Question 46

Parameters:

• Total data $= 10000\text{ bits}$
• Packet size $= 5000\text{ bits}$
• Number of packets $= \frac{10000}{5000} = 2$ packets: $P_1$ and $P_2$
• Link transmission rate $B = 10^7\text{ bps}$
• Transmission time per packet on each link: $T_t = \frac{5000}{10^7}\text{ s} = 5 \times 10^{-4}\text{ s} = 500\ \mu\text{s}$
• Propagation delay per link: $T_p = 20\ \mu\text{s}$
• Switch processing/forwarding delay: $T_{\text{proc}} = 35\ \mu\text{s}$

Detailed timeline of events:

• $t = 0\ \mu\text{s}$: Host 1 begins transmitting packet $P_1$.
• $t = 500\ \mu\text{s}$: Host 1 finishes transmitting $P_1$ and begins transmitting packet $P_2$.
• $t = 500 + 20 = 520\ \mu\text{s}$: The last bit of $P_1$ arrives at the packet switch.
• $t = 520 + 35 = 555\ \mu\text{s}$: The switch is ready to forward $P_1$ and begins transmitting $P_1$ on link 2.
• $t = 555 + 500 = 1055\ \mu\text{s}$: The switch finishes transmitting $P_1$ on link 2. Link 2 becomes idle.
• Meanwhile on link 1: Host 1 completes transmitting $P_2$ at $t = 500 + 500 = 1000\ \mu\text{s}$.
• $t = 1000 + 20 = 1020\ \mu\text{s}$: The last bit of $P_2$ arrives at the switch.
• With processing delay of $35\ \mu\text{s}$, $P_2$ is ready for transmission at $1020 + 35 = 1055\ \mu\text{s}$.

• At $t = 1055\ \mu\text{s}$, link 2 is freed by $P_1$, and $P_2$ is ready! Thus, transmission of $P_2$ on link 2 begins immediately at $t = 1055\ \mu\text{s}$.
• $t = 1055 + 500 = 1555\ \mu\text{s}$: The switch finishes transmitting $P_2$ onto link 2.
• $t = 1555 + 20 = 1575\ \mu\text{s}$: The last bit of $P_2$ arrives at Host 2.

The total elapsed time is $1575\ \mu\text{s}$.