A network using CSMA/CD has a bandwidth 10 Mbps. If the maximum propagation time (including delays in the devices and ignoring the time needed to send a jamming signal) is 25.6 micro seconds. What is the minimum size of the frame ?

Question

Accepted Answer

The correct option is (3): 64 bytes.

The core principle of the Carrier Sense Multiple Access with Collision Detection (CSMA/CD) protocol is that a transmitting station must continue to send its frame long enough to detect a collision from the most distant station in the network. The time required for a signal to travel to the farthest node and for a collision signal to return is called the *slot time*, which must be at least twice the maximum propagation delay ($2 \times T\_p$). In this case, the maximum propagation time ($T\_p$) is $25.6 \mu s$, so the minimum frame transmission time ($T\_{min}$) is $2 \times 25.6 \mu s = 51.2 \mu s$. The minimum frame size ($L\_{min}$) in bits is calculated by multiplying this time by the network's bandwidth (B): $L\_{min} = T\_{min} \times B$. Substituting the values: $L\_{min} = 51.2 \times 10^{-6} s \times 10 \times 10^6 bps = 512$ bits. Converting the result from bits to bytes ($512/8$) gives the standard minimum Ethernet frame size of 64 bytes, which is essential to ensure that any collision is detected before the transmission is complete.

The incorrect option is (1): 512 bytes. This value is a common distractor because it is the correct minimum frame size *in bits*. The calculation $51.2 \mu s \times 10 Mbps$ yields 512, which represents the number of bits that must be transmitted during the collision window. Since the options are given in bytes, this result must be divided by 8, thus 512 bytes is incorrect.

The incorrect option is (2): 1024 bytes. This option is twice the value of 512 bytes, which is the standard maximum size of a data field, not the minimum frame size. While a frame of 1024 bytes is valid, it does not represent the minimum required length to satisfy the CSMA/CD collision detection rule. The minimum size is determined by the physics of the network, specifically the bandwidth and the maximum propagation delay.

The incorrect option is (4): 32 bytes. A 32-byte frame is equivalent to 256 bits. At a 10 Mbps bandwidth, this frame would be transmitted in $256 / 10 \times 10^6 \approx 25.6 \mu s$. This transmission time is only equal to the *one-way* maximum propagation time ($T\_p$). This is insufficient because a collision occurring just as the first bit reaches the farthest station would only return to the sender after another $T\_p$ time, meaning the sender would have already completed its transmission and thus fail to detect the collision.