3.19) Three activities are candidates for crashing on a project network for a la
ID: 368288 • Letter: 3
Question
3.19) Three activities are candidates for crashing on a project network for a large computer installation (all are, of course, critical). Activity details are in the following table: a) What action would you take to reduce the critical path by 1 day? b) Assuming no other paths become critical, what action would you take to reduce the critical path one additional day? c) What is the total cost of the 2- day reduction? PREDE- NORMAL NORMAL C CRASH CRASH ACTIVITY CESSOR TIME COST TIME COST 7 days 6,000 6 days $6,600 4 days ,200 2 days 3,000 11 days4,000 9 days 6,000Explanation / Answer
Cost of Crashing per day for activity A
= ( Crash cost – Normal cost ) / ( Normal time – Crash time )
= ( $6600 - $ 6000) / ( 7 – 6 )
= $600
Cost of Crashing per day for activity B
= ( Crash cost – Normal cost ) / ( Normal time – Crash time )
= ( $3000 - $1200) / ( 4 -2 )
= $900
Cost of Crashing per day for activity C
= ( Crash cost – Normal cost ) / ( Normal time – Crash time )
= ( $6000 - $4000 ) / ( 11 – 9 )
= $ 1000
Answer to question a :
Since cost of crashing per day is least for activity A, activity A should be crashed by 1 day to reduce the critical path by 1 day
Answer to question b:
After crashing of activity A by 1 day, the next activity which has 2nd lowest cost of crashing per day is activity B at $900 / day
Therefore , activity B should be crashed by 1 day to reduce the critical path one additional day
Answer to question c :
Total cost of 2 day reduction ( as mentioned in answer #a and answer # b)
= Daly crashing cost of activity A + Daily crashing cost of activity B
= $ 600 + $900
= $1500
TOTAL COST OF 2 DAY REDUCTION = $1500
TOTAL COST OF 2 DAY REDUCTION = $1500
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