ISRO CSE 2016 PYQ

Detailed Explanation:

Time for 1-bit transmission = 1 / 10^7 $s = 0.1 \mu s$.

Detailed Explanation:

Class B has 16 host bits. 6 bits used for subnet => 10 bits remaining for hosts.

$Usable subnets = 2^6 - 2 = 62 (standard RFC 950 rule).$

Detailed Explanation:

Polynomial $x^3 + 1$ is 1001 (degree 3).

Append 3 zeros to message: 11001001000.

Modulo-2 division of 11001001000 by 1001 yields remainder 011.

Detailed Explanation:

Find the Transmission Time (\(T_{t}\))

Transmission time is the time taken to push all the bits of a single frame into the link:
\(T_{t}=\frac{\text{Frame\ Size}}{\text{Bandwidth}}=\frac{1000\text{\ bits}}{10^{6}\text{\ bps}}=10^{-3}\text{\ seconds}=1\text{\ ms}\)

Determine the Number of Frames in Transit

The problem asks to maximally pack the frames in transit within the link (one-way propagation).
• Propagation time (\(T_{p}\)) = \(25 \text{ ms}\)
• Transmission time for one frame (\(T_{t}\)) = \(1 \text{ ms}\)
The maximum number of frames that can physically reside inside the link at any given moment is:
\(\text{Max\ frames\ in\ transit}=\frac{T_{p}}{T_{t}}=\frac{25\text{\ ms}}{1\text{\ ms}}=25\text{\ frames}\)
(Alternatively, using the Bandwidth-Delay Product: \(\text{BDP} = 10^6 \text{ bps} \times 25 \text{ ms} = 25,000 \text{ bits}\). Dividing by the frame size gives \(\frac{25,000}{1000} = 25 \text{ frames}\).)