Question 1 |
A project consists of five activities (A, B, C, D
and E). The duration of each activity follows beta
distribution. The three time estimates (in weeks)
of each activity and immediate predecessor(s) are
listed in the table. The expected time of the project
completion is __________ weeks (in integer).
\begin{array}{|c|c|c|c|c|} \hline \text{Activity} & \text{Optimistic time (week)} & \text{Most likely time (week)} &\text{Pessimistic time (week)} &\text{Immediatepredecessor(s)} \\ \hline A& 4 & 5 &6 & None\\ \hline B& 1& 3 & 5 & A\\ \hline C& 1& 2 & 3 & A\\ \hline D& 2& 4 & 6 & C\\ \hline E& 3& 4 & 5 &B,D \\ \hline \end{array}
\begin{array}{|c|c|c|c|c|} \hline \text{Activity} & \text{Optimistic time (week)} & \text{Most likely time (week)} &\text{Pessimistic time (week)} &\text{Immediatepredecessor(s)} \\ \hline A& 4 & 5 &6 & None\\ \hline B& 1& 3 & 5 & A\\ \hline C& 1& 2 & 3 & A\\ \hline D& 2& 4 & 6 & C\\ \hline E& 3& 4 & 5 &B,D \\ \hline \end{array}
12 | |
15 | |
16 | |
18 |
Question 1 Explanation:
\begin{array}{|c|c|}
\hline
\text{Activity}& t_e=\frac{t_0+4t_m+t_p}{6}\\ \hline
A&5\\ \hline
B&3\\ \hline
C&2\\ \hline
D&4\\ \hline
E&4 \\ \hline
\end{array}
\begin{array}{|c|c|} \hline \text{Path}& \text{Duration}\\ \hline 1-2-4-5(A-B-E)&12\\ \hline 1-2-3-4-5 (A-C-D-E)&15\\ \hline \end{array}
Expected completion time of the project = critical path duration = 15 weeks
\begin{array}{|c|c|} \hline \text{Path}& \text{Duration}\\ \hline 1-2-4-5(A-B-E)&12\\ \hline 1-2-3-4-5 (A-C-D-E)&15\\ \hline \end{array}
Expected completion time of the project = critical path duration = 15 weeks
Question 2 |
Activities A to K are required to complete a project.
The time estimates and the immediate predecessors
of these activities are given in the table. If the
project is to be completed in the minimum possible
time, the latest finish time for the activity G is
_____ hours.
\begin{array}{|c|c|c|}\hline \text{Activity} &\text{Time (hours)} & \text{Immediate predecessors}\\ \hline A &2 &- \\ \hline B& 3 &- \\ \hline C& 2 &- \\ \hline D & 4 &A \\ \hline E & 5 & B\\ \hline F & 4 & B\\ \hline G & 3 & C\\ \hline H & 10 &D,E \\ \hline I & 5 & F\\ \hline J & 8 & G\\ \hline K & 3 &H,I,J\\ \hline \end{array}
\begin{array}{|c|c|c|}\hline \text{Activity} &\text{Time (hours)} & \text{Immediate predecessors}\\ \hline A &2 &- \\ \hline B& 3 &- \\ \hline C& 2 &- \\ \hline D & 4 &A \\ \hline E & 5 & B\\ \hline F & 4 & B\\ \hline G & 3 & C\\ \hline H & 10 &D,E \\ \hline I & 5 & F\\ \hline J & 8 & G\\ \hline K & 3 &H,I,J\\ \hline \end{array}
5 | |
10 | |
8 | |
9 |
Question 2 Explanation:
Activity 'G'
LFT of activity 'G' = 10 hours
Question 3 |
A PERT network has 9 activities on its critical path. The standard deviation of each activity on the critical path is 3. The standard deviation of the critical path is
3 | |
9 | |
27 | |
81 |
Question 3 Explanation:
In CPM,
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}
\begin{array}{l} \sigma=\sqrt{\text { sum of variance along critical path }} \\ \sigma=\sqrt{\sigma^{2}+\sigma^{2}+\ldots .+\sigma^{2}} \\ \sigma=\sqrt{9 \sigma^{2}}=\sqrt{9 \times 9}=9 \end{array}
Question 4 |
Activities A, B, C and D form the critical path for a project with a PERT network. The means and variances of the activity duration for each activity are given below. All activity durations follow the Gaussian (normal) distribution, and are independent of each other.
\begin{array}{|l|c|c|c|c|}\hline \text{Activity} & \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \text{Mean (days)} & \text{6} & \text{11} & \text{8} & \text{15} \\ \hline \text{Variance (days$^{2})$} & \text{4} & \text{9} & \text{4} & \text{9}\\ \hline \end{array}
The probability that the project will be completed within 40 days is _____ (round off to two decimal places).
(Note: Probability is a number between 0 and 1).
\begin{array}{|l|c|c|c|c|}\hline \text{Activity} & \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \text{Mean (days)} & \text{6} & \text{11} & \text{8} & \text{15} \\ \hline \text{Variance (days$^{2})$} & \text{4} & \text{9} & \text{4} & \text{9}\\ \hline \end{array}
The probability that the project will be completed within 40 days is _____ (round off to two decimal places).
(Note: Probability is a number between 0 and 1).
0.25 | |
0.5 | |
0.65 | |
0.85 |
Question 4 Explanation:
PERT-CPM
\begin{aligned} T_{S} &=40 \text { days, } T_{E}=6+11+8+15 \\ T_{E} &=40 \text { days, } \\ 2 &=\frac{T_{S}-T_{E}}{\sigma}=0 \rightarrow 50 \% \end{aligned}
Probability of completing project in expected time is always 0.5.
Question 5 |
Consider the following network of activities, with each activity named A-L, illustrated
in the nodes of the network
The number of hours required for each activity is shown alongside the nodes. The slack on the activity L, is ________ hours.
The number of hours required for each activity is shown alongside the nodes. The slack on the activity L, is ________ hours.
1 | |
2 | |
3 | |
4 |
Question 5 Explanation:
Time along path A-B-C-F-I-J-K = 42 hours
Time along path A-B-C-E-H-L = 31 hours
Time along path A-B-C-D-G-H-L = 40 hours
Slack for L = 42 - 40 = 2 hours
Time along path A-B-C-E-H-L = 31 hours
Time along path A-B-C-D-G-H-L = 40 hours
Slack for L = 42 - 40 = 2 hours
There are 5 questions to complete.