Scenario 2 has a well-defined set of traffic flows in a network whose links have fixed bandwidths. It should be possible to determine a set of rates for each flow which ensures the highest link utilisation, where possible, and also divides the link capacity fairly amongst the flows.
In Scenario 2, the bottleneck in the system is the 64kbps full-duplex link 2 3. We will assume that all acknowledgment rates are limited by their respective data rates by
(4) |
where denotes a data flow from source to its destination and denotes the corresponding acknowledgment flow in the reverse direction.
At time , the flows and commence. The optimum rates for both are:
Time | Optimum Rate (bps) | |
64,000 | 1706 |
At time , the flows and commence. As the link 2 3 is the bottleneck, we can write limiting equations for flows across the link 2 3 and across the link 3 2:
(5) | |||
(6) |
(2) is the limiting equation, as the data flows are substantially larger than the acknowledgment flows. By sharing bandwidth equally to the flows, we obtain the rates:
Time | Optimum Rate (bps) | |||
32,000 | 853 | 32,000 | 853 |
At time , the flows and commence. Here, the flows are in the opposite directions to the previous four flows. Again, we can write limiting equations for the flows:
(7) | |||
(8) |
Distributing bandwidth fairly, we have and . Assume both links are fully utilised and substituting for the acknowledgment rates using (1), we have:
(9) | |||
(10) | |||
(11) |
Substituting (8) into (6) gives , and the table:
Time | Optimum Rate (bps) | |||||
31,168 | 831 | 31,168 | 831 | 62,337 | 1,662 |
The six flows continue to transmit until time when the flows and commence. Again, distributing bandwidth fairly, we have and . Assume both links are fully utilised and substituting for the acknowledgment rates using (1), we have:
(12) | |||
(13) | |||
(14) |
Substituting (11) into (9) gives , and the table:
Time | Optimum Rate (bps) | |||||||
20,779 | 554 | 20,779 | 554 | 62,337 | 1,662 | 20,779 | 554 |
As flow only has 100 packets to send, it finishes quickly, and the rates return to the values before flow started:
Time | Optimum Rate (bps) | |||||
31,168 | 831 | 31,168 | 831 | 62,337 | 1,662 |
Similarly, flow has most of the bandwidth across the link 3 2, so it is the next to finish, and the remaining flows return to the rate values before flow started:
Time | Optimum Rate (bps) | |||
32,000 | 853 | 32,000 | 853 |
The flows , , and continue at these rates until one or both finish.