Question

Problem 17-7 Three recent college graduates have formed a partnership and have opened an advertising firm. Their first project consists of activities listed in the following table. Use Table B TIME IN DAYS Immediate Activity Predecessor Optimistic Most Likely Pessimistic 12 15 End E, G, I b.What is the probability that the project can be completed in 24 days or less? In 21 days or less? (Round your te and z values to 2 decimal places and "Standard deviation" to 3 decimal places. Round your final answers to 4 decimal places.) Probability Days 24 days or less 21 days or less 0.9686 c. Suppose it is now the end of the seventh day and that activities A and B have been completed while activity D is 50 percent completed. Time estimates for the completion of activity D are 5, 6, and 7. Activities C and H are ready to begin. Determine the probability of finishing the project by day 24 and the probability of finishing by day 21. (Round your intermediate calculations to 3 decimal places and final answers to 4 decimal places.) Probability Day 24 Day 21

Answer 1

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Please note the to is optimistic time

tm is most likely time

tp is pessimistic time

te is expected time and is calculated by the formula = \frac{t_{o}+4*t_{m}+t_{p}}{6}

and the standard deviation is calculated by = \frac{t_{p}-t_{o}}{6}

Variance = standard deviation ^2

It is calculated for each activity

Then we draw the network diagram of the project and as from the diagram it is evident that there are three paths to complete the project

A-C-E

B-H-I

D-F-G

Calculate the expected time of each path by adding the expected time of each path activity and variance by adding the variance

(** we cant add standard deviation simply to get the resultant we need a variance to add and then take the square root to calculate resultant standard deviation)

A-C-E

te= 6+8.17+6.33=20.50

variance = 0.111+0.694+0.444=1.249

St dev= sqrt(1.249) = 1.117

Path B-H-I

te= 8.50+4.17+6.83 = 19.50

variance = 0.25+0.028+0.25=0.528

std dev=0.726

Path D-F-G

te = 12+6+3.5= 21.50

variance = 1+0.111+0.694=1.805

std dev= 1.343

now we have to calculate the z values for each path and from the normal table we can get the probability

z=\frac{x-\mu}{\sigma}
for 24 days x=24

path A-C-E = z=\frac{24-20.5}{1.117} = 3.13 probability = NORMSDIST(3.13,TRUE) = 1.00

path B-H-I = z=\frac{24-19.5}{0.726} = 6.19 probabilty = 1.00

path D-F-G = z=\frac{24-21.5}{1.343} =1.86 probability = 0.97

So probability of fininshing the oriject 24 days or less = 1.00*1.00*0.97 = 0.9686

similarly for 21 days

path -A-C-E = z=\frac{21-20.5}{1.117} = 0.446 = prbobability = 0.67

path B-H-I = z= 2.06 prob = 0.98

path D-F-G = z= -0.37 prob= 0.36

so probability of completion 21 days or less = 0.67*0.98*0.36= 0.2363

***************************************************************************************************

A and B is completed by 7 th day so there expected time is 7

C and H are having no changes only D is

pathA- C-E

expected time= 7+8.17+6.33 = 21.5

variance = 41/36 = 1.138

path D-F-G

only D is changed its expected time will be = (5+4*6+7)/6 = 6 and variance is 0.111

expeted time for this path = 15.50

total duration = 7+15.5 = 22.5

variance = 33/36 = 0.916

(33 comes from addition of (tp-to)^2)

similarly for path B-H-I

total time =7+11= 18

variance= 10/36 = 0.278

Probabilty calcualtion for 24 days

path-A-C-E = z=\frac{24-21.5}{1.067} =0.99

path D-F-G = 0.94

path B-H-I = 1.00

so total probability = 0.99*0.94*1.00 =0.9306

similarly for 21 days or less

path A-C-E = 0.3192

path-D-F-G = 0.0582

pathB-H-I = 1.00

so total probability = 0.3192*0.0582*1= 0.0186

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