Question

5. A competitor of Kozar International, Inc. has begun marketing a new instant developing film project. Kozar has had a similar product under study in its R&D department but has not yet been able to begin production. Because of the competitor's action, top managers have asked for a speedup of R&D activities so that Kozar can produce and market instant film at the earliest possible date. The predecessor information and activity time estimates in months are shown in the next column. Assume that the project completion time is distributed normally Activity Pessimistic Time Immediate Predecessor A B с D E F G H Optimistic Time (weeks) 5 6 3 11 8 3 4 5 Most Probable Time 7 8 4 17 10 4 8 7 А B CE DE F 8 12 5 25 12 5 9 9 MGMT 3106: Management Science and Operations Management a. Calculate the expected time and variance for each activity. b. Develop an activity schedule for this project using early and late start and finish times, compute activity slack time, and define the critical activities. c. What is the probability of finishing the project in 30 weeks? (Show all calculations. "Guesses" are not accepted as an answer.) d. What is the probability of finishing the project in 35 weeks? (Show all calculations. "Guesses" are not accepted as an answer.)

Answer 1

Solution:
Expected time of each activity,

Activity	to	tm	tp	te=to+4⋅tm+tp/6	σ2=(tp-to/6)2
A	5	7	8	6.83	0.25
B	6	8	12	8.33	1
C	3	4	5	4	0.11
D	11	17	25	17.33	5.44
E	8	10	12	10	0.44
F	3	4	5	4	0.11
G	4	8	9	7.5	0.69
H	5	7	9	7	0.44

The earliest and latest expected time for each activity is calculated by considering the expected time te

Activity	Immediate Predecessors	Duration
A	-	6.83
B	-	8.33
C	-	4
D	A	17.33
E	B	10
F	C,E	4
G	D,E	7.5
H	F	7

Edge and it's preceded and succeeded node

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

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

2 D(17.33) 5 G(7.5) 8 A(6.83) d(0) H(7) B(8.33) 3 E(10) 6 H(7) C(4) do) H(7) C(4) C(4) 4 F(4) 7

Forward Pass Method
E1=0

E2=E1+t1,2 [t1,2=A=6.83]=0+6.83=6.83

E3=E1+t1,3 [t1,3=B=8.33]=0+8.33=8.33

E4=Max{Ei+ti,4}[i=1,6]

=Max{E1+t1,4;E6+t6,4}

=Max{0+4;18.33+0}

=Max{4;18.33}

=18.33

E5=Max{Ei+ti,5}[i=2,6]

=Max{E2+t2,5;E6+t6,5}

=Max{6.83+17.33;18.33+0}

=Max{24.16;18.33}

=24.16

E6=E3+t3,6 [t3,6=E=10]=8.33+10=18.33

E7=E4+t4,7 [t4,7=F=4]=18.33+4=22.33

E8=Max{Ei+ti,8}[i=5,7]

=Max{E5+t5,8;E7+t7,8}

=Max{24.16+7.5;22.33+7}

=Max{31.66;29.33}

=31.66


Backward Pass Method
L8=E8=31.66

L7=L8-t7,8 [t7,8=H=7]=31.66-7=24.66

L6=Min{Lj-t6,j}[j=5,4]

=Min{L5-t6,5;L4-t6,4}

=Min{24.16-0;20.66-0}

=Min{24.16;20.66}

=20.66

L5=L8-t5,8 [t5,8=G=7.5]=31.66-7.5=24.16

L4=L7-t4,7 [t4,7=F=4]=24.66-4=20.66

L3=L6-t3,6 [t3,6=E=10]=20.66-10=21.66

L2=L5-t2,5 [t2,5=D=17.33]=24.16-17.33=6.83

L1=Min{Lj-t1,j}[j=4,3,2]

=Min{L4-t1,4;L3-t1,3;L2-t1,2}

=Min{20.66-4;21.66-8.33;6.83-6.83}

=Min{16.66;13.33;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 is : 1-2-5-8 and critical activities are A,D,G

The total project time is 31.66
The network diagram for the project, along with E-values and L-values, is

G(7.5) 8 Eg = 31.66 Lg = 31.66 2 D(17.33) 5 E2 = 6.83 E5 = 24.16 22 = 6.83 L = 24.16 d(0) A(6.83) H(7) 1 B(8.33) 6 F = 18.33 26 = 20.66. H(7) F1 = 0 21 OI 3 E(10) Ez = 8.33 E 4 = 18.33 23 = 21.66 24 = 20.66 1 1 C(4) d(0) H7) = C(4) C(4) 4 F(4) 7 E = 22.33 27 = 24.66

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