Problem 2

Consider a path from node 0, through nodes 1, 2, ...,K, to node K+1. The channels, from node i to node i+1, for i = 0, 1, ..., K, have the same capacity and propagation delay:

C channel capacity in bits/second

p propagation delay in seconds

Consider packets, each of which has h header bits and L information bits. The delay of a packet from node 0 to node K+1 is defined to be the duration from when the last bit of the packet leaves node 0 to when the last bit of the packet arrives at node K+1. Suppose the delay also includes a processing time of q seconds in each of nodes 1, 2, ..., K. The processing time includes waiting time in queue.

a)

Draw a picture to show that the delay D of a packet from node 0 to node K+1 is given by the following formula, D = p + K [ (L+h)/C + p + q ]

Now, suppose that each packet is segmented into n cells (small packets), each of which has L/n information bits and h header bits. Each cell still requires a processing time of q seconds in an intermediate node. Suppose that the n cells of a packet are transmitted from one node to the next with no gap between consecutive cells, i.e., transmission of the last bit of a cell is immediately followed by transmission of the first bit of the next cell. (A node has multiple processors and can process the output of one cell and the input of another cell concurrently.)

Segmenting packets into cells has the advantage of pipelining, namely: as soon as a cell is completely received by node i, node i can process and transmit the cell to node i+1 without waiting for the rest of the packet. This observation is to be illustrated by the following analysis.

b)

Draw a picture to show that the delay of a packet that has been segmented into

cells is given by the following formula:

D(n) = p + K [ (L/n + h)/C + p + q ] , where D(n) is defined to be from when the last bit of the packet's last cell leaves node 0 until the last bit of the last cell arrives at node K+1.

c)

For p = q = 0.001 second

L = 10,000 bits

C = 1,000,000 bits/second

h = 100 bits

Perform the following computations:

1. Calculate delay for K = 10, n = 1

2. Calculate delay for K = 10, n = 20

3. Calculate delay for K = 4, n = 1

4. Calculate delay for K = 4, n = 20

d)

Plot the following:

1. For K = 10, plot delay as a function of n, for n = 1, 2, 3, ..., 40.

2. For n = 20, plot delay as a function of K, for K = 1, 2, 3, ..., 20.