# Pert estimate question

Here is the question i am confused how to calculate with the information gave:

A project manager has been asked by the client to meet the promise date of the project. The project manager analyzes the schedule before promising a date to the customer. The project manager uses the program evaluation and review technique to evaluate the project schedule. She decides that based on the PERT calculations she can promise a delivery date of June 30. The expected value of the project completion date is May 30. If the project manager is willing to accept a 5% probability that the project will be delivered later than June 30, what is the standard deviation of the duration of the activities on the critical path? Assume a five-day workweek.

a. Ten days
b. Fifteen days
c. One-half month
d. One month

This i found on the PM question bank, with no explanation.

Thank you all for the helpful info.

### Probability of failure to

Probability of failure to meet the dead line = Probability of slippage=5%=0.05

Probabilty of success to meet the dead line=1-Prob. failure = 1-0.05=95%

95% falls in the 2 sigma region from mean.

We have float of 1 month ( June 30 - May 30).

2 sigma = 1 month

1 sigma = 1/2 month

Chandra

### Pert estimate question

Thank you chandra, i know for sure you have the knowledge.

Question: how did you get the float 30 days?

### Expected Value of

Expected Value of completion= May 30

Promised Delivery Date to the customer: June 30

The difference is the Float

Chandra

### Reg Pert Estimate

Deviation =2 sigma since confidence is 95%

2SD= 4 weeks since (expected date+4 weeks = expected date+ SD*2 )

SD=4 weeks/2 = 2 weeks or 1/2 month so the answer is C

### Standard deviation

Preetham,

I am really having difficulty understanding the standard deviation solution you gave,

Is it possible, can you explain with little more steps.

I am looking at SD = (P-O)/6 as the formula to get the standard deviation.

Thank you very much for the helpful info.

### Look at Normal Distribution

Look at Normal Distribution ( Bell Shaped) curve.  The middle point is your Expected Value (Xbar).  Your data's mean falls at this location and the data's scatter gets on either side of this.  Suppose you want to machine a rod to 50 mm radius and on 10 samples you achieve 50.2, 48.3, 50.1 etc.  Ideally you want to be as close to mean as possible i.e. less scatter, less deviation from the mean and the mean as close as possible to the specification you are asked by customer/design drawing etc to get.  Now suppose your standard deviation is: 0.5.  If all the parts you produced follow the normal distribution, 68% of your measurements be :within 1 SD from mean. (X bar + 1 SD) If you are able to bring 95% of the parts within 50+/- 2 SD, If you are able to achieve 99.99% of parts within Xbar+/-3 SD ( Six Sigma), your process is robust.  Your rejection rate is very very low some 3 defects per 1 million parts you produced.

In this case you allow 5% slippage i.e. you want to meet the scheduled delivery date with a 95% confidence level which is 2 SDs from the mean(Xbar)

So Xbar+2 SD=X bar + 4 weeks ( From May 30 to June 30)

2 SD = 4 weeks

SD=2 weeks.

SD=(P-O)/6 works if you are given Pessimistic estimate and Optimistic Estimate.  Otherwise you have to assume Normal Distribution and use the 5% probability  for failure to meet the dead line.

Chandra

### Thanks a lot Chandra, now i

Thanks a lot Chandra, now i am getting it.

1 SD represents 68%

2 SD represents 95%

3 SD represents 99%

In this case 95%, it is 2SD = 4 weeks

sd = 4 weeks / 2 =  2 weeks.

sd = One half month.

Thank yoo,

Raj

### Imagine that you are in

Imagine that you are in the middle and you are moving to right.  SD is how far you have deviated from the mean .

ML+4 weeks= New estimate-->(1)

ML+2SD = New Estimate -->(2)                 [ 2SD =95% confidence]

Deducing (1) and (2)    --> SD = 2weeks

### I don't remember where I

Preetham, Thanks.

Is it possible can we exchange questions?

My email: miami3456@yahoo.com

Raj

### PERT Estimate

You should solve the Q by considering 2 sigma deviation from the mean.