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A network has a data transmission bandwidth of $20 \times 10^{6}$ bits per second. It uses **CSMA/CD** in the **MAC** layer. The maximum signal propagation time from one node to another node is $40$ microseconds. The minimum size of a frame in the network is __________ bytes.

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Since, CSMA/CD

Transmission Delay = RTT

Hence,

$L=B \times \text{ RTT}$

$\implies L=B \times 2 \times T_{\text{propagation delay}}$

$\implies L=(20 \times 10^6) \times 2 \times 40 \times 10^{-6}$

$\qquad =20 \times 2\times 40$

$\qquad =1600 \text{ bits}$

$\qquad = 200\text{ bytes}$

Hence, 200 Bytes** **is the answer.

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Collision detection condition of CSMA/CD protocol is

**Transmission Time >= 2 x Propagation Time + Transmission Time of Jam signal**

Here, we are not considering Jam signal. So, it’s transmission time will be considered as 0.

This collision detection condition puts a restriction on the size of frame being transmitted:

**Minimum Frame size =** **2 x Propagation delay x Bandwidth**

**Minimum Frame size =** 2*(40*10^-6)*(20*10^6) = 1600 bits = 200 Bytes.