CS455 Syllabus

### Maximum channel utilization of the Stop-and-Wait protocol

• Performance of Stop-and-Wait protocol

• Question:

• A sender transmits data frames to a receiver using the Stop-and-Wait protocol:

• What is the maximum channel utilization ?

 In other words: what fraction of time on the time line in the above figure is yellow (= data frames)

Note:

• The maximum channel utilization is experienced when there are no transmission errors

 When there are transmission errors, there will be extra overhead to recover from the error(s)

(So the question implies that we must assume that there are no transmission errors)

• Notations used in the performance analysis

• Notations:

• tdata = time to transmit the data frame

• tprop = time for the signal to propagate from the sender to the receiver

 This delay is called the propagation delay

• tACK = time to transmit the ACK frame

• NOTE:

• the ACK frame is usually very short compared to a data frame

• Therefore:

 ``` tACK ------ << 1 tdata ```

• It is often assumed that:

 ``` tACK ------ = 0 tdata ```

• Maximum channel utilization of the Stop-and-Wait protocol

• Maximum channel utilization of the transmission channel:

• The following pattern is repeated indefinitely when there are no transmission errors:

• The duration of each cycle is:

 ``` tdata + tprop + tACK + tprop = tdata + tACK + 2×tprop ```

• The amount of time in the cycle used to transmit data frames is:

 ``` tdata ```

• The fraction of time used to transmit data frames is:

 ``` tdata Channel efficiency = --------------------- tdata + tACK + 2 tprop tdata ~= --------------- tdata + 2 tprop ```

(tACK is very small and is usually discarded in the approximation)