CPA
Foundation Leval
Quantitative Analysis May 2018
Suggested solutions

Total profit (TP) = Total revenue (TR) - Total cost (TC)

Total Revenue (TR) = pq.

(100 - 5q)q

TR = 100q - 5q²

TC = q² + 4q + 300

TP = (100q - 5q²) - (q² + 4q + 300)

-6q² + 96q - 300

At maximum profit

Margin profit(MP) = 0

MP = dtp / dq

-12q = -96

q = -96 / -12 = 8 units

Hence, 8 electric components should be produced The maximum profit: so as to maximize profit

(ii) The maximum profit

TP_(Sh000) - 6q² + 96q - 300

q = 8 units

TP_(Sh000) = -6(8)² + 96(8) - 300

-384 + 768 - 300 = 84

Therefore, Maximum profit Sh.84,000

(iii) The maximum total revenue

TR_(Sh000) = 100q - 5q²

q = 8 units

TR_(Sh000) = 100(8) - 5(8)²

800 - 320 = 480

Therefore Maximum total revenue shs. 480,000

S = Steel dept
M = M vehicles
C = Costruction department



(ii) The total output of each department given that the LeontiePs inverse matrix is as provided below:



Hence total output of each dept









Hence, the total output in millions" of each department is 151.042, 265.104 and 159.375 for steel (a) motor vehicle (b) and construction (c) respectively.

(iii) The change in the total output of the construction department.

The vector matrix of final demand after incorporating changes:



The total output of each department after changes









168.75 - 159.375 = 9.375 Million

μ = x̄+ t_{α / 2,n-1} x δ / √n

Since the population standard deviation (δ) is not known, it can be estimated from the sample as



x̄ = ∑x / n = 584 / 9 = 64.8889

x	(x - x̄)2
64	0.7901
62	8.3457
65	0.0123
63	3.5679
68	9.6789
69	16.9011
65	0.0123
63	3.5679
65	0.0123
∑x = 584	42.8885

= 2.3154

α = 1 - 0.95 = 0.05

t_α/2,n-1 = t_0.025,8

t_0.025,8 = 2.306

μ = x̄ ± t_{α / 2,n-1} x δ/√n

μ = 64.8889 ± 2.306 x 2.3154/√9

μ = 64.8889 ± 1.7798

Formulating the hypothesis:

HO : μl - 2 - 0; i.e there is no significant difference in the monthly mean sales of the two outlets A and B

Hl : ul - u2 ≠01 i.e there is significant difference in the monthly measures of the two outlets A and B

The test statistic to be used in this case, it is a z-test for two sample means;

Where:

X̄₁ = 795 (sh000)
X̄₂ = 810 (sh000)
δ₁ = 50 (sh000)
δ₂ = 70 (sh000)
n₁ = 200
n₂ = 175



-15 / √40.5 = -2.357

α = 5% = 0.05

Z _α/2 = Z_0.025

Z_0.025 = + 1.96.

Illustration:



Decision
The observed Z = -2.357 is less than the critical Z = -1.96, and lies in the rejection region for Ho.The decision is therefore to reject Ho

Conclusion
There is significant difference in the monthly mean sales of the two outlets: A and B.

Contribution = Sp - Vc

Type of pen	Contribution per unit
	Sh
P1	200 - 100 = 100
P2	200 - 80 = 120
P3	200 - 60 = 140

π = Total contribution - Total fixed cost

Pay off table
Law(L)	Moderate(M)		High(H)
Decision alternative	(250,000)	(1,000,000)	(1,500,000)
P1(250,000)	23,000,000	-50,000,000	-100,000,000
P2(100,000)	26,800,000	116,800,000	60,000
P3(1,500,000)	29,000,000	134,000,000	204,000,000

W1

(P₁L)(250,000 × 100) - 2,000,000 = 23,000,000
(P₁M)(250,000 × 100 - 2,000,000) - (750,000 × 100 - 2,000,000) = -50,000,000
(PH)(250,000 × 100 - 2,000,000) - (1,250,000 × 100 - 2,000,0000) = -100,000,000

(P₂L) (250,000 × 120) - 3,200,000 = 26,800,000
(P₂ M) (1,000,000 × 120) - 3,200,000 = 116,800,000
(P₂H) [(1,000,000 × 120) - 3,200,000] - [500,000 × 120 - 3,200,000] = 60,000,000

(P₃L)(250,000 x 140 - 6,000,000) = 29,000,000
(P₃M)(1,000,000 x 140 - 6,000,000) = 134,000,000
(P₃H) (1,500,000 × 140 - 6,000,000) = 204,000,000

(ii) The type of pen to produce using the maximum criterion

The minimum of each row;

P1 = Sh-100,000,000
P2 = Sh.26,800,000
P3 = Sh.29,000,000

Therefore the maximum -Sh. 29,000,000 for P₃ Hence, the company should produce type 3 pen so as to maximize profit.


(iii) The type of pen to produce using the maximax criterion:

The maximum of each row:

P₁ = Sh 23,000,000
P₂ = Sh.116,800,000
P₃ = sh.204,000,000

Therefore the maximum of maximum is Sh 204,000,000; for P₃

Hence, the company should produce type 3 pen so as to maximize profit


(iv) The type of pen to produce using the minimax regret criterion

Constructing an opportunity loss table by deducting each payoff under each of the states of nature that is demand, from the maximum payoff under each of the given states of nature or demand. This gives an opportunity loss table as shown below

DEMAND
		Low	Moderate	High
Decision alternative	P1	6,000,000	184,000,000	304,000,000
	P2	2,200,000	17,200,000	144,000,000
	P3	0	0	0

The maximum request under each decision alternative

P1 = Sh. 304,000,000
P2 = Sh. 144,000,000
P3 = Sh. 0

Based on the minimax regret criterion, therefore the company should produce type3 pen so as to minimize the opportunity loss i.e maximum regret under the various states of demand.

Where;
x = Intelligence test score(%)
y = Weekly sales(sh.000)

X	Y 000	X2	Y2	XY
40	50	1,600	2,500	2,000
70	120	4,900	14,400	8,400
50	80	2,500	6,400	4,000
60	100	3,600	10,000	6,000
80	80	6,400	6,400	6,400
50	50	2,500	2,500	2,500
90	110	8,100	12,100	9,900
40	60	1,600	3,600	2,400
60	90	3,600	8,100	5,400
60	60	3,600	3,600	3,600
∑x=600	∑y=800	∑x2 38,400	∑y2 69,600	∑xy=50,600

n = 10



= 0.709

Interpretation

The correlation coefficient found above demonstrates a reasonably strong position association between the intelligence test result and the weekly sales made by the sales ladies.

The objective function (z);
Zmax = 2x + y

Subject to the constraints

Raw material RM1 constraint:

3x ≤ 30

Raw material RM2 constraint:

2y ≤ 30

Raw material RM3 constraint

x + y ≤ 20

Non negativity constraint

x ≥ 0 y ≥ 0

(ii) The optimum product mix using the simplex method

1st Simplex table
Solution Variable	Products		Slacks variable			Solution quantity
	x	y	S1	S2	S3
S1	3	0	1	0	0	27
S2	0	2	0	1	0	30
S3	1	1	0	0	1	20
Z	2	1	0	0	0	0

27 / 3 = 9, 20 / 1 = 20, 30 / 0 = 0

2nd Simplex table
Solution Variable	Products		Slacks variable			Solution quantity
	x	y	S1	S2	S3
X	1	0	1/3	0	0	9
S2	0	2	0	1	0	30
S3	0	1	-1/3	0	1	11
Z	0	1	-2/3	0	0	-18

9 / 0 = 0, 30 / 2 = 15, 11 / 1 = 11

3rd Simplex table
Solution Variable	Products		Slacks variable			Solution quantity
	x	y	S1	S2	S3
X	1	0	1/3	0	0	9
S2	0	0	2/3	1	-2	8
Y	0	1	-1/3	0	1	11
Z	0	0	-1/3	0	0	-29

Hence, the company should produce 9 units of product x and 11 units of product y so as to maximize the weekly profit

max z = 2(9) + 1(11) = 18 + 11

maximum profit = Sh. 29 per week

P(r) = e^-λ λ ^r / r!

where;

e = 2.7183

r = number of successes and r = 0,1,2,... n

入 = mean occurences

Therefore,

e = 2.7183

r = 0

入 = 0.4 per week



(ii) The probability that at most two transformers will be struck by lightning in a week.

p(r ≤ 2) = p(r = 0) + p(r = 1) + p(r = 2)

p(r = 0) = 0.6703





p(r ≤ 2) = 0.6703 + 0.2681 + 0.0536 = 0.992

Hence, the probability that most two transoformers be struck in week 0.992 or 99.2%

Expected/Average time (te)

Where;
a = most optimistic time estimate
m = most likely estimate
b = most perstimistic time estimate
Variance (δ²) = [(b - a) / 6]²

Activity	Expected/Average time(Weeks) te = a + 4m + b 6	Variance δ2 = [(b - a) / 6]2
A	5	2.7778
B	14	7.1111
C	9	0.1111
D	15	2.7778
E	8	0.4444
F	9	0
G	4	0.4444
H	5	0

(ii) The critical path

Is: A - D - G - H with a project duration of 29 weeks

(iii) The probability of completing the project within a 30 - week duration, (3 marks)

z = (t1 - te)/δ

δ = √(2.7778 + 2.7778+ 0.4444 + 0)

√6 = 2.4495 weeks

te = 29 weeks
t1 = 30 weeks

z = (30 - 29) / 2.4495 = 0.41

P(z ≤ 0.41) = 0.5 + 0.1591

0.6591 or 65.91%....

Illustration