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CPA
Foundation Leval
Quantitative Analysis May 2016
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1(a)

Q Explain four differences between the project evaluation and review technique (PERT) and the critical path analysis (CPA).
A

Solution


Differences Between PERT and CPA


  • Main Difference: PERT pertains to probabilistic time estimates, incorporating uncertainty in activity durations, while CPA relies on deterministic time estimates, assuming fixed durations.
  • Other Differences:
    • Pertains to Probabilistic Time Estimates
      • PERT is primarily used for projects with uncertain activity durations, incorporating probabilistic time estimates.
      • CPA relies on deterministic time estimates, assuming fixed durations for each activity.
    • Network Complexity
      • PERT involves the creation of a network diagram with multiple time estimates for each activity, considering optimistic, pessimistic, and most likely scenarios.
      • CPA uses a simpler network diagram, often represented as a Gantt chart, where activities are linked in sequence.
    • Focus on Critical Path
      • CPA emphasizes identifying the critical path, which is the sequence of activities that determines the project's minimum duration.
      • PERT provides a broader view, allowing for the analysis of the entire project network, including non-critical paths.
    • Activity Time Estimates
      • PERT uses three time estimates for each activity: optimistic (O), pessimistic (P), and most likely (M), to calculate expected durations and probabilities.
      • CPA uses a single deterministic time estimate for each activity, assuming a fixed duration for planning purposes.
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QUESTION 1(b)

Q A certain audit firm has two categories of employees, auditors and assistant auditors. The total monthly salary of I auditor and 5 assistant auditors amount to Sh.456.755 whereas the total monthly salary of 3 auditors and 9 assistant auditors amount to Sh.985,005. The firm has a total of 6 auditors and 14 assistant auditors. The employees contribute 12 per cent of their monthly salaries towards their sacco society.

Required:
(i) The monthly salary of an auditor and an assistant auditor, using matrix algebra,

(ii) The employees' total monthly contribution towards their sacco society.
A

Solution


(i) The monthly salary of an auditor and an assistant auditor, using matrix algebra,


Auditors salaries = A
Assisstant auditors salaries = B

1A + 5B = 456,755
3A + 9B = 985,005

(
1
3
5
9
)
(
A
B
)
=
(
456,755
985,005
)


A
=
Determinant
[
456,755
985,005
5
9
]
Determinant
[
1
3
5
9
]


A
=
(456,755 x 9) - (985,005 x 5)
(1 x 9) - (3 x 5)


-814,230 / -6 = Sh.135,705


B
=
Determinant
[
1
3
456,755
985,005
]
Determinant
[
1
3
5
9
]


B
=
(1 x 985,005) - (3 x 456,755)
(1 x 9) - (3 x 5)



-385,260 / -6 = Sh.64,210

ii)Employees total monthly contribution towards their sacco society


= (6 x 135,705) + (14 x 64,210)

= 814,230 + 898,940 = Shs 1,713,170

Total monthly pension contribution

12 / 100 x 1,713,170 = Sh. 205,580.40

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QUESTION 1(c)

Q Shujaa Limited deals in the manufacture of a product named "Zed". The product "Zed" is produced on order and the company does not keep inventory of the product. The demand and total cost functions (in thousands of shillings) of the company are given as follows:

P = 190 - q

and

TC = q² + 10q + 500

Where:

P is the unit selling price.

q is the quantity demanded in units.

TC is the total cost.

Required:
(1) The maximum profit of the company.

(ii) The output level that would maximise total revenue.
A

Solution


1)The maximum profit of the company


TR = PQ

= (190 - Q)Q = 190Q - Q²

Profit function (π) = TR - TC

190Q - Q² - (Q² + 10Q + 500)

= 190Q - Q² - Q² - 10Q - 500

190Q - 10Q - Q² - Q² - 500

π = 180Q - 2Q²-500

Δπ / ΔQ = 180 - 4Q = 0

4Q / 4 = 180/4

Q = 45 Units

Maximum profits of company

= (180 x 45) - (2 x 45² ) - 500

= Shs. 3,550 x 1000

= Shs 3,550,000

(ii) The output level that would maximize total revenue


ΔTR/ ΔQ = 190 - 2Q = 0

190 - 2Q = 0

2Q / 2 = 190 / 2

Q = 95,000 units

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QUESTION 2(a)

Q Distinguish between a "univariate function" and a "multivariate function".
A

Solution


Univariate vs. Multivariate Functions


  1. Definition:

    Univariate Function: A univariate function is a mathematical function that involves only one variable. It maps each value of that variable to a unique output.

    Multivariate Function: A multivariate function is a mathematical function that involves multiple variables. It maps combinations of values of these variables to unique outputs.

  2. Variables:

    Univariate Function: In a univariate function, there is only one independent variable. It is represented as f(x) where x is the single variable.

    Multivariate Function: In a multivariate function, there are two or more independent variables. It is represented as f(x, y, z, ) where x, y, z, are the multiple variables involved.

  3. Graphical Representation:

    Univariate Function: The graph of a univariate function is typically a curve in a two-dimensional coordinate system, where one axis represents the independent variable.

    Multivariate Function: The graph of a multivariate function exists in a multi-dimensional space, and visualizing it may require three or more dimensions, depending on the number of variables.

  4. Example:

    Univariate Function: f(x) = 2x + 3 where x is the independent variable.

    Multivariate Function: f(x, y) = 3x^2 + 2y - 5 where x and y are the independent variables.

  5. Dimensionality:

    Univariate Function: Univariate functions are one-dimensional, involving a single variable.

    Multivariate Function: Multivariate functions are multi-dimensional, involving two or more variables.

  6. Derivatives:

    Univariate Function: Derivatives in a univariate function are calculated with respect to a single variable, yielding a rate of change with respect to that variable.

    Multivariate Function: Partial derivatives in a multivariate function are calculated with respect to each variable independently, providing rates of change with respect to individual variables.

  7. Notation:

    Univariate Function: Represented as f(x) or y = f(x), where x is the independent variable.

    Multivariate Function: Represented as f(x, y) or z = f(x, y) , where x and y are the independent variables.

  8. Application:

    Univariate Function: Commonly used in modeling relationships involving a single variable, such as distance as a function of time.

    Multivariate Function: Applied in scenarios where multiple factors influence an outcome, such as the temperature as a function of both time and geographical location.

  9. Mathematical Representation:

    Univariate Function: Typically expressed as an equation involving a single variable, and solutions are points on a one-dimensional axis.

    Multivariate Function: Involves equations with multiple variables, and solutions are points in a multi-dimensional space.

  10. Optimization:

    Univariate Function: Optimization involves finding the maximum or minimum of the function with respect to a single variable.

    Multivariate Function: Optimization in a multivariate function requires finding the maximum or minimum considering multiple variables simultaneously.
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QUESTION 2(b)

Q The mean weight of 500 packaging tins from a production process are normally distributed with a mean weight of 151 grammes and a standard deviation of 15 grammes.

Required:
(i) The number of packaging tins that weigh between 120 grammes and 155 grammes.

(ii) The number of packaging tins that weigh more than 185 grammes.
A

Solution


(i) The number of packaging tins that weigh between 120 grammes and 155 grammes.




Z
=
x - μ
δ


Z
=
120 - 151
15
= 2.07


Z
=
155 - 151
15
= 0.27


Probability of a z score of 2.07 from normal table is 0.4808

Probability of a z score of 0.27 from normal table is 0.1064

P(120 ≤ x ≤155) = 0.1064 + 0.4808 = 0.5872

Number of packagi g tins weighing between 120 grammes and 155 grams

500 x 0.5872 = 293.6 tins

(ii) The number of packaging tins that weigh more than 185 grammes.




Z
=
x - μ
δ


Z
=
185 - 151
15
= 2.27


Probability from normal table of z score of 2.27 is

P(x ≥ 185) = 0.5 - 0.4884 = 0.0116

Components weighing above 185 grammes

(0.0116 x 500) = 5.8 tins
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QUESTION 2(c)

Q The following data were obtained from the records of Kiwandani Limited for the year 2015:



Month
January
February
March
April
May
June		 Total
overhead cost (y)
(Sh.)
16,250
15,000
15,000
14,500
15,250
15,750		 Director labour
hours (x)

1,056
736
840
800
880
1,008

Required:
(i) The least squares regression function relating direct labour hours and total overhead cost.

(ii) The coefficient of determination. Comment on your result.
A

Solution


(i) The least squares regression function relating direct labour hours and total overhead cost.


Y X XY X2 Y2
16,250 1,056 17,160,000 1,115,136 264,062,500
15,000 736 11,040,000 541,696 225,000,000
15,000 840 12,600,000 705,600 225,000,000
14,500 800 11,600,000 640,000 210,250,000
15,250 880 13,420,000 774,400 232,562,500
15,750 1,008 15,876,000 1,016,064 248,062,500
∑ y=91,750 ∑X=5,320 ∑XY=81,696,000 ∑2=4,792,896 ∑Y2=1,404,937,500


Y = a + bx

b
=
nΣxy - ΣxΣy
nΣx2 - (Σx)2


b
=
6 x 81,696,000 - 5,320 x 91,750
6 x 4,792,896 - (5,320)2


2,066,000 / 454,976 = 4.54

a
=
Σy
n
-
bΣx
n


a = 91,750 / 6 - 4.54 x 5,320 / 6

15,291.67 - 4,025.47 = Sh.11,266.2

Regression equation Y = 11,266.2 + 4.54x

(ii) The coefficient of determination. Comment on your result.


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


r2
=
6 x 81,696,000 - 5,320 x 91,750
√(6 x 4792896 - 5,3202) x (6 x 1,404,937,500 - 91,7502)


2,066,000 / 2,293,612.87 = 0.9008

90.08% variation in total over head cost is due to the variation in direct labour hours
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QUESTION 3(a)

Q Explain the following terms as used in game theory:

(i) Pay-off.

(ii) Value of a game.
A

Solution


(i) Pay-off


In game theory, a pay-off refers to the outcome or result that a player receives based on the combination of strategies chosen by all players in a game. It represents the utility, profit, or satisfaction that a player gains after the game is played. Pay-offs are typically represented numerically and are associated with each possible combination of strategies chosen by the players. The goal of each player is to maximize their own pay-off, and the pay-off matrix illustrates the various outcomes for different strategy combinations.

(ii) Value of a Game


The value of a game in game theory is a concept that indicates the expected outcome or average pay-off that a player can achieve over repeated plays of the game, assuming rational and optimal decision-making. It provides insight into the long-term prospects of a game for a player. The value of a game can be either positive, indicating a favorable situation for the player, or zero, suggesting a fair or balanced game where neither player has a significant advantage. The value of a game is calculated through various solution concepts, such as the minimax theorem or the Nash equilibrium, depending on the nature of the game.



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QUESTION 3(b)

Q Highlight eight steps followed in the simulation process.
A

Solution


Steps in the Simulation Process


  1. Problem Formulation:

    2. Clearly define the problem or system that needs to be studied through simulation. Identify the objectives, variables, and constraints that will be part of the simulation model.

  2. Model Development:

    3. Create a mathematical or logical representation (model) of the real-world system. Define the relationships and interactions among variables and components in the system.

  3. Data Collection:

    4. Gather relevant data to populate the simulation model. This may include historical data, statistical information, or experimental results necessary for accurate representation of the system.

  4. Model Validation:

    5. Verify the accuracy and reliability of the simulation model by comparing its output with real-world observations or data. Adjust the model if discrepancies are found.

  5. Experimentation:

    6. Conduct experiments with the simulation model to analyze the system's behavior under different conditions. This step helps in gaining insights, testing hypotheses, and making decisions.

  6. Sensitivity Analysis:

    7. Evaluate the sensitivity of the model to changes in input parameters. Identify critical factors that significantly influence the simulation results.

  7. Optimization:

    8. Fine-tune the simulation model or system parameters to achieve optimal performance or desired outcomes. This step may involve adjusting variables to improve efficiency or meet specific goals.

  8. Implementation:

    9. Implement the simulation model for practical use. This may involve integrating the model into decision-making processes or using it as a tool for planning and analysis.


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QUESTION 3(c)

Q The table below shows the actual sales and target sales of eight sales agents for the year 2015 in millions of shillings.

Sales agent

Actual sales (y)
Target sales (x)		 1

45
40		 2

41
27		 3

50
45		 4

56
38		 5

60
52		 6

42
35		 7

43
29		 8

52
44

Required:
The Spearman's rank correlation coefficient. Interpret your result.
A

Solution


Y Rank X Rank d d2
45 5 40 4 1 1
41 8 27 8 0 0
50 4 45 2 2 4
56 2 38 5 -3 9
60 1 52 1 0 0
42 7 35 6 1 1
43 6 29 7 -1 1
52 3 44 3 0 0
∑d=0 ∑d2=16


R
=
1 -
6d²
n(n² - 1)


Where:

R = Spearman's rank correlation coefficient

d = difference between the two ranks of each observation

n = number of observations

R
=
1 -
6 x 16
8(8² - 1)


R
=
1 -
6 x 16
512 - 8


1 - 0.1905 = 0.8095

Theres a strong positive correlation between actual sales and target sales

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QUESTION 3(d)

Q A cashier at a departmental store can serve on average 24 customers per hour. The arrival rate of customers averages 20 customers per hour. The departmental store applies a single channel queuing system.

Required:
(i) The probability that the cashier is idle.

(ii) The average number of customers in the queuing system.

(iii) The average time a customer spends in the queue waiting to be served.
A

Solution


(i) The probability that the cashier is idle.


P(cashier is idle)
=
1 -
λ
μ

Where;

λ = Arrival rate = 20
μ = mean = 24

1 - 20 / 24

1 - 0.8333 = 0.1667

(ii) The average number of customers in the queuing system.


Ls
=
λ
μ - λ


Ls
=
20
24 - 20
= 5 cuctomers


(iii) The average time a customer spends in the queue waiting to be served.


Wq
=
λ
μ(μ - λ)


=
20
24(24 - 20)
= 20 / 96


1hr = 60minutes
20 / 96 hrs = ?

20 / 96 x 60 = 12.5 minutes

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QUESTION 4(a)

Q Viwanda Limited deals in the production of a product named "Nguvu". The production cost of the product is Sh.500 per unit (excluding packaging cost). The product is sold at Sh.1,000 per unit. The company is considering the purchase of one out of three different packaging systems. The cost data for the three packaging systems are as follows:

Packaging system


A
B
C		 Purchase cost

Sh. "000"
100
200
400		 Variable cost per
unit of product
Sh. "000"
1.50
1.00
0.50		 Scrap value

Sh. "000"
10
20
40

All the three packaging systems have a useful life of one year after which they would be sold at their estimated scrap values. The probability distribution for the demand for product "Nguvu" is as provided below:

Demand (units)
100
200
400		 Probability
0.3
0.6
0.1

Required:
Recommend the packaging system that should be purchased by Viwanda Limited
A

Solution


Expected demand in units

(0.3 x 100) + (0.6 x 200) + (0.1 x 400)

30 + 120 + 40 = 190 units

Expected profit

A = (1,000𝑥 + 10,000) - (100,000 + 1,500𝑥 + 500𝑥)

1,000𝑥 + 10,000 - 100,000 - 1500𝑥 - 500𝑥

1,000𝑥 - 1,500𝑥 - 500𝑥 + 10,000 - 100,000 = -1,000𝑥 - 90,000

Assume you sell 190 units

∴

Expected profit A

=(-1,000 x 190) - 90,000 = (280,000)



Expected profit B

= (1,000𝑥 + 20,000) - (200,000 + 1,000𝑥 + 500𝑥)

1,000𝑥 - 1,000𝑥 - 500𝑥 + 20,000 - 200,000

-500𝑥 - 180,000

(-500 x 190) - 180,000 = shs(275,000)

Expected profit function C

= (1,000𝑥 + 40,000) - (400,000 + 500𝑥 + 500𝑥)

= 1000𝑥 - 500𝑥 - 500𝑥 + 40,000 - 400,000 = -360,000

Best recommendation is to buy system B because it will minimize losses made.

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QUESTION 4(b)

Q Farm Produce Limited is a producer and distributor of maize flour. The company owns milling plants in Eldoret, Nanyuki and Narok towns. The milling plants have not been able to meet the demand orders of the company's distribution offices located in Mombasa, Kisumu, Nairobi and Isiolo towns. The company is considering the construction of a new milling plant either in Nakuru town or Meru town, in order to expand its production capacity. The data below relate to the company's production and demand requirements.

Milling plant

Eldoret
Nanyuki
Narok		 Monthly
output (units)
30,000
12,000
28,000		 Unit production
cost (Sh.)
96
100
104

Distribution office
Mombasa
Kisumu
Nairobi
Isiolo		 Monthly demand (units)
20,000
24,000
30,000
18,000

Additional information:

  1. The estimated unit production costs in Nakuru and Meru towns are Sh.98 and Sh.106 respectively.
  2. The unit transportation costs (in shillings) from each milling plant to each distribution office are given as follows

    To
    From
    Eldoret
    Nanyuki
    Narok    		 Mombasa
    64
    56
    58    		 Kisumu
    36
    52
    42    		 Nairobi
    52
    44
    36    		 Isiolo
    58
    32
    50
  3. The estimated unit transportation costs (in shillings) from each of the proposed milling plants to each distribution office are as follows:

    To
    From
    Nakuru
    Meru    		 Mombasa
    60
    62    		 Kisumu
    46
    56    		 Nairobi
    40
    46    		 Isiolo
    52
    28
  4. Assume that the construction of one of the proposed milling plants would satisfy the demand deficiency

Required:
Using the Vogel's approximation method (VAM), advise the management of Farm Produce Limited on the best location to construct the milling plant.
A

Solution


When Plant is constructed in Meru


Workings




Eldoret to Mombasa
Eldoret to Kisumu
Eldoret to Nairobi
Eldoret to Isiolo
Nanyuki to Mombasa
Nanyuki to Kisumu
Nanyuki to Nairobi
Nanyuki to Isiolo
Narok to Mombasa
Narok to Kisumu
Narok to Nairobi
Narok to Isiolo
Meru to Mombasa
Meru to Kisumu
Meru to Nairobi
Meru to Isiolo		 Production cost
(Unit)
96
96
96
96
100
100
100
100
104
104
104
104
106
106
106
106		 Transportation cost
(Unit)
64
36
52
58
56
52
44
32
58
42
36
50
62
56
46
28		 Total

160
132
148
154
156
152
144
132
162
146
140
154
168
162
152
134

Demand deficit

Supply(Output) 30,000 + 12,000 + 28,000 70,000
Demand 20,000 + 24,000 + 30,000 + 18,000 92,000
Shortage 92,000 - 70,000 22,000

Using Vogels's Approximation Method(VAM) to determine best location to construct Milling plant.

Vogel's Approximation Method Steps


  1. Calculate the Penalty Matrix:
    • For each row and column in the cost matrix, find the two smallest costs. Subtract the smaller cost from the larger one. This gives you the penalty for each row and column.
    • Create a penalty matrix based on these calculated penalties.
  2. Identify the Cell with the Highest Penalty:
    • Locate the cell in the penalty matrix with the highest penalty. If there is a tie, you can choose any of the cells with the highest penalty.
  3. Allocate as much as Possible:
    • Allocate units to the cell with the highest penalty as much as possible, considering the supply and demand constraints.
  4. Update Supply and Demand:
    • After allocation, update the supply and demand values for the respective row and column. If the supply or demand becomes zero, eliminate the corresponding row or column.
  5. Recalculate the Penalties:
    • Recalculate the penalties for the remaining rows and columns based on the updated supply and demand values.
  6. Repeat Steps 2-5:
    • Repeat steps 2-5 until all supply and demand values are satisfied.

1st allocation


To
From
Eldoret
Nanyuki
Narok
Meru
Demand
Penalty
Mombasa
160
4,000
156
162
168
16,000
20,000
4
Kisumu
132
24,000
152
146
162
24,000
14
Nairobi
148
2,000
144
140
28,000
152
30,000
4
Isiolo
154
132
12,000
154
134
18,000
18,000 - 18,000 = 0
2
Supply
30,000
12,000
28,000
22,000 - 18,000 = 4,000
92,000
Penalty cost
A
16
12
6
18


2nd allocation


To
From
Eldoret
Nanyuki
Narok
Meru
Demand
Penalty A
Penalty B
Mombasa
160
156
162
168
16,000
20,000
4
4
Kisumu
132
24,000
152
146
162
24,000 - 24,000 = 0
14
14
Nairobi
148
2,000
144
140
28,000
152
30,000
4
4
Isiolo
154
132
12,000
154
134
18,000
0
2
0
Supply
30,000 - 24,000 = 6,000
12,000
28,000
4,000
74,000
Penalty cost
A
16
12
6
18
B
16
8
6
10


3rd allocation


To
From
Eldoret
Nanyuki
Narok
Meru
Demand
Penalty A
Penalty B
Penalty C
Mombasa
160
156
162
168
20,000
4
4
4
Kisumu
132
24,000
152
146
162
0
14
14
0
Nairobi
148
144
140
28,000
152
4,000
30,000 - 28,000 = 2,000
4
4
4
Isiolo
154
132
154
134
18,000
0
2
0
0
Supply
6,000
12,000
28,000 - 28,000 = 0
4,000
50,000
Penalty cost
A
16
12
6
18
B
16
8
6
10
C
12
12
22
16


4th allocation


To
From
Eldoret
Nanyuki
Narok
Meru
Demand
Penalty A
Penalty B
Penalty C
Penalty D
Mombasa
160
156
162
168
20,000
4
4
4
4
Kisumu
132
24,000
152
146
162
0
14
14
0
0
Nairobi
148
144
140
28,000
152
2,000
2,000 - 2,000 = 0
4
4
4
4
Isiolo
154
132
154
134
18,000
0
2
0
0
0
Supply
6,000
12,000
0
4,000 - 2,000 = 2,000
22,000
Penalty cost
A
16
12
6
18
B
16
8
6
10
C
12
12
22
16
D
12
12
0
16


5th allocation


Allocate the remaining units to Mombasa

To
From
Eldoret
Nanyuki
Narok
Meru
Demand
Penalty A
Penalty B
Penalty C
Penalty D
Mombasa
160
156
162
168
20,000
4
4
4
4
Kisumu
132
24,000
152
146
162
0
14
14
0
0
Nairobi
148
144
140
28,000
152
2,000
0
4
4
4
4
Isiolo
154
132
154
134
18,000
0
2
0
0
0
Supply
6,000
12,000
0
2,000
20,000
Penalty cost
A
16
12
6
18
B
16
8
6
10
C
12
12
22
16
D
12
12
0
16


Total Cost = (18,000 x 134) + (24,000 x 132) + (28,000 x 140) + (2,000 x 152) + (6,000 x 160) + (12,000 x 156) + (2,000 x 168)

12,972,000

Now calculate the cost for Nakuru following the same steps,compare the two costs and avice accordinmgly.
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QUESTION 5(a)

Q Outline three differences between the normal distribution and the t-distribution.
A

Solution


Differences Between Normal and t-Distribution


Characteristic Normal Distribution t-Distribution
Degree of Freedom Defined by variance and standard deviation; no specific degree of freedom. Dependent on the degree of freedom (df); df = n - 1, where n is the sample size. Applies when sample size is less than 30 items (n < 30).
Symmetry Symmetric around the mean. Normal distribution is peaked at the mean. Symmetric around the mean. t-Distribution is less peaked at the center.
Application Used when population parameters are known and data is normally distributed. Used when estimating the mean of a population based on a small sample. Distribution makes no assumption regarding data distribution.
Shape Shape is fully determined by mean and standard deviation. Shape varies with the degree of freedom; becomes more similar to the normal distribution as df increases. t-Distribution is leptokurtic and has higher kurtosis than the normal distribution.
Use Cases Applicable in situations with a known population variance and normal data distribution. Commonly used for small sample sizes or when population variance is unknown. t-Distribution is a family of distributions typically defined by degrees of freedom, whereas degrees of freedom for the normal distribution are infinite.
Zero Degrees of Freedom Not applicable; the normal distribution is always defined. Not defined; the t-distribution becomes undefined with zero degrees of freedom.


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QUESTION 5(b)

Q A certain project is expected to be completed within 18 weeks. The expected net revenue if the project is complete on time is Sh.1,120,000 but a penalty of Sh.484,000 will be imposed if the project is not completed on time. The cost of the project is Sh.459,000. The standard deviation of the project's duration is 2.08 weeks.

The table below is a summary of activities required to complete the project, the duration of the activities and their preceding activities.

Activity

A
B
C
D
E
F
G
H		 Duration
(weeks)
5
2
4
2
5
7
6
3		 Preceding activity

-
-
-
B
B,C
C
A,D
G, D, E, F

Required:
(i) A network diagram of the project.

(ii) The float times of activities B and D.

(iii) The critical path of the project.

(iv) A 95 per cent confidence interval of the expected completion time of the project.

(v) The expected profit from the project.
A

Solution


(i) A network diagram of the project.



(ii) The float times of activities B and D.


Total float = LFT - EST - D

Where:

LFT = Latest finish time

EST = Earliest start time

D = Duration

B = 3 - 0 - 2 = 1 Days

D = 5 - 2 - 2 = 1 Days

(iii) The critical path of the project.


Critical path A - G - H

(iv) A 95 per cent confidence interval of the expected completion time of the project.


Expected completion time = 14 weeks

Confidence interval = mean ± 1.968

14 ± 1.96 × 2.08

Upper limit

14 + 1.96 × 2.08 = 18.08

Lower limit

14 - 1.96 × 2.08 = 9.92

(v) The expected profit from the project.


P(𝑥 > 18 weeks)

Z
=
𝑥 - μ
δ


Z
=
18 - 14
2.08




Z = 1.92



μ = 14
𝑥 = 18 weeks


P(𝑥 > 18 week) = 0.5 - 0.4726 = 0.9726

Probability that the project is not completed on time

1 - 0.9726 = 0.0274

Expected profit

= (1,120,000 x 0.9726) - (0.0274 x 484,000) - 459,000 = 34,379.10

Shs.617,050.4
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