Greetings Nana,

Here is a solution to your question,

The probability P_{k}_{, }of a frame requiring exactly k transmissions is the probability of the first k - 1 attempts failing, p^{k-1}, times the probability of the k-th transmission succeeding,(1 - p). The mean number of transmission is then just

∞ ∞

∑ kP_{k}_{ = }∑ k(1-p)p^{k-1 = } 1 / (1-p)

k=1 k=1

You can find more information about, how to handle transmission errors here http://research.microsoft.com/en-us/um/people/costa/cn_slides/cn_03-handout.pdf.

**Note**: I would like to inform you that, this forum is for Windows Server 2008 Application compatibility and certification issues.

Hope the above information should help.

Regards,

Harish