Efficiency of Stop-and-Wait, Selective Reject and Go-Back-N ARQ when error probability is p

Question

Hi Guys,

Do you know formula for efficiency of  Stop-and-Wait, Selective Reject and Go-Back-N ARQ when error probability is p ?

PS: Although i am mentioning the link for answer but if you want to add some additional info.(or some more elegant proof for the formula)  then it will be really helpful for all readers. :)

Comments

Please refer ->

Check slide 11 - 14 (Notice efficiency can not be greater then 1. :)

https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0ahUKEwjB89CiiIDYAhXDL48KHbbsDEgQFghCMAQ&url=http%3A%2F%2Fwww.cs.nyu.edu%2Fcourses%2Ffall05%2FG22.2262-001%2Flectures%2Flec_5.ppt&usg=AOvVaw1OTCMST-xpUOywtbmBqOWM

http://www.sce.carleton.ca/faculty/lambadaris/courses/521/homeworks/solutions_2.pdf

http://www.eecs.yorku.ca/course_archive/2010-11/F/3213/CSE3213_11_FlowErrorControl_F2010.pdf

Mainly for G-back-N proof -->

Efficiency is defined as the ratio of the time required to transmit a packet, TRANSP, to the average time it takes for a packet to be transmitted, E(T),

Efficiency = TRANSP/ E(T)

From the above description we can calculate E(T) as follows:

E(T) = (1 − p)TRANSP + p(time-out + E(T)).

Answer 1

Let 'P' be the ERROR Probability

So (1-P) means no error

Efficiency in STOP n WAIT without error = 1 / (1+2a)

Efficiency in STOP n WAIT with error = (1-P) / (1+2a)

Efficiency in SR without error = W / (1+2a)

Efficiency in SR with error = W (1-P) / (1+2a)

Efficiency in GB-N without error = W / (1+2a)

Efficiency in GB-N with error = W (1-P) / (1+2a)                          if W>=(1+2a)

Efficiency in GB-N with error = W (1-P) / (1+2a) (1 - P + W*P)   if W<(1+2a)

Comments

What is w, window size ? And source of your formula ,than you !