Two hosts are connected via a packet switch with $10^7$ bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.

Given parameters:

• Bandwidth of each link $R = 10^7\text{ bps} = 10\text{ bits/}\mu\text{s}$

• Propagation delay per link $T_p = 20\ \mu\text{s}$

• Switch processing/queuing delay $T_{\text{proc}} = 35\ \mu\text{s}$

• Total data size = $10000\text{ bits}$

• Packet size $L = 5000\text{ bits}$

• Number of packets = $\frac{10000}{5000} = 2$ packets ($P_1$ and $P_2$)

Transmission time for each packet across a link:

Timeline of packet transfer from Host A to Host B via Switch S:

1. $t = 0\ \mu\text{s}$: Host A begins transmitting packet $P_1$.

2. $t = 500\ \mu\text{s}$: Host A finishes transmitting $P_1$ and immediately begins transmitting $P_2$.

3. $t = 500 + 20 = 520\ \mu\text{s}$: Switch S completely receives packet $P_1$.

4. $t = 520 + 35 = 555\ \mu\text{s}$: Switch S starts forwarding packet $P_1$ to Host B.

5. $t = 500 + 500 = 1000\ \mu\text{s}$: Host A finishes transmitting packet $P_2$.

6. $t = 1000 + 20 = 1020\ \mu\text{s}$: Switch S completely receives packet $P_2$.

7. $t = 555 + 500 = 1055\ \mu\text{s}$: Switch S finishes transmitting packet $P_1$ to Host B.

8. Packet $P_2$ is ready to be transmitted by Switch S at $\max(1020 + 35, 1055) = 1055\ \mu\text{s}$. Hence, Switch S starts transmitting $P_2$ immediately at $t = 1055\ \mu\text{s}$.

9. $t = 1055 + 500 = 1555\ \mu\text{s}$: Switch S finishes transmitting packet $P_2$.

10. $t = 1555 + 20 = 1575\ \mu\text{s}$: Host B completely receives packet $P_2$.

Thus, the total elapsed time is 1575 microseconds.