QUANTITATIVE ANALYSIS NOVEMBER 2020 - CPA SOLUTION|REVISION KITS|KASNEB

QUESTION 1a

Q (i) The profit function

(ii) The level of output that maximises pruth.

(iii) The breakeven output.

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

-b +- √b² - 4ac

-435 + -√435² - (4 x -2 x -8,750)

2 x -2

-435 +- √119,225

-4

-435 +- 345.29

-4

-435+345.29

-4

-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

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

Solution

N₁ = 13	x̄₁ = 45	S₁ = 4
N₂ = 15	x̄₂ = 55	S₂ = 5

S_p

√

(N₁ - 1)S₁² + (N₂ - 1)S₂²

N₁ + N₂ - 2

S_p

√

(13 - 1)4² + (15 - 1)5²

13 + 15 - 2

√

192 + 350

= 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.

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

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

1 -

6∑d ²

n(n² - 1)

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.

Solution

i). Service rate(μ)

μ = 4 patients per hour

ii). Length of queue

L_q

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

L_s

μ - λ

4 - 3

= 3 patients

iv). The time a patient takes being actually attended

W_s

μ - λ

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

P_{n >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.

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.

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

∑Y

- b

∑X

n∑xy - ∑x∑y

n∑x² - (∑x)²

(12 x 921,942) - (862 x 12,501)

(12 x 64,900) - 862²

287,442/35,756

8.039

∑Y

- b

∑X

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.

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.

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.

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.

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

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

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.

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.

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.

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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