Applied Test Series

Question

A link has a transmission speed of 500 × 10^6 bpsec. Assume acknowledgment has negligible transmission delay, and it's one way propagation delay is 2 sec. Also assumes that the processing delays at nodes are also negligible. If data packet size is 107 bits, then the efficiency of Go-Back-7 protocol is _______ (in %) [Correct upto two places of decimals].

Answer 1

Ans = 3.48 %

Given that –

Bandwidth = 500 * 10^6 bps

Packet size = 10^7 bits

Propagation delay = 2 sec.

We are also given that Processing delay and Acknowlgement delay as zero.

Therefore 

Tp/Tt = (2/10^7) 500 * 10^6

Tp/Tt = 100

$\eta = N/(1 + 2a)$ {a = Tp/Tt}

= 7/(1 + 2*100)

= 7/201

= 0.0348

= 0.0348 * 100

= 3.48 %

Answer 2

B =500*10^6 bits/sec

L=10^7 bits ,  Tt= L/B  =0.02sec

Tp=2 sec

efficiency= N*Tt/Tt+2Tp

put the value Tt & Tp

e=0.034* 100 =3.48%