This formula shows that partitioning the network decreases the total traffic if 1 N R (1 + p) < N R p < 1, i.e., if not all the packets have to go

Transcription

1 Chapter 3 Problem 2 In Figure 3.43 of the text every node transmits R bps, and we assume that both network partitions consist of 1 N nodes. So the total traffic generated by the nodes 2 of each Ethernet equals 1 N R. In addition, each Ethernet will receive a fraction 2 p of all bits transmitted on the other Ethernet. Hence the total traffic on each Ethernet equals 1N R + p 1N R = 1 N R (1 + p) bps

2 This formula shows that partitioning the network decreases the total traffic if 1 N R (1 + p) < N R p < 1, i.e., if not all the packets have 2 to go through both Ethernets, which is usually the case. From this, one could conclude that successive partitioning of Ethernets would be fruitful. However, there is some delay that is added to the normal processing time when a packet has to be processed by the connecting bridge. Clearly, the smaller is p, the more fruitful is this strategy.

3 Problem 7 Figure 3.1 shows the modified version of Figure 3.20 of the text to reflect the release after transmission protocol. Token Token Token T 1 T 2 T N node 1 PROP node 2 PROP node N 1-->2 2-->3 PROP N-->1 Figure 3.1: Timing diagram for release after transmission The efficiency, µ, can be calculated as follows: µ = T 1 + T T N T 1 + T T N + PROP = E[T i ] E[T i ]+PROP/N = 1 1+a/N. As we can see from the last equation, µ 1 as N

4 Problem 9 In this problem we want to calculate the asymptotic probability that N stations have m packets to transmit as N,p 0 in such a way that N p λ, a fixed value. The exact probability that m out of N stations have packets to send follows binomial distribution B(N,p) with p m = P {M = m} = ( N m) p m (1 p) N m. Taking the limit of the binomial distribution, and using the identities N = λ p and lim p 0(1 p) 1 p = e 1, we can write: lim N p 0 Np λ p m = lim N p 0 Np λ = lim N p 0 Np λ ( ) N p m (1 p) N m m (1 p) N = e λ 1 λm m! 1 λ λm = e m!, which is the Poisson distribution with mean λ. 1 (1 p) (Np)m N m m! N N 1 m +1 N N N

5 Problem 12 In this problem we analyze the timing of the FDDI MAC layer protocol in the case of asynchronous and synchronous traffic. (a) For the asynchronous traffic the maximum time between successive token arrivals in the FDDI network is TTRT + TRANSP. To show this we use induction. Initialize the induction at the beginning of the ring operation when no station has a packet to send. Therefore, since TTRT > TRT, each node gets the token in less than TTRT. Let node A receive the token at time Ta 1 and release the token so that node B receives it at time Tb 1. A and B will receive the token again at time Ta 2 and Tb 2, respectively. Assume that T a 2 Ta 1 TTRT + TRANSP, we want to show that,node B also receives the token in intervals less than TTRT + TRANSP, or Tb 2 T b 1 TTRT + TRANSP. If Ta 2 T a 1 TTRT, node A can transmit packages up to time TTRT + TRANSP (since FDDI protocol will allow a node to complete the transmission of a packet regardless of TTRT). Therefore, Tb 2 T a 1 TTRT + TRANSP, and we can calculate the bound on B s token interarrival time as

6 35 follows: Tb 2 T b 1 = Ta 2 T a 1 +(T b 2 T a 2 ) (T b 1 T a 1 ) Ta 2 T a 1 +(T b 2 T a 2 ) TTRT + TRANSP (b) However, if TTRT Ta 2 T a 1 TTRT+TRANSP, node A cannot transmit any package and send the token to B. Thus (Tb 2 T a 2)=0and T b 2 T b 1 TTRT + TRANSP. And our induction proof is completed. In the synchronous traffic, the protocol allocates TTRT time among all host requiring synchronous traffic. In a purely asynchronous traffic every node will see the token in TTRT time (assuming the packet transmission time is zero). When we have synchronous traffic, we should add to the asynchronous TTRT time the TTRT time used by the synchronous traffic. Therefore, every node will see the token in 2TTRT time. (c) The exact token arrival time to a node depends on the duration that each previous node kept the token and number of packets it had to transmit. Therefore, though the interarrival time is bounded, it is not periodic. There are two cases that can occur. Either all stations that have been assigned a fraction of TTRT for synchronous transmissions always have enough synchronous traffic to send, or some stations will cease to send their synchronous traffic at some time (which is the case with real systems). In the former case the token intervisit time can take any value between TTRT and 2 TTRT, hence the maximum deviation is TTRT. In the later case, the token intervisit time takes value between zero and 2TTRT, and the maximum deviation in the token inter-arrival time is 2 TTRT. Now, let us assume that a station has to transfer constant bit rate synchronous traffic (bit rate C). Also assume that the network allocates p TTRT for this traffic, and R is the bit rate of the FDDI network. Note that for stability we should have: R p TTRT 2 TTRT C Since the maximum token interarrival time is 2 TTRT, and during this time the station had p TTRT time allocation then the maximum buffer size the station needs is buffer size per station =(2 TTRT p TTRT) C

7 (d) The definition of fairness depends on what aspect of the resource one is interested. For FDDI we define fairness as follows: the network is fair if each node has the same average delay and have access to the same average transmission rate. Assume that all stations have packets to transmit. FDDI protocol treats all stations in a totally symmetrical manner independent of the location of the station on the ring. Also every station has the same average token interarrival time, and since they have the same average time to transmit, they have the same average transmission rate. (e) If all packets are placed in a single queue (a centralized queue) and then transmitted over the network, users with more packets to send receive a larger percentage of the network capacity. As previously mentioned the definition of fairness depends on the context. In the current context, if we define access to be fair if every user receives the portion of the network capacity in proportion to its transmission requirement, then FCFS is fair. However, if we accept the definition as all users should receive equal access to the network, then FCFS is not fair, since it allows one user to gain a larger share of the network (by transmitting or requesting a large file).

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