Stop and wait

Question

if packet size is 1KB and propogation time 15msec, channel capacity 10^9 b/s , find transmission time and sender utilization in stop and wait protocol ?

Answer 1

transmission time = 10^3*8/10^9 = 8*10^-6 s.

utilization = 8*10^-6/(8*10^-6+30*10^-3) =2.66*10^-4

Answer 2

Transmission time = Length of packet / Bandwidth =

Sender Utilization = Efficiency = 1/(1+2a) =

Answer 3

Transmission Time(Tx)= (Size of msg or packet or data)/ ( Brand width of channel or channel capacity)

So, transmission time=1KB/10^9 bps= 8*10^-6sec

Sender utilization means the utilization  of the channel by the sender 

n= 1/1+2a,a=Tp/Tx for stop and wait protocol

a=15 m sec/8*10^-3 m sec= 1.875 *10^3 

n= 1/1+2(1.875*10^3 )= 2.66*10^-4 sec

 

Comments

1) Link Utilisation is the time for which The Cable has been used

Sender utilises the cable which placing the data on cable i.e. during the transmission time

So, Sender Link Utilisation= Transmission Time/(Total Time)= 1/1+2a in  stop and wait protocol