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GATE CS 2016 Set 1 · Computer Networks

Question 65

NATPrinted as CS Q.552 marksNo negative markingMedium (estimate)

Question

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.

Official answer and solution
Official answer
2500
From
Official GATE 2016 answer key

Solution

Compute the total cycle time for transmitting one frame and its acknowledgment, then calculate throughput as payload divided by total cycle time.

Given parameters:

  • Data frame size Lf=1000 bytes=8000 bitsL_f = 1000\text{ bytes} = 8000\text{ bits}.
  • Sender transmission rate Rs=80 Kbps=80,000 bits/sR_s = 80\text{ Kbps} = 80,000\text{ bits/s}.
  • ACK frame size La=100 bytes=800 bitsL_a = 100\text{ bytes} = 800\text{ bits}.
  • Receiver transmission rate Rr=8 Kbps=8,000 bits/sR_r = 8\text{ Kbps} = 8,000\text{ bits/s}.
  • One-way propagation delay Tp=100 ms=0.1 sT_p = 100\text{ ms} = 0.1\text{ s}.

Calculate component delays:

  1. Transmission time of data frame:
Tt,frame=LfRs=8000 bits80,000 bits/s=0.1 s=100 msT_{t, frame} = \frac{L_f}{R_s} = \frac{8000\text{ bits}}{80,000\text{ bits/s}} = 0.1\text{ s} = 100\text{ ms}
  1. Propagation delay of data frame: Tp=100 msT_p = 100\text{ ms}.
  2. Transmission time of ACK frame:
Tt,ack=LaRr=800 bits8,000 bits/s=0.1 s=100 msT_{t, ack} = \frac{L_a}{R_r} = \frac{800\text{ bits}}{8,000\text{ bits/s}} = 0.1\text{ s} = 100\text{ ms}
  1. Propagation delay of ACK frame: Tp=100 msT_p = 100\text{ ms}.

Total cycle time for one frame:

Ttotal=Tt,frame+Tp+Tt,ack+Tp=100+100+100+100=400 ms=0.4 sT_{\text{total}} = T_{t, frame} + T_p + T_{t, ack} + T_p = 100 + 100 + 100 + 100 = 400\text{ ms} = 0.4\text{ s}

Calculate sender throughput in bytes/second:

Throughput=Data transferredTtotal=1000 bytes0.4 s=2500 bytes/s\text{Throughput} = \frac{\text{Data transferred}}{T_{\text{total}}} = \frac{1000\text{ bytes}}{0.4\text{ s}} = 2500\text{ bytes/s}
Final answer

2500

Common trap

Neglecting the transmission delay of the acknowledgement frame (100100 ms) or expressing throughput in bits/second rather than bytes/second.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2016 Computer Science and Information Technology (CS) Session 1
Question
65 of 65, printed as CS Q.55
Section
Computer Science and Information Technology

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