Efficiency and Throughput in Data Communication Systems

If N frames are transmitted continuously then the time spent in transmitting data will be N tf. The value of the efficiency in the absence of errors will depend on which of the two scenarios mentioned above applies, as follows: 1- If the send node is able to transmit continuously, the efficiency is given by: U = (time spent transmitting frames) / ( total time) =(N tf) / (N tf) = 1. 2- If the send window becomes full after a time (N tf,) when transmission stops until an acknowledgement is received after time (tf + 2td), the expression becomes: U = (time spent transmitting frames) / ( total time) = (N tf) / (tf + 2td) A window size of N = 1 corresponds to a stop-and-wait ARQ strategy which, although it never provides an optimum value of throughput, is adequate for relatively short transmission links in which the propagation delay is short.

A window size of 5 is adequate for most terrestrial links but in the case of links which have large propagation delays, such as satellite systems, a large window size is required to produce a satisfactory throughput (127 being a typical window size for such systems). Example 1: A frame oriented data communications system operates at a transmission rate of 512 kbps with a frame length of 512 bytes and 506 information bytes in each frame over a long-distance link which produces a propagation delay of 20 ms. A flow control system is required using a go- back-3 ARQ flow control (with a window size of 3), determine: (a) the efficiency, (b) the throughput obtained on an error-free link, (c) the throughput for a BER of 0.00003.

Solution: Frame length, n = 512 bytes. Information bits, k = 506 bytes. Delay time, td = 20 ms. Frame transmission time, tf = 8 ms. Window size, N = 3. (a) We know from the previous example in lecture 14, that the sending window becomes full before an acknowledgement is received (since N is less than 6) and the efficiency is not optimized. Under these circumstances, the efficiency is given by: U = (N) / (1 + 2a) where a = td / tf = 20 / 8 = 2.5 therefore: Efficiency (U) = (3) / (6) = 0.5 which has the percentage value of 50%.

(c) In the absence of errors, the throughput is given by: Throughput = (No. of Information bits transmission) / (Total time taken). = (Nk) / (tf + 2td) = (3 x 506 x 8) / (8 + 40) x 10-3 = 253 Kbps. (d) Bit error rate, E = 3 x 10-5 = 0.00003. Frame error probability is given by: P = 1 (1 E)n = 1 (1 0.00003)4096 = 0.11563. The effect of errors and the go-back-3 ARQ strategy will be for frames to be retransmitted. The average number of transmissions, m, is given as: m = (1 +NP) / (1 P) = (1 +(3 * 0.11563)) / (1 0.11563) = 1.52299 The total time taken by the data transfer will be increased by this amount, giving a value for throughput as follows: Throughput = (Nk) / m (tf + 2td) = (253) / (1.52299) = 166.12059 Kbps. The throughput has been reduced to under half the transmission rate.

Example 2: If the signal rate is 100 Mbps and frame length is 2000 bits, calculate the value ofThe factor (a = td / tf) for the following medium lengths: (a)250m. (b)4000m. Assume signal propagation, typical of twisted-pair cable, to be 2 108 m/s. Solution: (a) The factor a is given by: a = (Propagation delay) / (Frame transmission time) Propagation delay = (250 m) / (2 x 108 m/s) = 1.25 s. Frame transmission time = (No. of frame bits) / (Signal rate) = (2000 bits) / (100 Mbps) = 20 s. Therefore: a = td / tf = 1.25 / 20 = 0.0625. (b) If the medium length is changed from 250 m to 4000 m then a is increased by (4000 / 250) or 16 times. Hence a becomes 1, which would lead to a marked deterioration in throughput under heavy-load conditions.