380 views
2 votes
2 votes
A sender uses the stop and wait ARQ protocol for reliable transmissions of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps(1Kbps=1000bits/sec). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one way propogation delay is 100 milliseconds.

Assuming no frame is lost, the sender throughput is ______bytes/second.

1 Answer

Best answer
2 votes
2 votes
Throughput of sender = size of frame sent by sender/total time

Total time = t1+t2+t3+t4

Where

t1=transmission time of sender to place the frame of 1000B on the channel

t2=propagation time of the frame to reach the receiver

t3=transmission time of receiver to put the acknowledgement of 100B on the channel

t4=propagation time for the acknowledge to reach the sender

t2=t4=100ms (given)

t1=L/B,  L=1000B and B=80kbps

So t1=100ms

And t3=La/B' , La=100B and B'=8kbps

t3=100ms

Total time= 4*100ms =400ms

Throughput= 1000B/400ms = 2500B/s
selected by

Related questions

0 votes
0 votes
0 answers
1
amuchand47 asked Oct 15, 2017
321 views
Packets are being transmitted using GB5 and here every 4th packet is lost.How many packets need to be transmitted to transmit 10 packets?
2 votes
2 votes
2 answers
2
popo040 asked Jun 28, 2017
1,578 views
does packets in TCP follow the same route from the source to destination or different? And what is really a connection oriented and connectionless service?
1 votes
1 votes
1 answer
3
Nitesh Choudhary asked Jun 7, 2017
300 views
which edition i will follow for computer networks by tanenbaum?which is the current edition
0 votes
0 votes
1 answer
4
sup739 asked Feb 18
133 views
Is Vector Subspace, Span, Basis, Dimension part of the gate CSE engineering mathematics linear algebra syllabus ?