necessary for the completion of this project are listed in the following table T
ID: 385995 • Letter: N
Question
necessary for the completion of this project are listed in the following table Total Cost with Normal Time Crash Time Normal Activity (weeks) (weeks) Cost Crashing Immediate Predecessor(s) $2,000 $2,600 $2,100 $2,800 $600 $2.400 $2.700 $1,000 $1,450 S3,400 $4,900 $1,400 $1,900 $600 D, E a) Based on the given information regarding the activities for the project, the project length16 weeks b) The total cost required for completing this project on normal time S 12900 C) For reducing the duration of the project by one week, the activity that should be crashed first is activity D The cost of the project based on the first activity selected for crashing will increase by S 75 d) The maximum weeks by which the project can be reduced by crashingweeksExplanation / Answer
Answer to part D of first question :
The precedence diagram as follows :
The possible paths with their corresponding durations as follows :
A-D-G = 4 + 8 + 4 = 16 weeks
B-E-G = 2 + 6 + 4 = 12 weeks
C-F = 3 + 3 = 6 weeks
The critical path therefore is A-D-Gand the project length ( same as duration of critical path ) is 16 weeks
Maximum to which A-D-G can be crashed to = 3 + 4 + 2 = 9 weeks
Maximum to which B-E-G can be crashed to = 1 + 3 + 2 = 6 weeks
Maximum to which C-F can be crashed to = 3 + 2 = 5 weeks
The critical path ( i.e. path with longest duration ) continues to be A-D-G , but the revised project completion time = 9 weeks
The maximum weeks by which the project can be reduced by crashing9 weeks
Answer to part d of second question :
Expected completion time of the project ( i.e. expected completion time of the critical path ) = m = 20.84 weeks
Variance of the critical path = 5.36 weeks
Standard deviation of the critical path
= Sd
= Square root ( Variance)
= Square root ( 5.36)
= 2.31 days
Let the z value for the probability that the project will befinished in 21 weeks or less = Z1
Therefore ,
M + Z1 x Sd = 21
Or, 20.84 + Z1 x 2.31 = 21
Or, 2.31.Z1 = 0.16
Or, Z1 = 0.16/2.31
Or, Z1 = 0.0692 ( 0.07 rounded to 2 decimal places )
Corresponding probability for Z = 0.07 as per standard normal distribution table = 0.5279
Probability that the critical path will be finished in 21 weeks or less = 0.5279
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