## Stop-and-Wait-ARQ

 Question 1

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 (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds.

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

 A 2500 B 2501 C 2502 D 2503
Computer-Networks       Stop-and-Wait-ARQ       2016 set-01
Question 1 Explanation:
Given,
Frame size (L) =1000 bytes
Sender side bandwidth (BS) = 80 kbps = 10 * 103 bytes/sec
Acknowledgement size (LA) =100 bytes
Receiver side bandwidth (BR) = 8 kbps = 1 * 103 bytes/sec
Propagation delay (Tp) =100 ms
By formula:
Transmission delay (Tt ) = L/BS = 1000 bytes / 10 * 103 bytes/sec = 100 ms
Acknowledge delay (Tack ) = LA / BR = 100 bytes / 1 * 103 bytes/sec = 100 ms
Total cycle time = Tt + 2 * Tp + Tack = 100 ms + 2 * 100 ms + 100 ms = 400 ms
Efficiency (η) = Tt / Total cycle time = 100 ms / 400 ms = 1 / 4 = 0.25
Throughput = Efficiency (η) * Bandwidth (BS) = 0.25 * 10 *103 bytes/s = 2500 bytes/second
 Question 2
A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ___________.
 A 12 B 13 C 14 D 15
Computer-Networks       Stop-and-Wait-ARQ       GATE 2015 -(Set-2)
Question 2 Explanation:
Given, B=106bps
L=1000
η=25%
Tp=?
In stop-and-wait, η=1/1+2a
⇒1/4=1/1+2a⇒1+2a=4
2a=3;a=32
Tx=L/B=8×103/106=8ms
Tp/Tx=3/2;2Tp=3Tx
2Tp=24ms
Tp=12ms
There are 2 questions to complete.