P$6.$ This clementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B$,$ connected by a single link of rate $R$ bps$.$ Suppose that the two hosts are separated by $m$ meters, and suppose the propagation speed along the link is $s$ meters $\frac{?}{sec}.$ Host A is to send a packet of size $L$ bits to Host B$.$

a$.$ Express the pr$$pagation delay, $d_{\text{p}\text{r}\text{o}\text{p}},$ in terms of $m$ and $s.$

b$.$ Determine the transmission time of the packet, $d_{\text{t}\text{r}\text{a}\text{n}\text{s}},$ in terms of $L$ and $R.$

c$.$ Ignoring processing and queuing delays, obtain an expression for the endto$-$end delay.

d$.$ Suppose Host A begins to transmit the packet at time $t=0.$ At time $t=d_{\text{t}\text{r}\text{a}\text{n}\text{s}\text{,}}$ where is the last bit of the packet?

e$.$ Suppose $d_{\text{p}\text{r}\text{o}\text{p}}$ is greater than $d_{\text{t}\text{r}\text{a}\text{n}\text{s}}.$ At time $t=d_{\text{t}\text{r}\text{a}\text{n}\text{s}},$ where is the first bit of the packet?

f$.$ Suppose $d_{\text{p}\text{r}\text{o}\text{p}}$ is less than $d_{\text{t}\text{r}\text{a}\text{n}\text{s}\text{.}}.$ At time $t=d_{\text{t}\text{r}\text{a}\text{n}\text{s}},$ where is the first bit of the packet?

g$.$ Suppose $s=2\mathrm{.}5*{10}^8,L=120$ bits, and $R=56kbps.$ Find the distance $m$ so that $d_{\text{p}\text{r}\text{o}\text{p}}$ equals $d_{\text{t}\text{r}\text{a}\text{n}\text{s}}.$