operational management online chap 3

Complete this table of activity lengths and early and late start and finish times based on the indicated precedence requirements. All times are in days. Activity Predecessor Early Start Early Finish Late Start Late Finish Length A -- 0 5 B A 11 11 C A 9 14 D B 18 18 7 E B, C 11 15 F D, E 3

Activities A and F must be on the critical path, so their early and late start times must be identical and their early and late finish times must also be identical. Activity A must be 5 days long based on its early start of 0 and early finish of 5 days. Activity B must be critical since its early and late finish times are both 11. Its late start is 5 and late finish is 11, so its duration is 11-5=6. Activity C has an early start of 5 from activity A and an early finish of 9 from the table, so its duration must be 9-5=4, and its late start must be 14-4=10. Activity D has identical early and late finish times, so it is on the critical path. It has a late start of 11 since it follows B, it must also have an early start of 11 since it is critical, which can be confirmed by subtracting its length of 7 from the LS and LF values. Activity E has an early start of 11 and an early finish of 15, so its duration is 15-11=4. Its late finish must be 18, from D's late finish. Activity E has a late start of 18-4=14. Activity F has a duration of 3 and identical early and late starts of 18, so its early and late finish must be 18+3=21. Activity Predecessor Early Start Early Finish Late Start Late Finish Length A -- 0 5 5 0 5 B A 5 11 5 11 6 C A 5 9 10 14 4 D B 11 18 11 18 7 E B,C 11 15 14 18 4 F D,E 18 21 18 21 3

What is the probability that a project with a mean completion time of 23.9 days and a variance of 6 days will be finished in 26 days? 0.80 0.20 0.63 0.37

Correct Answer: 0.80. 26-23.9/sq root of 6 =.85. Use chart .3023+.5 =.80 INCORRECT. Use the other half of the normal curve.

Question 12. Activities A and B are both 6 days long and the only immediate predecessors to activity C. Activity A has ES=8 and LS=8 and activity B has ES=7 and LS=10. What is the ES of activity C? 14 13 16 15

Correct Answer: 14 8+6=14. 7+6=13. INCORRECT. Activity B is not on the critical path.

Question 10. An activity has an optimistic time of 15 days, a most likely time of 18 days, and a pessimistic time of 27 days. What is its expected time? 20 days 19 days 18 days 60 days

Correct Answer: 19 days. (15+4(18)+27)/6 INCORRECT. This is the sum of all estimates.

Question 11. An activity has an optimistic time of 11 days, a most likely time of 15 days, and a pessimistic time of 23 days. What is its variance? 4 2 15.6 16.33

Correct Answer: 4. ((23-11)/6)sqrd INCORRECT. This is the average of the three estimates.

Question 15. The EF of an activity is the -ES + Activity time. -Min{LS of all immediate following activities}. -LF-Activity time. -Max{EF of all immediate predecessors}.

Correct Answer: ES + Activity time. INCORRECT. This is the LF.

Which of the following is a direct responsibility of the project manager? -Calculating completion probabilities for all tasks in the project. -Drawing the network diagram. -Making sure that the people assigned to the project receive the motivation, direction, and information needed to do their jobs. -Performing all of the activities in the project.

Correct Answer: Making sure that the people assigned to the project receive the motivation, direction, and information needed to do their jobs. INCORRECT. The project manager is usually not obligated to do all of the work to complete the project.

Question 4. Which of the following is a basic assumption of PERT? -Only critical path activities in the network must be performed. -There is only one complete route from the start of a project to the end of a project. -No activity in the network must be repeated. -Activity completion times are known with certainty.

Correct Answer: No activity in the network must be repeated. There may be many routes from the start to end of a project.

Question 9. Which of the following is a limitation of PERT and CPM? -There is inherent danger of too much emphasis being placed on the critical path. -They are applicable to only a narrow variety of projects and industries. -They can be used only to monitor schedules. -The graphical nature of a network delays comprehension of the activity list's interrelationships.

Correct Answer: There is inherent danger of too much emphasis being placed on the critical path. INCORRECT. They can be used to monitor schedules and costs, which is an advantage.

Question 8. Dummy activities -cannot be on the critical path. -are used when two activities have identical starting and ending events. -are found in both AOA and AON networks. -have a duration equal to the shortest non-dummy activity in the network.

Correct Answer: are used when two activities have identical starting and ending events. INCORRECT. Dummy activities have no duration.

Question 2. With respect to PERT and CPM, slack -is a task or subproject that must be completed. -is the amount of time a task may be delayed without changing the overall project completion time. -is the latest time an activity can be started without delaying the entire project. -marks the start or completion of a task.

Correct Answer: is the amount of time a task may be delayed without changing the overall project completion time.

The critical path of a network is the Open Hint for Question 6 in a new window. -shortest time path through the network. -path with the fewest activities. -path with the most activities. -longest time path through the network.

Correct Answer: longest time path through the network. INCORRECT. Why would the shortest time path be critical?

Question 1. The three phases involved in the management of large projects are scheduling, designing, operating. scheduling, operating, evaluating. planning, scheduling, evaluating. planning, scheduling, controlling.

Correct Answer: planning, scheduling, controlling.

PERT analysis computes the variance of the total project completion time as - the variance of the final activity of the project. -the sum of the variances of all activities in the project. -the sum of the variances of all activities on the critical path. -the sum of the variances of all activities not on the critical path.

Correct Answer: the sum of the variances of all activities on the critical path. INCORRECT. This is a meaningless number; path variances are meaningful.

Describe at least four of the ethical issues faced in project management.

Project managers may be offered gifts from current contractors or contractors hoping to be selected for work. Other ethical issues include pressure to alter status reports to mask the reality of delays, false reports for charges of time and expenses, pressures to compromise quality to meet bonus or penalty schedules, outright bribery, and bid rigging (divulging confidential information to some bidders to give them an unfair advantage).

What is the probability that this project is completed in 28 days? All activity lengths are given in days. Activity Predecessor Optimistic Most Likely Pessimistic A -- 1 2 3 B A 2 4 6 C A 5 8 11 D A 3 4 5 E B 10 14 18 F D 6 8 10 G C, E, F 7 9 13

The routes through the network are ABEG=29.33 days; ACG=19.33 days; and ADFG=23.33 days. Route lengths are obtained by summing the expected times for each activity on the path. The variances for each route are ABEG=3.33; ACG=19.33; and ADFG=23.33. Path variances are computed similar to path lengths; sum the variance of each activity on the path to determine the overall path variance. You cannot add up activity standard deviations to get a path standard deviation, you must work with activity variances. To compute the probability of finishing within 28 days, compute a Z score for the critical path by subtracting the 29.33 from 28 and dividing the result by the standard deviation of path ADFG (the square root of the variance) which is 1.826. The resulting Z-score of -0.73 can be looked up in a normal table to reveal a probability of 0.23. The expected durations and variances for all activities are shown in the table below: Activity Mean Variance A 2 0.11 B 4 0.44 C 8 1 D 4 0.11 E 14 1.78 F 8 0.44 G 9.33 1

Use the crash costs, regular activity durations and precedence requirements in the table to determine the cheapest completion time and cost for this project if the project has a fixed cost of $1000 per day. All activity lengths are given in days. Activity Regular Time Minimum Time $/day crash fee Predecessor A 10 6 $500 -- B 6 5 $300 -- C 2 2 -- B D 4 2 $500 C E 6 3 $600 A F 8 5 $1200 D, E

The two routes through the network are AEF=24 days and BCDF=20 days; so with a duration of 24 days the initial project cost is $24,000. The first time/cost tradeoff should occur on the critical path AEF. The lowest cost per day is activity A at $500. Activity A can be shortened by 4 days for $2000, but this reduces the project by 4 days, saving $4000, for a total gain of $4000-$2000=$2000. Activity F costs more to crash than is saved in fixed costs, so it should not be expedited. Activity E costs $600 per day to expedite, but now the other path (BCDF) must be shortened equivalently. Activity B can be expedited for $300 per day, so the BE combination costs $900 to expedite while saving $1000 in fixed costs. Subsequent reductions in path BCDF in combination with activity E cost more than the $1000 per day fixed cost, so the shortest project is 19 days at a total cost of 19x$1000 (fixed) plus $2000 (activity A) plus $900 (activities B and E) for a total of $21,900.

Q5 Gantt charts are the best tool for showing interrelationships between activities and their resources. True False

Your Answer: False

Q6 The shortest completion time for a project is not the critical path. . True False

Your Answer: False

Q9 Early start (ES) and late start (LS) times are determined for each activity during the forward pass. True False

Your Answer: False

Question 10. The project's standard deviation is the sum of the standard deviations for all critical path activities. True False

Your Answer: False

Question 12. Microsoft Project uses activity on arc representation for network diagrams. True False

Your Answer: False

Question 2. The optimistic time is the greatest amount of time that could be required to complete an activity. True False

Your Answer: False

Question 3. The expected completion time of a PERT project is the sum of the most likely times of the activities on the critical path. True False

Your Answer: False

PERT requires three estimates of activity completion time, while CPM only requires a single estimate. True False

Your Answer: True

Q7 A network diagram drawn in AON style never needs a dummy activity. True False

Your Answer: True

Q8 There is never any slack on the critical path. True False

Your Answer: True

Question 11. The most cost-effective means of crashing a project would be to pay to expedite critical path activities. True False

Your Answer: True

Question 4. Work breakdown structure is the division of a project into more and more detailed components. True False

Your Answer: True

A project's critical path is composed of activities A, B, and C. Activity A has a standard deviation of 2, activity B has a standard deviation of 1, and activity C has a standard deviation of 2. What is the standard deviation of the critical path? 5 25 3 9

Your Answer: 3. 2sq+2sq+1sq=9. Square root of 9 is 3

Question 16. The ethical issue of divulging confidential information to some bidders to give them an unfair advantage is known as bid rigging. bribery. expense account padding. low balling.

Your Answer: bid rigging.

Q3 A dummy activity is required when Open Hint for Question 3 in a new window. -two or more activities have the same starting events. -the network contains two or more activities that have identical starting and ending events. -two or more activities have different ending events. -two or more activities have the same ending events.

Your Answer: the network contains two or more activities that have identical starting and ending events.