Chapter 9 Reducing Project Duration Chapter Outline 1. Rationale for Reducing Project Duration 2. Options for Accelerating Project Completion A. Options When Resources Are Not Constrained i. Adding Resources ii. Outsourcing Project Work iii. Scheduling Overtime iv. Establish a Core Project Team v. Do It Twice-Fast and Correctly B. Options When Resources Are Constrained i. Improve the Efficiency of the Project Team ii. Fast-Tracking iii. Critical Chain iv. Reducing Project Scope v. Compromise Quality 3. Project Cost-Duration Graph A. Explanation of Project Costs i. Project Indirect Costs Project Direct Costs 4. Constructing a Project Cost-Duration Graph A. Determining the Activities to Shorten B. A Simplified Example 5. Practical Considerations A. Using the Project Cost-Duration Graph B. Crash Times C. Linearity Assumption D. Choice of Activities to Crash Revisited E. Time Reduction Decisions and Sensitivity 6. What if Cost, Not Time, Is the Issue? A. Reduce Project Scope B. Have Owner Take on More Responsibility C. Outsourcing Project Activities or Even the Entire Project D. Brainstorming Cost Savings Options 7. Summary 8. Key Terms 9. Review Questions 10. Exercises 11. Case 9.1: International Capital, Inc.—Part B 12. Case 9.2: Whitbread World Sailboat Race 13. Case 9.3: Nightingale Project—A 14. Case 9.4: Nightingale Project—B 15. Case 9.5: The “Now” Wedding—Part A 16. Case 9.6: The “Now” Wedding—Part B Chapter Learning Objectives After reading this chapter you should be able to: LO 9.1 Understand the different reasons for crashing a project. LO 9.2 Identify the different options for crashing an activity when resources are not constrained. LO 9.3 Identify the different options for crashing an activity when resources are constrained. LO 9.4 Determine the optimum cost-time point in a project network. LO 9.5 Understand the risks associated with compressing or crashing a project. LO 9.6 Identify different options for reducing the costs of a project. Review Questions 1. What are five common reasons for crashing a project? Reasons given could include: Imposed deadline in which disfavor will be earned by not meeting superior’s deadline Time to market competitive advantage Realize benefits from incentive contracts To make up for lost time and avoid contract penalties Save extensive overhead costs Free up resources to work on other projects Exceed customer expectations. 2. What are the advantages and disadvantages of reducing project scope to accelerate a project? What can be done to reduce the disadvantages? Reducing the scope of the project can lead to big savings both in time and costs. It typically means the elimination of certain tasks. At the same time scaling down the scope may reduce the value of the project such that it is no longer worthwhile or fails to meet critical success parameters. The key is reassessing the project requirements to determine which are essential and which are optional. This requires the active involvement of all key stakeholders. More intense re-examination of requirements may actually improve the value of the project by getting it done more quickly and for a lower cost. 3. Why is scheduling overtime a popular choice for getting projects back on schedule? What are the potential problems for relying on this option? Scheduling overtime is popular because it often involves salary workers and no direct costs are added to the project. Even if it involves additional costs, such as hourly workers, you avoid Brook’s law and minimize additional coordination and training costs. The disadvantages are the additional time and half costs associated with hourly overtime and stress and fatigue that come with working long hours which can lead to accidents, inferior performance, and turnover. 4. Identify four indirect costs you might find on a moderately complex project. Why are these costs classified as indirect? Indirect (overhead) costs are costs that cannot be attributed to a specific activity or work package. Examples of indirect costs are supervision, consultants, debt interest charges, machinery common to several activities, accounting and information processing, public relations, penalties or incentives for early or late completion. In practice it is amazing how many project compression decisions are made without serious consideration of indirect costs. 5. How can a cost-duration graph be used by the project manager? Explain. A cost-duration graph is useful to the project manager for comparing alternatives. Any alternative that moves the project duration away from the optimum cost-duration point will increase costs. Additionally, incentives and penalties can be evaluated against the total, low cost point. 6. Reducing the project duration increases the risk of being late. Explain. Compressing the project duration means slack (float) on noncritical activities will be reduced. When slack of noncritical activities is reduced, the chance of new critical paths occurring increases; hence, the risk of the project becoming late increases. In addition, compressing will have the following other impacts on managing the project: Reduces flexibility by using slack Can increase number of critical activities Can increase interdependencies of paths Makes resource scheduling tighter (critical) May increase costs. 7. It is possible to shorten the critical path and save money. Explain how. The only way to shorten the critical path and save money is to have indirect costs which are greater than the additional direct costs of shortening the critical path one unit of time. The difference is a savings. Exercises 1. Use the information contained below to compress one time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost. Note: The correct normal project duration, critical path, and total direct cost are provided. Duration can be reduced from 12 time periods to eight. Total direct cost goes up from 1,000 to 1,300. Steps are shown below: The project has two paths. Path A-B-D-F takes 10 time periods and A-C-E-F takes 12 time periods. The A-C-E-F path is critical so it is the one to look at to reduce project time. Total direct cost is $1,000. The cheapest activity to reduce is A so we reduce it by its maximum reducion of one time unit to two time periods. The A-C-E-F path remains critical at 11 time periods and direct costs go up to $1,050 since it cost $50 to crash A. The next cheapest activity to reduce is C so we reduce it from four to three at a cost of $60. A-C-E-F remains critical, completion time goes down to ten time units and total direct cost goes up to $1,110 C can be reduced by two time units so we again reduce C by one time unit at a cost of $60. Completion time goes down to nine time units and both paths become critical. That means that further reductions will require either reducing the same time from an activity on both paths or finding an activity shared by both paths and reducing that activity. In this case, no shared activity can be reduced further. Finally, we can reduce D by one time period off of the first path at a cost of $60 and E by one day off of the second path at a cost of $70. That reduces the completion time to eight time periods and raises the total direct cost to $1,300. No further reductions are possible. 2. Use the information contained below to compress one time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost. Note: Choose B instead of C and E (equal costs) because it is usually smarter to crash early rather than late AND one activity instead of two activities Project duration is reduced from 13 time periods to ten. Total direct cost goes up from 1,000 to 1,240. The steps are shown below: The project has two paths, A-B-C-F, which takes 12 time periods, and A-B-D-E-F, which takes 13 time periods. This gives the project a duration of 13 time periods. The total direct cost is $1,000. The cheapest activity to reduce is D at a cost of $40 for one time period. That makes both paths critical. That means that further reductions will require either reducing the same time from an activity on both paths or finding an activity shared by both paths and reducing that activity. Total direct cost goes up from $1,000 to $1,040. Activity B is on both paths and can be reduced by one time period at a cost of $100. That lowers the completion time to 11 time periods and raises total direct cost to $1,140. Finally, we can reduce C on the first path at a cost of $50 and E on the second path at a cost of $50. That reduces the completion time to 10 time units and raises total direct cost to $1,240. No further reductions are possible. 3. Use the information contained below to compress one time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed and the adjusted total cost. Project duration is reduced from 16 to 12. Total direct cost increases from 1,400 to 1,640. The steps are shown below: The project has three paths, A-B-C-F-H, A-B-D-G-H (the critical path), and A-E-H. Completion time is 16 time periods and total direct cost is $1,400. We begin by reducing G by one time period at a cost of $20. That reduces the project completion time to 15 time units and raises total direct cost to $1,420. Next, we reduce D by one time period at a cost of $40. That reduces the project completion time to 14 and raises total direct cost to $1,460. Next, we reduce B by one time period at a cost of $80. That reduces the project completion time to 13 and raises total direct cost to $1,540. Finally, we reduce A by one time period at a cost of $100. That reduces the project completion time to 12 and raises total direct cost to $1,640. No further reductions are possible. 4. Given the data and information that follow, compute the total direct cost for each project duration. If the indirect costs for each project duration are $90 (15 time units), $70 (14), $50 (13), $40 (12), and $30 (11), compute the total project cost for each duration. What is the optimum cost-time schedule for the project? What is this cost? Total cost at the original duration of 15 time periods is 730 total direct cost plus 90 total indirect cost equals 820. That total goes to 810 for 14 time periods, 820 for 13 time periods, 1,000 for 12 time periods, and 1,180 for 11 time periods. Total costs are minimized at 14 time periods and 810. The steps are shown below: There are three paths, A-C-F-I, A-D-G-I, and B-E-H-I. The second, A-D-G-I, is critical at fifteen time periods. However, the first and third paths are 14 and 13 respectively, so other paths will rapidly become critical and will have to be monitored closely. Total direct cost is $730, indirect cost is $90, for a total cost of $820. Reducing D by one time period at a cost of $10 reduces the completion time to 14. The first path, A-C-F-I become critical. Total direct cost rises to $740, indirect cost falls to $70, and total cost falls to $810. Reducing A by one reduces the completion time to 13 and all three paths are now critical. Direct cost goes up to $770, indirect cost falls to $50, and total cost rises $820. Since total cost has started to rise, we know that 14 time periods is optimum. Several more crashes are shown below just to illustrate that total direct cost continues to rise faster than indirect costs fall so total cost continues to rise. However, it is reasonable for a student to stop at this point. With all three paths critical, each path must be reduced. It is cheaper to reduce B for $60, F for $100, and G for $30. Completion time does down to 12 and total cost goes up to $1,000. Reducing I by one time period, at a cost of $200, reduces all three paths. This reduces the completion time to 11 and raises total cost to $1,190. No further reductions are possible. 5. Use the information contained below to compress one time unit per move using the least cost method. Assume the total indirect cost for the project is $700 and there is a savings of $50 per time unit reduced. Record the total direct, indirect, and project costs for each duration. What is the optimum cost-time schedule for the project? What is the cost? Note: The correct normal project duration and total direct cost are provided. Total cost at 14 time units is 1,200 total direct cost plus 700 total indirect cost equals 1,900. This drops to 1,870 at 13 time units, 1,860 at 12 time units, and 1,910 at 11 time units. Total cost is minimized at 12 time units and total cost of 1,860. The steps are shown below: The project has three paths, A-B-D-G, A-C-E-I, and A-C-F-I. The second one, A C E I is critical at 14 time periods but the third, A-C-F-I, at 13 is almost critical. Direct cost is $1,200, indirect cost is $700, and total cost is $1,900. Reducing E by one at a cost of $20 reduces the time to 13 and makes the third path critical as well. Direct cost goes up to $1,220, indirect cost goes down to $650, and total cost goes down to $1,870. Reducing C by one at a cost of $40 lowers the completion time to 12. Direct cost goes up to $1,260, indirect cost goes down to $600, and total cost goes down to $1,860. Reducing E by one at a cost of $20 and F by one at a cost of $40 reduces completion time to 11. Direct cost goes up to $1,320, indirect cost goes down to $550, and total cost goes up to $1,870. Since total cost has started back up, we know that 12 time periods was optimal. 6. If the indirect costs for each duration are $300 for 27 days, $240 for 26 days, $180 for 25 days, $120 for 24 days, $60 for 23 days, and $50 for 22 days, compute the direct, indirect and total costs for each duration. What is the optimum cost-time schedule? The customer offers you $10 for every day you shorten the project from your original network. Would you take it? If so for how many days? With no incentive, day 27 has total direct cost of 300 and total indirect cost of 300 for a total of 600. Day 26 has total direct cost of 330, total indirect cost of 240, and an incentive of 10 for total cost of 560. (Remember, you subtract the incentive since the payment reduces cost.) Total cost at day 25 is 540, at day 24 it is 560, at day 23 it is 670, and at day 22 it is 830. Total cost is minimized at day 25. The steps and cost curve are shown below: The project has three paths, A-D-F (critical), B-F, and C-E-F. The third path is almost critical and will have to be monitored closely. Completion time is 27, direct cost is $300, indirect cost is $300, and the incentive is zero, for a total cost of $600. Reducing F by one at a cost of $30 reduces the completion time to 26. Direct cost goes up to $330, indirect cost goes down to $240, the incentive goes up to $10, and total cost goes down to $560. Reducing D by one at a cost of $50 reduces the completion time to 25 and makes the third path, C-E-F, critical as well. Direct cost goes up to $380, indirect cost goes down to $180, the incentive goes up to $20, and total cost goes down to $540. Reducing D by one at a cost of $50 and C by one at a cost of $40 reduces the completion time to 24. Direct cost goes up to $470, indirect cost goes down to $120, the incentive goes up to $30, and total cost goes up to $560. Since total cost is now going up, we know that 25 time periods is optimal. While 25 is optimal, several more crashes are shown below for illustration purposes. Reducing A by one at a cost of $80 and E by one at a cost of $100. Reducing A by one more time period at a cost of $80 and E by one more time period at a cost of $100. 7. Use the information contained below to compress one time unit per move using the least cost method. Assume the total indirect cost for the project is $2,000 and there is a savings of $100 per time unit reduced. Calculate the total direct, indirect, and project costs for each duration. Plot these costs on a graph. What is the optimum cost-time schedule for the project? Note: The correct normal project duration and total direct cost are provided. At the starting point of 20 time periods duration, total direct cost is 6,000 and total indirect cost is 2,000 for a total of 8,000. That total drops to 7,940 at 19 time periods, drops again to 7,890 at 18, stays 7,890 at 17, rises to 8,190 at 16, and rises again to 8,490 at 15. Total cost is minimized at either 17 or 18 time periods. The steps and cost curves are shown below: The project has four paths: A-B-C-E-F-H, A-B-C-E-G-H, A-B-D-E-F-H, and A B D E G H. The first and third are critical at 20 time periods and the second and third are near-critical at 19. Direct costs are $6,000, indirect costs are $2,000, and total cost are $8,000. Reducing F by one time period at a cost of $40 reduces the time to 19 and makes all four paths critical. Direct cost goes up to $6,040, indirect costs go down to $1,900, and total cost goes down to $7,940. Reducing B by one time period at a cost of $50 reduces the time to 18. Direct cost goes up to $6,090, indirect costs go down to $1,800, and total cost goes down to $7,890. Reducing E by one time period reduces the completion time to 17. Direct cost goes up to $6,190, indirect costs go down to $1,700, and total cost remains the same at $7,890. Reducing C and D each by one time period at a cost of $200 each reduces the completion time to 16. Direct cost goes up to $6,590, indirect cost goes down to $1,600, and total cost goes up to $8,190. We are now sure that total cost is on the rise and do not need to check further to minimize cost. Nevertheless, one more reduction is shown. Reducing C and D each again by one time period at a cost of $200 each reduces the completion time to 16. Direct cost goes up to $6,990, indirect cost goes down to $1,500, and total cost goes up to $8,490. A plot of the costs is shown below: 8. Use the information contained below to compress one time unit per move using the least cost method. Reduce the schedule until you reach the crash point of the network. For each move identify what activity(ies) was crashed, the adjusted total cost, and explain your choice if you have to choose between activities that cost the same. If the indirect cost for each duration are $1,500 for 17 weeks, $1,450 for 16 weeks, $1,400 for 15 weeks, $1,350 for 14 weeks, $1,300 for 13 weeks, $1,250 for 12 weeks, $1,200 for 11 weeks, and $1,150 for 10 weeks, what is the optimum cost-time schedule for the project? What is the cost? The project contains four paths: B-F-G-I, B-D-H-I (critical), C-D-H-I, and C-E-H-I. The third path is close to critical and must be monitored closely. Direct cost is $2,000, indirect cost is $1,500, and total cost is $3,500. Reducing D by one at a cost of $40 reduces the completion time to 16. Direct cost goes up to $2,040, indirect costs go down to $1,450, and total cost goes down to $3,490. Reducing H by one at a cost of $60 reduces the completion time to 15. Direct cost goes up to $2,100, indirect costs go down to $1,400, and total cost goes up to $3,500. Since total cost goes up, we know that 16 is optimal. Additional reductions are shown in order to construct a table and graph. Reducing H by one day at a cost of $60 causes completion time to go down to 14 and the first path to become critical as well. Direct cost goes up to $2,160, indirect costs go down to $1,350, and total cost goes up to $3,510. Reducing B by one day at a cost of $100 causes completion time to go down to 13 and the third path to become critical as well. Direct cost goes up to $2,260, indirect costs go down to $1,300, and total cost goes up to $3,560. Finally, reducing I by one day at a cost of $200 causes completion time to go down to 12. Direct cost goes up to $2,460, indirect costs go down to $1,250, and total cost goes up to $3,710. A summary table of the costs is shown below: A chart of the costs is shown below: Case 9.1 International Capital, Inc.—Part B Given the project network derived in Part A of the case from Chapter 7, Brown also wants to be prepared to answer any questions concerning compressing the project duration. This question will almost always be entertained by the accounting department, review committee, and the client. To be ready for the compression question, Brown has prepared the following data in case it is necessary to crash the project. (Use your weighted average times (te) computed in Part A of the International Capital case found in Chapter 7.) Using the data provided, determine the activity crashing decisions and best-time cost project duration. Given the information you have developed, what suggestions would you give Brown to ensure she is well prepared for the project review committee? Assume the overhead costs for this project are $700 per workday. Will this alter your suggestions? Part A suggests that the project would have to be compressed down to 61 days to reach the 95 percent chance of meeting the average. Compressing follows the steps listed below. Start at 73 workdays Cut A 3 days $1,500 70 days Cut K 6 days 6,000 64 days Cut J 1 day 1,000 63 days Cut H 1 day 2,000 62 days Cut D 1 day 3,000 61 days TOTAL $13,500 However, when these data are plotted with the indirect costs, the picture changes slightly. The optimum time/cost tradeoff is 70 days. See chart below. Clearly, the risk of being late has increased. Activities A&B are parallel and have added a critical path. Compressing D one day reduces slack for Activities E&F to one day. Brown and the project review committee have a tradeoff decision themselves. To get to 61 days will cost $5,700 ($159,200 – 153,500). Duration Project Overhead Normal Direct Costs Direct Crash Costs Total Direct Costs Total Costs 61 42,700 103,000 13,500 116,500 159,200 62 43,400 103,000 10,500 113,500 156,900 63 44,100 103,000 8,500 111,500 155,600 64 44,800 103,000 7,500 110,500 155,300 65 45,500 103,000 6,500 109,500 155,000 66 46,200 103,000 5,500 108,500 154,700 67 46,900 103,000 4,500 107,500 154,400 68 47,600 103,000 3,500 106,500 154,100 69 48,300 103,000 2,500 105,500 153,800 70* 49,000 103,000 1,500 104,500 153,500 71 49,700 103,000 1,000 104,000 153,700 72 50,400 103,000 500 103,500 153,900 73 51,100 103,000 - 103,000 154,100 74 51,800 103,000 - 103,000 154,800 Case 9.2 Whitbread World Sailboat Race Each year countries enter their sailing vessels in the nine-month Round the World Whitbread Sailboat Race. In recent years, about 14 countries entered sailboats in the race. Each year’s sailboat entries represent the latest technologies and human skills each country can muster. (Rest of case not shown due to length.) This is a fairly difficult case in which students have to create a project schedule and then use the time-cost method to determine whether it is possible to meet the 45 week deadline and $3.2 million budget limit. Students also have to factor in indirect costs in the form of a hammock for the additional cost of keeping vessels in service. Activity Normal Time Normal Cost Crash Time Crash Cost Slope A Design 6 $ 40 4 $ 160 60 B Build hull 12 1,000 10 1,400 200 C Install ballast tanks 2 100 2 100 - D Order mast 8 100 7 140 40 E Order sails 6 40 6 40 - F Order accessories 15 600 13 800 100 G Build deck 5 200 5 200 - H Coat hull 3 40 3 40 - I Install accessories 6 300 5 400 100 J Install mast and sails 2 40 1 80 40 K Test 5 60 4 100 40 L Sea trials 8 200 7 450 250 M Select crew 6 10 5 20 10 N Secure housing 3 30 3 30 - O Select crew equipment 2 10 2 10 - P Order crew equipment 5 30 5 30 - Q Routine sail/maintenance 15 40 12 130 30 R Crew maintenance Train 10 100 9 340 240 S Initial sail training 7 50 5 350 150 Total direct cost $2,990 Hammock (Indirect costs): Costs for keeping old vessel in service = $4,000/week for 25 weeks = $100,000 Cost for keeping new vessel in service for training = $6,000/week for 19 weeks = $114,000 Total hammock indirect costs = $214,000 Normal costs for 50-week plan: Normal direct costs $2,990,000 Indirect costs (hammock) 214,000 Total costs $3,204,000 Compressed to 45 Weeks: Costs for keeping old vessel in service = $4,000/ week for 21 weeks = $84,000 Cost for keeping new vessel in service for training = $6,000/week for 19 weeks = $114,000 Total hammock indirect costs = $198,000 Compressed Activities: A 2 weeks = 60,000+60,000 = $120,000 B 2 weeks = 200,000+200,000 = 400,000 (Reduce hammock 2 weeks) R 1 week = 240,000 = 240,000 $760,000 Normal direct costs 2,990,000 Indirect costs (hammocks) 198,000 (21x4,000=84,000) + (19x6,000=114,000) Total costs $3,948,000 Upon working the problem, students should advise Bjorn that yes it is possible to complete the project in 45 weeks but that the cost is estimated to be $3.948 million. This is $748,000 more than Bjorn’s $3.2 million. This begs the question of priorities. When asked what the priorities of the project are most students will rightly indicate: Whitbread Project Priority Matrix Time Performance Cost Constrain X X Enhance X Accept Students will point out that both time (45 weeks) and budget ($3.2 million) are constrained. However, based on their analysis it is impossible to meet both constraints. Given that performance (victory) is the primary goal, one should argue that Bjorn should seek additional funding and accept cost over-run. This is an important lesson. Often projects begin with similar constraints but are forced to make a trade off in the final analysis. Case 9.3 Nightingale Project—A You are the assistant project manager to Rassy Brown, who is in charge of the Nightingale project. Nightingale was the code name given to the development of a handheld electronic medical reference guide. Nightingale would be designed for emergency medical technicians and paramedics who need a quick reference guide to use in emergency situations. (Rest of case not shown due to length.) This case, along with Part B, is designed to have the students use the tools and concepts introduced in Chapters 6 and 9 to answer some basic questions concerning the schedule of the Nightingale Project. It is an excellent vehicle for training students to use project management software to create a network and assess alternative courses of action. At our college the students have access to MS Project and we have found that students are capable of completing this assignment by simply following the Tutorial contained within the program. Note: The answers provided are based on a January 1, 2017 start date which due to holidays means the project begins on January 2, 2017. We suggest you tell students to use this start date to insure consistency in answers. Students should submit a project schedule similar to the one presented in computer output below to support their answers. 1. Will the project as planned meet the October 25th deadline? No, the scheduled completion date is 12/13/17. 2. What activities lie on the critical path? Architectural decisions Feature specifications Database Review design Integration Procure prototype components Assemble prototypes Lab test prototype Field test prototypes Adjust design Order stock parts Assemble first production unit Test unit Produce 30 units Train sales representatives. 3. How sensitive is the network? The text defines sensitivity as the likelihood that the critical path may change during the course of the project. Sensitivity results from the number of critical paths and the amount of slack available to non-critical activities. Based on the information provided, this project is a fairly insensitive network. Most non-critical activities have between 20-37 days of slack, exceeding their estimated duration times. The lone exceptions are Price components which has only 5 days of slack and Order custom parts which has 8 days of slack. However, given the nature of these tasks it is unlikely that they will take more than 5 days beyond their estimated duration time to complete. It should also be noted that, after an initial burst of activities, there are only two non-critical activities (Document design and Order custom parts) during the second half of the project. Case 9.4 Nightingale Project—B Rassy and the team were concerned with the results of your analysis. They spent the afternoon brainstorming alternative ways for shortening the project duration. They rejected outsourcing activities because most of the work was developmental in nature and could only be done in-house. They considered altering the scope of the project by eliminating some of the proposed product features. After much debate, they felt they could not compromise any of the core features and be successful in the marketplace. They then turned their attention to accelerating the completion of activities through overtime and adding additional technical manpower. Rassy had built into her proposal a discretionary fund of $200,000. She was willing to invest up to half of this fund to accelerate the project, but wanted to hold onto at least $100,000 to deal with unexpected problems. After a lengthy discussion, her team concluded that the following activities could be reduced at the specified cost: (Rest of case not shown due to length.) Students should assess the suggestions presented in Part B and prepare a short memo to respond to the questions presented at the end of the case. 1. Is it possible to meet the deadline? Yes, it is possible. The revised schedule has a projected completion date of October 18th, right before MedCON. 2. If so, how would you recommend changing the original schedule (Part A) and why? While crashing activities will reduce project duration, introducing the suggested start-to-start lags has the biggest impact on schedule with the added bonus of no additional costs. The completion date was 11/10/17 after introducing all of the start-to-start lags. Implement: Document design could begin 5 days after the start of the review design. Adjust design could begin 15 days after the start of field test prototypes. Order stock parts could begin 5 days after the start of adjust design. Order custom could begin 5 days after start of adjust design. Training sales representatives could begin 5 days after the start of test unit and completed 5 days after the production of 30 units. To meet the October 25th deadline, three critical activities need to be crashed. Crash activities: Creation of database from 40 days to 35 days at a cost of $35,000. Procure prototype components from 20 days to 15 days at a cost of $30,000. Order stock parts from 15 days to 10 days at a cost of $20,000. These three activities are on the critical path and would reduce the project duration by 15 working days at a cost of $85,000. Neither External specifications, Document design, nor Voice recognition are on the critical path so, crashing these activities would have no impact on project duration. 3. What would the new schedule look like? Students should present a revised schedule similar to the table above. 4. What other factors should be considered before finalizing the schedule? The team needs to review the suggested revisions and make sure they are feasible. When crashing activities, there is always the danger that quality may be compromised; they need to take steps so that this does not happen. The team needs to assess the impact of these changes on the sensitivity of the network. A quick assessment of the slack available to noncritical activities suggests that the sensitivity of the network has not changed significantly, which enhances the chances of the project meeting the deadline. Introducing start-to-start lags may create resource conflicts if the same people are working on both activities. The team needs to check and make sure there are adequate resources available to avoid this problem. Finally, there is no slack between the estimated completion date and MedCON. The team should consider contingency plans if critical activities slip. One plan would be to have funds ready if it is necessary to crash production of the prototypes which occurs at the tail end of the project. Assignment Variation Part A could be assigned after completing Chapter 6 and Part B could be assigned for Chapter 9. Case 9.5 The “Now” Wedding—Part A On December 31 of last year, Lauren burst into the family living room and announced that she and Connor (her college boyfriend) were going to be married. After recovering from the shock, her mother hugged her and asked, “When?” The following conversation resulted: (Rest of case not shown due to length.) This is a great case for a class exercise. Students enthusiastically enter into the case and have fun. The case can be used several ways. First, the case can be used as a small team/group assignment. Second, the case can be used as suggested in the text—as an in-class exercise. Starting with the yellow sticky approach will get you past the many differences of opinion on how the network is developed. Careful reading of the case should bring the group around to the network suggested in the attached exhibits. Development of the original network suggests that the wedding cannot make the January 21 deadline. It would take until February 2 and be too late for Connor’s ship out date of January 30. However, tasks can be shortened at an additional cost and the deadline reached. Activity 8, Order material, can be cut from 8 days to 5 days at a cost of only $20. Activity 11, Sew dress, can be cut from 11 days to 9 days at a cost of $96 ($48x2). Activity 16, Mail invitations, must go out 10 days before the wedding. Reduce Activity 12, Order and receive invitations, from 7 days to 6 days. The cost is $20. Reduce Activity 13, Address invitations, from 3 days to 1 day. The cost is $80 ($40x2). Total costs = $216 The deadlines are met. The wedding can take place January 21 and have a seven-day honeymoon. Connor can leave on January 30. (Note: The project appears to end on January 28 if you look at row one in Project. That is because the honeymoon is included as activity 19. The wedding actually takes place on January 21.) Case 9.6 The “Now” Wedding—Part B Several complications arose during the course of trying to meet the deadline of January 20 for the “Now” Wedding rehearsal. Since Lauren was adamant on having the wedding on January 21 (as was Connor for obvious reasons), the implications of each of these complications had to be assessed. On January 1 the chairman of the Vestry Committee of the church was left unimpressed by the added donation and said he wouldn’t reduce the notice period from 14 to 7 days. Mother comes down with the three-day flu as she starts work on the guest list January 2. Bob’s Printing Shop press was down for one day on January 5 in order to replace faulty brushes in the electric motor. The lace and dress material are lost in transit. Notice of the loss is received on January 10. Can the wedding still take place on January 21? If not, what options are available? Part B presents some unknown risk events. Given these events, students will attempt to shorten Activity 11 and revise durations for Activity 7 and 12. Activity 11, Sew dress. Cut from 9 to 6 days at a cost of $144 ($48 x 3). Activity 7, Guest list. Add 3 days for the flu. Now duration is 7 days. Activity 12, Order and receive invitations. Add 1 day for press breakdown; now the duration is 7 days. Can the wedding take place on January 21? If not what options are available? Conclusion: Deadlines cannot be reached! The notice of the loss of dress material (Activity 8) in transit received January 10 kills the January 21 deadline. Reordering on January 11 will not do it! Although students are disappointed, give them an opportunity to suggest some alternatives. We find three suggestions almost always come up quickly. Buy the material locally and pay the price. Buy a ready-made wedding dress. Elope! Solution Manual for Project Management: The Managerial Process Erik Larson, Clifford F. Gray 9781259666094, 9780078096594
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