Question

This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Consider two hosts A andB, connected by a single link of rate R bps.suppose that the two hosts are separated by mmeters, and supose the propagation speed along the link is s meters/sec. Host A is to send a packet ofsize L bits to Host B.a. Express the propagation delay d(prop) interms of L and R.b. Determine the trnasmission time of the packet d(trans) in termsof L and R.c. Ignoring processing and queing delays obtain an expression forthe end to end delay.d. Supose Host A begins to transmit packet at time t=0 . At timet=d(trans) where is the last bit of the packet.e. Suppose d(prop)is greater than d(trans) . At time t= d(trans)where is the first bit of the packetf. Suppose d(prop)is less than d(trans) . At time t= d(trans) whereis the first bit of the packetg. Suppose s=2.5* 10power(8), L=100 bits and R=28 kbps. find thedistance m so tha d(prop) equals d(trans).

Answer 1

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, $d_{prop} = m/sec (s)$

(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,  $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$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$$\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time t=d_{\text {trans }}$ 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 $time t=d_{\text {trans }}$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time t=d_{\text {trans }}$ the bit has been transmitted from host A, but to condition (1),  the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time t=d_{\text {trans }}$, the first bit has reached destination on B

Here,$s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

        $\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$