CPM MCQ Quiz - Objective Question with Answer for CPM - Download Free PDF
Last updated on May 8, 2025
Latest CPM MCQ Objective Questions
CPM Question 1:
Which of the following conclusions is/are correct in relation to the given network diagram?
(i) Activity B is controlled by activity A.
(ii) Activity D is controlled by both activities A and C.
(iii) Activity B is independent of activity C.
Answer (Detailed Solution Below)
CPM Question 1 Detailed Solution
Explanation:
Activity A → B: Solid arrow means B can only start after A ends ⇒ B is dependent on A.
Activity C → D: Solid arrow means D can only start after C ends ⇒ D is dependent on C only.
Dummy C → B: Dummy arrow shows logical dependency ⇒ B is also dependent on C.
(i) Activity B is controlled by activity A – Correct
(ii) Activity D is controlled by both A and C – Incorrect (only C controls D)
(iii) Activity B is independent of activity C – Incorrect (dummy from C to B shows B depends on C)
CPM Question 2:
A critical activity in a project is estimated to take 15 days to complete at a cost of Rs. 30,000. The activity can be expedited to complete in 12 days by spending a total amount of Rs. 54,000. Consider the statements P and Q.
P: It is economically advisable to complete the activity early by crashing if the indirect cost of the project is Rs. 8,500 per day.
Q: It is economically advisable to complete the activity early by crashing, if the indirect cost of the project is Rs. 10,000 per day.
Which one of the following options is CORRECT?
Answer (Detailed Solution Below)
CPM Question 2 Detailed Solution
Explanation:
Given Data:
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Normal duration: 15 days
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Normal cost: Rs. 30,000
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Crashed duration: 12 days
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Crashed cost: Rs. 54,000
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Indirect cost: Rs. 8,500 per day (for P) and Rs. 10,000 per day (for Q)
Cost slope (Crashing cost per day) is given by:
\(\mathrm{C} / \mathrm{s}=\frac{54000-30000}{15-12}=8000 \mathrm{Rs} / \text { day }\)
Crashing is economically beneficial if the savings in indirect cost per day exceed the direct cost of crashing per day.
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If the indirect cost per day is Rs. 8,500 (Statement P):
Since Rs. 8,500 > Rs. 8,000, it is economically advisable to crash. So, P is true. -
If the indirect cost per day is Rs. 10,000 (Statement Q):
Since Rs. 10,000 > Rs. 8,000, it is also economically advisable to crash. So, Q is true.
CPM Question 3:
The direct cost of a project with respect to normal time will be
Answer (Detailed Solution Below)
CPM Question 3 Detailed Solution
Explanation:
In project management, normal time refers to the planned or standard duration needed to complete a project or an activity without expediting or crashing. During this period, the project operates under normal conditions, without any additional resources or acceleration.
The direct cost of a project typically includes labor, materials, and equipment costs. When a project is completed within the normal time, these costs are usually minimized because:
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There is no need for overtime or additional shifts.
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Standard resources are used without any premium charges.
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There is no need to expedite materials or equipment, which could increase costs.
However, if the project timeline is shortened (crashed), the direct costs increase due to additional expenses to speed up the work. Therefore, the direct cost of a project is at its minimum when completed within normal time.
CPM Question 4:
The time with which direct cost does not reduce with the increase of time is known as
Answer (Detailed Solution Below)
CPM Question 4 Detailed Solution
Explanation:
Direct Project cost:
(i) It is the cost that is directly dependent on the amount of resources involved for the completion of activities. It includes labour, materials, plants, and machining.
(ii) To get the same work done in less time, we have to increases the amount of labor, equipment, and time-saving material that too at extra charges which simply means increases indirect cost.
(iii) The project has the highest cost corresponding to the crash duration and has normal cost corresponding to the normal duration.
Normal time:
(i) Normal time is the standard time that an estimator would usually allow for an activity.
(ii) It is the time with which direct cost does not reduce with the increase in time.
Crash time:
(i) Crash time is the minimum possible time in which an activity can be completed, by employing extra resources.
(ii) Crash time is that time, beyond which the activity cannot be shortened by an amount of increases in the resources.
CPM Question 5:
Independent float of an activity (i, j) is denoted by IF (i, j). The earliest occurrence times of i and j are denoted by Ei and Ej respectively. The latest occurrence times of i and j are denoted by Li and Lj respectively. D (i, j) indicates the duration of the activity.
Answer (Detailed Solution Below)
CPM Question 5 Detailed Solution
Concept:
Float:
- Float indicates the time by which, starting or finishing of an activity can be delayed without completion time.
- Float is associated with an activity and is analogous to the term slack.
Independent Float:
- It is the amount of time by which an activity can be delayed when all the preceding activities are completed as late as possible and all succeeding activities started as early as possible.
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Independent float can also be defined as the excess of minimum available time over the required activity duration.
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Consider a partial network as below
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If activity c-d is under consideration, and activity b-c finishes as its latest possible time TcL and succeeding activity j-k start at its earliest time TiE.
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Then activity i-j can take up any duration from t" to ( TT) without affecting any activity of the project.
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The difference between (TjE-TiL) and t is known as Independent float for activity i-j Independent float for i-j,
FID = (TjE - TiL) - tij
FID = (TjE - tij ) - TiL [ FF = TjE - TiE - tij ]
FID = (FF + TiE) -TiL
FID = FF - (TiL - TiE) [Si = TiL - TiE]
FID = FF - Si
Solution:
FID = (TjE - TiL) - tij
FID = Ej – Li - D (i, j)
Additional Information
- The difference between maximum time available and the actual time required for the completion of the activity
- FT = (Maximum available Time) - Actual time required.
- Maximum available time we can get when the activity starts at Earliest start time and finish by latest finish time i.e.
- Maximum available time = LFT - EST
The actual time required for performing an activity is the estimated activity time
FT = LFT - EST - tij
Free Float (FF):
- It is defined as the amount of time by which an activity can be delayed without affecting the EST of the succeeding activity.
- In other words, it is that portion of total float that can be used by an activity without delaying any succeeding activity.
- Consider an activity i-j and successor activity j - k.
- EST of i-j = TiE, therefore EFT of i-j =TiE+tij
- EST of j-k =TjE, if TjE > (TiE+tij) then activity j-k can not start immediately after completion of activity i-j and the difference between TjE and (TiE+tij) is the free float for activity i-j.
- Free Float for i-j FF = TjE - (TiE + tij)
Interfering Float (FIT):
- It is defined as the difference between the total float and the free float of an activity.
- It is also equal to the slack of head event of head activity as shown below
FIT = FT - FF
Earliest occurrence time (TE):
- It is the earliest time at which an event can occur i.e. the time by which all the activities leading to an event under consideration are complete. It is also called as Earliest Event time.
- The earliest event time (TE) is analogous to the term earliest expected time used in PERT, except that the degree of uncertainty involved in the word expected is not there.
- Calculation of Earliest event time is done same as PERT analysis by forwarding pass rule i.e.
TjE = Maximum (TiE + tij)
Latest allowable occurrence time (TL):
- It is the latest (delayed) time by which an event must be completed to completion time is not affected.
- The latest allowable occurrence time is calculated the same as in PERT by the Backward pass rule. i.e. TiL= Minimum (TjL - tij)
- If the scheduled completion time (TS) of the project is given, then the latest event time of the final/end event will be equal to TS
- If the scheduled completion time (TS) is not specified, then TL, is taken equal to the earliest event time TE of the end/final event.
Latest Finish Time (LFT):
- It is the latest (or delayed) time which the activity can be finished, without delaying the project.
- It is equal to the latest occurrence time TL of the event at which the activity terminates.
Earliest Finish Time (EFT):
- It is the earliest time by which the activity can be completed. It is equal to the earliest start time
Top CPM MCQ Objective Questions
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
Answer (Detailed Solution Below)
CPM Question 6 Detailed Solution
Download Solution PDFConcept:
In CPM:
The standard deviation of critical path:
σcp = \(\sqrt {Sum\;of\;variance\;along\;critical\;path} \)
σcp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)
Where, σ1, σ2, ...., σ8, σ9 are the standard deviation of each activity on the critical path
Calculation:
Given:
σ1, σ2, ...., σ8, σ9 = 3
σcp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)
σcp = \(\sqrt {3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2} \)
σcp = \(\sqrt {9 \times 9} \) = 9
∴ the standard deviation of the critical path is 9.
Fulkerson’s rule is connected with the
Answer (Detailed Solution Below)
CPM Question 7 Detailed Solution
Download Solution PDFExplanation
- Fulkerson’s rule is used for numbering events involved in the project scheduling network.
- Types of network diagram
- Event on the node ( EON )
- Activity on the node ( AON )
Important Points
- Hungarian method – Assignment problems
- Johnson's rule – scheduling jobs
- Simplex method - linear programming
The difference between EST of succeeding activity and EFT of the activity under consideration is called
Answer (Detailed Solution Below)
CPM Question 8 Detailed Solution
Download Solution PDFEarly Start Time:
- The earliest point in the schedule at which a task can begin.
- EST = \(T_E^i\)
Early Finish Time:
- The earliest point in the schedule at which a task can finish.
- EFT = \(T_E^i\) + tij
Latest Start Time:
- The latest point in the schedule at which a task can start without causing a delay.
- LST = \(T_L^j\) - tij
Latest Finish Time:
- The latest point in the schedule at which a task can finish without causing a delay.
- LFT = \(T_L^j\)
Types of float:
1) Total float:
- It is the difference between the maximum time available and the actual time required for the completion of the activity. or It is the difference between the time available for an activity performance and the duration of the activity.
- FT = LST - EST = LFT - EFT = \(T_L^j - T_E^i - t_e^{ij}\)
2) Free float:
- It is the amount of time by which an activity can be delayed without affecting the EST of the succeeding activity.
- FT = \(T_E^j - T_E^i - t_e^{ij}\)
3) Independent float:
- It is the excess of minimum available time over the required activity duration
- FI = \(T_E^j - T_L^i - t_e^{ij}\)
4) Interfering float
- It is the difference between the total float and the free float of an activity.
- FIN = FT - FF
The time with which direct cost does not reduce with the increase in time is known as
Answer (Detailed Solution Below)
CPM Question 9 Detailed Solution
Download Solution PDFExplanation:
Direct Project cost:
(i) It is the cost that is directly dependent on the amount of resources involved for the completion of activities. It includes labour, materials, plants, and machining.
(ii) To get the same work done in less time, we have to increases the amount of labor, equipment, and time-saving material that too at extra charges which simply means increases indirect cost.
(iii) The project has the highest cost corresponding to the crash duration and has normal cost corresponding to the normal duration.
Normal time:
(i) Normal time is the standard time that an estimator would usually allow for an activity.
(ii) It is the time with which direct cost does not reduce with the increase in time.
Crash time:
(i) Crash time is the minimum possible time in which an activity can be completed, by employing extra resources.
(ii) Crash time is that time, beyond which the activity cannot be shortened by an amount of increases in the resources.
Which one of the following does have some limitations when applied to detailed engineering design work during early stages of a project?
Answer (Detailed Solution Below)
CPM Question 10 Detailed Solution
Download Solution PDFConcept
Dummy Activity: An activity that is used to maintain the predefined precedence relationship only during the construction of the project network is called a dummy activity.
Estimate: An approximate calculation or judgment of value
PERT is preferred when there are some uncertainties in the methodology, availability of materials, and the final answer is not clearly known.
Whereas CPM is applied when things are almost certain
For example in the case of a construction project, once the project has been finalized there is little doubt of men, material, and equipment required, time, etc. then PERT is used.
∴ CPM has limitations in the early stage of the project.
The excess of minimum available time over activity duration is called
Answer (Detailed Solution Below)
CPM Question 11 Detailed Solution
Download Solution PDFExplanation:
Float:
(i) It is associated with activity times.
(ii) It is analogous to the slack of events in PERT.
(iii) It is the range within which the start or finish time of activity may fluctuate without affecting the project completion time.
(iv) Floats are of the following types:
1. Total float: The time span by which starting or finishing of an activity can be delayed without delaying the completion of the project. Total float of activity affects total float of succeeding as well as preceding activities.
2. Free float: The delay can be made without delaying succeeding activities. It affects only preceding activities.
3. Independent Float: It is the minimum excess available time which exists without affecting any of succeeding or preceding activities.
4. Interfering Float: It is similar to head event slack.
Gantt charts indicate
Answer (Detailed Solution Below)
CPM Question 12 Detailed Solution
Download Solution PDFGantt charts are mainly used to allocate resources to activities. The resources allocated to activities include staff, hardware, and software. Gantt charts are useful for resource planning. A Gantt chart is a special type of bar chart where each bar represents an activity. The bars are drawn along a timeline. The length of each bar is proportional to the duration of time planned for the corresponding activity.
In a CPM network Earliest start time for an event i is 10 weeks. Activity i - j takes 4 weeks for Completion. Event J starts after 16 weeks. Total Float for activity i - j is
Answer (Detailed Solution Below)
CPM Question 13 Detailed Solution
Download Solution PDFConcept:
Total float: It is the amount of time that the completion time of an activity can be delayed without affecting project completion time.
Mathematically, Total Float = (Lj - Ei) - tij
Calculation:
∴ Total Float = (Lj - Ei) - tij = 16 - 10 - 4 = 2 week
Additional Information
Free float: It is the amount of time that the activity completion time can be delayed without affecting the earliest start time of the immediate successor activities in the network.
Mathematically, Free Float = (Ej - Ei) - tij
Independent Float: Amount by which an activity can be delayed without delaying the project; even if all predecessors are at Late Finish and all Successors are at Early Start.
Mathematically, Independent Float = (Ej - Li) - tij
In a bar chart the vertical axis represents
Answer (Detailed Solution Below)
CPM Question 14 Detailed Solution
Download Solution PDFBar Chart / Gantt Chart
First introduced by Hennery Gantt in 1910. Each bar represents a specific job or activity and length of the bar represent the time of completion of that activity. The starting and ending of the bar represent the starting and beginning of that activity.
CPM Network is updated
Answer (Detailed Solution Below)
CPM Question 15 Detailed Solution
Download Solution PDFConcept:
- Critical Path Methods (CPM) have been used for planning scheduling and control in construction project management.
- It is activity oriented method.
Sl. No. | CPM | PERT |
---|---|---|
1. | Critical path method | Program valuation and review technique |
2. | Deterministic | Probabilistic |
3. | Control time and cost | Control only time |
4. | Activity oriented | Event oriented |
5. | One time estimate | Three-time estimate |
6. | Use- Non research | Research |
CPM Network is updated whenever there is a difference in the planned and actual performance.