CEN 4007C Computer Networks Fundamentals

Instructor: Prof. A. Helmy Homeworks 3: Data Link Layer and Wireless Networks

Assigned: Nov 9, 2011. Due Date: Nov 22nd

, 2011 (beginning of lecture or to TA)

Chapter 5: Data Link Layer and MAC protocols

1. (4 points) If all the links in the Internet were to provide reliable delivery service, would

the TCP reliable delivery service be redundant? Why or why not?

Although each link guarantees that an IP datagram sent over the link will be received

at the other end of the link without errors, it is not guaranteed that IP datagrams will

arrive at the ultimate destination in the proper order. With IP, datagrams in the same

TCP connection can take different routes in the network, and therefore arrive out of

order. TCP is still needed to provide the receiving end of the application the byte

stream in the correct order. Also, IP can lose packets due to routing loops or

equipment failures.

2. (5 points) What are some of the possible services that a link-layer protocol can offer to

the network layer? Which of these link-layer services have corresponding services in IP?

In TCP?

Framing: there is also framing in IP and TCP; link access; reliable delivery: there is

also reliable delivery in TCP; flow control: there is also flow control in TCP; error

detection: there is also error detection in IP and TCP; error correction; full duplex:

TCP is also full duplex.

3. (3 points) Suppose nodes A, B, and C attach to the same LAN. If A sends frames to B

(addressed to the MAC address of B), will C's network card (adapter) process these frame

and will it pass the datagrams in the frames to its network layer?

How would your answers change if A sends frames with the MAC broadcast address?

C‟s adapter will process the frames, but the adapter will not pass the datagrams up the

protocol stack. If the LAN broadcast address is used, then C‟s adapter will both

process the frames and pass the datagrams up the protocol stack.

4. (3 points) Why is an ARP query sent within a broadcast frame? Why is an ARP

response sent within a frame with a specific destination MAC address?

An ARP query is sent in a broadcast frame because the querying host does not know

which adapter address corresponds to the IP address in question. For the response, the

sending node knows the adapter address to which the response should be sent, so

there is no need to send a broadcast frame (which would have to be processed by all

the other nodes on the LAN).

5. (4 points) Consider the information content of a packet is the bit pattern:

1010101010101011 and an even parity scheme is being used. What would the value of

the field containing the parity bits be for the case of a two-dimensional parity scheme?

[Your answer should be such that a minimum length checksum field is used.]

The rightmost column and bottom row are for parity bits.

1 0 1 0 0

1 0 1 0 0

1 0 1 0 0

1 0 1 1 1

0 0 0 1 1

6. Derivation of the efficiency of slotted ALOHA.

a. (4 points) With N active nodes, the efficiency of slotted ALHOA is Np(1-p)N-1

. Find

the value of p that maximizes this expression.

b. (3 points) Using the value of p found in (a), find the efficiency of slotted ALHOA by

letting N approach infinity. Hint: (1-1/N)N appraoches 1/e as N approaches infinity.

Thus

7. (6 points) Explain the effect of changing the number of stations, „N’, on the utilization

of CSMA/CD. Comment on the fact that „N’ does not appear in the expression for the

utilization [For CSMA/CD u=1/(1+5a)]

When N increases, the probability of collision increases, and hence the average time for

successful transmission increases (increasing in turn the wasted time in proportion to the

useful time), and the end result is the utilization decreasing with the increase of N.

The formula is „u = 1/(1+5a)‟ does not include N, so we will not be able to explain the

effect of N using this formula. In the derivation of the formula we used an approximation

that the average time taken to get a successful transmission (after possibly one or more

collisions) is ~5 prop. time. This approximation did not take into account that the

probability of collision is a function of N.

8. (5 points) For an Ethernet LAN the data rate was decreased from 100Mbps to 10Mbps.

Utilization (U) of this network will be more or less? [show your answer by calculating

„U‟] Suggest two ways in which we can return the utilization to what it was before

(Increasing or decreasing another parameter and by how much? Show your reasoning.)

- For Ethernet )51(

1

au

, where

onDelaytransmissi

nDelaypropagatio

Trans

Tpropa

With the decrease of the transmission rate from 100Mbps to 10Mbps there‟s an increase

in the transmission delay by 10 folds, hence there‟s a decrease in a by 10 folds and an

„increase‟ in the utilization. So the utilization becomes more.

To return the utilization to what it was before we can decrease the transmission delay by

decreasing the bits in a frame by 10 folds, or we can increase the propagation delay by

increasing the length of the LAN by 10 folds.

9. (total 13 points)

(5 points) Derive an expression for the utilization of the token ring release-after-

transmission algorithm. Comment on your result.

(3 points) What happens to the utilization with the increase in number of stations? Why?

(2 points) What happens to the utilization with the change in „a‟?

(3 points) Compare this result to utilization of token ring (release after reception) and

Ethernet.

For token ring release after transmission (RAT):

T1 T2 T3 TN

PROP1

2 PROP2

3 PROPN

1

Utilization

PROPT

T

PROPTTT

TTT

wastedUsefultimeTotal

timeonTransmissiUseful

i

i

iiN

N

121

21

.....

.....

)(

)( nTE

PROPa

where E(Tn) = expected (average) duration of node transmission N

i

i

N

T

So,

N

a

T

PROP

i

1

1

1

1

- As N increases there is less time wasted in propagating the token and utilization

increases. As a increases the utilization decreases.

- For token ring release after reception (RAR): a

1

1

For Ethernet (CSMA/CD): a51

1

For the same “a”, we see the u RAT > u RAR > u CSMA/CD

When „a‟ increases the utilization for all schemes drops, while at small „a‟ they exhibit somewhat

similar utilization.

[Extra: For token ring release after transmission (RAT), as N increases the utilization

increases.

For token ring release after reception (RAR), as N increases the utilization is mostly

unchanged.

For CSMA/CD, as N increases the utilization goes down due to higher chance of

collision.]

10. (5 points) A 2km long FDDI ring, with frame size of 10,000 bits in length and

multiple machines connected. Can you reason about a lower bound on the utilization of

such a network? [Note: FDDI is token ring (with release after transmission) with

100Mbps] [If not enough information is given please indicate so clearly in your answer,

pointing out what information is missing.]

- For token ring (release after transmission)

)1(

1

N

au

As N increases, so does the utilization. With multiple machines the least utilization we

get is when N=2, which represents the lower bound.

Get a for 2km and 10kbits at 100Mbps, a=0.1, and u=95.238%

With N > 2 we get u > 95.238%

Chapter 6: Wireless Communications and Networks

11. (4 points) As a mobile node gets farther and farther away from a base station what are

two actions that a base station could take to ensure that the loss probability of a

transmitted frame does not increase?

a) Increasing the transmission power

b) Reducing the transmission rate

12. (3 points) Describe the role of the beacon frames in 802.11?

APs transmit beacon frames. An AP‟s beacon frames will be transmitted over one

of the 11 channels. The beacon frames permit nearby wireless stations to discover

and identify the AP.

13. (4 points) Why are acknowledgements used in 802.11 but not in wired Ethernet?

In wired Ethernet collision detection is used to determine whether a packet was corrupt

during the link transmission. Collision detection is not used in 802.11 since it operates

over wireless channels (in which collision detection is not possible). Hence, to know

whether a packet was lost or not acknowledgements are used.

Furthermore, the probability of packet loss during to the physical medium characteristics

(i.e., BER) is quite low for Ethernet and is much higher for wireless channels, hence

there‟s a need for Acks to determine whether a frame was lost.

14. (3 points) Suppose the IEEE 802.11 RTS and CTS frames were as long as the

standard DATA and ACK frames. Would there be any advantage to using the CTS and

RTS frames? Why or why not?

No, there wouldn‟t be any advantage. Suppose there are two stations that want to

transmit at the same time, and they both use RTS/CTS. If the RTS frame is as

long as a DATA frames, the channel would be wasted for as long as it would have

been wasted for two colliding DATA frames. Thus, the RTS/CTS exchange is

only useful when the RTS/CTS frames are significantly smaller than the DATA

frames.

15. (5 points) What are the purposes of the HLR and VLR in cellular (e.g., GSM)

networks? What elements of mobile IP are similar to the HLR and VLR?

The home network in GSM maintains a database called the home location register

(HLR), which contains the permanent cell phone number and subscriber profile

information about each of its subscribers. The HLR also contains information

about the current locations of these subscribers. The visited network maintains a

database known as the visitor location register (VLR) that contains an entry for

each mobile user that is currently in the portion of the network served by the

VLR. VLR entries thus come and go as mobile users enter and leave the network.

The edge router in home network (or home agent) in mobile IP is similar to the HLR in

GSM and the edge router in foreign network (or foreign agent) is similar to the VLR in

GSM.

16. (6 points) Suppose there are two ISPs providing WiFi access in a particular cafe', with

each ISP operating its own AP and having its own IP address block.

a. Further suppose that by accident, each ISP has configured its AP to operate

over channel 11. Will the 802.11 protocol completely break down in this

situation? discuss what happens when two stations, each associated with a

different ISP, attempt to transmit at the same time.

b. Now suppose that one AP operates over channel 1 and the other over channel

11. How do your answers change?

a) The two APs will typically have different SSIDs and MAC addresses. A wireless

station arriving to the café will associate with one of the SSIDs (that is, one of the

APs). After association, there is a virtual link between the new station and the AP.

Label the APs AP1 and AP2. Suppose the new station associates with AP1. When

the new station sends a frame, it will be addressed to AP1. Although AP2 will

also receive the frame, it will not process the frame because the frame is not

addressed to it. Thus, the two ISPs can work in parallel over the same channel.

However, the two ISPs will be sharing the same wireless bandwidth. If wireless

stations in different ISPs transmit at the same time, there will be a collision. For

802.11b, the maximum aggregate transmission rate for the two ISPs is 11 Mbps.

b) Now if two wireless stations in different ISPs (and hence different channels)

transmit at the same time, there will not be a collision. Thus, the maximum

aggregate transmission rate for the two ISPs is 22 Mbps for 802.11b.

17. (4 points) Why are cells (and clusters of cells) used in cellular communication

systems instead of a powerful antenna covering the whole desired geographical area?

To resolve the issue of „spectral congestion‟, where a large number of users are

competing (or contending) for a small number of channels due to the limited frequency

spectrum. Using a single powerful antenna would cover a large geographical area, but

each channel can get used once (i.e., given a one user) to limit interference.

With cellular systems, the power of the antenna (or cell tower) is limited to a relatively

small geographical area. Hence the channels (using the same frequencies) can be used in

other cells (in different clusters) that are placed sufficiently distant to prevent

interference. Hence a much larger number of users can be supported at the same time.

18. (4 points) Why don't we use CSMA/CD in wireless networks?

Collision detection is not possible due to the difference in the physical medium

characteristics between Ethernet (in which CSMA/CD is used) and wireless channels:

1- the transmission power is high enough to distort reception at the sender

2- there could be little correlation between successful reception at the sender and the

receiver. For example, a receiver may have collision (by receiving enough signal power

from multiple senders), while a sender may not experience the collision and vice versa.

19. (4 points) Why does 802.11 use rate adaptation?

Signal to noise ratio (SNR) degrades fast as the mobile nodes move away from the access

point (or when the environment changes). This leads to much higher BER for the same

transmission rate (check the graph for SNR, BER and rate in the lecture slides or book).

Changing the modulation to a lower transmission rate achieves a much lower BER for the

same SNR.

Hence, reducing to a lower rate during bad SNR conditions improves performance.

On the other hand, increasing to a higher rate with favorable SNR conditions also boosts

performance.

Rate adaptation includes rules for both reducing the rate with bad SNR and increasing the

rate with good SNR.

20. (4 points) What the hidden terminal problem and how is it resolved?

When A can communicate with B, C can communicate with B, B can communicate with

both A and C, but A and C cannot communicate with each other (either due to a large

distance between them such that the signal received is so weak, or due to obstacles

between A and C). The problem occurs when both A and C want to transmit to B. Even if

A and C use RTS packets, this only prevents the nodes within their range from

transmitting at the same time, but does not prevent collision scenarios in which packets

from A and C arrive at B at similar times.

The problem is resolved using RTS/CTS exchanges, where the CTS is send by B in this

case, preventing the above collision (e.g., if A is sending RTS to B, B‟s CTS will be

heard by C and C will not transmit to B at the same time).

21. (4 points) In mobile IP, what effect will mobility have on end-to-end delays of

datagrams between the source and destination?

Because datagrams must be first forward to the home agent, and from there to the mobile,

the delays will generally be longer than via direct routing. Note that it is possible,

however, that the direct delay from the correspondent to the mobile (i.e., if the datagram

is not routed through the home agent) could actually be smaller than the sum of the delay

from the correspondent to the home agent and from there to the mobile. It would depend

on the delays on these various path segments. Note that indirect routing also adds a home

agent processing (e.g., encapsulation) delay.

22. (10 points: extra) While most conventional communication systems attempt to

conserve bandwidth, CDMA uses bandwidth that is much higher than that necessary to

transmit a signal. How does CDMA achieve such a high bandwidth and why? [clearly

mention the main operation and (5) advantages of using CDMA]

Operation: CDMA uses a pseudo noise (PN) sequence. The PN sequence has a much

higher frequency (hence much lower pulse duration) than the original signal. By

multiplying the PN sequence by the original signal, the resulting signal becomes of much

higher bandwidth than the original signal (i.e., the original signal is spread in bandwidth).

Each user uses a different PN sequence.

The advantages are:

1-secure: using the PN sequence, non-intended receivers with the wrong PN

sequence get noise

2- reduced fading effect . Fading affects only part of the frequency spectrum.

By spreading the signal over the whole spectrum only part of the signal is

affected, and can be in general recovered .

3- No frequency planning . All cells use the same frequency .

4- Soft handoff (no frequency change during handoff) . All cells use the same

frequency

5- Increased capacity: the number of users is not limited by the number of

channels anymore (there is no channel allocation), but is limited by the noise

floor. Typically the number of users supported by CDMA systems can go up

to 5 to 10 times more than other conventional cellular systems.