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Unformatted text preview: c. Packetization
delay (ms) Transmission
efficiency 2 1.0
N D32 0.9 4 0.8
0.6 8 16 32 64 128 Data field size of cell in octets
A data field of 48 octets, which is what is used in ATM, seems to provide a
reasonably good tradeoff between the requirements of low delay and high
3.9 a. The transmission time for one cell through one switch is t = (53 × 8)/(43 × 106 ) =
b. The maximum time from when a typical video cell arrives at the first switch (and
possibly waits) until it is finished being transmitted by the 5th and last one is 2 ×
5 × 9.86µs = 98.6µs.
c. The average time from the input of the first switch to clearing the fifth is (5 + 0.6
× 5 × 0.5) × 9.86µ s = 64.09µs.
d. The transmission time is always incurred so the jitter is due only to the waiting
for switches to clear. In the first case the maximum jitter is 49.3µs. In the second
case the average jitter is 64.09 – 49.3 = 14.79µs.
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This note was uploaded on 02/20/2012 for the course ENC 3250 taught by Professor Dr.timothycerner during the Fall '11 term at UNF.
- Fall '11