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
Question 6 |
The activities of a project, their duration and the precedence relationships are given in the table. For example, in a precedence relationship "X \lt Y \lt Z" means that X is predecessor of activities Y and Z. The time to complete the activities along the critical path is ______ weeks.
17 | |
21 | |
23 | |
25 |
Question 6 Explanation:
Critical path duration = 23 weeks
Question 7 |
The arc lengths of a directed graph of a project are as shown in the figure. The shortest path
length from node 1 to node 6 is _______.
7 | |
9 | |
11 | |
14 |
Question 7 Explanation:
Shortest path is
Shortest length = 7
It is the problem of shortest path which will be 7.
In this question do not confuse with critical path. Examinar ask shortest path. Critical path is the longest path which is not asked in this question.
Shortest length = 7
It is the problem of shortest path which will be 7.
In this question do not confuse with critical path. Examinar ask shortest path. Critical path is the longest path which is not asked in this question.
Question 8 |
Processing times (including setup times) and due dates for six jobs waiting to be processed at a work centre are given in the table. The average tardiness (in days) using shortest processing time rule is ___________ (correct to two decimal places).
4.26 | |
5.65 | |
6.33 | |
8.46 |
Question 8 Explanation:
By SPT Rule
Total tardiness = 38
Average tardiness per job =\frac{\text { Total tardiness }}{\text { No. of Jobs }}=\frac{38}{6}=6.33\text{days}
Total tardiness = 38
Average tardiness per job =\frac{\text { Total tardiness }}{\text { No. of Jobs }}=\frac{38}{6}=6.33\text{days}
Question 9 |
A project starts with activity A and ends with activity F. The precedence relation and durations of the activities are as per the following table:
The minimum project completion time (in days) is _______.
The minimum project completion time (in days) is _______.
19 | |
30 | |
39 | |
48 |
Question 9 Explanation:
There are two path,
\begin{aligned} A-B-D-F=30 \text { days } \\ A-C-E-F=24 \text { days } \end{aligned}
Critical path A-B-D-F and of 30 days and it is also minimum time required to complete project.
Question 10 |
The standard deviation of linear dimensions P and Q are 3 m and 4 m, respectively. When assembled, the standard deviation (in m) of the resulting linear dimension (P+Q) is ________
5 | |
6 | |
4 | |
3 |
Question 10 Explanation:
According to Root Sum Square or RSS model
\sigma_{a}=\sqrt{\sum_{i=1}^{n} \sigma_{i}^{2}}
where, \sigma_{a}= assembly tolerance standard deviation
\sigma_{i}=i^{\text {th }} component tolerance standard deviation
Here \sigma_{P}=3 \mu \mathrm{m} and \sigma_{Q}=4 \mu \mathrm{m}
\sigma_{a}=\sqrt{\sigma_{P}^{2}+\sigma_{Q}^{2}}=\sqrt{3^{2}+4^{2}}=5 \mu \mathrm{m}
\sigma_{a}=\sqrt{\sum_{i=1}^{n} \sigma_{i}^{2}}
where, \sigma_{a}= assembly tolerance standard deviation
\sigma_{i}=i^{\text {th }} component tolerance standard deviation
Here \sigma_{P}=3 \mu \mathrm{m} and \sigma_{Q}=4 \mu \mathrm{m}
\sigma_{a}=\sqrt{\sigma_{P}^{2}+\sigma_{Q}^{2}}=\sqrt{3^{2}+4^{2}}=5 \mu \mathrm{m}
There are 10 questions to complete.