Operations Research MCQ Quiz in मल्याळम - Objective Question with Answer for Operations Research - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation :

A critical path is a sequence of interdependent activities or tasks that must be finished before the project can be finished. It is the longest path (i.e. path with the longest duration) from project start to finish.

The path which moves along the activities having total float zero.

A. Critical Path Method
1. The critical path method (CPM) is a project modelling technique that's used by project managers to find important deadlines and deliver a project on time.

2. In a project, the critical path is the longest distance between the start and the finish, including all the tasks and their duration.

3. Once a critical path is determined, you'll have a clear picture of the project's actual schedule.

4. To find this, project managers use the CPM algorithm to find the least amount of time necessary to complete each task with the least amount of slack.
5. Once done by hand, nowadays, the critical path is calculated automatically by the project scheduling software. That makes the whole method a lot easier.

B. Critical Path Analysis
1. As mentioned, the purpose of a critical path is to find the least amount of time you'll need to complete a task.

2. Critical path analysis furthers your ability to make better estimates for scheduling because you're mapping out every important task that must be done for a successful project.

C. Critical Path Example

Explanation:

The project network is given along with the earliest and latest finish start time.

For activity '3':

Slack = Latest finish time – Earliest finish time

Slack = 11 - 9 = 2 hours

Explanation:

Total float: 

Total Float/Slack = Latest Finish time (LF) - Earliest Start time (ES) - (Earliest Finish time EF - Earliest Start time ES  )or Total Float/Slack = Latest Finish time (LF) - Earliest Finish time (EF) OR = Latest start time (LS) - Earliest start time (ES)

If we use the  formula with the given values, we get:

Total Float = LS - ES = 45 - 15 = 30 OR  45 - 15 = 30

Explanation:

Critical Path:

A critical path is a sequence of interdependent activities or tasks that must be finished before the project can be finished. It is the longest path (i.e. path with the longest duration) from project start to finish.

Critical Path Method:

• The critical path method (CPM) is a project modeling technique that’s used by project managers to find important deadlines and deliver a project on time.
• In a project, the critical path is the longest distance between the start and the finish, including all the tasks and their duration.
• Once a critical path is determined, you’ll have a clear picture of the project’s actual schedule.
• To find this, project managers use the CPM algorithm to find the least amount of time necessary to complete each task with the least amount of slack.
• For small projects, managers are often able to memorize and to coordinate all of the various tasks necessary for their completion.
• For larger projects, however, with numerous activities occurring simultaneously, remembering and coordinating these activities can prove much more difficult. CPM and related tools allow managers to determine which particular tasks most affect the total time of the project and enable managers to better schedule each task so that deadlines are met at the least possible cost

Explanation

Duality

• For every linear programming problem there exists a related unique linear programming problem involving the same data which also describe and solve the original problem.
• In Duality our aim is to find a Transpose matrix of the initial given problem known as Primal.
• It is done by transposing row and column and the final solution of both the problem will be the same.

Concept:

Referring to the graphs for given linear equations

The constraints x1 ≥ 0, x2 ≥ 0, and x1 + 2x2 ≤ 4 will be having feasible regions.

The corner points of the feasible region will give the best feasible solution for the objective function.

Calculation:

Referring to the corner points of feasible region,

The corner point of the feasible region are (0, 0), (0, 2) & (4, 0)

The value of the objective function at corner points,

⇒ Z (0,0) = 0,

⇒ Z (0,2) = 18

⇒ Z (4,0) = 12

∵ The maximum value of Z is at (0, 2)

⇒ Zmax = 18

Concept:

Slack time 

• It is the amount of time a task can be delayed before the project finish date is delayed. Thus, the slack is the difference between event times denoting the range within which an event time can vary.
• The Earliest Expected Time (TE) is the time when an event can be expected to occur earlier.
• The Latest allowable occurrence time (TE)is the latest time by which an event must occur to keep the project on schedule (without delaying the project).
• Event slack is defined as the difference between the latest event and the earliest event times. i.e. Slack = TL - TE

Positive slack: When TL > TE. It indicates the project is ahead of schedule meaning thereby the excess resources.

Zero slack: When TL = TE. It indicates that the project is going on schedule meaning thereby adequate resources.

Concept:

No. of customers in the system, \({L_S} = \frac{\lambda }{{\mu - \lambda }}\)

No. of customers in the queue, \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

Calculation:

Arrival rate λ = 10 minute/customer = 6 customers/hour

Service rate, μ = 6 minute/customer = 10 customers/hour

The average length of the queue, \({L_q} = \frac{{{\lambda ^2}}}{{\mu \left( {\mu - \lambda } \right)}}\)

\(\;{L_q} = \frac{{{6^2}}}{{10\;\left( {10\; - \;6} \right)}} = \;0.9 = \frac{9}{{10}}\)

Points to remember

• Average arrival time and the time spent in the system (Waiting time in system) = \({W_s} = \frac{1}{{\mu - \lambda }}\)
• Average arrival time and the time spent in the queue (before being served) (Waiting time in queue) = \({W_q} = \frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the system = \({L_s} = \lambda {W_s} = \lambda .\frac{1}{{\mu - \lambda }}\)
• Average number of customers in the queue = Average queue length = \({L_q} = \lambda {W_q} = \lambda .\frac{\lambda }{\mu }.\frac{1}{{\mu - \lambda }} = \frac{{{\lambda ^2}}}{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)
• Probability that there are k customers in the system = (ρ)k(1 - ρ) = \(\frac{\lambda }{\mu }\left( {\frac{1}{{\mu - \lambda }}} \right)\)

• Probability that there are more than k customers = \({\left( {\frac{\lambda }{\mu }} \right)^{k + 1}}\)

Concept:

Waiting time in the queue is given by,

Mean waiting time in the queue \(= \frac{λ }{{μ \left( {μ - λ } \right)}}\)

where, λ = arrival rate, μ = Service rate

Calculation:

Given:

λ = 20/hour, μ = 30/hour

Mean waiting time in the queue \(= \frac{λ }{{μ \left( {μ - λ } \right)}}\)

Mean waiting time in the queue \(= \frac{{20}}{{30\left( {30 - 20} \right)}}\)

Mean waiting time in the queue \(= \frac{{20}}{{300}} = \frac{1}{{15}}{\rm{\;hour}}\)

∴ Mean waiting time in the queue \(= \frac{1}{{15}} \times 60\)

∴ Mean waiting time in queue = 4 minutes

Explanation:

The given network diagram is given below:

As we can see, without any dummy activity we can draw the network diagram, so the sum of dummy activity required is zero.