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CPA
Foundation Leval
Quantitative Analysis November 2020
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1a

Q (i) The profit function

(ii) The level of output that maximises pruth.

(iii) The breakeven output.
A

Solution


(i) The profit function


AR = 150 - Q

TR = ARQ

Q(150 - Q)

= 150Q - Q²

Total variable cost = (Q - 285)Q
= Q² -285Q

Total cost = total fixed cost + total variable cost

= Q² - 285Q + 8,750

Profit function ( Π ) = TR-TC

1500 - Q² - (Q² - 285Q + 8,750)

= 150Q + 285Q - Q² - Q² - 8,750

= -2Q² + 435Q - 8,750

ii) The level of output that maximizes profit


Δπ / ΔQ = -4Q + 435

At profit maximizing output

Δπ / ΔQ = 0 = -4Q + 435
4Q / 4 = 435 / 4
Q = 108.75 = 109

iii). The breakeven output


At breakeven output profit = 0

-2Q² + 435Q - 8,750 = 0

Using Quadratic formula

Q
=
-b +- √b² - 4ac
2a


Q
=
-435 + -√435² - (4 x -2 x -8,750)
2 x -2


Q
=
-435 +- √119,225
-4


Q
=
-435 +- 345.29
-4


Q
=
-435+345.29
-4
or
-435-345.29
-4


Q = 22.43 or 195.07 units

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QUESTION 1b(i-iii)

Q (i) The optimum strategy for each player

(ii) The saddle point

(iii) The value of the game
A

Solution


i The optimum strategy for each player


Check for saddle point by using maxmin and minmax

Player B Strategies
Player A strategies
- A1 A2 A3 A4 A Minimax
B1 0 -4 1 2 4 -4
B2 -4 5 -1 1 9 -4
B3 13 5 3 11 9 3
B4 -2 8 -7 -1 -2 -7
Maximin 13 8 3 11 9 -


The game is a pure strategy game since it has a saddle point. Optimal strategy for player
A is A3 and optimal strategy for player B is B3

ii) The saddle point


A saddle point in game theory refers to an intersection point in a payoff matrix where the maximum value of one row coincides with the minimum value of one column. In other words, it is a specific cell in the matrix where a player can simultaneously achieve the best outcome against their opponent's strategy and the worst outcome for their opponent against their own strategy.

The saddle point = 3

iii) The value of the game


In game theory, the "value of the game" refers to the optimal expected outcome that a player can achieve in a given game. The value is often associated with the solution to the game, considering the best strategies for each player. One way to calculate the value of a game is through the concept of the minimax theorem.

For a two-player zero-sum game represented by a payoff matrix, where the gains of one player are balanced by the losses of the other, the value of the game (V) can be calculated using the minimax theorem.

The value of the game is 3

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QUESTION 1c

Q Determine whether there is any significant difference in the average test mark between the two classes at 5% level of significance
A

Solution


N1 = 13 x̄1 = 45 S1 = 4
N2 = 15 x̄2 = 55 S2 = 5
Sp
=
√
(N₁ - 1)S₁2 + (N₂ - 1)S₂2
N₁ + N₂ - 2

Sp
=
√
(13 - 1)42 + (15 - 1)52
13 + 15 - 2

=
√
192 + 350
26
= 4.57


∴ S X̄₁ = Sp / √N₁ = 4.57 / √13 = 1.27

SX̄₂ = Sp / √N₂ = 4.57 / √15 = 1.18

S(x̄₁ - x̄₂) = √Sx̄₁2 + Sx̄₂2

√1.272 + 1.182

1.73

T Calculated
=
S̄1-X̄2
Sx̄1-x̄2


=
45-55
1.73
= 5.78


From table 5% value with (n₁ + n₂ − 2)

(13 + 15 - 2) = 26 degrees freedom 2.056

Since calculated value greater than value, then there is a significant difference average test marks of the two classes

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QUESTION 2a

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

ii). Advise the management of the bakery on the best product to introduce into the unckot.
A

Solution



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




ii) Advice


Management should introduce white bread since it gives the highest expected monetary value of Sh. 576, 800


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QUESTION 2b(i - ii)

Q i) The rank correlation coefficient. Interpret your results.

ii). Coefficient of determination
A

Solution


i) The rank correlation coefficient. Interpret your results.


Contestant 1st Assessor Rank 2nd Assessor Rank d d 2
A 72 6 76 5 1 1
B 82 1 80 3 -2 4
C 79 3 78 4 -1 1
D 70 7 73 6 1 1
E 67 8 70 7 1 1
F 81 2 85 1 1 1
G 78 4 69 8 -4 16
H 75 5 83 2 3 9
I 65 9 68 9 0 0
34


R
=
1 -
6∑d 2
n(n2 - 1)


R
=
1 -
6 x 34
9(92 - 1)


1-204 / 720

1 - 0.283

0.7167

ii) Coefficient of determination


r2 = 0.71672 = 0.5137


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QUESTION 2c

Q i). Service rate

ii). Length of queue

iii). Length of the system.

iv). The time a parient takes being actually attended.

v). The probability that there are more than six patients in the outpatient hospital department.
A

Solution


i). Service rate(μ)


μ = 4 patients per hour

ii). Length of queue
Lq
=
λ2
μ(μ - λ)


λ = 3 patients per hour
μ = 4 patients per hour

∴
=
32
4(4 - 3)
= 2.25 ➫ 2 patients


iii). Length of the system


Ls
=
λ
μ - λ
=
3
4 - 3
= 3 patients


iv). The time a patient takes being actually attended


Ws
=
1
μ - λ
=
1
4 - 3
= 1 hour


v). The probability that there are more than six patients in the outpatient hospital department.


P(n >k) = (λ / μ)k+1

λ=3, μ=4, k=6

Pn >6 = (3 / 4)6+1 = 0.1335 .

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QUESTION 3a

Q The average contribution of Dolce Ltd. using Monte Carlo simulation with 10 simulations.
A

Solution


Selling price
S.P Probability Cummulative probability R.N
600 0.30 0.30 00-29
700 0.50 0.80 30-79
800 0.20 1.00 80-99


Variable cost
V.C Probability Cummulative probability R.N
300 0.40 0.40 00-39
400 0.50 0.90 40-89
500 0.10 1.00 90-99


Sales volume
S.V Probability Cummulative probability R.N
40,000 0.30 0.30 00-29
50,000 0.50 0.80 30-79
60,000 0.10 1.00 80-99


Simulation worksheet
Trials RN Selling price RN variable cost RN Sales volume Contribution
1 44 700 84 400 82 60,000 18,000,000
2 50 700 85 400 40 50,000 15,000,000
3 96 800 88 400 16 40,000 16,000,000
4 16 600 97 500 92 60,000 6,000,000
5 44 700 82 400 39 50,000 15,000,000
6 33 700 83 400 42 50,000 15,000,000
7 16 600 07 300 77 50,000 15,000,000
8 66 700 50 400 20 40,000 12,000,000
9 50 700 95 500 83 60,000 12,000,000
10 39 700 58 400 44 50,000 15,000,000
139,000,000

Average contribution

= 139,000,000 / 10 = 13,900,000

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QUESTION 3b(i)

Q Using the ordinary least squares method:

Formulate the indirect cost function.
A

Solution


Labour hours Indirect labour Cost
X2 X
"000"		 Y
"000"		 XY Y2
2,304 48 963 46,224 927,369
4,624 68 752 51,136 564,504
8,836 94 1,032 97,008 1,065,024
6,724 82 1,316 107,912 1,731,856
2,116 46 710 32,660 504,100
6,084 78 1180 92,040 1,392,400
9,216 96 1,456 139,776 2,119,936
3,600 60 770 46,200 592,900
5,184 72 1,004 72,288 1,008,016
3,844 62 1,211 75,082 1,466,521
7,744 88 917 80,696 840,889
4,624 68 1,190 80,920 1,416,000
∑X2 = 64,900 ∑X = 862 ∑Y=12,501 ∑XY = 921,942 ∑Y 2 = 13,630,615

a
=
∑Y
n
- b
∑X
n


b
=
n∑xy - ∑x∑y
n∑x2 - (∑x)2


b
=
(12 x 921,942) - (862 x 12,501)
(12 x 64,900) - 8622


287,442/35,756

8.039

a
=
∑Y
n
- b
∑X
n


12,501 / 12 - (8.039 x 862/12)

1,041.75 - 577.468

464.282

The indirect cost functions

Y = a + bx

Y = 464.282 + 8.039x


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QUESTION 3b(ii)

Q (ii) Compute the indirect labour cost for 120 labour hours.
A

Solution


Y = [464.282 + (8.039 x 120)]1,000

=(464.282 + 964.680)1,000

= Sh. 1,428,962

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QUESTION 3b(iii)

Q Calculate the coefficient of determination.
A

Solution


r2
=
nΣxy - ΣxΣy
√(nΣx2 - (Σx)2) x (nΣy2 - (Σx)2)

r2
=
(12 x 921,942) - (862 x 12,501)
√(12 x 64,900 - 8622) x (12 x 13,630,615 - 12,5012)


287,442 / 510,633.2378

0.5629

56.29% variation in indirect labour cost is due to change in direct labour hours


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QUESTION 4a(i-ii)

Q Explain the following terms as used in linear programming:

(i) Infeasibility.

(ii) Unboundedness.
A

Solution


i) Infeasibility


A linear programming problem is said to be infeasible if there is no solution that satisfies all the constraints. It represents a state of inconsistency in the state of constraints. Under the simplex method, the problem is said to have no feasible solution if at least one of the artificial variable remains in the final simplex table as basic variable with non-zero quantity

ii) Unboundedness


In unbounded solution of a linear programming problem is a situation where objective function is infinite. A linear programming problem is said to have unbounded solution if its solution can be made infinitely large without violating any of its constraints in the problem. For maximization problem, the values can increase indefinitely while in minimization the values can decrease to zero.

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QUESTION 4b(i)

Q Required:
The optimal assignment for these three topics.
A

Solution


The optimal assignment for three topics

Create dummy topic so that number of topics to be assigned is equal to number of lecturers.

T1 T2 T3 T4
L1 42 16 27 0
L2 48 40 25 0
L3 50 18 36 0
L4 58 38 60 0


1. Reduce each column by the largest figure in that column and ignore the resulting signs

T1 T2 T3 T4
L1 16 24 33 0
L2 10 0 35 0
L3 8 22 24 0
L4 0 2 0 0


2. Reduce each row by smallest figure in that row. Since smallest figure is zero for each row, then matrix remains as above

3. Cover zeros by minimum number of lines

T1 T2 T3 T4
L1 16 24 33 0
L2 10 0 35 0
L3 8 22 24 0
L4 0 2 0 0


4. Subtract 8 from any uncovered element and add 8 to any number where lines have intersected each other



T1 T2 T3 T4
L1 8 16 25 0
L2 10 0 35 8
L3 0 14 16 0
L4 0 2 0 8


Since number of lines covering all zeros is equivalent to number of allocations to be made, then allocation is optimal. Therefore, optimal assignment is:

T1 T2 T3 T4
L1 8 16 25 0
L2 10 0 35 0
L3 0 14 16 X0
L4 X0 2 0 8


Assignment

Topic Lecture Marks
T1 L3 50
T2 L2 40
T3 L3 60
Maximum score 150

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QUESTION 4b(ii-iii))

Q (ii) The maximum score.
(iii) The lecturer that will not be assigned any topic.
A

Solution


ii) Maximum score


150

(iii) The lecturer that will not be assigned any topic.


Lecturer 1
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QUESTION 4c(i)

Q 3 quarter moving average of the series.
A

Solution


Year Q Profits Three Quater moving avarage
2016 1 83
2 260 (83 + 260 + 2,150) / 3 = 186
3 215 (260 + 215 + 293) / 3 = 256
4 293 (215 + 293 + 105) / 3 = 204.33
2017 1 105 (293 + 105 + 383) / 3 = 260.33
2 383 (105 + 383 + 248) / 3 = 245.33
3 248 (383 + 248 + 553) / 3 = 394.67
4 553 (248 + 553 + 140) / 3 = 313.67
2018 1 140 (553 + 140 + 430) / 3 = 374.33
2 430 (140 + 430 + 323) / 3 = 297.67
3 323 (430 + 323 + 588) / 3 = 447
4 588 (323 + 588 + 168) / 3 = 359.67
2019 1 168 (588 + 168 + 503) / 3 = 419.67
2 503 (168 + 503 + 340) / 3 = 337
3 340 (503 + 340 + 755) / 3 = 532.67
4 755


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QUESTION 4c(ii)

Q The deseasonalised profit of the economic sector using the additive model.
A

Solution


Y T Y-T
260 186 74
215 256 -41
293 204.33 88.67
105 260.33 -155.67
383 245.33 137.67
248 394.67 -146.67
553 313.67 239.33
140 374.33 -234.33
430 297.67 132.33
323 447 -124
588 356.67 228.33
168 419.67 -251.67
503 337 166
340 532.67 -192.67


1 2 3 4
2016 0 74 -41 88.67
2017 -155.33 137.67 -146.67 239.33
2018 -234.33 132.33 -132.33 -124
2019 228.33 -251.67 166 -192.67
Total -161.33 92.33 -154 11.33
Average -53.78 23.08 -38.50 2.83
Adjustment factor 16.59 16.59 16.59 16.59
Final index -37.19 39.67 -21.91 19.42


Total = -53.78 + 23.08 - 38.50 + 2.83 = -66.37

Adjustment factor = 66.37 / 4 = 16.59

Year Quater Y Seasonal index Deseasonalised profit
2016 1 83 -37.19 120.19
2 260 39.67 220.33
3 215 -21.91 236.91
4 293 19.42 273.58
2017 1 105 -37.19 142.19
2 383 39.67 343.33
3 248 -21.91 269.91
4 553 19.42 533.58
2018 1 140 -37.19 177.19
2 430 39.67 390.33
3 323 -21.19 344.91
4 588 19.42 568.58
2019 1 168 37.19 205.19
2 503 39.67 463.33
3 340 21.91 361.91
4 755 19.42 735.58

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QUESTION 5a(i)

Q With reference to analysis of variances (ANOVA) tests:

Distinguish between one-way and two-way ANOVA tests.
A

Solution


Distinction between one-way and two-way ANOVA tests


A one way ANOVA test has one independent variable while a two way ANOVA test has two independent variables.


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QUESTION 5a(ii)

Q Outline four assumptions of two-way ANOVA tests.
A

Solution


Assumptions of two-way ANOVA tests


➢ Normality: That each sample is taken from a normally distributed population
➢Variance equality: That the variances of data in the different groups should be the same
Sample independence: Each sample has been drawn independently of the other sample
➢Two independent variables which should consist of two or more categorical independent groups
➢There should be no significant outliers
➢There should be independence of observations. This means that there should be no relationship between the observations in each group or between the group themselves
➢ Your dependent variable should be measured at the continuous level. They're interval or ratio variables.

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QUESTION 5a(iii)

Q Explain the difference between ANOVA tests and T-tests.
A

Solution


Difference between ANOVA tests and T-tests


➢ ANOVA is statistical technique that is used to compare the means of two populations while T-test hypothesistest that used compare of two populations
➢Test statistic for T-test is T-test whereas test statistic for ANOVA is ANOVA
➢ T-test tells us whether not two groups have the same mean. An ANOVA, other hand, tells whether or not three groups all have the same mean, but it doesn't tell which groups have means that are different from another.To find out which groups differ from one another, we would have to perform post hoc tests.

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QUESTION 5b(i)

Q In the context of critical path analysis (CPA) method:

Discuss two strengths and two weaknesses of CPA method.
A

Solution


Strengths and weaknesses of CPA method


Strengths of CPA
➤ Makes dependencies clear and transparent
➤ Identifies all critical activities that need attention
➤Helps to compare the planned with real status
➤ Saves time and helps in the management of deadlines
➤It defines the most important tasks
➤ The method visualizes projects in a clear graphical form.
It gives a fair and concise procedure of documenting project.
➤ It helps in optimization by determining the project duration
➤ An explicit and clear approach of communicating project plans schedules time and cost performance id developed.
➤It is extensively used in industry
➤It helps in determining the slack time
➤ CPA can strengthen team perception if it is applied properly.
It gives a practical and disciplined base which helps in determining how to reach objectives

Weaknesses of CPA method
➤The project still requires management because external factors may change
➤ Complex activities may be impossible to represent accurately on a network
Reliability of CPA largely based on accurate estimates and assumptions made
➤ Resources may not actually be flexible as management hope when they come to address the network float
➤Too many activities may make the network diagram too complicated. Activities might themselves have to be broken down into mini-projects.
➤The scheduling of personnel is not handled by CPA method
➤In CPA, it is difficult to estimate the completion time of an activity.
➤The critical path is not always clear in CPM
➤ It does not handle the scheduling of the resource allocation
In CPA, critical path needs to be calculated precisely

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QUESTION 5b(ii)

Q Explain three practical applications of CPA method.
A

Solution


Practical applications of CPA method


Construction of a Residential complex
Commercial complex
Petro-chemical complex
Ship building
Satellite mission development
Installation of a pipe line project etc

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