Final Review

Review

• The very basic wireless communication system• The bits are sent to the low pass filter to be turned into waveforms. The

low pass filter is also called the “pulse shaping filter.”• The impulse response of the filter is fixed. `1’ will excite a positive

waveform, `-1’ will excite a negative waveform. • Basically, you send a symbol to the low pass filter every symbol time. With

BPSK, the symbol is either `1’ or `-1’.• The impulse response waveform lasts more than one symbol time.

Remember its shape!

Review

• The final “baseband waveform” is the addition of the impulse responses generated at different times. – The impulse response is usually written as h(t). – Let the data symbols sent to the low pass filter be

x[n] at time n. x[n] will add a voltage of x[n]h(nT-t) to the baseband waveform at time t (assume current time is 0, current symbol is x[0]. The previous symbol is x[-1], the next symbol is x[1], and so on).

Review

• The signals we can actually send is the following, where I(t) and Q(t) are the baseband waveforms.

• In BPSK, the cosine is multiplied with a waveform, the sine is not. In QPSK, cosine and sine are multiplied with I(t) and Q(t), respectively.

Review

• Multi-path. Signals travel infinite number of paths to reach the receiver. The received signal is the addition of signals from all paths:

where I and are the is the attenuation and the delay of path i, respectively.

• After receiving the signal, we will first multiply it with sine and cosine wave, and pass it through the LPF, as explained before.

Review• Complex channels – because we can send both a sine and a

cosine wave, which can be conveniently represented as a complex wave.

• The sender and receiver will try to use the same frequency, but they cannot. Consider the cosine branch and consider a single path.

• So, after the low pass filter, it becomes

• Similarly, the sine branch will produce

• Therefore, the cosine and sine branch can be regarded as a complex number

Review• To establish the communication, we will first have to reproduce the

baseband waveform. Which means that we have to get rid of , a process called “carrier phase tracking.”

• We take samples from the received baseband waveform to get

• If the phase offset has been tracked perfectly,

• If is it not perfect but still okay, the sample will lead to a correct decision. Then, extract the phase error based on the sample and the constellation point, adjust the phase offset.

Review• Timing synchronization – how to

take samples at the correct time.• Using the shape of the impulse

response.

k k+1k-1

Review

• The current GNU SDR implementation uses the optimized mm algorithm. Assume a sample error of \delta. The sample at time n is

• The timing error is calculated as

• The reason is that

Review

• While

• The first line is the one that does the work. In BPSK, it becomes

Review

• Dealing with multipath.

Review

• Important thing to remember is that even with perfect carrier phase tracking and symbol timing, when taking a sample, it will contain some thing from the neighboring symbols.

• At time 0, the path with delay will contribute to this sample, but for k!=0, also

• And there are infinite number of paths.

Review

• The red curve represents the other path.

k k+1k-1

Review• We can use equalization. The basic idea is to

subtract the contributions of non-relevant samples from the current sample.

• A simple yet okay algorithm is LMS. Start with an arbitrary for all i, usually =1 and =0 for all other i.

• Let . Then , where

• Basically, if e< 0 while s_i>0, it means that c_i is not large enough…

• The idea is to minimize the mean square of the error.

Review

• A signal can also be represented in the frequency domain.

Review

• LTI system. – A linear system is a system such that

– Suppose . A linear system is “time invariant” if .

Review

• DFT:

• IDFT:

• Convolution:

CDMA

• Used in 3G networks.• Direct Sequence Spread Spectrum: spread a

data bit into multiple chips.• Each sender has a unique chip sequence, that

is *orthogonal* with other chip sequences.

Orthogonal Frequency-Division Multiplexing

• From the highest level, OFDM divides the communication bandwidth into a number of sub-channels, each occupying a fraction of the bandwidth.

• Each sub-carrier is modulated by BPSK, QPSK, or other schemes.

• Used in 802.11a and 802.11g.

OFDM

MIMO• First, having multiple receiving antennas means that you

can pick up more energy. • Also, when one antenna is having trouble receiving signal,

others are unlikely to be having the same problem. That is why commercial APs sometimes have multiple antennas also. It compares the received signal strength from different antennas and use the strongest one to decode the data. Called ``antenna diversity.’’

• As long as the antennas are sufficiently apart from each other, the signals are likely experiencing different fading. The space needs to be half of the carrier wavelength. If we are using 2.4GHz, the wavelength is about 10cm.

MIMO• Having multiple transmitting antennas does not necessarily mean

that you can send more energy, because the transmitting energy is determined by other issues, such as your battery.

• However, it does mean that you can have multiple paths between the sender and the receiver. With nt transmitting and nr receiving antennas, you have nt times nr paths that can be assumed to be independent. If one path is in trouble, i.e., there is someone in the blocking position right now, other paths are unlikely to be in the same situation at the same time. Much better than depending everything on only one path!

• Also, what makes MIMO possible is that the receiver antennas can operate in the linear range such that the received signal is the ADDITON of signals from multiple transmitting antennas.

SIMO

• Single Input Multiple Output.• Consider one transmitting antenna and two

receiving antennas.• Assume flat-fading, meaning that there is no

multi-path, i.e., the received sample is relevant only to the current data symbol. We write it as y[n]=x[n] + w[n].

• We can make this assumption because of OFDM.

SIMO

• With two receiving antennas, we will receive

that is, from the waveform received at each antenna, we can take a sample, and call it y1 and y2, respectively. Both samples are excited by x, but they are from different paths, therefore their channel coefficients (i.e., h1, h2) are different. One important thing to remember is that the noise from both antennas are usually assumed to be following the same distribution and have the same power and are independent from each other.

SIMO receiver

• The information from the strong channel is more valuable than the weak channel.

• The optimal -- Maximum Ratio Combining (Section 3.2.1 in the Tse book). We should weight the samples from the antennas according to the channel strength:

MISO• Now consider the case when the sender has

multiple antennas and the receiver has only one antenna.

• The sender has a power budget – the total transmitting power cannot exceed a threshold.

• Assume that all antennas are sending the same data symbol at any time, so the receiver will receive

where a1 and a2 represent the power allocated for antenna 1 and antenna 2, respectively.

MISO

• The problem is to maximize the magnitude of the received signal x(h1a1+h2a2) subject to the constraint that

MISO• Still maximum ratio combining. Define Lagrange

• Take the partial derivative of L over a1 and a2 :

• Means that a1 and a2 should be proportional to h1

and h2.

• But this requires the sender knows the channel – not always the case.

The Altamonte Scheme

• The key is that the transmitting antennas are NOT restricted to sending the same data symbols at the same time.

• The Altamonte Scheme (Tse book Section 3.3.2). Consider two data symbols to be sent in two consecutive symbol times, u1 and u2 . At time 1, ant1 transmits u1 and ant2 transmits u2.

At time 2, ant1 transmits –u*2 and ant2 transmits u*1.

The Altamonte Scheme

• (These two formulas are from the Tse book.) So,

• Rearrange it, we have

The Altamonte Scheme

• So, we have

• Note that

that is, the two vectors are orthogonal to each other. So, to recover u1 and u2, we can multiply

with the conjugate of either of the vectors.

The Altamonte Scheme

• So, the magnitude of the received signal is proportional to , even when the transmitter is not aware of the channel coefficient at all.

• If the transmitter simply sends the same symbol over two antennas at the same power, the received signal is proportional to h1 + h2 , and depends on the phase, they may cancel each other out!

2 by 2 MIMO

• Now consider we have two transmitting antennas and two receiving antennas.

• A simple scheme called ``V-BLAST:’’ Send independent data symbols over the transmitting antennas as well as over time.

MIMO

• MIMO receiver. Will receive two samples per time slot. hij: the channel coefficient from Tx ant j to Rx ant i.

• How to decode the data?

MIMO receiver

• The simplest receiver just do a matrix inversion:

• This is NOT the optimal decoder! The maximum likelihood decoder is better.

Error Correction Code

• Hamming code• Cyclic code• Both required!

Wireless LAN (802.11g)1. When a node gets a packet,

a. If its queue length is 0 at this time, i. if the medium is free, it waits for DIFS. If the medium is still free after DIFS (28us), it sends the packet. ii. If the medium is busy, it generates a random backoff counter and enters the WAIT_DIFS stateb. If the queue length is not 0 at this time, simply add this packet to the queue

2. When in the WAIT_DIFS state, maintain a medium_free counter. This counter is cleared to 0 every time the node finds the medium is busy. If the medium is free, the counter should be incremented every us. When the counter reaches DIFS, the node selects a backoff_timer and enters the WAIT_BACKOFF state. The backoff_timer is selected by picking a random number uniformly from [0, CW-1], then multiply this number by SLOT (9us).

3. When in the WAIT_BACKOFF state, the node maintains backoff_timer. If the medium is busy, it enters the WAIT_DIFS state. Otherwise, it decrements the backoff_timer every us. When the backoff_timer becomes 0, the node start sending the packet it wants to send if the queue is not empty. This packet should be the packet at the head of the queue.

4. When transmitting the packet, a node calculates the time needed for a packet transmission, including the packet transmission time, the SIFS, the ACK time, and stores it in a field in the packet. Other nodes, if heard this header of this packet, will backoff until this packet transmission is completed.

5. After the sending node finishes transmitting the packet, the receiving node checks whether it gets the packet. If yes, it sends an ACK packet back to the sending node SIFS (10us) after the packet transmission finishes. If it did not receive the packet, it does nothing.

6. The sending node of the packet waits for the ACK from the receiving node.a. If it did not receive the ACK before a timeout,, it doubles CW. If the packet has been transmitted several times (implementation dependent), it removes this packet from the queue. It then selects a backoff_timer and enters the WAIT_DIFS state.b. If it gets the ACK, it removes the packet from the queue. It then selects a backoff_timer and enters the WAIT_DIFS state. It resets CW=CWmin (CW is set to be CMmin = 16 at initialization).

802.11 channels

• In 802.11 b/g, there are 11 channels, starting at 2.412GHz at a spacing of 5MHz.

• Each channel owns a bandwidth of 22MHz.• So, only 3 non-overlapping channels, 1,6,11.• 802.11a has more channels and you may

check at http://www.moonblinkwifi.com/80211a_frequency_channel_map.cfm

802.11e

• Supporting QoS. The basic idea is to let higher priority packets have – Shorter IFS– Smaller contention window

• TXOP (Transmission Opportunity)– Everyone is competing for an equal share of air

time

Idle Sense

• Is doubling CW whenever not getting an ACK optimal?

• No, because not getting ACK could due to two reasons– Contention (CW should be doubled)– Random loss causing the packet not passing the

CRC check (shouldn’t double CW because there is no contention!)

Idle sense

• The basic idea is to sense the average duration of the idle period between two transmission attempts, and use it as an indicator of how many nodes want to transmit.

• If more nodes want to transmit, the idle period will be short; otherwise, it will be long.

• If too short, reduce the transmission probability. If too long, increase the transmission probability.

• Use AIMD to adjust for fairness.

WEP• The stations share a secret key.• Before the data transmission, a 24-bit random

Initialization Vector (IV) is generated by the sender. • The IV and the secret key are combined to make the

session key. • The data is encrypted with the session key by the RC4

stream cipher. Then the encrypted data is sent to the receiver along with the plaintext IV.

• The receiver can decrypt the data with the IV and the shared key.

• Different IV are used for each transmission.

Temporal Key Integrity Protocol(TKIP)

• A cipher suite based on old hardware (RC4 cipher)– A transmitter calculates a keyed cryptographic

message integrity code (MIC). TKIP appends the computed MIC. The receiver discards any frames with invalid MIC. MIC in TKIP is not perfect due to hardware constraints. Has a timeout countermeasure.

– Uses a transmit sequence counter (TSC) for every frame. Defend replay attack.

– Uses a key-mixing function to combine the temporal key, the transmitter address (TA), and the TSC into the seed (IV, key) for the RC4 stream cipher

TKIP Encapsulation

The figures are from http://standards.ieee.org/getieee802/download/802.11i-2004.pdf

Decpsulation

Format

Note: ICV is just the 4-byte checksum

Cellular Phone Networks• Compared to WLANs– Longer range– Less speed– Higher mobility– More for voice services (evolving!)

• Overview– Divides the area into cells.– Base stations in each cells.– User have cellular phones.– The phones talks to the base station directly in

wireless.

GSM

• Global System for Mobile (GSM)– Used by more than 3 billion people– 2nd Generation (2G), because everything is digital,

compared to the analog mobile phones which is 1G– Operates in 900MHz and 1800MHz band, or 850M

and 1900M bands in the US– Uplink and down link both 25MHz wide. In GSM900,

uplink band is 890-915M, downlink is 935-960M.– A band is divided into 124 channels with 200KHz

spacing. The data rate is about 270Kbps for a channel.

GSM• GSM continued

– The time in a channel is divided into 8 slots. Each slot is 577us and is allocated to a user.

– A mobile phone is allocated two channels, one for uplink and the other for downlink. Separated by 45MHz.

– The uplink and downlink slot numbers are separated by 3 – a phone never transmits and receives at the same time

– FDMA/TDMA (Frequency Division Multiple Access / Time Division Multiple Access).

– Different from In wireless LAN, in which a node is given the entire bandwidth for the time it needs to send a packet. Difference due to the nature of the application.

– The peak transmission power of a GSM phone is 2W in 900M band and 1W in 1800M band. In contrast, your wireless router transmits at 20dBm, which is 0.1W. Difference due to the distance.

Frequency Reuse• Adjacent cells should not use the same

frequency, because it will cause interference.• However, non-adjacent cells may use the same

frequency.• Due to frequency reuse, the actual number of

channels used in an area is larger than 124.• Given an area, reducing the size of cells (reducing

the transmit power) increases the frequency reuse, hence increasing the total capacity.

• Cell planning is a major issue.

Frequency Reuse

• Cell sizes in GSM:– Large microcell: 3-30km– Small microcell: 1-3km– Microcell: 0.1-1km– Picocell: 0.01-0.1km– Nanocell: 0.001-0.01km

• Typically, there are lots of small cells plus several large cells. The small cells carries the majority of the traffic. The large cells fills the holes.

Capacity

• Given the number of channels in a cell, the maximum number of users supported is fixed. – With GSM, if there are C channels, we can support

8C users at most• The number of users can be much larger than

the maximum number of slots. That is why some time you see “emergency calls only.”

• The hope is that not all users want to make calls at the same time.

http://elm.eeng.dcu.ie/~kaszubow/Biography/Lecture5.pdf

Management of GSM• Mobile System (MS)

– Mobile Equipment (ME)– Subscriber Identity Module (SIM)

• Base Station Subsystem– Base Transceiver Station (BTS)

• In charge of physical communication in the air. Has 1 to 16 transceivers– Base Station Controller (BSC)

• Controls hundreds of BTS• Network Switching Subsystem

– Mobile Switching Center (MSC)• Typical MSC supports up to 100,000 mobiles and 5000 simultaneous calls• MSC are connected with each other. • Gateway MSC connects the GSM system to external networks, e.g. PSTN. • Each MSC controls at least one Base Station System (BSS)

– Visitor’s Location Register (VLR)– Home Location Register (HLR). – Authentication Center (AuC). Holds different algorithms for authentication and encryption.– Operations and maintenance center (OMC)

HLR and VLR• HLR: database of all cellphones permanently registered in the system.

Stores– The address of the VLR currently associated with the phone– Encryption keys for data transmission and user authentication – Service type– …

• VLR: Each MSC connects to a VLR. The VLR is a data base with the information about cellphones temporarily located in the area served by particular MSC.

ME and SIM• ME, has the IMEI (International Mobile Equipment Identity)• SIM card, has

– Ki: Subscriber Authentication Key. 128 bit key shared by the subscriber and the operator. Stored in the SIM card and the HLR

– PIN: to protect the SIM card– IMSI: International Mobile Subscriber Identity– TMSI: Temporary Mobile Subscriber Identity. To prevent

eavesdropping, TMSI is used instead of IMSI. IMSI is used as rarely as possible. TMSI is randomly generated by the VLR.

– MSISDN: Mobile Station International Service Digital Network– LAI: Location Area Identification

GSM Security• When a mobile station needs to be authenticated,

1. The operator generates a random number, RAND (128 bit), and sends to the MS.

2. The MS and the operator both runs an algorithm, called the A3 algorithm, with Ki as the key, to produce SRES (32 bit) from RAND

3. The MS sends the SRES to the operator, and if SRES matches the operator’s SRES, consider passed authentication

4. RAND is passed to an algorithm called A8 as input with Ki as the key, to produce Kc (64 bit). Done by both the MS and the operator

5. Kc becomes the key for the A5 algorithm. A5 is a stream cipher for encrypting the data.

3G Overview• Use CDMA.• Generally, 3G will have a much better support for data

services. The numbers are different depending on the versions, but it will be about at least one order of magnitude higher than GRPS.

• Defines an air interface and maybe combined with the GSM/GRPS core network

• There are competing standards:– W-CDMA– CDMA2000– …

Power Control in CDMA Schemes

• The signal received at the base station are from multiple users at the same frequency

• If one user is transmitting at a high power, other users signal will be overshadowed

• CDMA schemes has to limit the transmitting power of the MS

• The BS may measure the signal strength and send instructions to the MS about increasing or decreasing the transmitting power.

W-CDMA

• Key features include– Radio channels 5MHz wide, both uplink and

downlink– Chip rate 3.84Mcps– Frame length 10ms– Adaptive power control updated 1500 times per

second– Cells not synchronized (synchronized in

CDMA2000)

Orthogonal variable spreading factor (OVSF)

• W-CDMA uses Orthogonal variable spreading factor (OVSF) to provide different data rates to different users

• The idea is that users may be assigned with codes of different lengths, but still orthogonal to each other.

• Because code length are different, a user assigned a shorter code will have a higher data rate

OVSF• Generation of OVSF code based on a simple binary tree

– Start with the root node {1}.– A node has two children. The upper and lower. If the node as code C,

the upper child is assigned code CC, and the lower child is assigned CC’ (C’ means inverting every bit in C).

– Repeat. • Two codes are orthogonal as long as no one is the prefix of the other• A major issue is how to assign codes

HSDPA

• Adaptive modulation and coding (AMC)– Depending on the channel state, send at different

data rates.– Use lower data rate if channel is weak– In wireless LAN, the rate adaptation

High-Speed Downlink Packet Access (HSDPA)

• Hybrid automatic repeat-request (HARQ) – When a data packet is received and found to be

corrupted, the receiver does not simply discard it, but saves it and combines it with the retransmissions

– When a packet is corrupted, the sender does not send the packet again, it sends some parity checking bits

– AMC is coarse grained, HARQ is fine grained

HSDPA

• Fast packet scheduling – Each user transmits to the base station the signal

quality – The base station determines which user to send to

for the next 2ms• Send to users with stronger channels • May send to multiple users simultaneously with the

channelization code• Must also ensure fairness

Multihop• Multi-hop wireless network has attracted lots attentions, because

it may offer interesting solutions to a number of problems– Wireless sensor networks: a large number of sensors are deployed in

an area. The sensors may collect data and then send data to the outside observer for processing. The transmission power of sensor is limited and cannot reach very far, therefore sensors have to reply other’s message. Applications include: scientific data gathering, military monitoring. Power is a main issue in wireless sensor networks because sensor nodes are battery-powered.

– Wireless Ad Hoc networks: wireless nodes organize themselves into a network with no infrastructure. Applications scenarios: emergency response.

– Wireless Mesh Networks: install mesh wireless routers to extend the one-hop to multi-hop. The routers forward the message from the nodes to the next mesh router in wireless. Comparing to installing the wired network such as Ethernet, this may offer faster, cheaper solutions for covering an area.

Routing• One of the key problems in multihop networks is routing.

– Provide connectivity• General approach: Link State (LS) and Distance Vector (DV)

– LS: nodes broadcast the link state information to all nodes in the network, and everyone computes the routing paths based on the same information.

– DV: a node only tells its neighbor the hop count to a certain destination node. A node will pick the shortest path from all neighbors.

• Challenges in wireless networks– Nodes may be mobile– Link is not binary but more a delivery ratio– Some networks, such as wireless sensor networks, can be very limited in

resources, e.g., computation, storage, battery, and periodical update of link state is a burden

– Other nodes can sometime overhear…

Proposed Protocols

• Dynamic Source Routing (DSR)– Source routing. The entire routing path specified in the

packet.– http://www.ietf.org/rfc/rfc4728.txt

• Ad-hoc On-demand Distance Vector Routing (AODV)– Hop-by-hop” protocol– http://www.ietf.org/rfc/rfc3561.txt

• Both on demand. Routing information discovered only when needed– Does not require periodic route updates– Suitable for mobile networks

DSR

• Route discovery. Invoked when the source needs to send packet to the destination but has no path. Basic idea:– Source broadcast Route Request message, with the

destination node’s ID and a unique packet ID– If a node overheard the request, it checks

• If it is the destination, it sends a packet back to the source. The routing path from the source to the destination is already in the packet.

• If it is not the destination, if it has not received the packet before, it appends its own ID to the packet and broadcast it. It does nothing if it received the packet before.

AODV

• Intermediate nodes stores the routing information in a lookup table. Needs only the destination address.

• Route discovery initiated when a node needs next hop information to a destination node

AODV

• Route Request– Source broadcast Route Request (RREQ) message

to a destination– Intermediate node broadcast the received RREQ

message. Creates a path entry for a reverse path to the source. Assumes bidirectional links. • S sends the RREQ message. A forwarded this message,

A knows the next hop to S is S. B forwards the message from A, B knows the next hop to S is A.

AODV

• Route Reply– Destination should reply the Route Reply message

(RREP).• This message will be forwarded along the reverse path

created when forwarding the RREQ message– An intermediate node creates next-hop entry for

destination as RREP is received. Also forwards along “reverse path” hop.

AODV• Each node will maintain a non-decreasing sequence number sent in RREQ and

RREP messages and increments it by one in every message.• The RREQ message contains the source node's ID, current sequence number,

broadcast ID. Nodes keep track of the RREQ's source ID and broadcast ID. If they receive a RREQ which they have already processed, they discard the RREQ and do not forward it.

• RREQ also contains the most recent sequence number for the destination of which the source node is aware. A node receiving the RREQ may send a route reply (RREP) if it is either the destination or if it has a route to the destination with corresponding sequence number greater than or equal to that contained in the RREQ. If this is the case, it unicasts a RREP back to the source. Otherwise, it rebroadcasts the RREQ.

• As the RREP propagates back to the source, nodes set up forward pointers to the destination. Once the source node receives the RREP, it may begin to forward data packets to the destination. If the source later receives a RREP containing a greater sequence number it may update its routing information for that destination and begin using the better route. If the sequence numbers are the same in two replies, pick the one with a smaller hop count.

Path Quality

• The path quality should be measured by the Expected Transmission Count (ETX). It is summation of the expected transmissions of all links on a path.

• Implemented LQSR on top of DSR, basically using ETX to select paths.

• http://www.cse.cuhk.edu.hk/~cslui/STUDY_GROUP/metrics.pdf

Geographical Routing• Karp, B. and Kung, H.T., Greedy Perimeter Stateless Routing for

Wireless Networks, in MobiCom 2000.• Using node’s geographical information to route.• Assumes:

– Nodes know their locations, e.g., by having GPS– Nodes know their one-hop neighbors, e.g., by asking every node to

broadcast beacon signals periodically, and one-hop ONLY– Knows the location of the destination, e.g., by overhearing or

broadcasting– The connectivity is a unit disc (arguable)

• Advantage– Nodes do not have to store too many information. Given a constant

amount of information in the packet, it can make routing decisions– AODV and DSR need to ask nodes to store hop-by-hop routing tables

or complete routing paths

Greedy Routing

• First idea:– At node A, if the destination is t, pick the next hop

neighbor that is closest to t among all neighbors of A

• To implement greedy routing, the packet needs to carry the destination’s location

Greedy routing

• As many greedy algorithms, greedy routing may run into problems.

Planar Graphs• Consider routing in planar graphs.

– A planar graph is a graph where there is no crossing edges.– The area closed by a set of edges is called a “face.” The outside

is also called a face.– Faces are connected.– There are ways to construct planar graphs by deleting edges.

Face Routing• Keep left hand on the wall, walk until hit the straight line

connecting source to destination.• Then switch to the next face.• The point is that every time you are switching to a face closer to the

destination.

Face routing• Face Routing may fail only if there is

no path to the destination.– Face routing will forward the packet

to the face– Then it will circle around the face,

without finding a point intersecting the line from s to t other than the last intersection point Lf.

– Eventually it will hit the first edge when entered this face, e0, and will realize that the destination is unreachable

– So, face routing needs to store in the packet header• Destination location• The starting point location• Lf• e0

Sample Problems• In an 802.11g wireless LAN, the maximum speed is

54Mbps. There are 5 nodes including the AP, where the AP is receiving traffic at 32Mbps and the nodes are receiving traffic at 16Mbps. Assume all nodes are operating at the highest data rate. Which of the following is true?– AP is having a larger throughput than the nodes.– AP is having the same throughput than the nodes.– AP is having a smaller throughput than the nodes.– The throughput of the AP is non-deterministic.

Sample problems• In 802.11g networks, suppose there are 3 nodes sending to the AP. The

AP has no traffic. The data rates of the nodes are 12Mbps, 24Mbps, and 48Mbps, respectively. What is the average data rate in the network (Neglecting all overhead such as ACK, DIFS, SIFS, backoff, and collision and loss)? – 24Mbps.– 12Mbps.– About 20Mbps.– None of the above.

Sample Problems

• Suppose an 802.11g LAN is in a 30 ft by 30 ft room, with 50 nodes and an AP. Which of the following statement is true?– The auto rate algorithm should be turned off because it has

negative effects.– The auto rate algorithm should be turned on because it has

positive effects.– Doesn’t matter, the auto rate algorithm has no effects.– None of the above.

•

Sample Problems

• In an 802.11 wireless LANs, if we modify the protocol as: if we received a packet header, we send an ACK. Suppose packet collision rarely occurs in the network.

• (1). Does this change the behavior of the network or not?

• (2). If it does, what kind of change? Good or bad, or both good and bad?

Sample Problems

• In cellular phone networks, a phone A may be not in its home network. If another phone B wishes to make a call to A, how can it find A?– With the phone number of A, the network can find the

HLR of A when find the current VLR A is registered to, then find the MSC.

– The MSC B is associated with remembers where A is.– The MSC B is associated with will broadcast a message

to find A’s location upon B’s request.– None of the above.

Sample Problems

• In the DSR routing protocol, why does the routing discovery packet contains a unique ID?– Because all packets have an unique ID– To avoid loops in the path– For security reasons– None of the above