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In multiple token operation, an idle token or the next busy token is generated immediately after the data frame leaves the source host. The effective service time is X/R, show below:
Average Transfer Delay in General
In general, the average transfer delay for multiple token operation is:
For fixed frame lengths:
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Average Transfer Delay for Fixed Frame Lengths
Average Transfer Delay for Exponentially Distributed Frame Lengths
Single Token Operation
In single token operation, there are a couple cases to consider – when the time to transmit a frame is greater than or equal to the ring latency and when the time to transmit a frame is less than the ring latency.
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In this case, the link is unavailable while the transmitter waits for the busy token to return. A gap in time occurs between the end of the data frame and the start of the subsequent idle token or busy token. During this time, the transmitter waits.
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The normalized ring latency, a', is similar to the normalized propagation delay, a = τ/(E[X]/R), in random access LANs. The normalized ring latency can be expressed as:
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The average transfer delay is expressed differently for different frame distributions. Let's consider fixed length frames and exponentially distributed frame lengths.
Fixed Length Frames Average Transfer Delay
If a' <= 1, T is given by the expression for multiple token operation with fixed length frames. If a' > 1, E = τ' for each frame, which implies for fixed length frames:
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Note that for a' > 1, stability is achieved only if Sa' > 1, i.e. if S < 1/a'. Also note that for a' > 1, W does not depend on E[X]; however, T does depend on E[X].
Exponentially Distributed Frame Lengths Average Transfer Delay
For some frames, X/R <= τ' and E = τ'. For other frames, X/R > τ' and E = τ'(X/R). X/R is an exponentially distributed RV with mean E[X]/R, so:
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In single frame operation, the idle or next busy token is generated immediately after the end of the data frame returns to the source host.
The
Average Transfer Delay in General
In general, the average transfer delay for single frame operation is:
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Note that the following must hold true for the system to be in a stable state:
Average Transfer Delay for Fixed Length Frames
The expression for T above can be simplified for single frame operation with fixed length frames:
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Average Transfer Delay for Exponentially Distributed Frame Lengths
The expression for T above can be simplified for single frame operation with exponentially distributed frame lengths:
Examples
Example 1
Assume:
- ring length of 1 km
- link rate of 4 Mbps
- average frame length of 1000 bits
- 40 hosts in ring
- Poisson arrivals at rate 10 frames/second per station
- host latency of 1 bit
- propagation velocity of 5 us/km
- single token operation
So:
- R = 4 Mbps
- E[X] = 1000 bits
- M = 40
- λ = 10 frames/second
- B = 1 bit
Thus we can find the ring latency:
the normalized ring latency:
and the normalized throughput:
We don't know the frame distribution, but we do know how to solve for fixed length and exponentially distributed. If frames are fixed length, then since a' < 1:
If frame lengths are exponentially distributed, then:
Example 2
Assume the same as Example 1, except the station latency, B = 10 bits. Then the ring latency, normalized ring latency, and normalized throughput can be computed as:
If frames are fixed length, then since a' < 1:
If frame lengths are exponentially distributed, then:
