Explanation:
(a)
Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec
Hence, propagation delay, [tex]d_{prop} = m/sec (s)[/tex]
(b)
Here, size of the packet is L bits
And the transmission rate of the link is R bps
Hence, the transmission time of the packet, [tex]d_{trans} = L/R[/tex]
(c)
As we know, end-to-end delay or total no delay,
[tex]\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}[/tex]
[tex]Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec[/tex][tex]\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }[/tex]
(d)
The expression, time [tex]time $t=d_{\text {trans }}$[/tex] means the\at time since transmission started is equal to transmission delay.
As we know, transmission delay is the time taken by host to push out the packet.
Hence, at [tex]time $t=d_{\text {trans }}$[/tex] the last bit of the packet has been pushed out or transmitted.
(e)
If [tex]\ d_{prop} >d_{trans}[/tex]
Then, at [tex]time $t=d_{\text {trans }}$[/tex] the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.
(f)
If [tex]\ d_{prop} <d_{trans}[/tex]
Then, at [tex]time $t=d_{\text {trans }}$[/tex], the first bit has reached destination on B
Here,[tex]s=2.5 \times 10^{8} \mathrm{sec}[/tex]
[tex]\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}[/tex]
It's given that [tex]\ d_{prop} =d_{trans}[/tex]
Hence,
[tex]\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}[/tex]
