1、Rec. ITU-R F.1192 1RECOMMENDATION ITU-R F.1192TRAFFIC CAPACITY OF AUTOMATICALLY CONTROLLED RADIO SYSTEMSAND NETWORKS IN THE HF FIXED SERVICE(Question ITU-R 147/9)(1995)Rec. ITU-R F.1192The ITU Radiocommunication Assembly,consideringa) the improvements available from the use of automatically controll
2、ed radio systems;b) the limited amount of spectrum available for the HF fixed service (FS);c) the demand for high reliability of HF FS;d) the demand for high traffic throughput;e) the demand for operation of networks in the HF FS;f) that automatic systems should be able to cope with a wide range of
3、traffic requirements and network size,recommends1 that, for automatically controlled HF radio networks, the methods described in Annex 1 are preferred for: the determination of the traffic throughput capacity; the estimation of the frequency requirements with respect to desired traffic throughput; t
4、he general planning of such systems; the comparison of the different types of system with respect to traffic throughput and frequency requirements.ANNEX 1Traffic capacity of automatically controlled radio systemsand networks in the HF FS1 IntroductionThere are three basic automatically controlled ra
5、dio systems. These are identified by the following acronyms:ACC : asynchronous common channelsSCC : synchronous common channelsSSCT : synchronous separate calling/traffic channelsThe ACC system asynchronously scans a set of pre-assigned channels in order to find a usable channel. Once a suitablechan
6、nel is found, the traffic is sent over that channel; i.e. the calling and traffic channels are common.The SCC system synchronously scans a set of pre-assigned channels in order to find a usable channel. Once a suitablechannel is found, the traffic is sent over that channel; i.e. the calling and traf
7、fic channels are common.2 Rec. ITU-R F.1192The SSCT system synchronously scans a set of pre-assigned “calling” channels in order to make the initial contact. Thecall offers a number of candidate “traffic” channels. Having established the call, the two stations move to the candidatetraffic channels a
8、nd select the best.There are two basic types of network which may utilize the above systems: multipoint and star.2 Multipoint networksThe analysis given in this section is for a multipoint network where every station is equally loaded.2.1 ACC system2.1.1 Calling sequenceThe ACC calling sequence is s
9、hown in Fig. 1.TLTsTmFIGURE 1ACC calling sequenceSelect candidatechannelListen to candidatechannelBusy?YesNoNoYesCallReply?Start messageD01FIGURE 1/1192.D01 = 3 CMIn the initial listening sequence, the station selects a candidate channel and checks that it is not occupied. There is aprobability, P1,
10、 that the channel is free. If busy, the station will select a second candidate channel and try again. Thenumber of attempts, on average, will be 1/P1.Rec. ITU-R F.1192 3When the station transmits the call on the selected (free) channel, there are three probabilities that the call will besuccessful:
11、the probability, P2, that another station is not transmitting on that same channel, the probability, P3, that the destination station is not already occupied, the probability, P4, that the propagation conditions are good enough.Expressions for P1, P2and P3have been derived in Appendix 1.The number o
12、f call attempts, i.e. transmissions, on average, is:1(P2P3P4)2.1.2 Traffic capacityLet C : number of scanned channelsN : number of propagating channelsS : number of stationsE : traffic capacity (E) relating to a network of N channels and S stationsTm: length of message (s)Ts: scanning (or calling) t
13、ime (s).The number of messages/hour/station, M, is:M = 3 600 ES (Ts/ (P2P3P4)+ 2 Tm)(1)Each station will transmit M messages and receive M messages in 1 h.There is a limit to how many messages an individual station can handle. Taking into account the amount of time that astation spends “listening” t
14、o check that the selected channel is free, and assuming that the maximum capacity of anindividual station is E max(E), the maximum number of messages/hour/station is:Mmax= 3 600 Emax(TL/ (P1P2P3P4)+ Ts/ (P2P3P4)+ 2 Tm)(2)where TLis the listening time (s)NOTE 1 2Tmis to take account of both received
15、and transmitted messages.NOTE 2 The appropriate value for E maxis the subject of further studies.2.2 SCC system2.2.1 Calling sequenceThe SCC calling sequence is shown in Fig. 2.The SCC system has been assumed to be one which arranges a different order of listening out frequencies for eachstation (us
16、ing an algorithm based on the station address).The station selects the candidate channel and then needs to wait (Tw) until the destination station selects that channel.Just before this occurs, the calling station must listen to ensure that the channel is free. Thus the station needs, onaverage, to w
17、ait: Tw/ P1(s).The calling sequence is similar to that of the ACC station except that Ts, the scanning time, is now much shorter as onlyone channel is “scanned”.If the system as described for the ACC is simply synchronized, the synchronized scanned channels effectively become asingle channel and sta
18、tions compete for it at the same time. This causes the system to lock-up at a certain point andtherefore the system described above for SCC is preferred.4 Rec. ITU-R F.1192TmTsTwFIGURE 2SCC calling sequenceSelect candidatechannelListen to candidatechannelBusy?YesNoNoYesCallReply?Start messageWaitD02
19、FIGURE 2/1192.D02 = 3 CM2.2.2 Traffic capacityThe number of messages/hour/station, M, is the same as that for an ACC system (equation (1).For an SCC system:Mmax= 3 600 Emax(Tw/ (P1P2P3P4)+ Ts/ (P2P3P4)+ 2 Tm)(3)2.3 SSCT system2.3.1 Calling sequenceThe SSCT calling sequence is shown in Fig. 3. The sy
20、nchronously scanned set of calling channels are equivalent to a“slotted Aloha” channel which has a capacity of 0.37 E. The probabilities P1and P2are taken into account in this way.However the probability P3must be considered.Rec. ITU-R F.1192 5When the destination station replies on the calling chan
21、nel, both stations move to the traffic channels. The calling stationsends a probe on each of the offered traffic channels and then waits for a reply on the preferred channel. This sequencewill take TTs.TsTTTmFIGURE 3SSCT calling sequenceYesNoReply?Call on calling channelSend probes on candidatetraff
22、ic channelsWait for reply Send messageD03FIGURE 3/1192.D01 = 3 CM2.3.2 Traffic capacityNumber of messages/hour/station, M, is:M = 3 600 0.37S Ts/ P3(4)and:Mmax= 3 600 EmaxTs/ P3+ 2 (TT+ Tm)(5)3 Star networksA star network consists of a central station with a number of outstations.3.1 OutstationsAs f
23、ar as an outstation is concerned, the analysis and the values for M and Mmaxare the same as for a multipoint network(where S is the number of outstations).3.2 Central stationIf there are S outstations then the central station will need to handle 2 M S messages in total. In order to achieve this, the
24、central station will need to be able to handle a number of simultaneous radio channels. If the outstations are operating atfull capacity, Mmax, then S simultaneous radio channels will be needed at the central station in order to achieve fullnetwork throughput.If the outstations are not at full capac
25、ity, M, then fewer simultaneous radio channels will be sufficient. The number ofchannels required at the central station, R, is given by:R = M SMmax(6)6 Rec. ITU-R F.1192For ACC and SCC systems:M = 3 600 ES Tm(7)where:Tm: time spent transmitting a messageTm= Ts/( P2P3P4) + TmFor a station operating
26、at full capacity:Mmax= 3 600 EmaxTtot(8)where Ttotis the total time that a station is occupied.For ACC systems:Ttot= TL/ (P1P2P3P4)+ Ts/ (P2P3P4)+ 2 Tm(9)For SCC systems:Ttot= Tw/ (P1P2P3P4)+ Ts/ (P2P3P4)+ 2 Tm(10)Thus:R = EEmax TtotTm(11)It should be noted that this is not strictly accurate as the
27、outstation will experience lower values of P1, P2and P3becausethey are addressing a central station which is running at full capacity. The result given by the above, however, ispractical.For an SSCT system:R = 0.37Emax (Ts/ P3+ 2 (Tt+ Tm)Ts/ P3(12)4 Basic assumptionsFor assessments of traffic capaci
28、ty and system performance, the following assumptions are appropriate.4.1 MessageA standard “page” of 20 lines of 69 characters per line, may be assumed (1 380 characters).The message time, Tm, may be calculated assuming 7 bit/character, 50% FEC and 10% repeats (i.e.1 380 7 10.5 1.1 = 21 252 bit/mess
29、age).It may also be assumed that every station in the network transmits and receives the same number of messages perhour, M.4.2 Probability of propagation, P4This is a measure of the success of the system in selecting the candidate channel in the ACC and SCC systems. A figureof 63% has been measured
30、 in test systems and this figure has been used for P4.Rec. ITU-R F.1192 74.3 Scanning time, Ts, ACCA scanning time Tsmay be assumed.Ts= t (2 C + 9) (13)where:t = 0.392 sC : number of scanned channels.4.4 Listening time, TL, ACCIt may be assumed that an ACC station listens for a time equal to the cha
31、nnel scanning time, in order to check that achannel is free.4.5 Waiting time, Tw, SCCIf the network has C allocated channels then, on average, the waiting time will be:Tw= C / 2 Tss (14)where Tsis the synchronous scanning time.4.6 Traffic channel set-up time, TT, SSCTFor an SSCT system which offers
32、five traffic channels in its initial call, the following procedure may be assumed. Callsare sent, sequentially, by the calling station which then listens, in turn, on each channel for a reply. The called stationreplies on the “best” channel but its algorithm is biased to select the first offered cha
33、nnel which is the most favouredchannel according to the calling station. On average the calling station would wait 2.5 Ts(s) for the reply, but the biascauses this to be 1.5 Tsin practice.Therefore:TT= (2.5 2 + 1.5) Ts= 6.5 Tss5 The results of traffic calculationsFrom the results of calculations whi
34、ch considered the full range of diurnal, seasonal and solar cycle variations inpropagation conditions, and a wide range of data rates and network sizes, the following conclusions were drawn.5.1 ACCAs the number of propagating channels is increased the performance of the network does not necessarily
35、improve. Forexample, for a data rate of 1 200 bit/s, 6 channels perform better than 14 if the network consists of less than 30 stations.Ten channels give a better performance than 14 channels up to 100 stations. This effect tends to be more pronounced forthe higher data speeds.For small networks of
36、10 or less stations, the choice of the number of allocated channels is quite complicated but there isno advantage in allocating too many.Increasing the length of the message to two pages does not change the relative performance of the network with respectto the channels. The system is slightly more
37、efficient with the longer messages in that just over half the number ofmessages are transmitted and received per station compared with the one page message length.8 Rec. ITU-R F.11925.2 SCCSimilar to the ACC results, the SCC system also shows that the performance for 10 channels generally exceeds th
38、atof 14, and that 6 channels is often better than 14. Also for networks of 10 or less stations, it is wasteful to allocate toomany channels as the performance tends to be as good with fewer channels.Increasing the length of the message to two pages does not change the relative performance of the net
39、work with respectto the channels. The system is slightly more efficient with the longer messages in that just over half the number ofmessages are transmitted and received per station compared with the one page message length.5.3 SSCTThe performance of the SSCT system becomes better as the data speed
40、 is increased, the message becomes longer and thenumber of scanning channel sets is increased.The effect of longer messages is particularly interesting because, as the system is capable of setting up a particularnumber of calls, determined only by the scanning rate, then the number of messages/hour/
41、station, M, is almost the samefor one and two page messages.The effect of two calling channel sets is to double the throughput.5.4 DiscussionFor networks consisting of a few stations, say 15 or less, there is no particular advantage in any of the systems. As thenetwork increases in size the SCC and
42、the SSCT systems show a marked improvement over the ACC.At the lower data speeds the SSCT system shows a better performance than SCC but the 2 systems show similarperformance at the higher data speeds (SCC with 10 propagating channels). The SSCT is significantly better as themessage length increases
43、 or if 2 calling channel sets are used.Both ACC and SCC systems need careful planning as the number of scanned channels is of prime importance.Frequency management is very necessary as if too many channels are scanned the performance can suffer. In planningsystems based on SCC and ACC the future exp
44、ansion possibilities need to be carefully considered. An increase in thenumber of stations could change the number of frequency channels required for optimum performance.6 ConclusionThe formulas and results derived in this Annex can be used as a planning tool. It is possible to predict the expectedp
45、erformance of an automatically controlled radio system and also to realise the limitations of a such a system.APPENDIX 1TO ANNEX 1Calculations of probabilities relating to clashing and occupied channels1 Probability P1that selected channel is free ACC and SCCIn 1 h the total number of messages/chann
46、el/s:m = M S3 600 N(15)Rec. ITU-R F.1192 9where:M : messages/hour/stationS : No. of stationsN : No. of channels.Let the time that a message occupies a channel be Tm(s).The probability of K new messages during time interval t is given by the Poisson distribution:P (K) = (mt)Ke mt / K!K 0 (16)where m
47、is the average message rate.A channel is free as long as no other station began a message Tm(s) previously, or starts one within the next Tm(s).Thus, the probability of there being no messages ( K = 0) in time 2Tmis the probability that the channel is free.From equation (16):P(0) = exp 2 m Tm= exp 2
48、 M S Tm3 600 N(17)Now:M = 3 600 ES Tm(18)where E (E) is the total network traffic capacity.Thus:P1(0) = exp ( 2 E / N ) (19)1.1 Case of ACC stationWhen an ACC station is at full capacityMmax= 3 600 Emaxt1+ ts+ 2 Tm(20)where:t1: total listening time (Ts/ ( P1P2P3P4)ts: total scanning timets= Ts(2 C +
49、 9)(P2P3P4)(21)C : No. of channelsTm: time of the message.Now:Tm= ts+ Tmand:P1= exp 2 Emax(ts+ Tm)SN (tw+ ts+ 2 Tm)(22)10 Rec. ITU-R F.11921.2 Case of SCC stationWhen an SCC station is at full capacity:Mmax= 3 600 Emaxtw+ Ts+ 2 Tm(23)where:tw: total waiting time ( Ts/ ( P1P2P3P4)and:P1= exp 2 Emax(Ts+ Tm)SN (tw+ Ts+ 2 Tm)(24)2 Probability P2that no clash occurs on a selected channel ACC/SCCThe station listens before attempting to transmit on the selected channel. Thus a channel is
