Operations Research MCQ Quiz - Objective Question with Answer for Operations Research - Download Free PDF

Explanation:

Critical Path Method (CPM):

• The Critical Path Method (CPM) is a project management technique used to determine the sequence of activities that directly affect the project completion time. It identifies the longest path of dependent activities in the project schedule, known as the critical path. The duration of this path determines the shortest possible project completion time.

Critical Path and Slack:

• The critical path in a project is defined as the sequence of tasks where any delay in one task would directly result in a delay in the overall project completion. The slack (or float) is the amount of time an activity can be delayed without affecting the project’s completion date. For activities on the critical path, the slack is zero because there is no flexibility to delay these activities without impacting the entire project.

Why Slack on the Critical Path is Zero:

• The critical path represents the longest duration path in the project network. If any activity on this path is delayed, the entire project completion time will also be delayed.
• Slack is calculated as the difference between the latest allowable finish time (LF) and the earliest finish time (EF) of an activity:
  Slack = LF - EF
• For activities on the critical path, the LF equals EF because they determine the project’s end date. Therefore, their slack is zero.

Explanation:

Critical Path Method (CPM):

Definition: The Critical Path Method (CPM) is a project management technique used to analyze and schedule tasks within a project. It identifies the longest sequence of dependent tasks (known as the critical path) that determines the shortest possible project duration. Any delay in the tasks on the critical path will directly impact the overall project completion time.

How to Identify the Critical Path:

• List all the tasks required to complete the project.
• Define dependencies between tasks (i.e., which tasks must be completed before others can start).
• Determine the duration of each task.
• Calculate the earliest start (ES) and finish (EF) times for each task by performing a forward pass through the network diagram.
• Calculate the latest start (LS) and finish (LF) times for each task by performing a backward pass through the network diagram.
• Identify the tasks with zero slack (i.e., tasks where ES = LS and EF = LF). These tasks form the critical path.

To identify the critical path, we analyze the given network diagram and follow these steps:

• Step 1: List all paths in the network diagram and calculate the total duration for each path.
• Step 2: Identify the path with the longest duration. This is the critical path because it governs the minimum time required to complete the project.

Based on the given data:

• Path 1: 1-2-3-7
  • Total duration = Sum of durations of tasks along this path.
• Path 2: 1-2-4-5-6-7
  • Total duration = Sum of durations of tasks along this path.
• Path 3: 1-2-4-5-6
  • Total duration = Sum of durations of tasks along this path.
• Path 4: 1-2-4-7
  • Total duration = Sum of durations of tasks along this path.

From the calculations, Path 2 (1-2-4-5-6-7) has the longest duration. Therefore, it is the critical path.

Concept:

• In order to find the maximum value of the objective function, the constraints of the objective function are drawn and the region formed by the constraints is the feasible region.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

Calculation:

Given:

• Following cases are observed by the region formed by the constraints

Calculation:

Given:

• The objective function to maximize is,

Z = 2X₁ + 3X₂

• Which is subjected to the constraints,

2X₁ + X₂ ≤ 6

X₁ – X₂ ≥ 3

X₁, X₂ ≥ 0

• If the given constraints are drawn graphically then,

• The feasible region is a point here, therefore the given LPP has an optimal solution and it will occur at point (3, 0).

Concept:

• The corner points of the feasible region are the points for which we need to check the value of the objective function.
• The maximum value of the objective function for these corner points of the feasible region will be the optimal feasible point(s).​
• In case if the objective function Z = mx + ny has the same maximum value on the two corner points of the feasible region and these two corner points are lying on the same line segment of the constraint then the number of points at which Zmax occurs will be infinite.

Calculation:

Given: The objective function Z = mx + ny has the same maximum value on the two corner points of the feasible region.

• Referring to the given example of the feasible region drawn with the feasible region shaded as,

• The corner points of the feasible region are:

C(15, 15), B(5, 5), M(10, 0) and N(60, 0)

• There is no change in corner points occurs due to extra constraints.

• The maximum value of x + 3y will occur at two points C and N.

• Now check whether there is a possibility of multiple solutions.
• For that join the points C and N. If the points C and N are lying at the same line segment CN so for all the points on that line segment CN, the value of the objective function will be a maximum of 60.
• Since corner points C and N are lying on the same line segment CN so , at every point on the line segment CN, we will get the maximum value of the objective function.

• So, the correct answer is option 4.

Solution:

Converting constraints into equations we get,

x + y = 6; 

2x + 3y = 3;

and x = 3, y = 3

these are represented in the following diagram:

The above diagram represents the boundary of the constraints

For first constraint,  x + y ≤ 6

The solution region is below the line as Origin satisfies the condition,

For second constraints,  2x + 3y ≥ 3

The solution region is away from the origin as the origin doesn't satisfy the constraint.

For x ≥ 3 and y ≥ 3, 

The solution region is away from the origin,

Hence, the Common region of all the constraints together is only one point i.e., (3, 3)

 z = 5x + 7y = 15 + 35 = 50

Since only one point is in the feasible solution, Hence, we can't say it is the maximum or minimum.

Concept:

In CPM:

The standard deviation of critical path:

σ_cp = \(\sqrt {Sum\;of\;variance\;along\;critical\;path} \)

σ_cp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

Where, σ₁, σ₂, ...., σ₈, σ₉ are the standard deviation of each activity on the critical path   

Calculation:

Given:

σ1, σ2, ...., σ8, σ9 = 3

σcp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

σ_cp = \(\sqrt {3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2} \)

σ_cp = \(\sqrt {9 \times 9} \) = 9

∴ the standard deviation of the critical path is 9.

Explanation:

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.

To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.

PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.

• PERT (Project Evaluation and Review Technique) approach takes account of the uncertainties. In this approach, 3-time values are associated which each activity. So it is probabilistic.

• CPM (Critical Path Method) involves the critical path which is the largest path in the network from starting to ending event and defines the minimum time required to complete the project. So it is deterministic.

Difference between PERT and CPM (Critical Path Method)

Concept:

Project Evaluation and Review Technique (PERT) is probabilistic in nature and is based upon three-time estimates to complete an activity.

Optimistic Time (to): It is the minimum time that will be taken to complete an activity if everything goes according to the plan.

Pessimistic Time (tp): It is the maximum time that will be taken to complete an activity when everything goes against the plan.

Most likely time (tm): It is the time required to complete a project when an activity is executed under normal conditions.

Average or most expected time is given by \({t_E} = \left( {\frac{{{t_p}\; + \;4{t_m}\; + {t_o}}}{6}} \right)\)

The variance gives the measure of uncertainty of activity completion. The variance of the activity is given by 

Variance, \(V = {\left( {\frac{{{t_p} - {t_0}}}{6}} \right)^2}\)

Standard duration, \(\sigma = \sqrt {variance} \)

Calculation:

Given:

t_p = 10 days, t_o = 4 days

\({\rm{V}} = {\left( {\frac{{{{\rm{t}}_{\rm{p}}} - {{\rm{t}}_{\rm{o}}}}}{6}} \right)^2} = {\left( {\frac{{10 - 4}}{6}} \right)^2} = 1\)

The variance of the activity is 1.

Concept:

In a transportation problem with m supply points and n demand points

Number of constraints = m + n

Number of variables = m × n

Number of equations = m + n - 1

Calculation:

Given:

m = 4, n = 5

Number of constraints = m + n = 4 + 5 = 9

The correct answer is Kolkata.

Key Points

• Indian Railways is divided into 18 zones and 73 divisions.
• A Divisional Railway Manager (DRM) heads the division and he/she reports to General Manager (GM).
• A Railway Division is the smallest administrative unit of Railways.
• North Zone is the largest zone.

Given below is the list of all railway zones and their headquarters:

Explanation

Slack or Event Float

• Slack corresponds to the event in PERT.
• Float corresponds to activity in CPM.

Slack

• It is defined as the amount of time by which an event can be delayed without delaying the project schedule.
• Slack of an event = Latest Start Time – Earliest Start Time OR Latest Finish Time – Earliest Finish Time

There are three types of floats.

Concept:

For a system of equation, the number of possible basic solution is calculated by - \({n_C}_m\)

n = number of variables.

m = number of equations.

Inequalities must be converted into equalities.

Calculation:

Given:

X1 + X2 + X3 ≤ 5

X₁ + X₂ + X₃ + S₁ + 0S₂ = 5     (1)

2X1 + X2 + 3X3 ≤ 10

2X1 + X2 + 3X₃ + 0S₁ + S₂ = 10      (2)

n = number of variables = 5

m = number of equations = 2

∴ number of basic solution = \({n_C}_m ⇒ {5_C}_2\)

∴ \(\frac{5!}{2!\;\times\;(5-2)!}\Rightarrow10\)

Explanation:

PERT stands for "Program Evaluation and Review Technique". This network model is used for project scheduling.

Difference between PERT and CPM (Critical Path Method)

PERT does take uncertainties involved in the estimation of times, therefore three-time estimates have been taken for the calculation project duration. They are optimistic (t_o), pessimistic (t_p), and most likely (t_m).

\(T_e=\frac{t_o\;+\;4t_m\;+\;t_p}{6}\)

Therefore, the option 2 is the incorrect statement among the given options.

Explanation:

If \(x_{ij}\ge 0, \)  is the number of units shipped from i^th source to j^th destination, then the equivalent LPP model will be

Minimize \(Z = \sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {{c_{ij}}} } {x_{ij}}\)

Subjected to:

\(\begin{array}{l} \sum\limits_{i = 1}^m {{x_{ij}}} \le {b_i}\,\,(demand)\\ \sum\limits_{j = 1}^n {{x_{ij}}} \le {a_i}\,\,(\sup ply) \end{array}\)

If total supply = total demand then it is a balanced transportation problem otherwise it is called an unbalanced transportation problem.

There will be (m + n - 1) basic independent variables out of (m x n) variables.

Explanation:

PERT stands for Program Evaluation and Review Technique and was developed to address the needs of projects for which the time and cost estimates tend to be quite uncertain.

It has a probabilistic approach and hence suitable for the projects which are to be conducted for the first time or projects related to research and development.

PERT uses 3 cases:

• Optimistic time ⇒ estimates the shortest possible time required for the completion of the activity.
• Most likely time ⇒ estimates the time required for the completion of activity under normal circumstances.
• Pessimistic time ⇒ estimates the longest possible time required for the completion of the activity.