Stop and Wait

Question

Consider a wireless link, where the probability of packet error is 0.6. To transfer data across the links, Stop and Wait protocol is used. The channel condition is assumed to be independent from transmission to transmission. The average number of transmission attempts required to transfer x packets is 500. The value of x is _______.

Comments

X is 200

500 = x/1-p

500 = x/1-0.6

x=200

Answer 1

number of packet be x(given)

probability of error=0.6

the average number of retransmission number=500(given)

500=x+0.6*x+0.6*0.6*x+0.6*0.6*0.6*x+.............................................infinite GP

500=x/1-0.6

x=200 packet

Answer 2

Packet without error prob=0.4

In 100 packets 40 packets are reaching safely

To reach 500 safely 500/40 = 8

Total attempts 8*100=800

Comments

what will be right ans...?

I got 200