Register

Finding maximum throughput in Stop and Wait

• edited by
3,346 views
0 0 votes

The problem:

Given:
$B=$  64 kbps satellite channel
$L=$ 512 B data frames
Round Trip Time (RTT) = 64 ms
What is the maximum throughput for window size of 1?

1 Answer

4 4 votes

Throughput=Efficiency * Bandwidth

Efficiency= Transmission Time/(Transmission Time+ Round Trip Time)

Tt=Length of Pkt/Bandwidth

Tt=512*8/(64*10^3)=64 msec

Therefore, Efficency=64/(64+64)=0.5

Thus, Throughput=0.5*64 kbps=32 kbps.

Voting & Accepting an answer helps improve the community
← Previous · DigitalNext → · Time to build a heap
← Previous in category · Virtual Gate Test Se...Next in category → · Virtual Gate Test Se...
Position:
Show:

Related questions

1 1 vote
1 1 answer
3.0k
3.0k views
iarnav asked Nov 8, 2018
2,970 views
#CN Stop and Wait ARQ Numerical about finding efficency?
Consider the use of 10 K-bit size frames on a 10 Mbps satellite channel with 270 ms delay. What is the link utilization for stop-and-wait ARQ technique assuming P = 10-3?...
0 0 votes
1 answers 1 answer
2.3k
2.3k views
iarnav asked Nov 3, 2018
2,334 views
#CN Stop And Wait Numerical Question - Finding Throughput of Receiver?
A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1 Kbps...
8 8 votes
3 answers 3 answers
6.0k
6.0k views
Aakash Das asked Nov 4, 2016
5,992 views
Packet loss numerical in stop and wait
In a communication link out of p packets one packet will be lost. If stop and wait protocol is used then expected number of retransmissions for a packet?(A) P/(1-P)(B) P(...
0 0 votes
2 answers 2 answers
2.6k
2.6k views
vishwa ratna asked Jan 18, 2017
2,620 views
Stop and wait
A channel has a capacity of 256 Mbps, maximum packet size is 1024 bytes and RTT is 200 μsec. So what is efficiency of sender (Assume that channel uses stop & wait protoco...