Question

You are a junior manager with expertise in project management and a member of a project team of a company that just won the bid to develop a new internet banking management system for a major bank in South Africa. The project will entail the following activities along with their respective precedence, duration and cost.

Activity Code	Activity	Immediate Predecessor	Normal Time(Weeks)	Crashing Time(weeks)	Normal Cost	Crashing cost
1	A	–	12	8	$100000	$104000
2	B	A	17	15	90000	140000
3	C	–	11	7	44000	60000
4	D	C	17	14	220000	280000
5	E	C	18	16	230000	256000
6	F	C	12	10	14000	20000
7	G	D,E	15	13	264000	312000
8	H	F	10	10	90000	–
9	I	H	12	11	300000	330000
10	J	B	16	14	130000	175000
11	K	D,B	15	13	125000	180000

The project requires a crane that rents for $1300 per week and a crew of general labour that cost $5200 per week. Ahead of the next project team meeting, the project manager has tasked you to develop the project schedule along with relevant analysis on possible cost savings related to completing the project ahead of schedule.

For the questions that follow, use the activity codes in place of the activity. For instance, activity B should be written as activity 2. That is, 2 stands for B. Thus, a critical path of say, CBJ must be written as 3210 in the space provided.

Note: All answers must be in numeric form, This is why activities are given number codes.

i. The latest start time for activity G in order to complete the project on time is week  Blank 1. Fill in the blank, read surrounding text. .

ii. The number of weeks for which activity J could be delayed without delaying the project is  Blank 2. Fill in the blank, read surrounding text. weeks.

iii. The project completion time is  Blank 3. Fill in the blank, read surrounding text. weeks.

iv. The critical path of the project (in terms of the activity codes, and in the order from left to right) is  Blank 4. Fill in the blank, read surrounding text. . Note: In case of multiple critical paths, give only the one for which the sum of the codes for the activities is the minimum.

v. The total cost of the project within the normal time frame is $  Blank 5. Fill in the blank, read surrounding text.

vi. If the project manager would want to save cost by taking advantage of the weekly indirect cost, then the earliest permissible time the project would be completed is in  Blank 6. Fill in the blank, read surrounding text. weeks.

vii. The total cost of the project if the project manager undertakes the action in question (vi) is   $  Blank 7. Fill in the blank, read surrounding text. .

viii. If the project has to be completed in 41 weeks, then the activity/activities to crash when on week 42 in order to achieve the 41 weeks (using activity codes, and listing the smallest first) is  Blank 8. Fill in the blank, read surrounding text. . For example, if the activities to crash to get to week 40 when on week 41 is C,I, provide the answer as 39.

ix. The minimum total cost of the project if it has to be completed in 41 weeks through crashing is $  Blank 9. Fill in the blank, read surrounding text. .

Answer 1

Answer:

Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost

A	-	12	100000	8	104000
B	A	17	90000	15	140000
C	-	11	44000	7	60000
D	C	17	220000	14	280000
E	C	18	230000	16	256000
F	C	12	14000	10	20000
G	D,E	15	264000	13	312000
H	F	10	90000	10	0
I	H	12	300000	11	330000
J	B	16	130000	14	175000
K	D,B	15	125000	13	180000

Indirect cost = 6500

Solution:
The given problem is

Activity	Immediate Predecessors	Normal Time	Normal Cost	Crash Time	Crash Cost
A	-	12	100000	8	104000
B	A	17	90000	15	140000
C	-	11	44000	7	60000
D	C	17	220000	14	280000
E	C	18	230000	16	256000
F	C	12	14000	10	20000
G	D,E	15	264000	13	312000
H	F	10	90000	10	0
I	H	12	300000	11	330000
J	B	16	130000	14	175000
K	D,B	15	125000	13	180000

Edge and it's preceded and succeeded node

Edge	Node1 → Node2
A	1→2
C	1→3
B	2→4
D	3→5
E	3→6
F	3→7
d	4→5
J	4→9
d	5→6
K	5→9
G	6→9
H	7→8
I	8→9

The network diagram for the project, along with activity time, is

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=12]=0+12=12

E3=E1+t1,3 [t1,3=C=11]=0+11=11

E4=E2+t2,4 [t2,4=B=17]=12+17=29

E5=Max{Ei+ti,5}[i=3,4]

=Max{E3+t3,5;E4+t4,5}

=Max{11+17;29+0}

=Max{28;29}

=29

E6=Max{Ei+ti,6}[i=3,5]

=Max{E3+t3,6;E5+t5,6}

=Max{11+18;29+0}

=Max{29;29}

=29

E7=E3+t3,7 [t3,7=F=12]=11+12=23

E8=E7+t7,8 [t7,8=H=10]=23+10=33

E9=Max{Ei+ti,9}[i=4,5,6,8]

=Max{E4+t4,9;E5+t5,9;E6+t6,9;E8+t8,9}

=Max{29+16;29+15;29+15;33+12}

=Max{45;44;44;45}

=45

Backward Pass Method
L9=E9=45

L8=L9-t8,9 [t8,9=I=12]=45-12=33

L7=L8-t7,8 [t7,8=H=10]=33-10=23

L6=L9-t6,9 [t6,9=G=15]=45-15=30

L5=Min{Lj-t5,j}[j=9,6]

=Min{L9-t5,9;L6-t5,6}

=Min{45-15;30-0}

=Min{30;30}

=30

L4=Min{Lj-t4,j}[j=9,5]

=Min{L9-t4,9;L5-t4,5}

=Min{45-16;30-0}

=Min{29;30}

=29

L3=Min{Lj-t3,j}[j=7,6,5]

=Min{L7-t3,7;L6-t3,6;L5-t3,5}

=Min{23-12;30-18;30-17}

=Min{11;12;13}

=11

L2=L4-t2,4 [t2,4=B=17]=29-17=12

L1=Min{Lj-t1,j}[j=3,2]

=Min{L3-t1,3;L2-t1,2}

=Min{11-11;12-12}

=Min{0;0}

=0

The critical path in the network diagram has been shown. This has been done by double lines by joining all those events where E-values and L-values are equal.
The critical path of the project are :
(1) 1-2-4-9 and critical activities : A,B,J

(2) 1-3-7-8-9 and critical activities : C,F,H,I

The total project time is 45
The network diagram for the project, along with E-values and L-values, i

Critical activity	Crash cost per week (Rs)
A (1 - 2)	104000-10000012-8=1000
C (1 - 3)	140000-9000011-15=-12500
B (2 - 4)	60000-4400017-7=1600
D (3 - 5)	280000-22000017-14=20000
E (3 - 6)	256000-23000018-16=13000
F (3 - 7)	20000-1400012-10=3000
d (4 - 5)	-
J (4 - 9)	175000-13000016-14=22500
d (5 - 6)	-
K (5 - 9)	180000-12500015-13=27500
G (6 - 9)	312000-26400015-13=24000
H (7 - 8)	-
I (8 - 9)	330000-30000012-11=30000

Total cost = Direct normal cost + Indirect cost for 45 weeks
=1607000+45×6500=1899500


The crashing activity J in the path 1-2-4-9 (A,B,J), activity I in the path 1-3-7-8-9 (C,F,H,I), each by 1 week

E-values and L-values for next crashed network

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=12]=0+12=12

E3=E1+t1,3 [t1,3=C=11]=0+11=11

E4=E2+t2,4 [t2,4=B=17]=12+17=29

E5=Max{Ei+ti,5}[i=3,4]

=Max{E3+t3,5;E4+t4,5}

=Max{11+17;29+0}

=Max{28;29}

=29

E6=Max{Ei+ti,6}[i=3,5]

=Max{E3+t3,6;E5+t5,6}

=Max{11+18;29+0}

=Max{29;29}

=29

E7=E3+t3,7 [t3,7=F=12]=11+12=23

E8=E7+t7,8 [t7,8=H=10]=23+10=33

E9=Max{Ei+ti,9}[i=4,5,6,8]

=Max{E4+t4,9;E5+t5,9;E6+t6,9;E8+t8,9}

=Max{29+15;29+15;29+15;33+11}

=Max{44;44;44;44}

=44

Backward Pass Method
L9=E9=44

L8=L9-t8,9 [t8,9=I=11]=44-11=33

L7=L8-t7,8 [t7,8=H=10]=33-10=23

L6=L9-t6,9 [t6,9=G=15]=44-15=29

L5=Min{Lj-t5,j}[j=9,6]

=Min{L9-t5,9;L6-t5,6}

=Min{44-15;29-0}

=Min{29;29}

=29

L4=Min{Lj-t4,j}[j=9,5]

=Min{L9-t4,9;L5-t4,5}

=Min{44-15;29-0}

=Min{29;29}

=29

L3=Min{Lj-t3,j}[j=7,6,5]

=Min{L7-t3,7;L6-t3,6;L5-t3,5}

=Min{23-12;29-18;29-17}

=Min{11;11;12}

=11

L2=L4-t2,4 [t2,4=B=17]=29-17=12

L1=Min{Lj-t1,j}[j=3,2]

=Min{L3-t1,3;L2-t1,2}

=Min{11-11;12-12}

=Min{0;0}

=0

The critical path in the network diagram has been shown. This has been done by double lines by joining all those events where E-values and L-values are equal.
The critical path of the project are :
(1) 1-2-4-5-6-9 and critical activities : A,B,d,d,G

(2) 1-2-4-5-9 and critical activities : A,B,d,K

(3) 1-2-4-9 and critical activities : A,B,J

(4) 1-3-6-9 and critical activities : C,E,G

(5) 1-3-7-8-9 and critical activities : C,F,H,I

The total project time is 44
The network diagram for the project, along with E-values and L-values, is

Total cost = Direct normal cost + Indirect cost for 44 weeks
=1607000+1×22500+1×30000+44×6500=1945500

Since total project cost for 44 weeks is more than the cost for 45 weeks. So further crashing is not desirable.
Hence, project optimum time is 45 weeks and cost is 1899500.

Crashing schedule of project

Project duration	Crashing activity and time	Direct Normal Cost	Direct Crashing Cost	Total (Normal + Crashing)	Indirect Cost	Total Cost
45		1607000		1607000	45×6500=292500	1899500
44	J;I=1	1607000	1×22500+1×30000=52500	1659500	44×6500=286000	1945500

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