What is the minimum cost of crashing the following project that Roger Solano manages at Slippery Rock University by 4 days?
Normal Crash Time
Activity Time (days) (days)
A
6
5
B
8
6
С
5
4
D
6
3
E
6
4
0 000
Normal
Cost
5800
$300
5600
$750
$1.200
Total Cost Immediate
with Crashing Predecessor(s)
$1,100
$600
$650
$1,500
$1.650
Il co
By how many days should each activity be crashed to reduce the project completion time by 4 days? Fill in the table below (Enter your responses as whole numbers)
Each Activity Should be
Reduced BY
Activity
(days)
А
B
С
mo00
D
E
The total cost of crashing the project by 4 days is $(Enter your response as a whole number)

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adioabiola adioabiola · Answer 1 · 2022-02-02T07:53:50+01:00

The minimum cost of crashing the project that Roger Solano manages at Slippery Rock University by 4 days is $1100

Critical path of the project = Longest path of the project = CE

= 4 + 8

= 12 days

Other Paths = AD

= 6 + 5

= 11 days

In order to reduce project time by 4 days , we need to crash

so as to have total project duration to be 8 days

Cost of crashing = Crashing cost of A, C, D and E

= (1000 - 900) + (600 - 500) + (1200 - 900) + (1600-1000)

= 100 + 100 + 300 + 600

= $1100

Cost to crash per period = (Crash Cost - Normal Cost) / (Normal Time - Crash Time)

Cost to crash per period of Activity A = ($1,000 - $900) / (6 - 5)

= $100

Cost to crash per period of Activity C = (600 - 500) / (4 - 3)

= $100

Cost to crash per period of Activity D = (1200 - 900) / (5 - 3)

= $150

Cost to crash per period of Activity E = (1600 - 1000) / (8 - 5)

= $200

Total = $100 + $50 + $100 + $150 + $200

= $600

The total cost of crashing the project by 4 days is $550

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