Test by Bikram | Mock GATE | Test 1 | Question: 52

Question

Two peer processes $A$ (sender) and $B$ (receiver) use stop-and-wait $ARQ$ to send packets over a single link with capacity $C$.  All packets have the same length of $100$ bits. The round-trip time (which is the time until $A$ receives an acknowledgment for a sent packet) is equal to $2$ seconds.

Assume that no packets or $ACK's$ are dropped and that all packets and $ACK's$ arrive error-free. Furthermore, assume that the capacity $C$ is equal to $100,000$ bits per second.

The average transmission rate (in bits per seconds) with which process $A$ sends data to process $B$ is ________.

Answer 1

Here transmission time = length of packet / bandwidth of channel

=100 bits /100000 bits/sec = 10^-3 sec

as given RTT =2 sec

as we know RTT = transmission time of packet + propagation delay of packet + processing delay+ quening delay + ack transmission delay + ack propagation delay = 2 sec

so in 2 sec only utilization is 10^-3 sec because of STOP and WAIT .

so efficiency = 10^-3 / 2
so effective bandwidth = 10^5 * 10^-3 /2 =100 /2

=50 bits / sec

correct me if i am wrong .

Comments

As it is stop and wait, sender will be sending a packet and waiting for the ack to reach to send the next packet. this takes 2S and in that time, All it could send was 100Bits. which means 50Bits/sec effectively. Is this correct ?

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yes, correct @Junk_mayavi

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round trip time=2*tp   ??? is this after approximation???

actual RTT=Tp+ Tt+ Tq + Tr    ???

Answer 2

If you ever get confused regarding what is what, look at the units and verify.

capacity C is equal to 100,000 bits per second.

This is the Bandwidth.

transmission rate (in bits per seconds) ...

They're asking for Sender Throughput.

 

Throughput = efficiency * bandwidth.

 

Efficiency in Stop-And-Wait = $\frac{1}{1+2a}$

$a = \frac{T_p}{T_t}$

=> $\frac{1s}{100bits*seconds/100000bits}$ (Notice the units cancel each other perfectly. This indicates we're going right)

=> $a=1000$

 

So, $\frac{1}{1+2a}$

= $\frac{1}{2001}$

=> efficiency = $ 0.00049975012 $

 

Finally, Throughput = $ 0.00049975012 * 100000bps=49.975 bps$