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GATE CS 2014 Set 2 · Computer Networks

Question 36

MCQPrinted as CS Q.262 marksWrong answer: −2/3Medium (estimate)

Question

Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 10610^6 bytes / sec. A user on host A sends a file of size 10310^3 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

Store and forward packet switched network connecting host A to host B via routers R1 and R2
Store-and-forward network topology from Host A to Host B
Choose one option
Official answer and solution
Official answer
D
From
Official GATE 2014 answer key

Solution

Calculate the end-to-end transmission time across the 3 links for a single packet versus pipelined transmission of multiple packets.

  1. System Parameters:

    • Link bandwidth B=106B = 10^6 bytes/sec = 11 byte/μ\mus.
    • Topology: A→R1→R2→BA \to R1 \to R2 \to B consists of 3 links (3 hops).
    • User data = 103=100010^3 = 1000 bytes.
    • Header per packet = 100100 bytes.

  2. Case 1 (Single packet):

    • Data size = 10001000 bytes, Header = 100100 bytes   ⟹  \implies Packet size = 11001100 bytes.
    • Transmission delay per link = 1100106=1.1\frac{1100}{10^6} = 1.1 ms.
    • Total time across 3 hops (store-and-forward):
      T1=3×1.1=3.3 ms.T_1 = 3 \times 1.1 = 3.3 \text{ ms}.

  3. Case 2 (10 packets):

    • Data per packet = 100010=100\frac{1000}{10} = 100 bytes.
    • Packet size = 100+100=200100 + 100 = 200 bytes.
    • Transmission delay per link = 200106=0.2\frac{200}{10^6} = 0.2 ms.
    • Due to pipelining, the first packet reaches Host B after 3 hops (3×0.2=0.63 \times 0.2 = 0.6 ms), and each of the remaining 9 packets arrives every 0.20.2 ms thereafter:
      T2=3×0.2+(10−1)×0.2=0.6+1.8=2.4 ms.T_2 = 3 \times 0.2 + (10 - 1) \times 0.2 = 0.6 + 1.8 = 2.4 \text{ ms}.

  4. Case 3 (20 packets):

    • Data per packet = 100020=50\frac{1000}{20} = 50 bytes.
    • Packet size = 50+100=15050 + 100 = 150 bytes.
    • Transmission delay per link = 150106=0.15\frac{150}{10^6} = 0.15 ms.
    • Total pipelined time:
      T3=3×0.15+(20−1)×0.15=0.45+2.85=3.3 ms.T_3 = 3 \times 0.15 + (20 - 1) \times 0.15 = 0.45 + 2.85 = 3.3 \text{ ms}.

  5. Comparison:

    T1=3.3 ms,T2=2.4 ms,T3=3.3 ms.T_1 = 3.3 \text{ ms}, \quad T_2 = 2.4 \text{ ms}, \quad T_3 = 3.3 \text{ ms}.

    Therefore, T1=T3T_1 = T_3 and T3>T2T_3 > T_2, which matches option (D).

Final answer

D

Common trap

Forgetting that in store-and-forward switching with multiple packets, transmission across subsequent links overlaps in a pipeline rather than multiplying total packets by 3.

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

Source

Paper
GATE 2014 Computer Science and Information Technology (CS) Session 2
Question
36 of 65, printed as CS Q.26
Section
Computer Science and Information Technology

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