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A.
Interpretation: A network diagram representing the activities of the project is to be constructed.
Concept Introduction: Network diagram is a graphical representation of tasks and events happening in a project. It depicts the work flow and is used to compute the total duration of the project.
A.
Explanation of Solution
Given information: Theactivities to be taken up for the construction of a gas station and the time duration for each are given in the table.
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
The activity times and the precedence relationship are shown in the table:
| Activity | Time | Predecessors |
| A | 4 | |
| B | 2 | A |
| C | 3 | B |
| D | 1 | C |
| E | 1 | D |
| F | 2 | B |
| G | 3 | F |
| H | 1 | G |
| I | 5 | B |
| J | 3 | I |
| K | 4 | C, F, I |
| L | 2 | K |
| M | 1 | L |
The network diagram showing the work flows is as follows:
B.
Interpretation: The critical path is to be identified by computing the earliest and the latest start and finish times.
Concept Introduction: Network diagram is a graphical representation of tasks and events happening in a project. It depicts the work flow and is used to compute the total duration of the project.
B.
Answer to Problem 26AP
The critical path is A − B − I − K − L − M with completion time as 18 weeks
Explanation of Solution
Given information:
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
The earliest and latest start and finish times are calculated using the formulae:
Earliest Start (EST) = Maximum EFT of the predecessor activity
Latest Start (LST) = LFT - Duration
Earliest Finish (EFT) = EST + Duration
Latest Finish (LFT) = Minimum LST of the immediate successor
Slack = LFT − EFT (or) LST − EST
| Activity | Time | EST | EFT | LST | LFT | Slack |
| A | 4 | 0 | 4 | 0 | 4 | 0 |
| B | 2 | 4 | 6 | 4 | 6 | 0 |
| C | 3 | 6 | 9 | 8 | 11 | 2 |
| D | 1 | 9 | 10 | 16 | 17 | 7 |
| E | 1 | 10 | 11 | 17 | 18 | 7 |
| F | 2 | 6 | 8 | 9 | 11 | 3 |
| G | 3 | 8 | 11 | 14 | 17 | 6 |
| H | 1 | 11 | 12 | 17 | 18 | 6 |
| I | 5 | 6 | 11 | 6 | 11 | 0 |
| J | 3 | 11 | 14 | 15 | 18 | 4 |
| K | 4 | 11 | 15 | 11 | 15 | 0 |
| L | 2 | 15 | 17 | 15 | 17 | 0 |
| M | 1 | 17 | 18 | 17 | 18 | 0 |
The critical path includes the activities with zero slack. So, here the critical path is A − B− I − K − L − M. The total time taken to complete the project is 18 weeks.
C.
Interpretation: A Gantt chart of the
Concept Introduction: Gantt chart is a project management tool that is a type of bar chart. It depicts the project schedule and is used for resource allocation.
C.
Answer to Problem 26AP
A Gantt chart for the project has been drawn.
Explanation of Solution
Given information:
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
Activity times and early start times:
| Activity | EST | Time |
| A | 0 | 4 |
| B | 4 | 2 |
| C | 6 | 3 |
| D | 9 | 1 |
| E | 10 | 1 |
| F | 6 | 2 |
| G | 8 | 3 |
| H | 11 | 1 |
| I | 6 | 5 |
| J | 11 | 3 |
| K | 11 | 4 |
| L | 15 | 2 |
| M | 17 | 1 |
The Gantt chart depicting the project based on the EST:
D.
Interpretation: The impact of replacement of the air compressor which takes two weeks on the completion of the project is to be determined.
Concept Introduction: Network diagram is a graphical representation of tasks and events happening in a project. It depicts the work flow and is used to compute the total duration of the project.
D.
Explanation of Solution
Given information:
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
Installing the air compressor is represented by activity G.
| Activity | Time | EST | EFT | LST | LFT | Slack |
| A | 4 | 0 | 4 | 0 | 4 | 0 |
| B | 2 | 4 | 6 | 4 | 6 | 0 |
| C | 3 | 6 | 9 | 8 | 11 | 2 |
| D | 1 | 9 | 10 | 16 | 17 | 7 |
| E | 1 | 10 | 11 | 17 | 18 | 7 |
| F | 2 | 6 | 8 | 9 | 11 | 3 |
| G | 3 | 8 | 11 | 14 | 17 | 6 |
| H | 1 | 11 | 12 | 17 | 18 | 6 |
| I | 5 | 6 | 11 | 6 | 11 | 0 |
| J | 3 | 11 | 14 | 15 | 18 | 4 |
| K | 4 | 11 | 15 | 11 | 15 | 0 |
| L | 2 | 15 | 17 | 15 | 17 | 0 |
| M | 1 | 17 | 18 | 17 | 18 | 0 |
The latest start time of the activity G is 14th week. If the failure is identified before installation, the project will not be delayed as the replacement time (2 weeks) is less than the slack (6 weeks).
If the failure is found during testing of the air compressor, represented by activity H, still the project would not be delayed, as again there is a slack of 6 weeks.
E.
Interpretation: The activities that can be completed by the end of the 15th week without delaying the project completion are to be determined.
Concept Introduction: Network diagram is a graphical representation of tasks and events happening in a project. It depicts the work flow and is used to compute the total duration of the project.
E.
Answer to Problem 26AP
The activities A, B, C, F and I have to be completed by the end of the 15th week to guarantee project completion without delay.
Explanation of Solution
Given information:
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
The table shows the Earliest and Latest times of the activities
| Activity | Time | EST | EFT | LST | LFT | Slack |
| A | 4 | 0 | 4 | 0 | 4 | 0 |
| B | 2 | 4 | 6 | 4 | 6 | 0 |
| C | 3 | 6 | 9 | 8 | 11 | 2 |
| D | 1 | 9 | 10 | 16 | 17 | 7 |
| E | 1 | 10 | 11 | 17 | 18 | 7 |
| F | 2 | 6 | 8 | 9 | 11 | 3 |
| G | 3 | 8 | 11 | 14 | 17 | 6 |
| H | 1 | 11 | 12 | 17 | 18 | 6 |
| I | 5 | 6 | 11 | 6 | 11 | 0 |
| J | 3 | 11 | 14 | 15 | 18 | 4 |
| K | 4 | 11 | 15 | 11 | 15 | 0 |
| L | 2 | 15 | 17 | 15 | 17 | 0 |
| M | 1 | 17 | 18 | 17 | 18 | 0 |
The latest finish times of the activities A, B, C, F and Iam less than 15 weeks. Therefore, these activities have to be completed by the end of the 15th weeks to guarantee project completion without any delay.
F.
Interpretation: The optimal solution for the problem is to be determined using linear programming.
Concept Introduction: Linear programming is a mathematical technique used to achieve the best outcome, given a list of parameters usually restrictions on the resources.
F.
Answer to Problem 26AP
The optimal solution for the problem is obtained using linear programming.
Explanation of Solution
Given information:
| Activity | Time |
| A | 4 |
| B | 2 |
| C | 3 |
| D | 1 |
| E | 1 |
| F | 2 |
| G | 3 |
| H | 1 |
| I | 5 |
| J | 3 |
| K | 4 |
| L | 2 |
| M | 1 |
The various paths of the project and the durations are:
The critical path is A − B − I − K − L − M with the completion time of 18 weeks.
Using linear programming, the EST is calculated.
Objective function = =SUM (B2:M2)
| Nodes | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| EST |
The constraints
| Activity | Time | ||
| A | 0 | > | 4 |
| B | 0 | > | 2 |
| C | 0 | > | 3 |
| D | 0 | > | 1 |
| E | 0 | > | 1 |
| F | 0 | > | 2 |
| G | 0 | > | 3 |
| H | 0 | > | 1 |
| I | 0 | > | 5 |
| J | 0 | > | 3 |
| K | 0 | > | 4 |
| L | 0 | > | 2 |
| M | 0 | > | 1 |
| P1 | 0 | > | 0 |
| P2 | 0 | > | 0 |
| P3 | 0 | > | 0 |
Solver Solution:
| Objective Function |
| 120 |
EST for the nodes:
| Nodes | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| EST | 0 | 4 | 6 | 9 | 8 | 11 | 11 | 11 | 10 | 15 | 17 | 18 |
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Chapter 10 Solutions
Production and Operations Analysis, Seventh Edition
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