### Computer Network | Capacity of Channel and Pipelining

#### Capacity of Channel

Capacity can be measured by counting number of packets of data embedded in a snapshot of a channel while it is transferring maximum amount of data it can hold. Which means the holding capacity of channel during transmission.

If a half duplex channel can send $$n$$ bits per second(Band Width) the it has a propagation delay($$T_p$$) of $$p$$ seconds, channel will have a capacity $$n*p$$ bits.

$$Capacity\ of\ channel = Band\ Width( BW ) * Transmission\ dealy( T_p )$$

Channels with high capacity is called thick channels/thick pipes and those with lower capacity is called thin pipes. Since the sender has to wait until receive acknowledgement from the receiver, in efficient usage of available space on thick pipes makes it not suitable for stop and wait protocol. We have efficiency( $$\eta$$ ) of stop and wait protocol, which can be rewrite as,

$$\eta =\frac{1}{1 +2\ a} =\frac{1}{1 +\frac{2\ T_p}{T_t}} \\ \\ \eta =\frac{1}{1 +\frac{2\ T_p\ BW}{L}} \\ \\ \eta =\frac{1}{1 +\frac{2\ T_p\ BW}{L}}=\frac{1}{1 +\frac{ ChannelCapacity }{L}}$$

For a full duplex communication link, Channel Capacity will be twice that of half duplex $$[Channel\ Capacity= 2\ T_p\ BW]$$ , since it can communicate to and fro at same time with out collision.

$$\Rightarrow$$ Efficiency of stop and wait protocol is inversely proportional to channel capacity. So it will give poor performance on a channel with higher capacity, this can be improved by a method called pipelining.

#### Pipelining/ Sliding window protocol

Let $$T_t = 1 \mu s$$ and $$T_p = 1.5 \mu s$$

If I apply stop and wait protocol for data transmission, $$\ \ efficiency\ \ \eta = \frac{1}{1+2a}= \frac{1}{1+2* \frac{1.5}{1}} = \frac{1}{4}$$

The efficiency is very less. The sender has used only small portion of the total time, because it has to wait during the to and fro transmissions until the acknowledgment from the receiver reaches at sender.

In order to increase the efficiency of the transmission this waiting time should be used for the sending of more number of packets and each of the packets transmitted from the sender has to be copied to the window buffer at the sender to keep track of its success rate of transmission, until its acknowledgment messages reaches at sender.

Data which are already acknowledged are stored pops out from the window, and inside the window we keep only those which are already transmitted and did not acknowledged till that time. So the number of such data packets defines the size of the window and which is a function of propagation and transmission delays.

By observing the timing diagram it is found that sender window size $$W_s = 1+ 2a$$

In order to identify data packet stored in window each of them should have a sequence number, which can be in increasing order of of natural numbers. But as the number of data packets grew up, the space to store sequence number will also become high. So it is a more convenient to use sequence number as minimum as possible with lower size. To do that sequence number last packet whose acknowledgement had received can be used.

$$Window\ size\ W_s = min( 1 + 2a, 2^N) \\ \\ N-\ Number\ of\ bits\ available\ in\ sequence\ number\ field \\ \\ Minimum\ sequence\ number\ required\ = 1 + 2a \\ \\ Bits\ used\ for\ number\ of\ sequence\ number\ = \left \lceil log_2(1+ 2a) \right \rceil$$

Q) Tt = 1 ms and Tp = 56.5 ms What should be sender window size to get maximum efficiency?

$$\\ a = \frac{T_p}{T_t} = \frac{56.5}{1}= 56.5 \\ \\ N = \left \lceil log(1+ 2a) \right \rceil = \left \lceil log_2(1+ 2*56.5) \right \rceil = \left \lceil log_2(114) \right \rceil = 7 \\ \\ Window\ size\ W_s = min( 1 + 2a, 2^N) = min( 114, 2^7) = min( 114, 128) = 114 \\$$

Notes:
Half Duplex
A half duplex communication medium is a channel which can be used either for sending data or receiving data at an instant, which means even though it provides bidirectional data path, the transmission time and receival time will be different.

Full Duplex
Full duplex channels are those transmission mediums where sending and receiving can be done simultaneously. Which means, the channel will have the capability to transport data from receiver to sender and from sender to receiver on same time.

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