CHAPTER 3. MAC SCHEME– PROBABILISTIC QUOTA PLUS CREDIT
3.2 B ANDWIDTH A LLOCATION - P ROBABILISTIC Q UOTA D ETERMINATION
Assume that there are S server-nodes (S-node 1 to S-node S) in the network dividing itself into S sections (sections 1 to S). Each section contains more than one node including the server node of the section. To simplify the illustration, the S-node for a section is placed in the most downstream location in that section. Namely, section 1 is preceded by S-node S in section S; and section k is preceded by S-node
k-1 in section k-1, for k=2 to S. More specifically, for a network with N nodes, we
have , where N
1 S
k k
N N
=
=
∑
k is the total number of nodes in section k.Moreover, a slot passed by an S-node is considered as either Available
Bandwidth (AB) if the slot is empty or erased, or used bandwidth (UB) if the slot is
non-empty and cannot be erased (have not been read) (see Figure 5). Thus, the quota for a node can be computed as the mean value of the total amount of AB observed by a section divided by the total number of nodes in the section. For instance, in an observed section (referred to as section b), we attainQ
b =AB N
b/ b, where Qb denotesthe quota to be allocated to any node in section b,
AB the mean value of the total
b amount of AB passed down by S-node b-1, and Nb is the total number of nodes in section b. Notice that the value ofAB is relevant to the traffic destination
b distribution; and S-nodes are possible to receive more traffic than O-nodes.Accordingly, we derive in the sequel a closed form for
AB under two different
b destination distributions.Case 1
In this case, traffic is uniformly distributed to all nodes. Moreover, for simplicity we consider a prevailing case in which S-nodes are evenly located in the network, namely Nk =N/S, where k=1 to S. Accordingly, each node is given the same AB and quota Q, where
Q
=AB N S
/( / ), for all nodes in the network.The value of AB can be computed as the mean total bandwidth (total number of slots in a cycle) minus UB. Thus, in the sequel we analyze the UB value passed through S-node b-1. The analysis is presented in two parts: one is to consider the transmissions from any section to itself, and the other is to consider the transmissions from a section to the other sections. For the first part, since the total amount of traffic (slots) generated from any section (take section k as an example) is Q·(N/S), thus the traffic amount from section k to section k itself is Q·(N/S)/S. Within this amount of traffic, the proportion Q·(N/S)/2S will be erased by the most downstream node (S-node) of the section. Notice that erasable traffic corresponds to the traffic sent from upstream to downstream nodes within this section. Therefore, the remaining traffic,
Q·(N/S)/2S, which is sent from downstream to upstream nodes within this section, will
Free
Will be erased at Si Pass-Through Traffic
Available Bandwidth
Used Bandwidth
AB
UB Si
Section i Section i+1
W
C C
Before Si After Si
Figure 5. Quota Determination.
be passed through the entire ring and seen by the section as UB.
For the second part, take section b+2 as an example. The traffic that is sent from section b+2 and passed through the entire ring and seen by S-node b-1 and section b as UB is the sum of the traffic destined to sections b and b+1. Thus, the UB is equal to 2·Q·(N/S)/S. Finally, with all sections taken into account by summing all UB which passes through S-node b-1, and given that the total bandwidth in a cycle is C·W, where C is the total number of slots in a cycle and W the number of wavelength channels, we obtain AB as the following equation
1
In this case, S-nodes are to receive additional traffic amount than O-nodes. We derive a closed form for
AB under an assumption that a probability p
b A of destination traffic is uniformly distributed among all nodes (including the S-nodes), while the remaining probability 1–pA (=pS) of traffic is additionally destined to all S-nodes. Clearly in the case of pS=0, destination traffic is uniformly distributed among all nodes in the network. Regarding the value of pA, it can be obtained through periodic traffic monitoring via network management protocols, which are beyond the scope of this thesis.Similar to case 1, to compute the value of
AB , we analyze the
b UBvalue passedthrough S-node b-1. The value of
AB can be computed as the mean total bandwidth
b (total number of slots in a cycle) minus the mean UB. Thus, in the following we analyze the mean UB passed through S-node b-1. The analysis is presented in two parts: one is to consider the transmissions from any section to itself, and the other is to consider the transmissions from a section to the other sections. For the first part, since the total amount of traffic (slots) generated from any section (take section k as an example) is Qk·Nk, thus the traffic amount from section k to section k itself is (Qk·Nk·pS/S)+(Q
k·Nk·pA·Nk/N). Of this amount of traffic, the fraction
(Qk·Nk·pS/S)+(Q
k·Nk·pA·Nk/2N), will be erased by section k’s most downstream node,
i.e., S-node k. Notice that the second term corresponds to the traffic sent from upstream to downstream nodes within section k. Therefore, the remaining traffic,Q
k·Nk·pA·Nk/2N, which is sent from downstream to upstream nodes within section k,
will be passed through the entire ring and seen by section b as UB.For the second part, let’s take section b+2 as an example. The traffic that is sent from section b+2, passed through the entire ring, and seen by S-node b-1 and section
b as UB, is the total amount of traffic destined to sections b and b+1. Thus, the mean UB becomes Q
b+2·Nb+2·((ps/S)+(pA·Nb/N)) + Qb+2·Nb+2·((ps/S)+(pA·Nb+1/N)). Finally, with all sections taken into account by summing all UB which pass through S-nodeb-1, and given that the total bandwidth in a cycle is C·W (where C is the total number
of slots in a cycle and W the number of data channels), we obtain A Bb as the following equation1
, where 2
S
A k
b k k k
k
AB C W Q N p N U
=
N
⎧ ⎛ ⎞⎫
= ⋅ −
∑
⎨⎩ ⎜⎝ + ⎟⎠⎬⎭1
Notice that Equation (3) cannot be solved because there is more than one unknown variable (
AB and Q
b k’s) in the equation. In the following, we solve the equation under a prevailing case in which S-nodes are evenly located in the network, namely N1=N2=N3=…=NS=N/S. In this case, due to the same behavior of S-nodes, we obtain the same quota Q, whereQ AB N S
= /( / ), for all nodes in the network. With this additional equation and the simplified version of Equation (3) as1
we can attain the closed form solution for Q, as 2
With the quota determined, we are now ready to obtain the probabilistic quota, denoted as PQ. Given the total number of packets currently in the queue as npqueue, PQ
can simply be expressed as
min ( , queue)
Q