GATE CSE 2016 Set 1 | Question: 55

Question

A sender uses the Stop-and-Wait $\text{ARQ}$ protocol for reliable transmission of frames. Frames are of size $1000$ bytes and the transmission rate at the sender is $80\;\textsf{Kbps} (1\textsf{Kbps} = 1000\;\text{ bits/second}).$  Size of an acknowledgment is $100$ bytes and the transmission rate at the receiver is $8\;\textsf{Kbps.}$ The one-way propagation delay is $100$ milliseconds.

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

Comments

Thats why we say, never remember the formula

always remember the logic

then you will always derive the correct useful time

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how bw is calculated plz some one help

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GO to the Root of Concept: Video Explanation with timestamp 15 Questions on Stop Wait Protocol| GATE PYQs| MIT, Berkeley |Efficiency & Throughput| With NOTES

Answer 1

Sender's transmission time (Tt) = 1000B/10kBps = 0.1 second.

Receiver's transmission time (Tt) = 100B/1kBps= 0.1 second.

In stop and wait sender sends a frame, it reaches the receiver ( sender's transmission time + sender's propagation time), receiver process it  and send back acknowledgement to receiver ( receiver's transmission tiE + receiver's propagation time), 

So total time for a frame = sender's transmission time + sender's propagation time + receiver's transmission time + receiver's propagation time.  (Processing delay at receiver is neglected here)

Throughput = efficiency * Bandwidth

Efficiency = Usefull time / Total time 

(Always transmission time of the sender is the  Usefull time)

So, efficiency = 0.1/(0.1+0.1+0.1+0.1) = 0.1/0.4 = 0.25

Throughput = efficiency * bandwidth

 = 0.25* 10kBps = 0.25*10⁴Bps = 2500 byte/second

 

 

Answer 2

As we know, Throughput = efficiency * Bandwidth

 Now we have to calculate throughput at the sender, so first calculate the efficiency as we already know the Bandwidth (i.e Transmission rate at sender) 

Efficiency = Tt (sender) / (Tt) sender + Tp + Tp + (Tt) ack

= 100ms/100ms+100ms+100ms+100ms

=100/400

=¼

Now, Throughput at sender = 1/4 * 80Kbps = 2500 B/sec

Note : Divide by 8 to get Byte per sec from bits per sec. 

Answer 3

The 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 = 1000 bits/second). The size of an acknowledgment is 100 bytes, and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds.

In this scenario, the throughput can be calculated by considering the time it takes for a frame to be sent and acknowledged. This includes the transmission time of the sender, the propagation delay, and the transmission time of the receiver.

The transmission time for the sender is the size of the frame divided by the transmission rate, which is 1000 bytes * 8 bits/byte / 80000 bits/second = 0.1 seconds.

The transmission time for the receiver (the acknowledgment) is the size of the acknowledgment divided by the transmission rate, which is 100 bytes * 8 bits/byte / 8000 bits/second = 0.1 seconds.

The total time to send a frame and receive an acknowledgment is the sum of these times and twice the propagation delay (for the frame to reach the receiver and the acknowledgment to reach the sender). This is 0.1 seconds + 0.1 seconds + 2*0.1 seconds = 0.4 seconds.

The throughput is then the size of the frame divided by this total time, which is 1000 bytes / 0.4 seconds = 2500 bytes/second.