Assignment #7

Activity	A	B	C	D	E	F	G	H
Duration	4	10	5	15	12	4	8	7
Early Start	0	0	4	10	10	25	22	30
Early Finish	4	10	9	25	22	29	30	37
Late Start	2	0	6	11	10	26	22	30
Late Finish	6	10	11	26	22	30	30	37
Total Slack	2	0	2	1	0	1	0	0

The critical path is B-E-G-H with duration: 37
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• 107/8
• a)
 

Activity	Optimistic	Most Likely	Pesimistic	Average	variance
A	5	8	11	8	1.00
B	4	8	11	7.83	1.36
C	5	6	7	6	0.11
D	2	4	6	4	0.44
E	4	7	10	7	1.00


Path        Total Expected Time
A-C           8+6 = 14.00
A-D-E     8+4+7 = 19.00
B-E           7.83 + 7 = 14.83

The critical path is A-D-E since it is the longest path. Expected project duration is 19.00 (sum of the expected durations of critical activities)
 

• b) Probablity of completing the project in 21 days: Variance of the project duration is 2.44 (sum of the variances of critical activities)
  
                                                                            

       z = (T - TE)/ standard deviationof project duration  =  (21 - 19) / (2.44^0.5) = 1.28
Assuming normal distribution applies, the probability that the projectcan be completed within 21 days is about 90%.
 

• c)  z = (T - TE)/ standard deviation of project duration  = (17 - 19) / (2.44^0.5) = -1.28 . Probablity of finishing projectwithin 17 days is 10%


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• 108/10




Activity	Start	A	B	C	D	E	F	Finish
Duration	0	5	3	2	5	4	7	0
ES	0	0	0	5	3	8	8	15
EF	0	5	3	7	8	12	15	15
LS	0	4	0	9	3	11	8	15
EF	0	9	3	11	8	15	15	15
Slack	0	4	0	4	0	3	0	0

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The critical Path is Start-> B -> D -> F -> Finish
Project Duration is 15
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• 110/17




   Critical path, using normal durations, is A-D-G with a duration of 29. Direct cost of this schedule is $13,050.
 

	start	A	B	C	D	E	F	G	H	I	Finish
Dur	0	12	13	18	9	12	8	8	2	4	0
ES	0	0	0	0	12	12	13	21	24	18	29
EF	0	12	13	18	21	24	21	29	26	22	29
LS	0	0	6	7	12	15	19	21	27	25	29
LF	0	12	19	25	21	27	27	29	29	29	29
S	0	0	6	7	0	3	6	0	3	7	0

Note: Since the project is rather simple we could have calculated the critical path by enumerating all the paths to select the longest one as the critical path.
A-D-G       29
A-E-H       26
B-F-H       23
C-I            22
Clearly, the path A-D-G must be reduced by four time units, A-E-H one to bring all the paths to within 25 time units.


Cost analysis:
 

Activity	Crash Cost/Day	Maximum Crash
A	600	1
B	112.5	4
C	750	2
D	250	4
E	225	2
F	350	1
G	200	2
H	200	1
I	900	2


 

 

Trial	Crash  Activity	Resulting  Critical Path	Time reduction	Project Duration	Crash Cost
0	-	A-D-G	-	29	-
1	G	A-D-G	2	27	400
2	D, H	A-D-G  A-E-H	2, 1	25	500 +200

Note:  In trial 2 if we only crash D for 2 days the project durationwill go down to 26, since A-E-H has a normal length of 26.  To get the project length down to 25, A-E-H must be shortened to 25 days-- the cheapest activity among A,E,H is H which is crashed for 1 day addinga cost of $200. Thus the project cost is now 13,050 + $700 +400 = $14,150
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