# I need like this answe r Example for answer Direct Costs Activity Slope Maximum Crash time Normal…

I need like this answe r Example for answer Direct Costs Activity Slope Maximum Crash time Normal Time Normal Cost Crash Time Crash Cost A (70-50)/1 = 20 1 3 50 2 70 B (160-80)/2 = 40 2 6 80 4 160 C (90-60)/1 = 30 1 10 60 9 90 D (150-50)/5 = 25 4 11 50 7 150 E (160-100)/2 = 30 2 8 100 6 160 F (70-40)/1 = 30 1 5 40 4 70 G na 0 (cant crash) 6 70 6 70 Initial total direct costs = \$450 Direct costs section of table are given = estimated by accounting Calculation of Slope = (CC-NC)/(NT-CT) Slope is just the cost per time period to reduce the time it may be given directly Calculation of maximum

I need like this answe r Example for answer Direct Costs Activity Slope Maximum Crash time Normal Time Normal Cost Crash Time Crash Cost A (70-50)/1 = 20 1 3 50 2 70 B (160-80)/2 = 40 2 6 80 4 160 C (90-60)/1 = 30 1 10 60 9 90 D (150-50)/5 = 25 4 11 50 7 150 E (160-100)/2 = 30 2 8 100 6 160 F (70-40)/1 = 30 1 5 40 4 70 G na 0 (cant crash) 6 70 6 70 Initial total direct costs = \$450 Direct costs section of table are given = estimated by accounting Calculation of Slope = (CC-NC)/(NT-CT) Slope is just the cost per time period to reduce the time it may be given directly Calculation of maximum crash time = NT-CT This tells how many days each activity can be crashed Initial analysis is to determine critical path with our usual method, given the Normal times Critical path is A-D-F-G, project length of 25 days Crashing Critical activities are crashed to reduce the total project length. They should be crashed in order of cost – least costly to crash (lowest slope) should be crashed first. You should generally do crashes one time period at a time, so you dont miss the emergence of new critical paths and you should redraw the network each time. As you crash critical activities, the slack is reduced in non-critical activities and when new critical paths emerge, you will have to crash both the original and the new paths. Look at slopes for critical activities A, D, F, G = 20, 25, 30, 0 Cant crash G because NT = CT Activity A has the lowest slope = lowest crash cost per day – crash it one day Direct cost is increased by \$20 = \$450+\$20 = \$470, project length is decreased by 1 day Redo the network and critical path with new time on A A cant be crashed any further – check D and F for next lowest slope D=\$25, F = \$30 Reduce D by one day to 10, project length goes to 23 direct cost up by \$25 to \$495 Redraw & resolve the network Now there are two critical paths A-C-F-G and A-D-F-G both must be crashed in order to shorten the project time further We have two ways to crash both paths Crash C and D simultaneously cost would be \$30 + \$25 = \$55 Or Crash F cost would be \$30 The cheapest way to crash now is to crash F, since it is on both paths, both will be crashed Crashing F will take the project to 22 days, an additional direct cost of \$30, total direct costs now up to \$525 and another critical path arises. Now we have three critical paths – the only way to reduce to 21 days is to crash activities D, C and E – all can be crashed the total increase in direct cost is \$25+\$30+\$30 = \$85, for a total direct cost of \$610, project length of 21 After this reduction to 21 days we can no longer crash any activity and reduce the project length so we STOP Summary Costs by Duration Project duration Direct costs Added direct costs from crash Indirect costs Total costs 25 450 400 850 24 450 20 350 820 23 450 25 300 795 22**** 450 30 250 775**** 21 450 85 200 810 Once project direct costs are determined for each possible project duration, we have to determine indirect costs – often there is an assumed time per day for indirect costs, available from accounting department Note in this example: Indirect costs per day appear to be \$50 but this doesnt work if you do 400/25 = 8, obviously there is some minimum value and then \$50 per day is added to that. i.e. the minimum indirect cost for a 20 day project is \$150 any time above that is \$50 per day The book does not explain this issue for this example in table 9.5 but it is showing that we save \$50 per each day the project is shortened When the crash cost per day goes above the savings per day, we should stop. The final crash to 21 days wouldnt make economic sense it saves \$50 but costs \$85. ****Determine optimal project time = 22 days ****Direct costs + Indirect costs = Total costs = \$775

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