# Stop and Wait retransmission of packets?

In a communication link out of p packets one packet will be lost. If stop and wait protocol is used then expected number of retransmissions for a packet?

(A) P/(1-P)

(B) P

(C) 1/(1-P)

(D) 1/P

I am getting option a) as answer. Anyway to solve this?

• Is P same as the number of p packets? If so, it doesn't make sense since 1-P is a negative number. Nov 3 '18 at 15:02
• No, P is total packets and p packets out of these P we are sending! Nov 3 '18 at 15:52
• I become more confused after your reply to @BanghuaZhao. Either I am crazy or ... Nov 3 '18 at 17:08

Suppose that the probability that a packet is lost is $$1/p$$. Then the number of packets you need to send until the packet is sent successfully is a geometric random variable with success probability $$1-1/p$$. Hence the expected number of packets you need to send is $$1/(1-1/p) = p/(p-1)$$. The expected number of retransmissions is $$p/(p-1) - 1 = 1/(p-1)$$. So none of the answers is correct.