GB2417852A - Network data flow control - Google Patents

Abstract

A method of determining the blocking probability of traffic, Pm at a server M, said server having at least two inputs, at least one being Poisson, and one of which is from one or more nodes which are non-Poisson, (non-Poisson node(s)) but whose input is/are Poisson comprising the steps of:: viewing the arrangement as a network having a single Poisson input with incoming traffic equal to the sum of the Poisson traffic into the 1<st> set of nodes and the Poisson traffic into server M; and for each of said non-Poisson nodes determining the sum of the blocked customers/second and using equivalence to determine Pm.

Description

24 1 7852

NETWORK DATA FLOW CONTROL

The invention has a wide variety of applications including, but not limited to, telephone networks, computer networks, optical networks (e.g., optical burst switching network), wireless networks production lines and manufacturing systems and traffic classes. The term "network" as described hereinafter should be interpreted as such. In general the network comprises a number of interconnected nodes (these may be e.g., processors in a telecommunication system), where information flows between the nodes by links, e.g., in packets along wires. The term "pathway" comprises several network nodes joined by links.

The invention relates to a method of determining blocking probabilities in the nodes receiving non-Poisson traffic in merge networks. Merge networks are those networks like the one at the top of figure 2. When one moves from source (left of the figure) to destination (right) there are only network branches which converge (or merge) into other branches. It is like a tree structure starting from the top of the tree to the bottom (to the root).

Poisson traffic, referred to throughout this specification, means that the packet arrivals are described by a Poisson Arrival Process. This is the same as saying that the inter-arrival time between consecutive data packet arrivals is exponentially distributed.

Note that by saying that we have Poisson traffic we determine probabilistically the time epochs at which packets arrive but we say nothing about their size. The packet size distribution has nothing to do with the question whether the packet arrival process is Poisson (i.e., have Poisson traffic) or not. Since the volume of traffic is affected by the packet size distribution.

In each network node blocking might take place. Thus, a packet that.goes through one of these nodes is blocked with a certain blocking probability. The blocking probability of a pathway is the probability that a packet is blocked somewhere in the pathway, that is, that the packet is blocked in the first node of the pathway OR in the second OR.... OR in the last node of the pathway. Node blocking probability is the probability that a certain packet is blocked in a node. For instance, with a blocking probability of 0.1 of every 100 packets that go through a node, 10 will be blocked.

Physically seen if we have two or more packets that arrive at a node and wish to be sent through the same output port (this is called contention), only one of them can get through it. The other packets could wait in a buffer until the output port is once more free, but since there are no buffers in networks which are not store and forward, they will be directly blocked (i.e., dismissed, thrown away). The blocking probability measures the probability that this happens.

The Erlang B formula is widely known and has proved to be extremely reliable lo and accurate for calculating blocking probabilities when incoming traffic to the node is Poisson. The problem however, is that once a Poisson traffic stream has become blocked, it is not Poisson any more, and the Erlang B formula does not work.

A link load of one Erlang means that you offer to the link in average exactly as much bits per second (bps) as it can carry. For instance, if the average packet arrival rate is 1 packet/second, and the average packet size is 1 Mbyte/packet, the offered load to a link is their product: 1 Mbyte/second. If the link has a capacity of 1 Mbyte/second then we have a link load of 1 Erlang. If the link has a capacity of 0.5 Mbytes/second, then we have a link load of 2 Erlangs, and so on.

The problem has been partially solved using only approximated solutions which have been given to the problem of calculating the blocking probability in open networks with traffic classes. In adopt the nature of these solutions is iterative and recursive which means long calculation times, as they are only valid for certain restricted topologies.

The invention comprises a method of determining the blocking probability of traffic, Pm at a server MJ said server having at least two inputs, at least one being Poisson, and one of which is from one or more nodes which are non-Poisson, (non- Poisson node(s)) but whose input is/are Poisson comprising the steps of: i) viewing the arrangement as a network having a single Poisson input with incoming traffic equal to the sum of the Poisson traffic into the lS' set of nodes and the Poisson traffic into server M; ii) for each of said non-Poisson nodes determining the sum of the blocked customers/second and using equivalence to determine Pm.

With this method, the blocking probabilities of any merge network can be calculated. The advantages of the method are that it provides for the first time a direct, simple and exact solution, it is valid for any split or merged cut-through open network combining blocked and/or non-blocked Poisson traffic, It is valid for any arrival rate values and It is valid for any service time distribution. It is also applicable for nodes with one or multiple servers, solving networks with several aggregation stages (tandem), all circuit switched networks (since all of them are cutthrough) and those packet switched networks that directly forward the information without storing it in the nodes (i.e., cut- through).

The invention will now be described by means of example, and with reference to the following figures: Figure 1 shows a generalized network Figure 2a shows a one stage sequence network Figure 2b shows an equivalent of figure 2a Figures 3a, 3b and 3c show examples of a network and its equivalents Figure 4 shows a flow diagram of the method step to calculate blocking probability where there are different priority classes Figure S. 6 and 7 illustrate a worked example Figure 8 shows how the process can be repeated when there are several stages Figure I shows a network 1 having a number of outer nodes Al, B1... Gl, and a plurality of inner nodes A, B... G. At the outer nodes (depicted by a disc. the blocking probability can be determined by the Erlang B formula. Within the inner ring of outer nodes the traffic is non Poisson and no exact solution for the blocking probability can be readily determined. The thick arrows show the flow of all packets coming from node Al. No, nodes Al to Fl are NOT physically different to nodes from A to G. The only difference is that in the first nodes you can use the Er]ang B formula, but the second ones not. The thin dines, (in fact all lines) are bidirectional, so data flows in both directions.

Figure 2a shows a one stage merge network and figure 2b shows an equivalent network. They are equivalent in the sense that they block in average the same number of packets per second when they are offered the same input traffic load.

In a large network parts of it that look like the network at the top of figure 2 are identified and they are then considered as figure 2b. By doing this substitution the blocking behavior of the big network is not changed since due to the theorem of 0 equivalence, both networks in figure 2 block the same number of packets per second.

The substitution simplifies the larger network successively until a network of one node is obtained as in figure 2b, which blocks in average the same amount of packets per second as the whole big and complicated original network. When you look at a large network from source to destination therefore, the invention involves identifying components that look like the top of figure 2 and substituting by their equivalent networks that look like the bottom of these figures. All these nodes that you substitute by just one node represent one stage of the network. Once you have made the substitution you go again from source to destination identifying equivalent networks, that is substituting the next network stage by its equivalent. The terms arriving customer: is a packet that arrives (customer = network packet) and service time: relates to the packet size (service time = packet size/link speed). . .

For the Figure 2 representation, the equivalence which is the core of the . invention, can be stated mathematically as follows: A' P + [amp (} P) + 365- pM = (3, + 26) Ps Equation a ' 25 Where l, *A is the average number of blocked customers per second in the server i,(+ ++2B)*PB is the average number of blocked customers per second in the server B and [2,-(1-P,)+2B]-PM is the average number of blocked customers per second in the server M. In equation 1, the blocking probabilities PE,Nli (l,...,n) and PB can be calculated directly with the Erlang B formula, since those nodes receive Poissonian traffic. Therefore, it is possible to express PM as a function of them: (2 + JIB) P. I,; P PM = n Equation Z2. (I P. )+ SIB with: P' = Em ('l, ,u)\li (1, ..., n) PB Em ((I + JIB) flu) Equation y I=] Where En'(p)is the Erlang B formula for m servers and a load of lo p =*,Uerlangs.

The case in figure 2 also contemplates the possibility of having several Poisson sources going to the optical serve M. Due to the additive property of Poisson processes, they are equivalent to a single source of a rate ilB equal to the addition of all their arrival ..

. . . rates.

. 15 Finally, the results obtained in the networks of figures 2 and 3 are generalized for the case of networks with several stages (tandem networks). The traffic coming out ace..

from both networks in figure 2 and from both networks in figure 3 has exactly the same statistical properties. This will allow us to substitute in figure 2 and figure 3 the complex network above by the simple network below respectively, in the case that the ë.

network above is a section belonging to a bigger network. Equation 2 can be then applied successively to the different stages of a tandem network in order to solve the different blocking probabilities of nodes receiving non-Poissonian traffic. Therefore in summary the method step comprises: calculating the blocking probability with the Erlang B formula in every merge network that receives exclusively Poisson traffic, applying on each stage equation 2 to solve each particular group of the stage (see figure 4), and substituting e in the stage every group by its equivalent simplified network (the s network at the bottom of figure 2 and figure 3), move one stage to the right and repeat the whole process again.

The invention is to realises that the two networks in figures 2 and 3 produce an output traffic statistically identical. In particular they block in average exactly the same number of customer per second.

Example 1

Figure 3a shows an example of a packet switched network, which is an optical burst switching (OBS) networks.Four different traffic sources (src_O, src_2, src_3, src_4) generate Poisson traffic of intensity = 1 packet/seg. The optical switches ql, q2, q3, q4 have an average service time of,u = 10-3 seg/packet. The blocking probabilities in ql and q2 are calculated according to the traditional Erlang B formula.

The blocking probability in the nodes ql and q2 can be calculated with the Erlang B formula: Node ql: Pb' = 0.0099 Node q2: Pb2 = 0.0099 Applying the method to the node q3, the network can be substituted by the following equivalent, as shown in Figure 3b. ëa a

And the blocking probability can be calculated according to equation 2: . a a a 2s Pd3 = = 0.0227 . a a . a

Further enhancement for priority classes.

The term "differentiated services" is a way of introducing priority classes in traffic flows of packets. Imagine several packets that go through a node. Introducing traffic classes in the fashion of differentiated services means that you have, for instance, s two different classes of packets: the high priority packets and the low priority ones.

Packets with a high priority have a special bit field in the packet header set to 1 whereas the others have this bit field set to 0 (for instance). When contention takes place in a node using a differentiated services scheme, high priority packets (with the bit set to 1) will always be chosen to be sent through the output port before low priority packets 0 (with the bit set to 0). That is, high priority packets always win contentions against low priority packets. One can say that high priority packets do not see low priority packets, because they are not affected. As a result of all of this, high priority packets experience a lower blocking probability that low priority packets. This blocking probability is however, not zero, since when contention between two high priority packets take place, one of them has to be blocked, so high priority packets can also be blocked by other packets of the same priority. We can easily extend all these schemes for more than two priority classes. The term "blocking probability per priority class" is the blocking probability for the packets belonging to each one of the traffic classes. In our example above, it would be the blocking probability for the high priority packets and the blocking probability for the low priority packets.

The conservation law relates to the following: imagine that you have a certain node and a flow of packets arriving at it. We calculate the blocking probability for this flow of packets: Pb. If one has the same node and the same flow of packets, but a division is made and say that some packets have higher priority than others. If in total 2s there are a number of n traffic classes (priority distinctions), the blocking probability for the packets belonging to each traffic class: Phi, with I ranging from 1 to n, is calculation.

The conservation law says that the sum of the blocking probabilities for each traffic class is equal to the blocking probability of the original low without a division of priorities: Pbl + Pbl + ... + Pbn = Pb.

General Methodology for Multiple Priority Classes If we have a case illustrated in figure 2 with a total number of traffic classes, 1 Poisson stream and N non-Poisson traffic stream.

The case in figure 2 also contemplates the possibility of having several Poisson sources going to processor M. Due to the additive property of Poisson processes, they are equivalent to a single source of a rate OR equal to the addition of all their arrival rates.

The problem consists on solving the blocking probabilities Pi' Eli {1,..., N jJ{M}, \/k {1,..., C}, in each processor i for every priority class k.

Consider that the priority of class i is higher than that of j when iSi. According to a blocking scheme as described, priority class i is perfectly isolated from any priority classy if iSi. This isolation means that the blocking probability for packets with priority i is not affected by packets with a lower priority. In particular, following the same principle the subset of priority classes S = {I,...p,' is isolated from the subset of priority classes S2 = {p+l' ..., C), Alp {1..., C}. Using this property, the method to calculate the blocking probabilities per priority class in a network like figure 2 is described below.

Let P,s be the blocking probability in the processor i of the group of priority classes S. STEP 1 Initialise A = S to be the total set of priority classes (1, 2, ..k, , K) - K = total number of priority classes, the set of priority classes to be considered in the network.

The blocking probability for the whole set of priority classes A in processors from I to N I in figure 2 can be calculated according to the well-known furlong B formula for N servers and p Erlangs of load: Em(p). The exact expression to calculate Pin Vi {1,..., N} is: P.! A = Empty,' IN) brie {1' IN} Equation] keA The blocking probability for the whole set of priority classes A in processor M in figure 2 is calculated by:

N N

it, k + FIR, k PB, A- PE,, ill, k PM A = =i ken keA =i keA (1 BE,.) k + ilB, k Equation 2 =1 keA keA PER = Ems k Ill I, vi (19 N) PBA = Ems Eli k + zA B, ) /U) Equation 3 STEP 2 If the subset S has elements, separate the last element {k = K,' of the set of the lo priority classes S. and makes S = fl, ..., K-l}. The set of priority classes S is now isolated from the priority class k = K. Therefore, we can proceed as if there were only two traffic classes, one representing the subset S and other representing the priority class n. In addition define the set of priority classes of which solution has already been calculated as R = {n+l, ..., C} (in the first loop of the algorithm R = d> ).

STEP 3 : . . Calculate -,5 it' e {1, ., N} i is not an equivalent network introduced in a a. previous step. PiS can be calculated with the Er]ang B formula since it is isolated from the set of priority classes. That is, packets with a priority belonging to S are not ". 20 interfered of affected by packets with priority belonging to {n} u R. Pl. = Ems, k it, Hi (1, , N)/ i is not an equivalent network Equation 4 l..e STEP 4 Calculate from the conservation law the blocking probability for priority class n in each processor: P,e{1,...,N} ' is not an equivalent network introduced in a previous step.

Nit, e {1, , N}, is not an equivalent network Equation 5 p = (keA) ',A 5 ok P. k 2,n STEP 5 Calculate for the processor M the blocking probabilities of the set of priority lo classes S. according to: . (,k+2Bk) PB S PF-'k p 5 ==1 keS keS =1 keS Z' (1 PEI S)Z, 11 k + Equation 6 =1 keS keS e.e..

P,,s = Em (ok ')7 Fit (17...7 N) keS a PB,S Em ((I k + 2 B,k) /) Equation 7 =1 keS INS ...s . STEP 6 Calculate from the conservation law the blocking probability PM,n for priority class n in processor M according to: Equation 8 ( i] k (1 Pi A)) PMA (I (I PIA) 2t k) PMS (Z (1 P.! A) k PMk) on (1 P[.A) t=/ STEP 7 If S has more than one element, then jump to step 2. Before doing so k is to be decremented by 1. Otherwise, if S has only one element, then the all the blocking probabilities for this stage have been obtained, and we can proceed to follow step 8.

STEP 8 In this case, the calculated P., s b,, {1,..., N} and PM,S are in fact the blocking probabilities for the priority class 1: Pi,,, {l,...,N} and PM,I- At this point all the blocking probabilities P. I, {1,...,N}U{M} ,VkG {1,...,C}, in each processor i 0 and for every priority k have been calculated.

DESCRIPTIVE SUMMARY

Imagine we have 3 priority classes, and priority 1 is higher than 2, which is .

. higher than 3. So we have a set of A = {1, 2, 3} priority classes and we want to calculate their respective blocking probabilities. We first calculate the blocking probability of lowest priority class (3). What we make in step 2 is to say that we want to calculate the block. Probability of priority class 3 making n = 3, and that we are going to take into account the interference of the rest of the higher priority classes S = { 1, 2} as one single higher priority class. Packets from a priority class x are not affected by packets with a lower priority class y (y > x) because when there is a conflict with lower class packets, they always get the resources. On the other hand, packets from a priority class x are affected by packets with a higher priority class z (z < x) because when there is a conflict with higher class packets, they are always blocked (the higher priority packets get the resources). So at this point we consider how the higher priority classes S = { 1, 2} affect the blocking of our lower priority class K (=3) but without taking into account which higher priority class in particular (1 or 2) is actually blocking our lower priority class 3. With steps 2-7 we calculate the blocking probability for the lower priority class 3 in all nodes of the network.

Now we want to calculate the blocking probability of priority class 2. Since it has a higher priority than priority class 3, it is actually not affected by it, so we just forget about priority class 2 and remove it from our calculations. This is made in step 2, where we say that this time k = 2 (this is the blocking probability that we want to calculate now) and that S = { 1} this is the set of priority classes higher than our priority class 2. Note that S is being reduced in size very time we are in step 2 (that's the mathematical trick), because we are intentionally forgetting in our analysis the lower priority classes which we have already calculated. SO, we go again from steps 2-7 and we calculate all the blocking probabilities for priority class 2. But this time S has only lo one element (step 7), and as a secondary result from our calculations in steps 2-7 we also calculated the blocking probabilities for the set of priority classes S. Since this time this set has only one element, we have just calculated "by-the-way" the blocking probabilities for the highest priority class (1). So in step 8 we just "copy" these results from our intermediate calculations and put them in the right variables.

Example 2

The case of split cut-through networks is very simple. In a cut-through network the nodes have no possibility to store arriving customers before forwarding them. This implies that the average service times in each processor are the same in the whole network. According to this, all the blocked customers in figure 3 will be blocked in processor M, and therefore the blocking probabilities for any priority class in processors 1, ..., n are zero. The blocking probabilities for any priority class in processor M can be calculated with the conservation law and the Erlang B formula.

This is illustrated in figure 4.

Example 3

If we consider the network shown in figure 5 with four priority classes. Three different Poisson sources send traffic with four different priority classes. The traffic has the following characteristics: l' = \2 = \3 = = I customer/sec, and the service time is /1 = 10 2 SeC / CRStOmer. We wish to calculate the blocking probabilities per priority class in each processor: Pl,l, Pl,2, P2,* P2,4, P3,1, P3,2, P3.4, P4,1, P4,2, P4,3 and P44.

The network in figure 5 has two stages. In order to apply the method described in figure 3, we concentrate on the first stage; illustration of figure 6.

s We begin to follow the method described in figure 3. note that N = 2.

STEP 1 A=S= {1, ...,4}.

P. A = El i l fin p,2} k=l Therefore P1,A = P2,A = E(2*10 2)=0.0196.

P 4 PB A-2 - PE,,4-2 - PE2,4 0 0192 -2 (1-PEI,4) + 2 (1 PE2 4) ë where, ..

PEI.4 = PE2,4 = El 2-10)= 0.0196 PB A = E(4 -10)= 0.0385 STEP2 A. I, . n=4,S=1, ,3},R=q). i...e

STEP 3 from equation 4 it follows: Pl,S = El (2 10 2) = 0.0196 P2,S = El (1 10 2) = 0.0099 STEP 4 One of the solutions of equation 5 (Ply = 0) is trivial, since there is no priority 4 traffic in processor 1. The other solution is one of the desired blocking probabilities: P (2) P2. A-P2.s 0 0293 STEP 5 P 3 PB S-2 PEAS-PEAS 00130 2 (1-PEI 5)+2 (1-PE2 5) where PE, 5 = El(2 - 10-2)= 0.0196 PE25 = El(2- 10-2)= 0.0099 PB S = E'(3-10-2)= 0.0291

I

. STEP 6 P L2 (1-Pl.A)+2(1-P2. A)3 PM A-L2 (1-PI A)+(1-P2. A)A PM S O 0386 STEP 7 Since S has more than one element, jump to step 2.

STEP 2 n=3,S={1,2},R={4} STEP 3 from equation 4 it follows: Pl,S= El (2 10-2) = 0.0196 P2S= E/ (1 10-2) = 0.0 STEP 4 One of the solutions of equation 5 (Pl,3 = 0) is trivial, since there is no priority 3 traffic in processor 1. The other solution is one of the desired blocking probabilities: P (2). P2 A-P2 4 0 0099 l s STEP 5 PMS = 0 STEP 6 P3 3 = L2 (1 P' A)+ 2 (1 P2 A); PM A (1-P2 4)- P3 4 O. 0383 STEP 7 Since S has more than one element, jump to step 2. STEP2

n=2,S=(l},R=(3,4} e. .

STEP 3 . e.

from equation 4 it follows: PlS=El(I 102)=o.oogg P2,s=E'(O 10-2)=0 STEP 4 One of the solutions of equation 5 (P2,2 = 0) is trivial, since there is no priority 2 traffic in processor 2. The other solution is one of the desired blocking probabilities: P.} 2 = (2) = 0.0293 STEP 5 PM,S = 0 STEP 6 p L2 (1-Pi, A)+ 2 (1-P2, A)] PM, A L2 (1 P2,3)+ (1 P2 4)1 P3 4 = 0 (1 - PI,2) STEP 7 S has one element.

lo STEP 8 Pl,l = Pl,S, P2,1 = P2,S, and P3,/ = PITS STEP 9 Substitute network in figure 6 by its equivalent in figure 2. This is equivalent to lea. 15 the network of figure 7. ease, C

We begin to follow the method described in figure 3. note that N = 1.

STEP 1 A=S=1, ,4} PI, A = E'(4 10-2) = 0.0385 and PB A-4 PEI 4 4 (1-PEE)+ 1 where 2s PE,,4 = E'(4 10-2) = 0.0385 PB,A = E1(S 10)= 0.0476 STEP2 n=4,S={1, .

,3},R=(P STEP3 Since processor 3 is an equivalent network, do nothing (see equation 4). STEP4..DTD: Since processor 3 is an equivalent network, do nothing (see equation 5).

lo STEPS P. 4 PB S-3 PEI S O 0170 3 (1-PE, S)+ 1 where PE,. S = El(2 1 0-2) = 0.029 1 tale eee PE S = El(3 10)= 0.0385 acre 15 t t) STEP6 L4 (1-Pl A)+ 1 PM A L3 (1 Pl A)+ 1 P = 00190 : (1-PI, A j ..: e STEP7 Since S has more than one element, jump to step 2. STEP2

n=3,S={1,2},R=4} 2s STEP3 Since processor 3 is an equivalent network, do nothing (see equation 4). STEP4

Since processor 3 is an equivalent network, do nothing (see equation 5).

STEPS

PM,S = 0 STEP6 p L4 (1-P! A) + 13 PM A-(1-P.! A) P4 4 O. 03370 (1-Pi A)+1 lo STEP7 Since S has more than one element, jump to step 2. STEP2

.'. n=2,S= {1},R= {3,4} .. STEP3

Since processor 3 is an equivalent network, do nothing (see equation 4). STEP4 I.

20 Since processor 3 is an equivalent network, do nothing (see equation 5).

STEPS

PMS = 0 STEP6 P4.2=0 STEP7 S has one element.

STEP 8 P41 = PMS, = 0 s Summarising the simulation and analytical results, we observe that they lead to the same blocking probabilities: Blocking probability (simulation) Block probability (calculation) | P. 0.00994 0. 0099 Pll2 0.0294 0.0293 P2,3 0.00995 0.0099 P24 0.0294 0.0293 P3.1 O O P3, 2 o 0 P3.3 0.0391 0.0383 P3,4 0.0390 0.0386 P4,} O O P4,2 O P4,3 0.0342 0. 0337 P4,4 0.0195 0.0190 Table 1: Comparison of the simulation and analytical results

Claims (10)

1. In a method of determining the blocking probability of traffic, Pm at a server M, said server having at least two inputs, at least one being Poisson, and one of which is s from one or more nodes which are nonPoisson, (non-Poisson node(s)) but whose input is/are Poisson comprising the steps of: i) viewing the arrangement as a network having a single Poisson input with incoming traffic equal to the sum of the Poisson traffic into the is' set of nodes and the Poisson traffic into server M; ii) for each of said non-Poisson nodes determining the sum of the blocked customers/second and using equivalence to determine Pm.

2. A method as claimed in claim 1, wherein step ii) comprises: . (a) determining for the non-Poisson input, or where there are a plurality of c 15 non-Poisson inputs, the blocking probability at the or each nonPoisson node; eea (b) determining the blocking using the equation: a ä-e a n n it + ilk PBI PE, : PM =- ,=1 I=! I, 22 (1-PEI) + ilB =!

3. A method as claimed in claim 1, wherein step i) uses the Erlang B formula.

4. A method as claimed in claim 3 where -Ps,=._(l 10, I) i= a) 1 - 1 PEI = EHI(;l] ,U)Vi (1,..., n) ( n) )

5. A method where there are more than one input which is Poisson, but these several Poisson inputs are regarded as a single Poisson input.

6. A method as claimed above, wherein there are a plurality of different traffic classes having different priorities, of determining Pm for each different priority class.

A method as claimed above comprising the steps of i) determining PEA (PA) for each processor i (all processors 1 to N) for all lo priority classes as an aggregate, ii) determining PM A the Probability for processor M for all priority classes,

the classes being determine as an aggregate, iii) setting S= to be the set of classes (1, 2, . K-1) where K is total number of classes :., 15 iv) determining P,,s determining for each processors P,,s where Pi, S is the P for each processor for all priority classes of set S as an aggregate and : v) determining P,,k for each process for priority class k, vi) determining PM S where P is the blocking probability for processor M, for all priority classes as an aggregate in the set S Vii) determining from the above PM,k viii) decrementing S by 1 and repeating steps until iv to vii until priority class k required is determined

7. A method as claimed in claim 5, wherein step i) uses the formula: P,A = Em( k il) \]i {1 N}

8. A method as claimed in claim 5, wherein step ii) uses the formula:..CLME: N N

il,k+2Bk PB,A-PE.,',k PM. A = - !=1 keA keA i=l keA I, (1-PE! )E il' k + IB, k =1 keA keA and PE! A = En( k U) 7i (1 N) keA e,.

PB, A = Em(( 2' k + iB k) ) =1 keA keA

9. A method as claimed in claim 5, wherein step iv) uses the formula: : * P.' s = Em Zi k U 7 i (17 7 N)/ i keS

10. A method as claimed in claim 5, wherein step v) uses the formula: (1, 7is not an equivalent network ( ,k,n) F:,A Prs 2, k,k p = ke A ke S ,n A method as claimed in claim 5, wherein step vi) uses the formula:

N N

il, k + iB.k PB, S- PEI, s h, k n =1 keS keS r=1 keS rM,S = N (1-PEI _ 5) , k + lB, k =1 keS keS and P,,s = Em (2,,k 'p), i (1,..., N) keS

N

PB,S Em ((IJ 2, k + iB,k) ) =l keS keS ë 12. A method as claimed in claim 5, wherein step vii) use the formula: e se.. 10 e C e. _ _ N k PMS- 1-Pl A k PMk k 1 Pl A PMA 1 PIA z ' ( ')) (( )z) (( )E ) I =I keA l =I keS l =I kR PM,r = N itl,n (1- Pl,A) 1=1 e eeeee e