UGC NET CSE | January 2017 | Part 3 | Question: 26

Question

Station $A$ uses $32$ byte packets t transmit messages to station $B$ using sliding window protocol. The round trip delay between $A$ and $B$ is $40$ milliseconds and the bottleneck bandwidth on the path between $A$ and $B$ is $64$ kbps. The optimal window size of $A$ is

• A. $20$  
• B. $10$  
• C. $30$
• D. $40$

Comments

option B is correct

Answer 1

Total data 40x64x10³ x10^-3 =  =320 bytes.

No. of packets 320/32  = 10

ANS: 2

Answer 2

2) 10

OWS=2a

a=Tp/Tt

2a=40/4

2a=10

Comments

the optimal window size in sliding window protocol if considered accordingly

then Go-Back N = N+1

and selective repeat = 2N

where N is the window size , so to avoid ambiguity we use formula

 

window size= (Bandwidth∗RTT ) / Packet size

Answer 3

Length=32*8bits

Bandwidth=128*10^3bps

Transmission time =frame length/ Bandwidth

So ,Transmission time=4ms

RTT=40ms

No of frames to achieve 100% utilization= RTT/Transmission time

So 40/2=20

20 is the optimal window size

Answer 4

window size= $\frac{Bandwidth*RTT}{packet size}$

packet size = 32 bytes

RTT = 40 millisecond

B.W = 64Kbps

W.S=$\frac{(64*10^{3})*(40*10^{-3})}{32*8}$ = 10