΣy
nΣx² - (Ex)²


b
=
(10 x 20,110,000) - (66 x 2,420,000)
(10 x 514) - 66²


41,380,000/784

Shs. 52, 780.61

a
=
∑y
n
- b
∑x
n


2,420,000/10 - 52,780.61 x 66/10

242,000-348,352.026

=-106,352.026

ye= -106.352.026 + 52,780.61x

Mean monthly sales for a salesman with 15years of experience

∴
= -106,352.026 + (15 x 52,780.61)

= -106,352.026 + 791,709.15

= Shs. 685,357.124

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QUESTION 5a

Q Outline four merits of using the project evaluation and review technique (PERT) to plan and analyse a project in an organisation.
A

Solution


Merits of using the project evaluation and review technique (PERT) to plan and analyse project in an organisation


1. With the help of PERT, large scale project planning and project business planning can be easily done.

2. Visibility of critical paths: The PERT method will display critical paths in a well-defined way

3. Examination of activity - The PERT networks are used to analyze the activity and occurrences. Both of these are examined separately and jointly. This will provide information about the project's budget and likelihood of completion.

4. It’s useful even if there is little or no previous schedule data.

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QUESTION 5b(i))

Q The expected completion time and variance of each of the activities.
A

Solution


Expected time
=
0 + 4ML + P
6


Activity Expected time
A (7 + 4 x 17 + 27)/6 = 17days
B (5 + 4 x 11 + 23)/6 = 12days
C (3 + 4 x 8 + 19)/6 = 9days
D (23 + 4 x 31 + 47)/6 = 32days
E (9 + 4 x 21 + 39)/6 = 22days
F (9 + 4 x 11 + 25)/6 = 13days
G (2 + 4 x 5 + 14)/6 = 6days
H (9 + 10 x 4 + 17)/6 = 11days


Variance
=
[
(Pessimistic time-Optimistic time)
6
]
2


Expected Variance of each activity

A
=
(
27 - 7
6
)
2
= 11.11


B
=
(
23 - 5
6
)
2
= 9


C
=
(
19 - 3
6
)
2
= 7.11


D
=
(
45 - 23
6
)
2
= 13.44


E
=
(
39 - 9
6
)
2
= 25


F
=
(
25 - 9
6
)
2
= 7.11


G
=
(
14 - 2
6
)
2
= 4


H
=
(
17 - 9
6
)
2
= 1.78



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QUESTION 5b(ii)

Q The total float of each activity.
A

Solution


The total float of each activity


Total float = Latest finish time - Earliest start time - Activity duration

Activity
A 17 - 0 - 17 = 0
B 29 - 17 - 12 = 0
C 52 - 29 - 9 = 14
D 61 - 29 - 32 = 0
E 74 - 38 - 22 = 14
F 74 - 61 - 13 = 0
G 85 - 61 - 6 = 18
H 85 - 74 - 11 = 0


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QUESTION 5(iii-iv)

Q (iii) The expected completion time and variance of the project.

(iv) The 95% confidence interval for the project's completion time.
A

Solution


(iii) The expected completion time and variance of the project


Expected project completion = 85 days

Variance of the project

= 11.11 + 9 + 13.44 + 7.11 + 1.78 = 42.44

(iv) The 95% confidence interval for the project's completion time


μ
=
t̄+
z
δ
√n

t̄ = 85days

z+= 1.96

n = number of activities = 8

δ = √ 42.44

Time interval μ

=
85̄+
1.96 x
√42.44
√8


85+1.96 x 2.30

85+4.508

At 95% confidence level project completion time will lie between 80.49 days to 89.51 days

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