GATE CS 2015 Set 3 Q36 — Packet, Circuit and Virtual Circuit Switching

To find the total time elapsed, we first calculate the transmission delay ($T_t$) for a single packet of size $5000$ bits:

We are given:

• Propagation delay per link ($T_p$) = $20\ \mu\text{s}$
• Switch forwarding delay ($T_s$) = $35\ \mu\text{s}$
• Total data to transmit = $10000$ bits, which corresponds to $2$ packets ($P_1$ and $P_2$) of $5000$ bits each.

Let's trace the timeline of transmission starting at $t = 0$:

1. $t = 0\ \mu\text{s}$: Host $H_1$ starts transmitting packet $P_1$.
2. $t = 500\ \mu\text{s}$: $H_1$ finishes transmitting $P_1$ and immediately starts transmitting $P_2$.
3. $t = 520\ \mu\text{s}$: $P_1$ is completely received at the switch $S$ ($500 + T_p$).
4. $t = 555\ \mu\text{s}$: Switch $S$ begins forwarding $P_1$ (after the $35\ \mu\text{s}$ processing delay).
5. $t = 1000\ \mu\text{s}$: $H_1$ finishes transmitting $P_2$.
6. $t = 1020\ \mu\text{s}$: $P_2$ is completely received at the switch $S$.
7. $t = 1055\ \mu\text{s}$: Switch $S$ finishes transmitting $P_1$ ($555 + 500$). At this exact moment, the switch is also ready to forward $P_2$ (since $1020 + 35 = 1055\ \mu\text{s}$), so it immediately starts transmitting $P_2$.
8. $t = 1555\ \mu\text{s}$: Switch $S$ finishes transmitting $P_2$ ($1055 + 500$).
9. $t = 1575\ \mu\text{s}$: $P_2$ is completely received at the destination host $H_2$ ($1555 + T_p$).

Thus, the total time elapsed is 1575 microseconds.