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Official PYQ · 2015Set 3 · Q6Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsMCQ1 mark

GATE CS 2015 Set 3 Q6 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Consider a CSMA/CD network that transmits data at a rate of 100  Mbps  (108  bits100\;\textsf{Mbps}\; (10^8\;\text{bits} per second) over a 1  km1\;\textsf{km} (kilometre) cable with no repeaters. If the minimum frame size required for this network is 1250  bytes,1250\;\text{bytes}, What is the signal speed (km/sec)\textsf{(km/sec)} in the cable?

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Answer: D

In a CSMA/CD network, to ensure that a transmitting station can detect a collision before it finishes transmitting, the transmission time (TtT_t) of the minimum-sized frame must be at least twice the propagation delay (TpT_p) of the cable:

Tt≥2⋅TpT_t \ge 2 \cdot T_p

For the limiting case of the minimum frame size, we have:

Tt=2⋅TpT_t = 2 \cdot T_p

1. Calculate the Transmission Time (TtT_t):

  • Minimum frame size, L=1250 bytes=1250×8 bits=10,000 bitsL = 1250 \text{ bytes} = 1250 \times 8 \text{ bits} = 10,000 \text{ bits}
  • Transmission rate, B=100 Mbps=108 bits/secB = 100 \text{ Mbps} = 10^8 \text{ bits/sec}
Tt=LB=10,000108=10−4 secondsT_t = \frac{L}{B} = \frac{10,000}{10^8} = 10^{-4} \text{ seconds}

2. Express the Propagation Delay (TpT_p):

  • Cable length, d=1 kmd = 1 \text{ km}
  • Let the signal speed be vv in km/sec\text{km/sec}
Tp=dv=1v secondsT_p = \frac{d}{v} = \frac{1}{v} \text{ seconds}

3. Solve for Signal Speed (vv):

Substitute TtT_t and TpT_p into the CSMA/CD condition:

10−4=2⋅(1v)10^{-4} = 2 \cdot \left(\frac{1}{v}\right) v=210−4=20,000 km/secv = \frac{2}{10^{-4}} = 20,000 \text{ km/sec}

Thus, the signal speed in the cable is 20,000 km/sec20,000 \text{ km/sec}.

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