Project Management Time-Cost Model Tutorial Objectives and Information Step 1. Normal duration and cost Step 2. What to crash Step 3. Crash the activity Step 3. Repeat Step 4. The all crash duration and cost Step 5. The tradeoff graph Developed by: Alex J. Ruiz-Torres, Ph.D.

Objectives and Information

Objectives and Information The goal of this process is the development of alternative project plans. Each project plan has a different cost and duration. As the project duration is reduced the cost increases. The idea is similar to the options for shipping via UPS,.. 5 days costs $10, 3 days $16, 2 days costs $22, and overnight $40.

Objectives and Information The project information that must be already defined: Set of activities and precedence relationships. Activity duration when its performed at its “normal” speed. Called the Normal Time (NT) Shortest possible duration. Called the Crash Time (CT) Activity cost when performed at its normal speed. Called the Normal Cost (NC) Activity cost when performed at its shortest duration. Called the Crash Cost (CC)

Objectives and Information EXAMPLE   Months Millions $ CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e

Step 1. Normal Duration and Cost

Step 1. Normal Duration and Cost We start by considering all activities are performed at their normal time. No activity is being expedited/ crashed.   Months Millions $ CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e

Step 1. Normal Duration and Cost   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We determine the duration and CP for the project with all NT’s 4 9 9 17 0 4 b d 17 20 a f c e Duration = 20 Months CP = a – b – d – f 9 15 11 17 4 8 5 9

Step 1. Normal Duration and Cost   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e 4 9 9 17 0 4 b d 17 20 a f Given all activities are at their NT, the total cost is the sum of Normal Costs = $36 Million c e 9 15 11 17 4 8 5 9

Step 1. Normal Duration and Cost Normal Case Current Duration = 20 Months Current Costs = $36 Million

Step 2. What to Crash?

Step 2. What to Crash? Objective: to determine what activity to crash in order to reduce the project’s duration. This is done one time unit at a time. In this example, one month at a time. For each activity we calculate the crash rate CR = (CC – NC) / (NT – CT) The crash rate is the cost to reduce that activity by one time unit For this example = $million/reduce one month

Step 2. What to Crash? crash rate CR = (CC – NC) / (NT – CT) Months   Months Millions $ $M/mo. CT NT NC CC Rate Pred a 2 4 8 18 5 b 3 13 c 9.1 1.1 d 7 6 8.5 2.5 b,c e 14 3.5 f 1 d,e

Step 2. What to Crash?   mo.  mo. $M $M  $M/mo. CT NT NC CC Rate Pred a 2 4 8 18 5 b 3 13 c 9.1 1.1 d 7 6 8.5 2.5 b,c e 14 3.5 f 1 d,e Activity c has the smallest rate (therefore smallest cost increase), BUT This activity is not in the CP. Therefore reducing the time of activity c will not reduce the project’s duration

Step 2. What to Crash? b d a f c e If the duration of activity c is three months (instead of 4 as the original ), the project still takes 20 months 4 9 9 17 0 4 b d 17 20 a f c e 9 15 11 17 4 7 6 9 new duration = 3

Step 2. What to Crash? We will crash the activity in the CP with the smallest rate. Activity d. CP = a – b – d – f SMALLEST   mo.  mo. $M $M  $M/mo. CT NT NC CC Rate Pred a 2 4 8 18 5 b 3 13 c 9.1 1.1 d 7 6 8.5 2.5 b,c e 14 3.5 f 1 d,e

Step 3. Crash the activity

Step 3. Crash the activity The duration of Activity d is reduced by one time unit duration = 7 months (instead of 8) We will next determine the project’s completion time.

Step 3. Crash the activity   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We calculate the completion time and critical path with d = 7 9 9 16 new duration = 7 0 4 b d 19 a f Duration = 19 Months Thus crashing d by one time unit reduced the project by one time unit. c e 15 8

Step 3. Crash the activity When expediting activity d by one month the project cost increases by the crash rate for that activity Therefore the project cost at 19 months is $38.5 Million (36 + 2.5)

Step 4. Repeat This means steps 2 – 3 will be repeated X times or until no alternatives are available. Back to step 2. What to crash?

Step 2. What to Crash? b d a f c e CP = a – b – d – f (The CP did not   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We need to determine the critical path 4 9 9 16 0 4 b d 16 19 a f c e CP = a – b – d – f (The CP did not CHANGE) 9 15 10 16 4 8 5 9

Step 2. What to Crash? The critical path is CP = a – b – d – f d cannot be crashed again. Is at its minimum possible duration (already at 7 months). The CT is the smallest possible duration.. Next smallest is b. available LOWEST   mo.  mo. $M $M  $M/mo. CT NT NC CC Rate Pred a 2 4 8 18 5 b 3 13 c 9.1 1.1 d 7 6 8.5 2.5 b,c e 14 3.5 f 1 d,e

Step 3. Crash the activity   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We calculate the project’s duration with b = 4 months 4 8 8 15 0 4 b d 15 18 a f c e Duration = 18 Months 8 14 8

Step 3. Crash the activity Activity b had a crash rate of 4 Reducing the time of an activity increases the cost of the project by its crash rate. Project cost at 18 Months is 38.5 + 4 = $42.5 Million

Step 4. Repeat Back to step 2. What to crash?

Step 2. What to Crash? b d a f c e We determine the critical path   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We determine the critical path 4 8 8 15 all activities but e have 0 slack 0 4 b d 15 18 a f Duration = 18 Months CP = a – b – d – f CP = a – c – d – f (we now have 2 CPs) c e 8 14 9 15 4 8

Step 2. What to Crash? We now have 2 CPs CP = a – b – d – f CP = a – c – d – f To reduce the project to 17 months we could do one of the following Crash a Crash both b and c Crash f Lowest cost = a   $M/mo. Rate Pred a 5 b 4 c 1.1 d 2.5 b,c e 3.5 f 6 d,e

Step 2. What to Crash? b d a f c e Why not just c? If we only crashed c, the project’s completion time would not be reduced. Same if we only crashed b. They are in parallel, so both must be reduced simulateneosly to change the project’s completion time. 4 8 8 15 0 4 b d 15 18 a f CP = a – b – d – f CP = a – c – d – f c e 8 14 9 15 4 7 5 8 29

Step 3. Crash the activity Activity a is selected. Now has a duration of 3 months. 3 7 7 14 0 3 b d 14 17 a f c e Duration = 17 Months 7 13 3 7 30

Step 3. Crash the activity Activity a had a crash rate of 5 Reducing the time of an activity increases the cost of the project by its crash rate. Project cost at 17 Months is 42.5 + 5 = $47.5 Million Step 4.We will stop here, X times = 3 times

Step 5. The all crash duration and cost

Step 5. The all crash duration and cost   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e We calculate the completion time and critical path with all durations at crash time 2 5 5 12 0 2 b d 12 13 a f c e Duration = 13 Months 5 9 8 12 2 5

Step 5. The all crash duration and cost   CT NT NC CC Pred a 2 4 8 18 b 3 5 13 c 9.1 d 7 6 8.5 b,c e 14 f 1 d,e As we assumed all the activities are crashed to their maximum, the project’s cost is the sum all the crash costs. Project Cost = $76.6 Million. This is a rough estimate. For example, activity e can be un-crashed back to 6 months and the project would still have a duration of 13 months.

Step 6. The Tradeoff Plot

Step 6. The Tradeoff Plot Alternatives Generated Duration Cost 20 36 19 38.5 18 42.5 17 47.5 13 76.6 Alternatives Generated