F. Theoretical Rate Calculation for Scenario 2

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 $\leftrightarrow$ 3. We will assume that all acknowledgment rates are limited by their respective data rates by

$$A_x = R * D_x,~where~R = \frac{ack~pkt~size}{data~pkt~size} = \frac{40}{1,500}$$ (4)

where $D_x$ denotes a data flow from source $x$ to its destination and $A_x$ denotes the corresponding acknowledgment flow in the reverse direction.

At time $t=0$, the flows $D_5$ and $A_5$ commence. The optimum rates for both are:

Time	Optimum Rate (bps)
	$D_5$	$A_5$
$t \ge 0$	64,000	1706

At time $t=12$, the flows $D_7$ and $A_7$ commence. As the link 2 $\leftrightarrow$ 3 is the bottleneck, we can write limiting equations for flows across the link 2 $\rightarrow$ 3 and across the link 3 $\rightarrow$ 2:

$$D_5 + D_7 \leq 64,000$$ (5)

$$A_5 + A_7 \leq 64,000$$ (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)
	$D_5$	$A_5$	$D_7$	$A_7$
$t \ge 12$	32,000	853	32,000	853

At time $t=18$, the flows $D_8$ and $A_8$ commence. Here, the flows are in the opposite directions to the previous four flows. Again, we can write limiting equations for the flows:

$$D_5 + D_7 + A_8 \leq 64,000$$ (7)

$$A_5 + A_7 + D_8 \leq 64,000$$ (8)

Distributing bandwidth fairly, we have $D_5=D_7$ and $A_5=A_7$. Assume both links are fully utilised and substituting for the acknowledgment rates using (1), we have:

$$2D_5 + R * D_8 =64,000 = 2R * D_5 + D_8$$ (9)

$$2D_5 (1 - R) = D_8 (1 - R)$$ (10)

$$2D_5 = D_8$$ (11)

Substituting (8) into (6) gives $D_5=31,168$, $D_8=62,337$ and the table:

Time	Optimum Rate (bps)
	$D_5$	$A_5$	$D_7$	$A_7$	$D_8$	$A_8$
$t \ge 18$	31,168	831	31,168	831	62,337	1,662

The six flows continue to transmit until time $t=140$ when the flows $D_6$ and $A_6$ commence. Again, distributing bandwidth fairly, we have $D_5=D_7=D_6$ and $A_5=A_7=A_6$. Assume both links are fully utilised and substituting for the acknowledgment rates using (1), we have:

$$3D_5 + R * D_8 =64,000 =3R * D_5 + D_8$$ (12)

$$3D_5 (1 - R) = D_8 (1 - R)$$ (13)

$$3D_5 = D_8$$ (14)

Substituting (11) into (9) gives $D_5=20,779$, $D_8=62,337$ and the table:

Time	Optimum Rate (bps)
	$D_5$	$A_5$	$D_7$	$A_7$	$D_8$	$A_8$	$D_6$	$A_6$
$t \ge 140$	20,779	554	20,779	554	62,337	1,662	20,779	554

As flow $D_6$ only has 100 packets to send, it finishes quickly, and the rates return to the values before flow $D_6$ started:

Time	Optimum Rate (bps)
	$D_5$	$A_5$	$D_7$	$A_7$	$D_8$	$A_8$
$t \ge Off_6$	31,168	831	31,168	831	62,337	1,662

Similarly, flow $D_8$ has most of the bandwidth across the link 3 $\rightarrow$ 2, so it is the next to finish, and the remaining flows return to the rate values before flow $D_8$ started:

Time	Optimum Rate (bps)
	$D_5$	$A_5$	$D_7$	$A_7$
$t \ge Off_8$	32,000	853	32,000	853

The flows $D_5$, $D_7$, $A_5$ and $A_7$ continue at these rates until one or both finish.