What is Channel allocation problem? Explain CSMA/CD protocol. A network with CSMA/CD has 10 Mbps bandwidth and 25.6ms

Channel Allocation Problem

The channel allocation problem is the problem of assigning channels to users in a multi-user system. There are a number of different channel allocation algorithms, each with its own goal. Some channel allocation algorithms aim to maximize the total throughput of the system, while others aim to minimize the delay or jitter experienced by users.

Carrier Sense Multiple Access with Collision Detection (CSMA/CD)

CSMA/CD is a variation of CSMA that uses collision detection to further reduce the number of collisions. If a station detects a collision, it immediately stops transmitting and waits for a random amount of time before trying again

A  network with CSMA/CD has 10 Mbps bandwidth and 25.6 μs maximum  propagation delay. What is the minimum frame size? (This solution is for micro second not mili second)

Data:

L = size of frame

BW → bandwidth = 10 Mbps = 10 × 10^6 b/s

Tp → propagation time = 25.6 μs = 25.6 × 10^-6 s

Tt → transmission time

Formula:

For CSMA/CD,

Tt ≥ 2 × Tp

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

Calculation:

\(\frac{{\rm{L}}}{{{\rm{BW}}}} ≥ 2 × T_p\)

L ≥ 2× T_p × BW

L ≥ 2 × 25.6 × 10^-6 × 10 × 10^6

L ≥ 512 bits

 Lmin = 64 bytes

Therefore, the minimum size of a frame in the network is 64 bytes.

Minimum frame size = 2 × propagation delay × bandwidth = 512,000 bits (64,000 bytes) for 25.6 ms. (If the delay is 25.6 µs, as in the usual textbook version, it is 512 bits = 64 bytes.)

Channel allocation problem: when many stations share one broadcast channel (e.g. a LAN cable or radio channel), the network must decide who gets to use the channel and when. Static allocation (FDM/TDM) divides the channel into fixed parts and wastes capacity for bursty traffic; dynamic allocation lets stations compete for the channel on demand, using multiple-access protocols such as ALOHA, CSMA and CSMA/CD.

CSMA/CD (Carrier Sense Multiple Access with Collision Detection), used in classic Ethernet:

1. Carrier sense: a station listens; if the channel is busy it waits.
2. If idle, it starts transmitting and keeps listening while sending.
3. If a collision is detected, it stops at once and sends a short jam signal so all stations know.
4. It waits a random time using binary exponential backoff (after the k-th collision, a random number of slots from 0 to \(2^k - 1\)) and tries again.

For a collision to be detected, a station must still be transmitting when the collision signal returns, i.e. the frame must last at least one round-trip time:

\[T_{fr} \ge 2T_p \Rightarrow L_{min} = 2T_p \times B\]

Calculation: \(B = 10\ \text{Mbps} = 10^7\ \text{bit/s}\), \(T_p = 25.6\ \text{ms}\)

\[L_{min} = 2 \times 25.6\times10^{-3} \times 10^{7}\]

\[= 512{,}000\ \text{bits} = 64{,}000\ \text{bytes}\]

With \(T_p = 25.6\ \mu\text{s}\): \(L_{min} = 2 \times 25.6\times10^{-6} \times 10^{7} = 512\) bits = 64 bytes, which is the standard Ethernet minimum frame.