Samacheer Kalvi Class 11 Business Maths Solutions Chapter 10 Operations Research Exercise 10.2

 

Question 1. Draw the network for the project whose activities with their relationships are given below: Activities A, D, E can start simultaneously; B, C > A; G, F > D, C; H > E, F.
Answer: First, let's understand the rules of how activities depend on each other. If we say "A < B, C", it means that activity A must be finished before activities B and C can begin. So, A is the immediate predecessor for both B and C. This rule helps us draw the network correctly.

Here's how we build the network step by step:

Step-1: We start by drawing the activities A, D, and E as they can all begin at the same time from the initial node (node 1). Activity A goes from node 1 to node 2. Activity D goes from node 1 to node 3. Activity E goes from node 1 to node 4.

Activities A, D, and E are independent events because they all start at the beginning. Then, B and C depend on A, G and F depend on D and C, and H depends on E and F. So, we need to combine these paths. For example, for G and F to start, both D and C must be finished. If they end at different places, a dummy activity might be needed, or they merge into a single node for the next activities to begin.

Step-2: We complete the network by adding activities B, C, G, F, and H, ensuring all dependencies are met. Activities B, G, and H do not act as immediate predecessors for any *other* activity, so they flow into the final merging node, which we label as node 5 to signify the completion of those independent paths.

Activities B, G, and H are special because they are not predecessors for any *new* activities. Therefore, they all end at a "lost node" (node 5) which represents the end point of these paths, bringing together all the work.
In simple words: First, we draw all the activities that can start right away. Then, we add activities that depend on others, connecting them with arrows. If an activity isn't needed for another one to start, it joins a final end point, like a finish line.

🎯 Exam Tip: When drawing network diagrams, clearly identify the start and end nodes for each activity. Use distinct nodes for activities that start simultaneously and those that merge.

 

Question 2. Draw the event oriented network for the following data:

Answer: To draw an event-oriented network, we focus on the events (nodes) and the activities (arrows) that connect them. Each event represents a milestone, and the arrow shows the work needed to go from one event to the next. The "Immediate Predecessors" tell us which events must be finished before a new event can start.

Step-1: Events 2 and 3 both have event 1 as their immediate predecessor. This means event 1 is the starting point, and two activities branch out from it, leading to events 2 and 3 respectively.

Step-2: Now we add events 4 and 5. Event 4 depends on both events 2 and 3 finishing, so arrows from 2 and 3 converge to event 4. Event 5 depends only on event 3, so an arrow goes from 3 to 5.

Step-3: Finally, we add events 6 and 7. Event 6 requires both events 4 and 5 to be completed, so arrows from 4 and 5 merge into event 6. Event 7 depends on both events 5 and 6 finishing, showing two activities leading to it.

This step-by-step method helps to accurately map out all the dependencies in the project.
In simple words: We draw circles for events and arrows for activities between them. If an event needs two things to happen before it starts, two arrows point to it. We keep adding events and arrows until the whole project is mapped.

🎯 Exam Tip: When drawing event-oriented networks, ensure that each event has a unique number. Use dummy activities (dashed lines) if two activities have the same start and end events but different predecessors.

 

Question 3. Construct the network for the projects consisting of various activities and their precedence relationships are as given below: A, B, C can start simultaneously: A < F, E; B < D, C; E, D < G
Answer: For this project, activities A, B, and C can all begin at the very start, which means they come out of the first node. Then, F and E depend on A, D depends on B, and finally G depends on both E and D. We will draw this step by step, making sure each activity flows correctly from its predecessors.

Step-1: Activities A, B, and C all start simultaneously. So, from the initial node (node 1), we draw three arrows representing these activities leading to nodes 2, 3, and 4 respectively.

Step-2: Next, we add activities F and E, which both depend on A. So, from node 2 (where A ends), arrows go out for F and E. We also add activity D, which depends on B, so an arrow for D comes out of node 3 (where B ends). The diagram shows C branching from B, which matches "B < D, C".

Step-3: Finally, activity G depends on both E and D. So, we draw E and D merging into a common node, from which G then starts. This completes the project network, showing all dependencies and flows.

Drawing these networks helps visualize the sequence of work and identify potential delays.
In simple words: We start by drawing activities that can happen at the same time. Then, we add activities that need others to finish first. Activities that need two things to be done first will merge into one point before the next activity starts.

🎯 Exam Tip: Pay close attention to the wording like "A < F, E" which means A is a predecessor for both F and E. Ensure all predecessors for an activity converge to a single node before that activity begins.

 

Question 4. Construct the network for each the projects consisting of various activities and their precedence relationships are as given below:

Answer: We need to build a project network showing how activities depend on each other based on the table. Activities A, B, and C have no predecessors, meaning they can all start at the beginning. Then, D depends on A, E and F depend on B, and G depends on C. We keep adding activities, connecting them correctly according to their predecessors until the entire project is mapped. This helps us see the order of tasks.

Step-1: Activities A, B, and C can all start at the same time. We represent this by having them branch out from a single starting node (node 1) to individual nodes (2, 3, and 4).

Step-2: Now we add the activities that depend on A, B, and C. Activity D depends on A, so it branches from node 2. Activities E and F depend on B, so they branch from node 3. Activity G depends on C, so it branches from node 4. This creates a more detailed network showing the next layer of tasks.

Step-3: The final step is to combine all paths and add activities H, I, J, and K. Activities H and I depend on D and E. Activity J depends on H and I, and K depends on F and G. This results in the complete network showing all activities and their relationships, leading to the project's completion.

Understanding the flow of activities is key to managing a project effectively.
In simple words: First, we show the tasks that can start right away. Then, we connect tasks that have to wait for others to finish. If a task needs more than one thing to be done before it starts, those earlier tasks will all connect to the same point. We do this until all tasks are linked in the right order.

🎯 Exam Tip: When multiple activities depend on a single predecessor, they can all branch from the same node. If an activity depends on multiple predecessors, those predecessors must all converge to a single node before that activity can begin.

 

Question 5. Construct the network for the project whose activities are given below. Calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity. Determine the critical path and the project completion time.

Answer: First, we construct the project network using the given activities and their durations. Then, we perform forward and backward pass calculations to find the earliest and latest times for each event. Finally, we identify the critical path and the total project completion time.

The network diagram helps visualize the project flow:

Forward pass: This step calculates the earliest event times (E values) by moving from the start of the project to the end. We always pick the maximum value if an event has multiple incoming activities. \( E_1 = 0 + 3 = 3 \) \( E_2 = E_1 + t_{12} = 3 + 8 = 11 \) \( E_3 = 0 + 12 = 12 \) (The source calculation `E3 = 3 + 12 = 15` is based on E1 for node 3, but activity 0-3 also leads to it, with duration 12. Assuming 0-3 means 1-3, and E1 is the EST of node 1. So E3 is max(E1+t13, E0+t03) = 15 or 12. But 0 is the start node. The diagram shows 1-3. So \( E_3 = E_1 + t_{13} = 3 + 12 = 15 \). This aligns with the source calculation.) \( E_4 = \text{max}(E_2 + t_{24}, E_3 + t_{34}) = \text{max}(11 + 6, 15 + 3) = \text{max}(17, 18) = 18 \) \( E_5 = E_2 + t_{25} = 11 + 3 = 14 \) \( E_6 = E_3 + t_{36} = 15 + 8 = 23 \) \( E_7 = \text{max}(E_4 + t_{47}, E_5 + t_{57}, E_6 + t_{67}) = \text{max}(18 + 5, 14 + 3, 23 + 8) = \text{max}(23, 17, 31) = 31 \) The project will take 31 weeks to complete.
Backward pass: This step calculates the latest event times (L values) by moving from the end of the project back to the start. We always pick the minimum value if an event has multiple outgoing activities. \( L_7 = 31 \) \( L_6 = L_7 - t_{67} = 31 - 8 = 23 \) \( L_5 = L_7 - t_{57} = 31 - 3 = 28 \) \( L_4 = L_7 - t_{47} = 31 - 5 = 26 \) \( L_3 = L_6 - t_{36} = 23 - 8 = 15 \) \( L_2 = \text{min}(L_4 - t_{24}, L_5 - t_{25}) = \text{min}(26 - 6, 28 - 3) = \text{min}(20, 25) = 20 \) \( L_1 = \text{min}(L_2 - t_{12}, L_3 - t_{13}) = \text{min}(20 - 8, 15 - 12) = \text{min}(12, 3) = 3 \) \( L_0 = 0 \)

The critical path is the longest path in the network, which determines the minimum time needed to complete the project. Activities on the critical path have zero slack (EFT = LFT and EST = LST).
Critical path is 0-1-3-6-7 and the duration is 31 weeks.
In simple words: First, we draw a map of all the tasks and how long they take. Then, we find the earliest and latest times each task can start and finish without delaying the whole project. The longest path of tasks that cannot be delayed is called the critical path, and its total time is how long the project will take.

🎯 Exam Tip: Remember to calculate Earliest Start Time (EST) and Earliest Finish Time (EFT) using the forward pass, and Latest Start Time (LST) and Latest Finish Time (LFT) using the backward pass. The critical path consists of activities where EST = LST and EFT = LFT.

 

Question 6. A project schedule has the following characteristics: Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Answer: We begin by drawing the network diagram using the given activities and their durations. Then, we calculate the earliest start and finish times (E values) using the forward pass, and the latest start and finish times (L values) using the backward pass. These calculations help us find the critical path and the total time required to complete the project.

The network diagram with E and L values for each node is shown below:

Forward pass: Here, we calculate the earliest time each event can occur. We always choose the maximum value if an event is reached by more than one path. \( E_1 = 0 \) \( E_2 = E_1 + t_{12} = 0 + 4 = 4 \) \( E_3 = E_1 + t_{13} = 0 + 1 = 1 \) \( E_4 = \text{max}(E_2 + t_{24}, E_3 + t_{34}) = \text{max}(4 + 1, 1 + 1) = \text{max}(5, 2) = 5 \) \( E_5 = E_3 + t_{35} = 1 + 6 = 7 \) \( E_6 = E_5 + t_{56} = 7 + 4 = 11 \) \( E_7 = E_5 + t_{57} = 7 + 8 = 15 \) \( E_8 = \text{max}(E_6 + t_{68}, E_7 + t_{78}) = \text{max}(11 + 1, 15 + 2) = \text{max}(12, 17) = 17 \) \( E_9 = \text{max}(E_4 + t_{49}, E_2 + t_{29}) = \text{max}(5 + 5, 4 + 5) = \text{max}(10, 9) = 10 \) \( E_{10} = \text{max}(E_9 + t_{9,10}, E_8 + t_{8,10}) = \text{max}(10 + 7, 17 + 5) = \text{max}(17, 22) = 22 \) The total project duration is 22 days.
Backward pass: This calculates the latest time each event can occur without delaying the project. We pick the minimum value if an event leads to multiple paths. \( L_{10} = 22 \) \( L_9 = L_{10} - t_{9,10} = 22 - 7 = 15 \) \( L_8 = L_{10} - t_{8,10} = 22 - 5 = 17 \) \( L_7 = L_8 - t_{78} = 17 - 2 = 15 \) \( L_6 = L_8 - t_{68} = 17 - 1 = 16 \) \( L_5 = \text{min}(L_6 - t_{56}, L_7 - t_{57}) = \text{min}(16 - 4, 15 - 8) = \text{min}(12, 7) = 7 \) \( L_4 = L_9 - t_{49} = 15 - 5 = 10 \) \( L_3 = \text{min}(L_4 - t_{34}, L_5 - t_{35}) = \text{min}(10 - 1, 7 - 6) = \text{min}(9, 1) = 1 \) \( L_2 = \text{min}(L_4 - t_{24}, L_9 - t_{29}) = \text{min}(10 - 1, 15 - 5) = \text{min}(9, 10) = 9 \) \( L_1 = \text{min}(L_2 - t_{12}, L_3 - t_{13}) = \text{min}(9 - 4, 1 - 1) = \text{min}(5, 0) = 0 \)
Now, let's find the critical path by examining paths where \( E_i = L_i \).

The critical path is 1-3-5-7-8-10 as it takes 22 units to complete the project. This path shows the sequence of activities that must be followed exactly on schedule to ensure the project finishes on time.
In simple words: We draw the project tasks on a map with how long each takes. We figure out the earliest and latest times each task can start and end. The 'critical path' is the longest route through these tasks, which tells us the shortest possible time to finish the whole project.

🎯 Exam Tip: To correctly identify the critical path, always calculate the total duration of all possible paths through the network. The path with the maximum duration is the critical path, and any delay on this path will delay the entire project.

 

Question 7. Draw the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Answer: We will first draw the network diagram based on the given jobs and their durations. Then, we will calculate the earliest and latest start and finish times for each activity using forward and backward passes. Finally, we will identify the critical path, which shows the sequence of tasks that determines the project's overall length.

The network diagram with calculated E and L values for each node is as follows:

Forward pass: This step calculates the earliest event times (E values) by moving from the start of the project to the end. \( E_1 = 0 \) \( E_2 = E_1 + 6 = 0 + 6 = 6 \) \( E_3 = E_1 + 5 = 0 + 5 = 5 \) \( E_4 = \text{max}(E_2 + 10, E_3 + 3) = \text{max}(6 + 10, 5 + 3) = \text{max}(16, 8) = 16 \) \( E_5 = \text{max}(E_4 + 6, E_3 + 4) = \text{max}(16 + 6, 5 + 4) = \text{max}(22, 9) = 22 \) \( E_6 = \text{max}(E_5 + 9, E_4 + 2) = \text{max}(22 + 9, 16 + 2) = \text{max}(31, 18) = 31 \) The total project duration is 31 days.
Backward pass: This step calculates the latest event times (L values) by moving from the end of the project back to the start. \( L_6 = 31 \) \( L_5 = L_6 - 9 = 31 - 9 = 22 \) \( L_4 = \text{min}(L_5 - 6, L_6 - 2) = \text{min}(22 - 6, 31 - 2) = \text{min}(16, 29) = 16 \) \( L_3 = \text{min}(L_4 - 3, L_5 - 4) = \text{min}(16 - 3, 22 - 4) = \text{min}(13, 18) = 13 \) \( L_2 = L_4 - 10 = 16 - 10 = 6 \) \( L_1 = \text{min}(L_2 - 6, L_3 - 5) = \text{min}(6 - 6, 13 - 5) = \text{min}(0, 8) = 0 \)

The Earliest Finish Time (EFT) and Latest Finish Time (LFT) are the same for activities 1-2, 2-4, 4-5, 5-6. This indicates these activities form the critical path, which means they have no slack.
The critical path is 1-2-4-5-6 and the duration to complete the project is 31 days.
In simple words: We draw a map of all project tasks and their times. We then calculate the earliest and latest times each task can begin and end. The 'critical path' is the longest route of tasks where any delay will make the whole project late. In this case, it takes 31 days.

🎯 Exam Tip: To identify the critical path, look for activities where the Earliest Start Time (EST) equals the Latest Start Time (LST) AND the Earliest Finish Time (EFT) equals the Latest Finish Time (LFT). These activities have zero float.

 

Question 8. The following table gives the activities of a project and their duration in days. Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the critical path of the project and duration to complete the project.

Answer: We begin by drawing the network diagram for the project, showing all activities and their durations. Then, we use forward and backward passes to determine the earliest and latest start and finish times for each task. This helps us find the critical path, which is the longest sequence of activities that determines the overall project completion time.

The network diagram with calculated E and L values for each node is:

Forward pass: Here, we calculate the earliest time each event can occur. We always choose the maximum value if an event is reached by more than one path. \( E_1 = 0 \) \( E_2 = E_1 + 5 = 0 + 5 = 5 \) \( E_3 = \text{max}(E_1 + 8, E_2 + 6) = \text{max}(0 + 8, 5 + 6) = \text{max}(8, 11) = 11 \) \( E_4 = \text{max}(E_2 + 7, E_3 + 5) = \text{max}(5 + 7, 11 + 5) = \text{max}(12, 16) = 16 \) \( E_5 = \text{max}(E_3 + 4, E_4 + 8) = \text{max}(11 + 4, 16 + 8) = \text{max}(15, 24) = 24 \) The total project duration is 24 days.
Backward pass: This calculates the latest time each event can occur without delaying the project. We pick the minimum value if an event leads to multiple paths. \( L_5 = 24 \) \( L_4 = L_5 - 8 = 24 - 8 = 16 \) \( L_3 = \text{min}(L_4 - 5, L_5 - 4) = \text{min}(16 - 5, 24 - 4) = \text{min}(11, 20) = 11 \) \( L_2 = \text{min}(L_3 - 6, L_4 - 7) = \text{min}(11 - 6, 16 - 7) = \text{min}(5, 9) = 5 \) \( L_1 = \text{min}(L_2 - 5, L_3 - 8) = \text{min}(5 - 5, 11 - 8) = \text{min}(0, 3) = 0 \)

The Earliest Finish Time (EFT) and Latest Finish Time (LFT) are the same for activities 1-2, 2-3, 3-4, and 4-5. This means these activities are on the critical path.
So, the critical path is 1-2-3-4-5, and the time taken to complete the project is 24 days.
In simple words: We map all the project activities and their timings. Then we find the earliest and latest times each task can start and finish. The "critical path" is the longest sequence of tasks in this map, and its total time tells us how long the project will actually take.

🎯 Exam Tip: When determining the critical path, remember to check activities where EST=LST and EFT=LFT. If a dummy activity is part of the critical path, it indicates a critical dependency, even if it has no duration.

 

Question 9. A project has the following time schedule Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the critical path of the project and duration to complete the project.

Answer: First, we will construct the network diagram for the project, mapping out all activities and their durations. After that, we perform forward and backward pass calculations to determine the earliest and latest possible start and finish times for each activity. This process allows us to find the critical path and the total project completion time.

The network diagram with calculated E and L values for each node is:

Forward pass: This step calculates the earliest event times (E values) by moving from the start of the project to the end. \( E_1 = 0 \) \( E_2 = E_1 + 7 = 0 + 7 = 7 \) \( E_6 = E_1 + 6 = 0 + 6 = 6 \) \( E_3 = E_2 + 14 = 7 + 14 = 21 \) \( E_4 = E_2 + 5 = 7 + 5 = 12 \) \( E_7 = E_6 + 11 = 6 + 11 = 17 \) \( E_5 = \text{max}(E_3 + 11, E_4 + 7) = \text{max}(21 + 11, 12 + 7) = \text{max}(32, 19) = 32 \) \( E_8 = \text{max}(E_5 + 4, E_7 + 18) = \text{max}(32 + 4, 17 + 18) = \text{max}(36, 35) = 36 \) The total project duration is 36 days.
Backward pass: This step calculates the latest event times (L values) by moving from the end of the project back to the start. \( L_8 = 36 \) \( L_7 = L_8 - 18 = 36 - 18 = 18 \) \( L_5 = L_8 - 4 = 36 - 4 = 32 \) \( L_6 = L_7 - 11 = 18 - 11 = 7 \) \( L_4 = L_5 - 7 = 32 - 7 = 25 \) \( L_3 = L_5 - 11 = 32 - 11 = 21 \) \( L_2 = \text{min}(L_3 - 14, L_4 - 5) = \text{min}(21 - 14, 25 - 5) = \text{min}(7, 20) = 7 \) \( L_1 = \text{min}(L_2 - 7, L_6 - 6) = \text{min}(7 - 7, 7 - 6) = \text{min}(0, 1) = 0 \)

The Earliest Start Time (EST) and Latest Start Time (LST), and Earliest Finish Time (EFT) and Latest Finish Time (LFT) are the same for activities 1-2, 2-3, 3-5, 5-8. These activities form the critical path.
Therefore, the critical path is 1-2-3-5-8, and the time to complete the project is 36 days.
In simple words: We draw a map of all the tasks and how long they take. We then find the earliest and latest possible times each task can start and finish. The "critical path" is the longest sequence of tasks where no delays are allowed, and its total time tells us the shortest possible time to finish the entire project.

🎯 Exam Tip: When all four time values (EST, EFT, LST, LFT) are equal for an activity, it signifies that the activity is on the critical path and cannot be delayed without impacting the overall project duration. Clearly mark these activities.

 

Question 10. The following table gives the activities of a project and their duration in days. Construct the network for the project, calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity, and find the critical path. Compute the project duration.

Answer:

Forward pass:
\( E_1 = 0 \)
\( E_2 = 0 + 22 = 22 \)
\( E_3 = (0 + 27) \) or \( (22 + 12) = 34 \). We choose the maximum, so \( E_3 = 34 \).
\( E_4 = (22 + 14) \) or \( (34 + 6) = 40 \). We choose the maximum, so \( E_4 = 40 \).
\( E_5 = 40 + 12 = 52 \)

Backward pass:
\( L_5 = 52 \)
\( L_4 = 52 - 12 = 40 \)
\( L_3 = 40 - 6 = 34 \)
\( L_2 = (40 - 14) \) or \( (34 - 12) = 22 \). We choose the minimum, so \( L_2 = 22 \).
\( L_1 = (22 - 22) \) or \( (34 - 27) = 0 \). We choose the minimum, so \( L_1 = 0 \)

The Earliest Start Time (EST) and Latest Finish Time (LFT) are the same for activities 1-2, 2-3, 3-4, and 4-5. This means these activities are critical. Therefore, the critical path is 1-2-3-4-5. The total time to complete the project is 52 days.
In simple words: First, we draw a diagram showing all the tasks and how long they take. Then, we calculate the earliest and latest times each task can begin and end. The path where the earliest and latest times are the same for all tasks is the "critical path," which shows the longest time needed to finish the whole project. In this case, it's 52 days.

🎯 Exam Tip: When finding the critical path, remember to use the maximum value for forward pass calculations and the minimum value for backward pass calculations to correctly determine the earliest and latest event times. The critical path consists of activities where EST = LST and EFT = LFT.