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CPA
Foundation Leval
Quantitative Analysis May 2017
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1(a)

Q Describe four types of sets as used in set theory.
A

Solution


Types of Sets in Set Theory


1. Finite Set:


A set that contains a specific number of elements.

Example: A = {1, 2, 3}

2. Infinite Set:


A set that contains an infinite number of elements.

Example: B = {1, 2, 3, ....}


3. Equal Sets:


Two sets that have exactly the same elements.

Example: C = {a, b, c} and D = {c, a, b} are equal sets.


4. Null Set (Empty Set):


A set that contains no elements.

Denoted by ∅ or {}.

Example: E = ∅


5. Singleton Set:


A set that contains only one element.

Example: F = {5}


6. Subset:


If every element of set A is also an element of set B, then A is a subset of B, denoted as A ⊆ B.

Example: If A = {1,2} and B = {1,2,3,4}, then A is a subset of B.


7. Power Set:


The set of all subsets of a given set.

If A = {a,b}, then the power set of A is P(A) = {∅,{a},{b},{a,b}}.

Where:

∅ is the empty set.

{a} is the subset containing only the element 'a'.

{b} is the subset containing only the element 'b'.

{a,b} is the subset containing both 'a' and 'b'.


8. Universal Set:


The set that contains all the elements under consideration in a particular discussion or problem. Denoted by U .


Example: If discussing natural numbers, the universal set might be U = {1, 2, 3, .....}


9. Complement of a Set:


The set of elements not in a given set A with respect to a universal setU. Denoted by A' or A

Example: If U = {1,2,3,4,5} and A = {1,2}, then the complement of A is A′ is = {3,4,5}.


10. Disjoint Sets:


Sets that have no elements in common. Example: If X = {a, b} and Y = {c, d} , then X and Y are disjoint sets.


11. Equal and Equivalent Sets:


Equal sets have the same elements. Equivalent sets have the same cardinality (number of elements).


Example: {1, 2, 3} and {3, 1, 2} are equal sets, while {a, b} and {1, 2} are not equal but equivalent sets.





QUESTION 1(b)

Q Explain the following terms as used in Markov analysis

(i) Transition probability.

(ii) Absorbing state.
A

Solution


Markov Analysis:


Markov analysis, specifically in the context of Markov Chains, is a mathematical model used to describe a sequence of events where the probability of each event depends only on the state attained in the previous event. Two key terms in Markov analysis are:

1. Transition Probability:


In a Markov Chain, a transition probability is the probability of moving from one state to another in a single step. It quantifies the likelihood of transitioning from one state to another in the next iteration of the process. Transition probabilities are often organized into a matrix known as the transition probability matrix.


For a Markov Chain with n states, the transition probability matrix P is an n × n matrix, where Pij represents the probability of transitioning from state i to state j. The sum of probabilities in each row of the matrix is always 1, as the system must transition to some state.


Mathematically, the transition probability matrix is represented as

P
=
[
P11
P21
.
.
.
Pn1
P12
P22
.
.
.
Pn2
.
.
.
.
.
.
.
.
.
.
.
.
P1n
P2n
.
.
.
Pnn
]

Transition probabilities play a crucial role in understanding and predicting the behavior of Markov Chains over time.


2. Absorbing State:


An absorbing state in a Markov Chain is a state from which, once entered, the system cannot leave. In other words, an absorbing state is a state in which the system remains indefinitely once it reaches that state. Once the system enters an absorbing state, it will stay in that state with a probability of 1 in all subsequent time steps.


In the context of a transition probability matrix, an absorbing state is characterized by having a diagonal element Pii = 1 and all other elements in that row and column equal to 0. The absorbing states are often represented by rows and columns where all elements are 0 except for the diagonal element.


The concept of absorbing states is significant in various applications, such as modeling absorbing barriers in stochastic processes, analyzing the long-term behavior of a system, and predicting the ultimate fate of a Markov Chain.





QUESTION 1(c)

Q (i) The proflt function.

(ii) The level of output that would maximise profit.
A

Solution


(i) The proflt function.


Tp = Total revenue(Tr) - Total cost(Tc)

Tp = Tr - Tc

Tr = (Ar)X

(650 - 15X)X = 650x - 15X2

Tc = ∫(Mc)X

9X2+1 / 3 - 14X1+1/2 + (180X / 1) + F.C

3X3 - 7X2 + 180X + F.C

∴ F.C = 25

Tc = 3x3 - 7x2 + 180x + 25

Tp = (650x - 15x2) - (3x3 - 7x2 + 180x + 25)

Tp = -3x3 - 8x2 + 470x - 25

(ii) At maximum profit


Margin profit(MP) = 0

Mp = dtp / dx = - 9x2 - 16x + 470

-9x2 - 16x + 470 = 0

X2 + (16 / 9)x = 470 / 9

Use completing the square method ↓ .
Completing the square is a method used to manipulate a quadratic expression into a perfect square trinomial, making it easier to solve or analyze. The process involves the following steps:

  1. Start with the Quadratic Expression: Begin with a quadratic expression in the form ax² + bx + c.
  2. If Coefficient of x² (a) is Not 1: If a ≠1, divide the entire expression by a to make the coefficient of x² equal to 1.
  3. Rearrange the Terms: Move the constant term (c) to the other side of the equation if it's not already there.
  4. Add and Subtract (b / 2)²: Take the coefficient of x (b), divide it by 2, square the result, and add and subtract it to the expression.
  5. Factor the Perfect Square Trinomial: Express the squared binomial as (x + b / 2)²
  6. Solve for x: If solving for x, take the square root of both sides and solve for x.


(x + 8 / 9)2 = 470 / 9 + (8 / 9)2

470 / 9 + 64 / 81 = 4,294 / 81

X + 8 / 9 = + √4,294 / 81 = +65.5286 / 9

X = -8 / 9 + 65.5286 / 9

-8.1698 or 6.3921

= 6.3921





QUESTION 1(d)

Q (i) The least squares regression function relating the monthly income and expenditure incurred by the households. Interpret your results.

(ii) The expenditure incurred by a household whose monthly income is Sh.30,000.
A

Solution


(i) The least squares regression function


y = a + bx

where;

x = monthly income
y = expenditure

a and b are constants

a
=
Σy

n
-
bΣx

n


b
=
nΣxy - ΣxΣy

nΣx2 - (Σx)2


X"000" Y X2 XY
15 2,000 225 30,000
6 200 36 1,200
9 500 81 4,500
3 500 9 1,500
20 2,500 400 50,000
11 800 121 8,800
14 1,500 196 21,000
10 1,500 100 15,000
12 1,600 144 19,200
100 11,100 1,312 151,200


n = 9

b
=
9(151,200) - 100(11,100)

9(1,312) - (100)2


250,800 / 1,808 = 138.717

a
=
11,100

9
-
138.717(100)

9


-307.967

y = 138.717x - 307.967

(ii) The expenditure incurred by a household whose monthly income is Sh.30,000.


y = 138.717x - 307.967

x = 30

= 138.717(30) - 307.967 = 3,853.543





QUESTION 2(a)

Q Highlight four properties of a binomial experiment.
A

Solution


Properties of a Binomial Experiment


1. Fixed Number of Trials:


  • The experiment consists of a fixed number of identical trials or observations.
  • Each trial can result in one of two possible outcomes: success or failure.

2. Independent Trials:


  • The trials are independent, meaning the outcome of one trial does not affect the outcome of any other trial.
  • Independence implies that the probability of success or failure remains constant from trial to trial.

3.Constant Probability of Success:


  • There are only two possible outcomes for each trial: success and failure.
  • The probability of success (denoted as p ) remains constant for each trial.

4. Discrete Random Variable:


  • The variable of interest is the number of successes in a fixed number of trials.
  • This random variable follows a binomial probability distribution.



QUESTION 2(b)

Q (i) The probability that exactly four machines break down in a given day. using poisson distribution.

(ii) The probability that exactly four machines break down in a given day, using binomial distribution.

(iii) Comment on the results obtained in (b)(i) and (b)(ii) above.
A

Solution


(i) The probability that exactly four machines break down in a given day. using poisson distribution.


Pr
=
e λr

r!

where;

λ = 0.015 x 30 = 0.45 Machines breaking down per day

e = 2.7183
r = 4

P(r = 4)
=
2.7183-0.45 - 0.454

4!


= 0.00109

(ii) The probability that exactly four machines break down in a given day, using binomial distribution.


P(r) = nCr x P rqn-r

Where;

n = 30
r = 4
p = 0.015
q = 1 - p = 1 - 0.014 = 0.985

P(r = 4) = 30C4 x 0.0154 x 0.98526

= 0.00094

(iii) Comment on the results obtained in (b)(i) and (b)(ii) above.


The probabilities calculated in (i) and (ii) are extremely low, suggesting a minimal likelihood of all four machines experiencing a breakdown on any given day.



QUESTION 2(c)

Q (i) Expected profit.

(ii) Minimising risk.
A

Solution


(i) Expected profit.


Expected project E(X̄)

For small scale production

E(X̄) = 0.25(40) + 0.45(140) + 0.30(180) = 127,000,000

For large scale production:

E(x) = 0.25(5) + 0.45(90) + 0.30(280) = Sh. 125,750,000

Hence ABC ltd should adopt small scale production because it's more profitable.

(ii) Minimising risk.


Using coefficient of variation (CV):

CV
=
Standard deviation [S.D]

Expected value: E[x]


S.D = Σp(x - x)²

S.D = 0.25(40 - 127)2 + 0.45(140 - 127)2 + 0.30(180 - 127)2

S.D = 2,811

Sh.53.02 million

CV = 53.02 / 127 x 100% = 41.75%

For large scale prodction

S.D = 0.25(5 - 125.75)2 + 0.45(90 - 125.75)2 + 0.30(280 - 125.75)2

S.D = 11358.1875 = 106.57 million

CV = 106.57 / 125.75 x 100% = 84.75%


Therefore, ABC Ltd ought to consider implementing small-scale production, as it involves lower risk when compared to large-scale production.




QUESTION 3(a)

Q Test at a 5 per cent level of significance, whether there is a significant difference between the sample mean and the population mean.
A

Solution


Since, population standard deviation is unknown and sample size; "n" is very small one n < 30; the test is t-test:
Where;

t
=
x̄ - μ

s / √n


S
=
∑(X̄ - X̄)2

n - 1


X (X-X)2
4,400 2,500
4,800 202,500
3,700 422,500
3,900 202,500
5,500 1,322,500
4000 122,500
3,700 422,500
4,100 62,500
4,000 122,500
5,400 1,102,500
43,500 3,985,000


X̄ = ∑x / n

43,500 / 10 = 4,350 hrs

S
=
3,985,000

10 - 1


= 665.4155

t
=
x̄ - μ

s / √n


n = 10
x = 4350 hrs
μ = 4000 hrs
s = 665.4155 hrs

t
=
4350 - 4000

665.4155 / √10


1.6633

Formulating Hypotheses:
Ho: u = 4000 ie There is no significant difference between sample mean and the population mean.

HI: μ ≠ 4000 ie There is significant difference between the sample mean and the population mean.

α = 5% or 0.05.

From HI: The test if a two tailored test.

t 0.025,9 = 2.262

With an observed t-value of 1.6633, which is below the critical t-value of 2.262, we proceed to accept the null hypothesis (Ho). This leads us to the conclusion that there is no noteworthy difference between the sample mean and the population mean concerning the lifetime of the electric bulbs.




QUESTION 3(b)

Q (i) Formulate the mathematical model for the linear programming problem (4 marks)

(ii) The maximum contribution of C1 and C2,

(iii) Explain the effect on contribution of the availability of additional plastic waste and machine time.

(iv) Explain the sensitivity of the model to changes in contribution per unit of Cl and C2.

(v) The increase in contribution of Green Furniture Limited assuming that the management overcomes the plastic waste constraint.
A

Solution


(i) Formulate the mathematical model for the linear programming problem


Let;

C₁=Number of units of plastic chairs c1 produced
C₂ =Number of units of plastic chair c2 production.

Objective
Maximize contribution (z)

Objective function ➫ Zmax = 555C₁ + 490C₂

Subject to:
Plastic wastage constraint: ➫ 4C₁ + 4C₂ ≤ 34000

Machine time x constraint: ➫ 0.5C₁ + 0.3C₂ ≤ 3400

Machine time y constraint: ➫ 0.4C₁ + 0.45C₂ ≤ 3840

The -ve constraints

C₁ ≥ 0
C₂ ≥ 0

(ii) The maximum contribution of C₁ and C₂


Zmax = 555C₁ + 490C₂

C₁ = 4250
C₂ = 4250

Max.contribution (z) = 4250(555) + 4250(490)

2,358,750 + 2,082,500 = Sh. 4,441,250

(iii) Explaining the effect on contribution of the availability of additional plastic waste and machine time.


An augmentation of one unit in plastic waste will lead to a rise in the total contribution by shs 1733.33. Conversely, a reduction of one unit in plastic waste will cause a decrease in contribution by shs 6800.


For every increase in one unit of machine hours for variable x, there will be a corresponding increase in total contribution by shs 850. Conversely, a reduction of one unit in machine x hours will result in a decrease in total contribution by shs 850.


(iv) The sensitivity of the model to changes in contribution per unit of C₁ and C₂


For additional unit of C₁, produced will increase the total contribution by shs. 261.70
while a decrease in unit of C₁ will decrease the total contribution by shs. 65.00
For every additional unit of C₂ produced, the total contribution will increase by shs. 65.00
which a decrease in every unit of C₂ produced, decreases the total contribution by shs. 157.00

(v) The increase in contribution of Green Furniture Limited assuming that the management overcomes the plastic waste constraint.


An increase in one unit of the plastic waste will increase the total contribution of shs.1,733.33




QUESTION 4(a)

Q Outline five assumptions of game theory.
A

Solution


Assumptions of Game Theory


➢ No player is aware of his opponents' choice until he decides his own.

➢ The overall outcome for all players would be zero at the end of the game.


➢ A game is played when each player chooses a course of action (strategy) out of available strategies.


➢ Each outcome determines the gain or loss of each player.


➢ The outcome of the play depends on every combination of the courses of action.


➢ Each player has a finite list of courses of action or strategy.


➢ The number of players is finite.





QUESTION 4(b)

Q The rank coefficient of correlation. Interpret your result.
A

Solution


r
=
1 -
6∑d2

n( n2 - 1 )

or

r
=
1 -
6( ∑d2 + ( t3- ε ) / 12 )

n(n 2 - 1)


Student Mathematics score Rank English score Rank d d2
1 31 7 56 3 4 16
2 26 5.5 46 5 0.5 0.25
3 44 4 66 2 2 4
4 28 8 46 5 3 9
5 56 3 36 7 -4 16
6 76 2 26 8 -6 36
7 36 5.5 46 5.5 0.0 0.25
8 96 1 76 1 0 0
0 81.5


where;

∑d2 = 81.5
n = 8
t = 2 + 3 = 5

r
=
1 -
6( ∑81.5 + ( 53- 5)/12 )

8( 82 - 1 )


1 - ( 549 / 504 ) = -0.0893

Interpreting the result obtained above:

The obtained coefficient of correlation is both negative and exceedingly low, suggesting an extremely weak correlation or association between the scores achieved in mathematics and English.




QUESTION 4(c)

Q Using eight trials, simulate the average profit of Pure Grain Society.
A

Solution


Selling price per tonne Probability Cummulative probabilitv Range
240 0.18 0.18 00-17
250 0.29 0.47 18-46
260 0.31 0.78 47-77
270 0.14 0.72 78-91
280 0.08 1.00 92-99


Yield per acre Probability Cummulative probabilitv Range
70 0.09 0.09 00-08
75 0.16 0.25 09-24
80 0.24 0.49 25-48
85 0.38 0.87 49-86
90 0.13 1.00 87-99


Cost per acre
"000"

Probability
Cummulative
probabilitv

Range
12,000 0.14 0.14 00-13
14,000 0.22 0.36 14-35
16,000 0.36 0.72 36-71
18,000 0.26 0.98 72-97
20,000 0.02 1.00 98-99


Trial
Number

Range
Sp
(1)

Range
Yield
(2)

Range
Cost
(3)
Profit
(4)=(1x2)-3
1 03 240 69 85 03 12,000 8,400
2 91 280 85 91 18,000 5,800
3 38 250 75 38 16,000 2,750
4 55 260 75 55 16,000 3,500
5 17 240 85 17 14,000 6,400
6 46 250 90 46 16,000 5,500
7 32 250 80 32 14,000 6,000
8 43 250 80 43 16,000 4,000
43,350


Average profit = 43,350,000/8 = Sh.5,418,750 per acre



QUESTION 5(a)

Q Enumerate five characteristics of a simple queuing system.
A

Solution


Characteristics of a Simple Queuing System


  1. Arrival Process: Describes how entities arrive at the queue, following a specific distribution.
  2. Service Process: Describes how entities are served or processed, with service times following a specific distribution.
  3. Queue Discipline: Specifies the rules governing the order in which entities are served, such as FCFS or LCFS.
  4. Queue Capacity: Represents the maximum number of entities the queue can hold, being limited or unlimited.
  5. Balking and Reneging: Balking occurs when entities decide not to join the queue upon arrival, and reneging occurs when entities leave before being served.
  6. Queue Configuration: May include single or multiple queues serving a single or multiple service stations.
  7. Service Channels: Indicates the number of parallel servers available to serve entities, with options for a single or multiple channels.
  8. Queue Length: Represents the number of entities currently in the queue, indicating system congestion.
  9. Waiting Time: The time entities spend waiting in the queue before being served, influenced by arrival and service processes.
  10. Utilization Factor: Measures the proportion of time the service facility is busy, calculated as the ratio of total service time to the total time.
  11. System Efficiency: Represents the effectiveness of the queuing system in utilizing resources, influenced by the balance between arrivals and service rates.
  12. Stability: A stable system is one where the queue length and waiting times do not grow indefinitely, influenced by the relationship between arrival and service rates.



QUESTION 5(b)

Q The optimal strategy for each company.
A

Solution


Company Bs strategies
Company As strategies
0
A1
A2
A3
A4
Maxmin
[
B1
7
4
3
1
7
B2
4
2
-1
5
5
B3
1
0
-2
-3
1
]
Minmax
1
0
-1
-3


Maxmin = Minmax

Thus, a saddle point exists in the game, representing a pure strategy. The game's value is 1. Consequently, the optimal strategy for company A is A1, while the optimal strategy for company B is B3.



QUESTION 5(c)

Q (i) A network diagram for the project.

(ii) The minimum cost of the project.

(iii) Determine whether or not Ujenzi Limited should accept the bonus offer.
A

Solution


(i) A network diagram for the project.



(ii) The minimum cost of the project.


Total indirect cost + total direct cost

Total indirect cost (sh. Million)

= 100 + 75 + 135 + 70 + 160 + 255 + 30 + 90 + 55 = 970

Total direct cost (Shs. Million)

5 x 24 months = 120

Total cost = Sh. (970 + 120) million

Shs. 1,090 Million

(iii) Determine whether or not Ujenzi Limited should accept the bonus offer.


Critical path

A - D - Dummy - F::Project duration 24 days

Critical activity Cost slope(million per month) Duration(months)
A (125 - 100) / 2 = 12.5 2
D (125 - 100) / 2 = 12.5 2
F (125 - 100) / 2 = 12.5 2


If project is completed within (24 - 6) = 18 months

Total cost Normal cost + direct cost + additional cost - Bonus

970 + (18 x 5) + (12.5 x 2 + 7.5 x 2 + 10 x 2) - 25

= 970 + 90 + 60 - 25 = Sh. 1095 Million

This is more than the normal cost of completing the project within 24 months

If project is completed within 19 months

Total cost = 970 + (19 x 5) + (7.5 x 2 + 10 x 2 + 12.5 x 1) - 25

970 + 95 + 47.5 - 25 = Sh.1,087.5 Million

Therefore, it is advisable for Ujenzi Ltd to accept the bonus offer and finalize the project within 19 months, resulting in a savings of Sh. 2.5 million (1,090 - 1,087.5).



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