# Solution for HW1

(5 pts)

(5 pts)

(10 pts)

i.: 0.00001753
ii.: 0.730749

## Problem 2

### Part I

(10 pts)

Consider each packet, the length of which is L+h, so the transmission delay for each packet is t=(L+h)/C.
At each node i, where i=1, 2, ..., K, the transmission delay is t and the propagation delay is p. Hence the total transmission and propagation delay from node 1 to node K+1 are added up to K * (t+p).
Also, at each intermediate node i, where i=1, 2, ..., K, q seconds are required to process each packet and so the total processing time is K*q.
Therefore, the total delay time from node 0 to node K+1 is given by the following formula after adding the extra propagation delay p from node 0 to node 1:

D = p + K (t + p) + K*q = p + K [ (L + h)/C + p + q]

### Part II

(10 pts)

Consider the last cell, which has the length of (L/n+h) and has the transmission delay of tc = (L/n+h)/C. For the same reason shown in Part I, the last cell takes p + K[(L/n+h)/C+p+q] seconds to be transmitted from node 0 to node K+1.

So the delay is given by the following formula:

D(n) = p + K [ (L/n+h)/C + p + q]

### Part III

(10 pts)

For K = 10, n = 1, D = 0.122s
For K = 10, n = 20, D = 0.027s
For K = 4, n = 1, D = 0.0494s
For K = 4, n =20, D = 0.0114s

(10 pts)