Chapter 3

Problem 2

In Figure 3.43 of the text every node transmits R bps, and we assume that bothnetwork partitions consist of 1

2N nodes. So the total traffic generated by the nodes

of each Ethernet equals 12N ×R. In addition, each Ethernet will receive a fraction

p of all bits transmitted on the other Ethernet. Hence the total traffic on eachEthernet equals 1

2N × R + p× 1

2N × R = 1

2N × R× (1 + p) bps.

29

30 CHAPTER 3. PACKET SWITCHED NETWORKS

This formula shows that partitioning the network decreases the total trafficif 1

2N × R × (1 + p) < N × R ⇔ p < 1, i.e., if not all the packets have

to go through both Ethernets, which is usually the case. From this, one couldconclude that successive partitioning of Ethernets would be fruitful. However,there is some delay that is added to the normal processing time when a packethas to be processed by the connecting bridge. Clearly, the smaller is p, the morefruitful is this strategy.

Problem 3

An ethernet switch forwards packets from one LAN segment to another dependingon MAC address. It therefore, operates upto data link layer. A router, on the otherhand, forwards packets depending on their IP address. It therefore, operates uptothe network layer. Hosts on the other hand, have applications running which sendor receive packets. The flow of information is thus, as shown in the diagram.

Problem 4

(a) (1) Ethernet hubs are mere physical layer repeaters. Thus, the whole net-work shown is one LAN segment and thus, one collision domain.

(2) Node A’s ARP table contains entries for B and C since they are on thesame LAN.

(3) Node A uses node C’s MAC address listed in its ARP table.

(4) Since, nodes A, B, and C are on the same LAN, node C can see thecommunication between nodes A and B.

(b) (1) Ethernet switches forward packets depending on the MAC address ofthe packet. They segregate the LAN into segments. There are thus,three collision domains separated by the switches.

(2) Yes.

(3) Same as in (a) (3).

(4) No, because the switch connecting node B and node C’s LAN seg-ments do not forward packets meant for middle LAN segment ontorightmost LAN segment.

32 CHAPTER 3. PACKET SWITCHED NETWORKS

Problem 7

Figure 3.1 shows the modified version of Figure 3.20 of the text to reflect the“release after transmission” protocol.

PROP1-->2

PROP2-->3

PROPN-->1

T1 T2 TN

Token Token Token

node 1 node 2 node N

Figure 3.1: Timing diagram for release after transmission

The efficiency, µ, can be calculated as follows:

µ =T1 + T2 + · · ·+ TN

T1 + T2 + · · ·+ TN + PROP

=E[Ti]

E[Ti] + PROP/N

=1

1 + a/N.

As we can see from the last equation, µ→ 1 as N →∞

Problem 8

In this problem we want to analyze the efficiency of a token bus network. A node(say node i) after receiving the token, transmits its packets and then the tokenindicating that the token is intended for the next node (node i + 1). The situationhere is very similar to the situation in the previous problem. A node does not needto wait any time after it has received the token before it sends out its own data.

The difference comes in the propagation time, PROP . If the nodes are num-bered consecutively, i.e., node 1 is at one end of the bus, node N is at the otherend, and the node numbers increasing monotonically, then PROP in the expres-sion of Problem 4 should be replaced by the amount of time it would take a signalto go from node 1 to node N and back to node 1. So the efficiency will be

1

1 + b/N,

33

where b =2PROPN,1

E[Ti]. This is the best arrangement for the token bus.

On the other hand, if the nodes were clustered at the two ends of the bus, withthe odd numbered nodes at one end and the even numbered nodes at the other end,then b =

NPROPN,1

E[Ti], which is the worst case set up of the nodes on the bus.

Problem 9

In this problem we want to calculate the asymptotic probability that N stationshave m packets to transmit as N →∞, p→ 0 in such a way that N × p → λ, afixed value. The exact probability that m out of N stations have packets to sendfollows binomial distribution B(N, p) with pm = P{M = m} =

(Nm

)pm(1 −

p)N−m. Taking the limit of the binomial distribution, and using the identities

N = λp

and limp→0(1− p)1p = e−1, we can write:

limN → ∞p → 0

Np → λ

pm = limN → ∞p → 0

Np → λ

(N

m

)pm(1− p)N−m

= limN → ∞p → 0

Np → λ

(1− p)N × 1

(1− p)m× (Np)m

m!× N

N· N − 1

N· · · N −m+ 1

N

= e−λ × 1× λm

m!× 1

= e−λλm

m!,

which is the Poisson distribution with mean λ.

Problem 10

The probability that a successful data packet tranmission takes place in any timeslot is (

4

1

)p(1− p)3(1− peD)

where p is the probability that a node transmits in a given time slot, and peD is

the probability of a data packet being in error. The probability that the ACK is

34 CHAPTER 3. PACKET SWITCHED NETWORKS

received error-free in subsequent time slots is (1−peA) where peA is the probability

of an ACK being in error. Thus, the probability that a transmission is successfulin two time slots is

Psuccess =

(4

1

)p(1− p)3(1− peD)(1− peA)

Therefore, throughput is

λ = 2p(1− p)3(1− peD)(1− peA) packets/time slot.

Problem 11

The answer to this question varies. In general the CPU and internal buses are morethan capable to fill up the Ethernet capacity. The data rate is usually limited bydisk access rate.

Problem 12

In this problem we analyze the timing of the FDDI MAC layer protocol in the caseof asynchronous and synchronous traffic.

(a) For the asynchronous traffic the maximum time between successive tokenarrivals in the FDDI network is TTRT + TRANSP. To show this we useinduction. Initialize the induction at the beginning of the ring operationwhen no station has a packet to send. Therefore, since TTRT > TRT, eachnode gets the token in less than TTRT.

Let node A receive the token at time T 1a and release the token so that node

B receives it at time T 1b . A and B will receive the token again at time

T 2a and T 2

b , respectively. Assume that T 2a − T 1

a ≤ TTRT + TRANSP, wewant to show that ,node B also receives the token in intervals less thanTTRT + TRANSP, or T 2

b − T 1b ≤ TTRT + TRANSP.

If T 2a − T 1

a ≤ TTRT, node A can transmit packages up to time TTRT +TRANSP (since FDDI protocol will allow a node to complete the trans-mission of a packet regardless of TTRT). Therefore, T 2

b − T 1a ≤ TTRT +

TRANSP, and we can calculate the bound on B’s token interarrival time as

35

follows:

T 2b − T 1

b = T 2a − T 1

a + (T 2b − T 2

a )− (T 1b − T 1

a )

≤ T 2a − T 1

a + (T 2b − T 2

a )

≤ TTRT + TRANSP

However, if TTRT ≤ T 2a −T 1

a ≤ TTRT+TRANSP, node A cannot transmitany package and send the token to B. Thus (T 2

b − T 2a ) = 0 and T 2

b − T 1b ≤

TTRT + TRANSP. And our induction proof is completed.

(b) In the synchronous traffic, the protocol allocates TTRT time among all hostrequiring synchronous traffic. In a purely asynchronous traffic every nodewill see the token in TTRT time (assuming the packet transmission timeis zero). When we have synchronous traffic, we should add to the asyn-chronous TTRT time the TTRT time used by the synchronous traffic. There-fore, every node will see the token in 2TTRT time.

(c) The exact token arrival time to a node depends on the duration that each previ-ous 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 as-signed a fraction of TTRT for synchronous transmissions always have enoughsynchronous traffic to send, or some stations will cease to send their syn-chronous traffic at some time (which is the case with real systems). In theformer case the token intervisit time can take any value between TTRT and2× TTRT, hence the maximum deviation is TTRT. In the later case, the to-ken intervisit time takes value between zero and 2TTRT, and the maximumdeviation in the token inter-arrival time is 2× TTRT.

Now, let us assume that a station has to transfer constant bit rate syn-chronous 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 thatfor stability we should have:

R× p× TTRT2× TTRT

≥ C

Since the maximum token interarrival time is 2×TTRT, and during this timethe station had p× TTRT time allocation then the maximum buffer size thestation needs is

buffer size per station = (2× TTRT− p× TTRT) · C

36 CHAPTER 3. PACKET SWITCHED NETWORKS

(d) The definition of fairness depends on what aspect of the resource one is in-terested. For FDDI we define fairness as follows: the network is fair ifeach node has the same average delay and have access to the same averagetransmission rate.

Assume that all stations have packets to transmit. FDDI protocol treats allstations in a totally symmetrical manner independent of the location of thestation on the ring. Also every station has the same average token interar-rival time, and since they have the same average time to transmit, they havethe same average transmission rate.

(e) If all packets are placed in a single queue (a centralized queue) and thentransmitted over the network, users with more packets to send receive alarger percentage of the network capacity. As previously mentioned thedefinition of fairness depends on the context. In the current context, if wedefine access to be fair if every user receives the portion of the networkcapacity in proportion to its transmission requirement, then FCFS is fair.However, if we accept the definition as all users should receive equal accessto the network, then FCFS is not fair, since it allows one user to gain a largershare of the network (by transmitting or requesting a large file).

Problem 13

In this problem we want to analyze the efficiency of DQDB protocol. There areN equally spaced station on the network. The stations are indexed 1, 2, · · · , N ,arranged from left to right. The propagation time between adjacent stations isPROP , and T is the transmission time for a 53-byte frame.

(a) F idle frames are transmitted by the head station. If T seconds are needed totransmit a frame, then the value of F in terms of T is

F =1

T.

(b) To show that the utilization of bus A in Figure 3.26 is 100 percent, we shouldshow that every frame carries data. Consider a station that has informationto send to its right. There are three possibilities for a passing frame:

1. The frame is already busy;