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CPA
Foundation Leval
Quantitative Analysis November 2016
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Quantitative analysis
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QUESTION 1(a)

Q Explanation of the following terms as used in Linear programming

(i) Infeasibility.

(ii) Unboundedness.

(iii) Alternate optimality.
A

Solution


(i) Infeasibility:


Definition: Infeasibility refers to a situation where there is no feasible solution to the linear programming problem. Feasible solutions are those that satisfy all the constraints of the problem.

Explanation: The constraints in a linear programming problem define a region in the solution space where valid solutions must lie. If this region is empty, meaning there is no point that satisfies all the constraints simultaneously, the problem is considered infeasible. Infeasibility could result from contradictory or incompatible constraints.


(ii) Unboundedness:


Definition: Unboundedness occurs when there is no finite optimal solution to the linear programming problem. In other words, the objective function can be improved indefinitely without reaching an optimal solution.

Explanation: Linear programming aims to optimize (maximize or minimize) a linear objective function subject to linear constraints. If there is no upper or lower limit on the objective function within the feasible region, the problem is considered unbounded. This situation might arise when the feasible region extends infinitely in one or more directions, and the objective function can improve indefinitely without reaching an optimal value.


(iii) Alternate Optimality:


Definition: Alternate optimality, also known as degeneracy, occurs when there are multiple optimal solutions to the linear programming problem.

Explanation: Ideally, a linear programming problem has a unique optimal solution. However, in some cases, there might be more than one combination of decision variables that yield the same optimal value for the objective function. These different solutions are considered alternate optimal solutions. Degeneracy can occur when the feasible region has redundant constraints or when the objective function is parallel to one of the constraint boundaries.





QUESTION 1(b)

Q (i) The probability that product "X" contains no defect.

(ii) The probability that product "X" contains only one of the three defects.
A

Solution


(i) The probability that product "X" contains no defect.


P(A and B) = P(A) x P(B) = 0.15 x 0.14

0.021

Probability that product x only contains defect A or B

P(A or B) = P(A) + P(B) - P(A and B)

(0.15 + 0.14) - 0.021 = 0.269

Probability of no defect

Ā = 1 - 0.15 = 0.85 Probability of no defect

B̄ = 1 - 0.14 = 0.86 P(A and B̄) = 0.15 x 0.86 = 0.129

P(B and Ā) = 0.14 x 0.85 = 0.119

P(A and B̄) or P (B and A) = 0.129 + 0.119 = 0.248

Probability no defect (A&B)

P(Ā and B̄) = 0.86 × 0.85 = 0.731

Probability of no defect c when we have no defect A&B = 1 - 0.3 = 0.7

Therefore P(Ā and B̄ and C̄) = 0.7 x 0.731 = 0.5117

(ii) The probability that product "X" contains only one of the three defects.


Probability of defect A but no defect B

B = 0.731 x 0.3 = 0.2193

Probability of either A or B but no defect

= 0.248 x 0.8 = 0.1984

So probability of only one deficit type

0.2193 + 0.1984 = 0.4177




QUESTION 1(c)

Q Using the paired t-test, determine whether the advertisement was a success at a 5 per cent level of significance.
A

Solution


College No. of students after advert Pre advert students d (d-d̄)2
1 170 165 5 2.25
2 141 140 1 6.25
3 142 143 -1 20.25
4 167 160 7 12.25
5 168 162 6 6.25
6 157 154 3 0.25
21 47.5


=
∑d

n
=
21

6
= 3.5


Sd
=
(d-d̄)2

n - 1


Sd
=
47.5

6 - 1
= 3.08


tcalculated
=
d̄ - μd

Sd

√n
t0.05,5 = 2.571


Since t calculated is greater than t critical we reject null hypothesis and concluded advertising was a success.



QUESTION 2(a)

Q Highlight four applications of Markov analysis in business.
A

Solution


Applications of Markov Analysis in Business


1. Employee Turnover:


Markov analysis models and predicts employee movement within different job categories or employment states over time. This aids in workforce planning and optimizing recruitment strategies.

2. Customer Loyalty and Retention:


Markov analysis is used to model customer behavior, predicting transitions between different states of loyalty or engagement. This helps in designing effective customer retention strategies.


3. Financial Portfolio Management:


Markov models analyze and predict changes in financial states of investment portfolios, assisting in risk assessment, investment decisions, and asset allocation strategies.


4. Supply Chain Management:


Markov analysis can model the transitions between different states in a supply chain, optimizing supply chain efficiency, reducing bottlenecks, and improving overall performance.


5. Quality Control and Manufacturing:


Markov models can be applied to analyze and predict transitions between different quality states of products during production. This helps in identifying potential defects and improving overall product quality.


6. Project Management:


Markov analysis is applied to project management to model the progression of projects through different phases or states. This helps in predicting project completion times and optimizing project schedules.


7. Marketing and Sales Funnel Analysis:


Markov models can be employed to analyze customer journeys through the marketing and sales funnel, optimizing marketing strategies and improving the sales process.


8. Insurance and Risk Assessment:


Markov analysis is used to model transitions between different risk states or insurance coverage states, aiding in pricing policies, assessing risk exposure, and optimizing insurance product offerings.


9. Credit Scoring and Risk Management:


Markov models are applied in credit scoring to predict transitions between different credit risk states over time. This helps in assessing creditworthiness and optimizing lending practices.


10. Healthcare Management:


Markov analysis is applicable in healthcare for modeling patient transitions between different health states, aiding in healthcare planning, resource allocation, and optimizing patient care.





QUESTION 2(b)

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

(ii) The probability that a student is enrolled for all the three courses.

(iii) The probability that a student is enrolled for Accounting or Statistics but is not enrolled for Information Technology.
A

Solution


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


Let A represent number of students doing accounting
Let I represent number of students doing information technology
Let S represent students doing statistics

Number of students doing all the three courses (A)

A ∩ I ∩ S + A + I + S - A ∪ S - A S - I ∪ S + AIS

500 = 329 + 186 + 295 - 83 - 217 - 63 + AIS

500 = 447 + AIS

AIS = 500 - 444 = 53

Students doing accounting and information technology but not statistics = 83 - 53 - 30

Students doing accounting and statistics but not IT = 217 - 53 = 164

Students doing statistics and information technology but not accounting = 63 - 53 - 10

Venn diagram

A(Accounting)
I(IT)
A-82
E-53
B-30
I-93
D-10
F-164
S-68
S(Statistics)


(ii) Probability student enrolled in all three courses


53 / 500 = 0.106

(iii) Probability student enrolled for accounting or statistics but not ICT


(82 + 164 + 68) / 500

314 / 500 = 0.628




QUESTION 2(c)

Q (i) The missing values of X1 ,X2 , X3, X4, X5, X6 and X7.

(ii) A 95 per cent confidence level of the labour hours worked,
A

Solution


(i) The missing values of X1 ,X2 , X3, X4, X5, X6 and X7.


X1 = 0.029 / 0.04 = 0.725

X2 = √0.725 = 0.8515

X3 = 23 - 22 = 1

X4 = 0.04 - 0.029 = 0.011

X5 = 0.029 / 0.000455 = 63.74

X6 = 0.077 / 11.328 = 0.00679732

Statistic = Coefficient / Standard error

X7 = 0.826 / 0.103 = 8.0194

(ii) A 95 per cent confidence level of the labour hours worked,


tcalculated 8.0194

tcalculated = (1- α / 2) at dƒ= n - 1

tcalculated = (1 - 0.5 / 2) = 0.2069

Since tcalculated 8.0194 is greater than t critical 2.069 so labour hours are statistically signigificant




QUESTION 3(a)

Q Formulate a linear programming model to maximise TOC Limited's revenue, assuming that only transport cost is variable.
A

Solution


Assume;

A = Daily production of diesel at refinery 1

B = Daily production of petrol at refinery 1

C = Daily production of diesel at refinery 2

D = Daily production of petrol at refinery 2

Contribution = selling price per unit - variable cost per unit

Contribution per litre of petrol refinery 1 = 160 - 80 = Sh80

Contribution per litre of petrol of refinery 1 = 160 - 100 = Sh.60

Contribution per litre of diesel refinery 2 = 170 - 80 = Sh.90

Contribution per litre of diesel refinery 2 = 170 - 100 = Shs.70

Objective function

90A + 80B + 70C + 60D

Subject to;

Crude oil 1 : 3A + 1B ≥ 12000

Crude oil refinery 2: 3C + 2D ≤ 15000

Furnace time refinery 1: 2a + 18 ≤ 10

Furnace time refinery 2: C = d ≤ 4

Mixer: 8A + 6B + 7C + 5D ≤ 80000

A ≥ 0, B ≥ 0, D≥0




QUESTION 3(b)

Q (i) The adjusted seasonal component for the four quarters using the additive model.

(ii) The deseasonalised production data for each quarter.

(iii) Explain the significance of the deseasonalised data.
A

Solution


(i) The adjusted seasonal component for the four quarters using the additive model.


Y 4CMT 4CMTA CMACT Y/T Y-T
250
200 930 232.5
180 1010 252.5 242.5 0.742 62.5
300 1090 272.5 262.5 1.143 37.5
330 1170 292.5 282.5 1.168 47.5
280 1250 312.5 302.5 0.926 -22.5
260 1330 332.5 322.5 0.806 -62.5
380 1420 355 343.75 1.105 36.25
410 1,500 375 365 1.123 45
370 1580 395 365 0.961 -15
340 1648 412 403.5 0.843 -63.5
460
478


Multiplicative adjusted seasonal component

Year Q1 Q2 Q3 Q4
2012 0.742 1.143
2013 1.168 0.926 0.806 1.105
2014 1.123 0.961 0.843
2015 =
Total 2.291 1.887 2.391 2.248
Average 1.146 0.994 0.797 1.124 0.4011


Addictive adjusted seasonal component

Year Q1 Q2 Q3 Q4
2012 -62.50 37.50
2013 47.50 -22.50 -62.50 36.25
2014 45.00 -15.00 -63.50
Total 92.50 -37.50 -188.50 -73.75
Average 46.25 -18.75 -62.83 36.88 1.55
Adjustment factor -0.39 -0.39 -0.39 -0.39 -1.55
ASI 45.86 -19.14 -63.22 36.49


(ii) The deseasonalised production data for each quarter.


Year Quater Y ASI Deseasonalised data
2012 1 250 45.86 204.14
2 200 -19.14 219.14
3 180 -63.22 243.22
4 300 36.49 263.51
2013 1 330 45.86 284.14
2 280 -19.14 299.14
3 260 -63.22 323.22
4 380 -63.22 323.22
2014 1 410 45.86 364.14
2 370 -19.14 389.14
3 340 -63.22 403.22
4 460 36.49 423.51
2015 1 478 45.86 432.14


(iii) Explain the significance of the deseasonalised data.


The deseasonalised data can be used in least square method to come up with a regression equation



QUESTION 4(a)

Q Enumerate four limitations of linear programming models.
A

Solution


Limitations of Linear Programming Models


1. Linearity Assumption:


Linear programming models assume a linear relationship between decision variables and the objective function as well as constraints.

2. Additivity and Proportionality:


Linear programming assumes additivity and proportionality, which may not accurately represent complex real-world relationships.


3. Assumption of Certainty:


Linear programming assumes that all parameters, coefficients, and data are known with certainty, neglecting uncertainties and variability.


4. Continuous Decision Variables:


Linear programming assumes continuous decision variables, while in reality, some decisions may be discrete.


5. Static Nature:


Linear programming models are generally static, not accounting for dynamic or time-dependent changes over time.


6. Limited to Linear Objectives:


Linear programming is designed for linear objective functions, limiting its applicability in situations with non-linear objectives.


7. Sensitivity to Parameter Changes:


Linear programming solutions can be sensitive to changes in coefficients and parameters, impacting the robustness of the model.


8. No Consideration of Time Value of Money:


Linear programming does not inherently account for the time value of money, crucial in many decision-making scenarios.


9. Assumption of Independence:


Linear programming assumes independence between decision variables, which may not hold true in real-world interdependent scenarios.


10. Computational Complexity:


Solving large-scale linear programming problems can be computationally demanding, especially with a large number of variables and constraints.


11. Difficulty Handling Non-Quantifiable Factors:


Linear programming is primarily quantitative and may struggle to handle qualitative or non-quantifiable factors effectively.





QUESTION 4(b)

Q Summarise four decision criteria used in decision making under uncertainty.
A

Solution


Decision Criteria in Decision-Making Under Uncertainty


1. Maximax Criterion:


Description: Select the alternative that maximizes the maximum possible payoff.

Decision Rule: Choose the alternative with the highest possible maximum payoff.


2. Maximin Criterion:


Description: Select the alternative that maximizes the minimum possible payoff.

Decision Rule: Choose the alternative with the highest minimum payoff.


3. Minimax Regret Criterion:


Description: Focuses on minimizing the regret associated with each decision.

Decision Rule: Minimize the maximum regret, where regret is the difference between the payoff of the chosen alternative and the best possible payoff for each state of nature.


4. Hurwicz Criterion:


Description: A compromise between optimism and pessimism, using an optimism coefficient (α) to weigh the minimum and maximum payoffs.

Decision Rule: Calculate the weighted average for each alternative and choose the one with the highest value.


5. Expected Monetary Value (EMV):


Description: Calculates the average payoff for each alternative, considering the probabilities of different states of nature.

Decision Rule: Choose the alternative with the highest expected monetary value.


6. Expected Opportunity Loss (EOL):


Description: Measures the average opportunity loss for each alternative.

Decision Rule: Choose the alternative with the lowest expected opportunity loss.


7. Expected Value of Perfect Information (EVPI):


Description: Measures the value of acquiring additional information to eliminate uncertainty.

Decision Rule: Compare the expected payoff with perfect information to the expected payoff without additional information.


8. Decision Trees:


Description: Utilizes a graphical representation of decision alternatives, probabilities, and payoffs to analyze complex decision scenarios.

Decision Rule: Evaluate decision nodes, probability nodes, and terminal nodes to identify the optimal decision path.


9. Real Options Analysis:


Description: Applies financial options concepts to evaluate the value of flexibility in decision-making.

Decision Rule: Incorporate the value of deferring, expanding, or abandoning a decision based on changing circumstances.


10. Bayesian Decision Making:


Description: Applies Bayesian probability to update beliefs and make decisions as new information becomes available.

Decision Rule: Update probabilities based on prior knowledge and new evidence to revise decision choices.





QUESTION 4(c)

Q (i) The price and quantity that would maximisc profit.

(ii) The maximum profit.
A

Solution


(i) The price and quantity that would maximisc profit.


Total cost of function = 10,000 + 65x

Q = a - bp

3100 = a - 95b

2,900 = a - 105b

200 / 10 = 10b / 10

b = 20

3100 = a - (95 x 20)

a = 5000

x = 5,000 - 20p

20p / 20 = 5,000 / 20 - x/20

P = 250 - 0.05x

Total revenue = P.Q

x(250 - 0.05x) = 250x - 0.05x²

Profit function

π=(250x - 0.05x²) - (10,000 + 65x) = 250x - 65x - 0.05x² - 10,000 = 185x- 0.05x² - 10,000

Δπ/Δx = 185 - 0.1x = 0

0.1x / 0.1 = 185 / 0.1

x = 1850

Profit maximizing price

= 250 - (0.5 x 1850)

= 250 - 92.5

= Sh. 157.50

(ii) Maximum profit


Profit function = 185x - 0.05x2 - 10,000

= (185 x 1850) - (0.05 x 18502) - 10,000

= 342,250 - 171,125 - 10,000 = Sh. 161,125




QUESTION 4(d)

Q (i) The average number of hair cuts that will be completed by the barber each hour.

(ii) The average time each customer will spend at the barber shop.
A

Solution


(i) The average number of hair cuts that will be completed by the barber each hour.


12 minutes = 1 customer
60 minutes = ?

60 minutes/12 minutes x 1 customer = 5 customers per one hour.

(ii) The average time each customer will spend at the barber shop.


λ = 20 customers per hour
μ = 5 customers per hour
K = Available seats = 10

P = λ / μ = 20 / 5 = 4

Probability of barber being idle

=
1 - P

1 - Pk+1
=
1 - 4

1 - 410+1
=
-3

-4,194,303
= 0.0000007153


Arrival rate of customers

λ = μ(1 - P idle) = 5(1 - 0.0000007153) = 4.999 ≈ 5

Number of customers in barbership at any given time

L
=
P

1-P
-
(k + 1)Pk+1

1 - Pk+1


L
=
4

1 - 4
-
11 x 411

1 - 411


L = (4 / -3) - (-11)

L = 9.67 customers

Expected time to be spent in barber shop

W
=
1

λ
L
=
1

5
x 9.67 = 1.934 hrs




QUESTION 5(a)

Q Outline five limitations of game theory.
A

Solution


Limitations of Game Theory


1. Assumption of Rationality:


Limitation: Game theory assumes that players are rational and always make decisions that maximize their utility. In reality, individuals may not always act in a perfectly rational manner due to emotions, cognitive biases, or incomplete information.

2. Complete Information:


Limitation: Many game models assume that players have complete information about the game, including the payoffs and strategies of all players. In practice, information may be incomplete or asymmetric, leading to different outcomes.


3. Static Nature:


Limitation: Game theory often models situations as one-shot games or in discrete time steps. In dynamic and evolving environments, where interactions occur continuously over time, traditional game theory may not capture the complexity of strategic decisions.


4. Limited Number of Players:


Limitation: Game theory becomes more complex as the number of players increases. Analyzing games with a large number of players becomes computationally challenging, and finding equilibrium solutions may become impractical.


5. Assumption of Common Knowledge:


Limitation: Game theory assumes that players have common knowledge of the rules, strategies, and payoffs. Achieving common knowledge in real-world scenarios can be challenging, and assumptions of common knowledge may not hold.


6. Nash Equilibrium:


Limitation: While Nash equilibrium is a central concept in game theory, not all games have unique equilibria, and some games may have multiple equilibria. The predictive power of Nash equilibrium relies on the uniqueness and stability of equilibria.


7. Mixed Strategies:


Limitation: In certain situations, players may find it difficult to randomize their strategies, especially if they have a strong preference for one particular strategy. The assumption of players employing mixed strategies may not always hold.


8. No Consideration of Cooperation:


Limitation: Traditional game theory assumes that players act independently and do not cooperate. In reality, cooperation and collusion among players are common, and game theory might not fully capture cooperative strategies.


9. Sequential Rationality:


Limitation: In extensive-form games, players are expected to make sequentially rational decisions. However, in situations of imperfect information or changing circumstances, players may deviate from their intended strategies.


10. Behavioral Considerations:


Limitation: Game theory often overlooks behavioral aspects, such as emotions, trust, and social norms, which can significantly influence decision-making in strategic interactions.


11. Dynamic and Adaptive Players:


Limitation: Game theory assumes fixed strategies, but in reality, players may adapt their strategies based on feedback and learning from previous interactions. Dynamic and adaptive behavior is not always adequately captured.


12. External Factors:


Limitation: Game theory may not account for external factors, such as regulatory changes, economic shocks, or technological advancements, that can impact the strategies and payoffs of players.





QUESTION 5(b)

Q (i) A network diagram of the project.

(ii) The expected completion time of the project

(iii) The probability that the project will be completed between 13 weeks and 17 weeks.
A

Solution


(i) A network diagram of the project.


Expected time = (0 + 4ML + P) / 6

Where;

0 -> Optimistia time
ML -> Most Likely time
P -> Pessimistical time

Activity Expected time
A [1.5 + (4 x 2) + 2.5] ÷ 6 = 2
B (2 + 2.5 x 4 + 6 ) ÷ 6 = 3
C (1 + 2 x 4 + 3 ) ÷ 6 = 2
D (1.5 + 2 x 4 + 2.5) ÷ 6 = 2
E (0.5 + 1 x 4 + 1.5) ÷ 6 = 1
F (1 + 2 x 4 + 3) ÷ 6 = 2
G (3 + 3.5 x 4 + 7) ÷ 6 = 4
H (3 + 4 x 4 + 5) ÷ 6 = 4
I (1.5 + 2 x 4 + 2.5) ÷ 6 = 2




Critical path A - B - G - H - I

(ii) Expected project completion time of project


Expected project completion time 15 days

(iii) The probability that the project will be completed between 13 weeks and 17 weeks.


δ2= ((Persimistic time-optimistic time) / 6)2

δ2 = ((2.5 - 1.5) / 6)² + ((6 - 2) / 6)² + ((7 - 3) / 6)² + ((5 - 3) / 6)² + ((2.5 - 1.5) / 6)²

= 0.028 + 0.444 + 0.444 + 0.111 + 0.028 = 1.055

Standard deviation

√1.055 = 1.027

δ = 1.207


X = 13
X̄ = 15
X = 17



Z
=
X - X̄

δ


Z
=
13 - 15

1.027
= 1.95


Probability from normal table = 0.4744

Therefore Probability P(13 ≤ X ≤ 17)




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