Decision Analysis
Case Problem RC Coleman
Activity Description Immediate Predecessor
A Determine equipment needs -
B Obtain vendor proposals -
C Select vendor A, B
D Order System C
E Design new warehouse layout C
F Design warehouse E
G Design computer interface C
H Interface computer D,F,G
I Install system D,F
J Train system operators H
K Test system I,J
(a) (m) (b) Time
Activity Optimistic Most Probable Pessimistic
A 4 6 8
B 6 8 16
C 2 4 6
D 8 10 24
E 7 10 13
F 4 6 8
G 4 6 20
H 4 6 8
I 4 6 14
J 3 4 5
K 2 4 6
Where mean (expected time), t = (a + 4m + b) / 6
tA = (4 + 4(6) + 8) / 6 = 6 tB = (6 + 4(8) + 16) / 6 = 9 tC = (2 + 4(4) + 6) / 6 = 4 tD = (8 + 4(10) + 24) / 6 = 12 tE = (7 + 4(10) + 13) / 6 = 10 tF = (4 + 4(6) + 8) / 6 = 6 tG = (4 + 4(6) + 20) / 6 = 8 tH = (4 + 4(6) + 8) / 6 = 6 tI = (4 + 4(6) + 14) / 6 = 7 tJ = (3 + 4(4) + 5) / 6 = 4 tK = (2 + 4(4) + 6) / 6 = 4
for Network Diagram, refer to attachment.
Critical Path for activities = B, C, E, F, H, J, K = 43
Question 1: R.C. Coleman’s top management established a required 40 week completion time for the project. Can this completion be achieved? Include probability information in your discussion. What recommendations do you have if the 40 week completion time is required?
To calculate the probability to complete in 40 weeks:-
Z = (T –TE) / ((v)^2)^0.5
V = ((b-a)/6)^2
vB = ((16 – 6) / 6)^2 = 2.77
vC = ((6-2)/6)^2 = 0.44
vE = ((13-7)/6)^2 = 1
vF = ((8-4)/6)^2 = 0.44
vH = ((8-4)/6)^2 = 0.44
vJ = ((5-3)/6)^2 = 0.11
vK = ((6-2)/6)^2 = 0.44
Z = 3 / ((2.77)+(0.44)+(1)+(0.44)+(0.44)+(0.11)+(0.44))^0.5
= 3 / (5.64)^0.5
= 3 / 2.374
= 1.26
From the Normal Distribution Chart:-
Z (1.26) = 0.8962
= 89.62 %
Therefore; the probability
Activity Description Immediate Predecessor
A Determine equipment needs -
B Obtain vendor proposals -
C Select vendor A, B
D Order System C
E Design new warehouse layout C
F Design warehouse E
G Design computer interface C
H Interface computer D,F,G
I Install system D,F
J Train system operators H
K Test system I,J
(a) (m) (b) Time
Activity Optimistic Most Probable Pessimistic
A 4 6 8
B 6 8 16
C 2 4 6
D 8 10 24
E 7 10 13
F 4 6 8
G 4 6 20
H 4 6 8
I 4 6 14
J 3 4 5
K 2 4 6
Where mean (expected time), t = (a + 4m + b) / 6
tA = (4 + 4(6) + 8) / 6 = 6 tB = (6 + 4(8) + 16) / 6 = 9 tC = (2 + 4(4) + 6) / 6 = 4 tD = (8 + 4(10) + 24) / 6 = 12 tE = (7 + 4(10) + 13) / 6 = 10 tF = (4 + 4(6) + 8) / 6 = 6 tG = (4 + 4(6) + 20) / 6 = 8 tH = (4 + 4(6) + 8) / 6 = 6 tI = (4 + 4(6) + 14) / 6 = 7 tJ = (3 + 4(4) + 5) / 6 = 4 tK = (2 + 4(4) + 6) / 6 = 4
for Network Diagram, refer to attachment.
Critical Path for activities = B, C, E, F, H, J, K = 43
Question 1: R.C. Coleman’s top management established a required 40 week completion time for the project. Can this completion be achieved? Include probability information in your discussion. What recommendations do you have if the 40 week completion time is required?
To calculate the probability to complete in 40 weeks:-
Z = (T –TE) / ((v)^2)^0.5
V = ((b-a)/6)^2
vB = ((16 – 6) / 6)^2 = 2.77
vC = ((6-2)/6)^2 = 0.44
vE = ((13-7)/6)^2 = 1
vF = ((8-4)/6)^2 = 0.44
vH = ((8-4)/6)^2 = 0.44
vJ = ((5-3)/6)^2 = 0.11
vK = ((6-2)/6)^2 = 0.44
Z = 3 / ((2.77)+(0.44)+(1)+(0.44)+(0.44)+(0.11)+(0.44))^0.5
= 3 / (5.64)^0.5
= 3 / 2.374
= 1.26
From the Normal Distribution Chart:-
Z (1.26) = 0.8962
= 89.62 %
Therefore; the probability