INTRODUCTION:
INTRODUCTION The critical path provides us the minimum possible project completion time under normal conditions. But sometime, the project manager may want to achieve project completion earlier than the completion time given by the critical path In that case ,the critical path will have to be shortened or reduced. This can be done by reducing completion time of some or all of the critical activities. To achieve this we will need to employ extra recourses . This process of critical path to achieve earlier completion of the project is called project crashing.TIME – COST TRADE OFF:
TIME – COST TRADE OFF Direct cost Indirect costPowerPoint Presentation:
3 TIME-COST RELATIONSHIP Crashing costs increase as project duration decreases Indirect costs increase as project duration increases Reduce project length as long as crashing costs are less than indirect costs Time-Cost Tradeoff cost time Direct cost Indirect cost Total project cost Min total cost = optimal project timePROJECT CRASHING:
PROJECT CRASHING Crashing reducing project time by expending additional resources Crash time an amount of time an activity is reduced Crash cost cost of reducing activity time Normal time -- it is the activity duration under normal circumstances. Normal cost --This is the minimum direct cost required for performing that activityPowerPoint Presentation:
darla/smbs/vit 5 ACTIVITY CRASHING Activity cost Activity time Crashing activity Crash time Crash cost Normal Activity Normal time Normal cost Slope = crash cost per unit timeCOST SLOPE:
COST SLOPE The cost slope of an activity is the increase in direct cost per day of crushing . Formula – Crash cost- Normal cost Cost slope = Cost slope represents the increased in direct cost of an activity if we crash the activity by one day. Normal time – Crash timeOPTIMAL COST AND OPTIMAL TIME:
OPTIMAL COST AND OPTIMAL TIME Optimal cost - It is the point of minimum total cost Optimal time - Optimal project completion time is that duration for which total project cost is minimum. It is the project completion time corresponding with optimal cost.CRASHING PROCEDURE:
CRASHING PROCEDURE Draw the Network diagram Find critical path and Normal project completion time Calculate total normal cost Prepare a table showing for each activity Crashing Cost computation Parallel crashing Continue the crashing procedure till we reach the point of optimal total cost.EXAMPLE:
EXAMPLE Activity 1-2 1-3 2-4 2-5 3-4 4-5 Normal Time (Days) 8 4 2 10 5 3 Normal cost(Rs.) 100 150 50 100 100 80 Crash Time(Days) 6 2 1 5 1 1 Crashed cost(Rs.) 200 350 90 400 200 100 Indirect Cost Rs. 70 per daySOLUTION:
SOLUTION Network Diagram : 1 2 3 4 5 A B 8 4 D 10 c 2 E 5 F 3PowerPoint Presentation:
2. Path in Network : a) A-D : Duration = 18 days b) A-C-F : Duration = 13 days c)B-E-F : Duration = 12 days 18 DaysPowerPoint Presentation:
3. Cost slope and Rank Table : Activity Maximum Crashable limit(days) Cost slope Rs./Day Rank A (1-2) 2 50 4 B(1-3) 2 100 6 C(2-4) 1 40 3 D(2-5) 5 60 5 E(3-4) 4 25 2 F(4-5) 2 10 1 Rank is given in ascending order of cost slope. Minimum cost slope = 10 Rs. /day = Rank 1.PowerPoint Presentation:
4. Crashing : Principle of Crashing: a) Only critical activities are crashed. b) Crashing is started with the critical activity having minimum cost slop i.e minimum rank. c) Critical path should always remain longest. Hence, we will start crashing from critical activity A. Project completion duration is shown in box e.g 18PowerPoint Presentation:
A-D A – C – F B – E – F Direct Indirect Total 13 12 580 1260 1840 13-1 =12 12 580 + (50 x 1) =630 17 x 70 = 1190 1820 12-1 = 11 12 630 + (50 x 1) =680 16 x 70 =1120 1800 11 12 680 + (60 x 1) =740 15 x 70 = 1050 1790 11 12 740 + ( 60 x 1) =800 14 x 70 = 980 1780 11 12 800 + ( 60 x 1) =860 13 x 70 = 910 1770 11 12 860 + (60 x 1) =920 12 x 70 =840 1760 11-1 = 10 12-1 = 920 + ( 60 x 1) =990 11 x 70 = 770 1760 11 12