Question

The list of activities of a data analysis project, their precedence relationships, and their activity times (in number of days) are given in the table below. Activity ID Predecessors Activity Time (Days) A 3 B А 4 с B 2 D с 5 E B 7 F с 2 G EF 2 H D G 3 All numeric answers should be integers. Do NOT use comma in your numeric answers. Find the earliest Start Time (EST), Earliest Completion Time (ECT). Latest Start Time (LST), Latest Completion Time (LCT), and Slack Time of each activity and fill them in the table below. Assume that activity A starts at T = 0. Activity ECT LST LCT Slack Time ID EST А B с D E F H

If you would like to reduce the overall completion time of the project by 1 day, which activity (A, B, C, D, E, F, G, or H) should be crashed?

Answer 1

First I have prepared network Diagram based on the above table:

Network Diagram 9 D 14 21512 7clg 21212 9 31 B 17 4 0 3 7 16 16 Hig ol3 0 16 19 11 11 0 A 3 o o 91F11 3213 12 14 TE/14 0 710 17 14 3 149/16 0/210 16 14

Legend:

Earliest Completion time or ECT=Earliest Start time + Duration Earliest Start time or Activity D EST=highest Earliest finish time of all the predecessor activities Slack=LST-EST Duration Latest start time or LST=Latest completion time - Duration Slack=LCT-ECT Latest completion time= lowest Latest start time of all the follower activities.

Current project completion time is 19 days

ECT Activity ID A B с EST 0 3 7 9 LST 0 3 9 11 7 9 14 LCT 3 7 11 16 Slack 0 0 NINI O E 7 14 7 14 9 F G 11 16 12 14 14 16 3 0 14 H 16 19 6 19 0

we will reduce project completion time by 1 day, hence, we will crash critical path activities and lowest cost activities, here, we can crash A or G, we will crash A as Crash cost per day for activity A=300 that is lowest one, hence, we will crash A

Crash cost per day Activity ID A B Normal Time 3 4 Normal Cost 600 700 Crash time 2 2 Crash Cost 900 1200 1 с D 2 5 500 1000 600 1200 3 E 7 1500 2 4000 (900-600/(3-2)=300 (1200-700)/(4- 2)=250 (600-500)/(2-1)=100 (1200-1000)/(5- 3)=100 (4000-1500)/(7- 2)=500 (4000-2000)/(2- 1)=2000 (1000-600)/(2- 1)=400 (1800-900)/(3- 1)=450 2000 1 4000 N F G 2 600 1 1000 H 3 900 1 1800