Data and Computer Comm Midterm

What is the time to send a message of 100,000 bits using packet switching with packet size 10,000 bits over 4 nodes? Each link has a latency of 10 us and a bandwidth of 1Gbps. Assume that other times such as packet processing time at switches are negligible.

- Num packets = message size/packet size = 100,000 bits/10,000 bits = 10 packets - Prop delay = Latency x (num of nodes - 1) = 10 us x 3 = 30 us - Transmission delay: __Time for all packets to get to 1st: ____Latency x num packets = 10us x 10 = 100us __Time for last packet to get to 2nd: ____ + latency = 100us + 10us = 110us __Time for last packet to get to 3rd: ____ + latency = 110us + 10us = 120us - *Total time = Prop delay + Tran Delay = 30us + 120us = 150us* _____________________________ - TD = 10,000 / 1Gbps = 10us, for 9 its 90us. - Last packet: (TD + PD) x (num of nodes - 1) - Total time: ((num of packets - 1) x TD) + ((TD + latency) x (num of nodes - 1)) - (9 x 10us) + ((10us + 10us) x 3) = 90us + 60us = 150us

Understand how the signal in baseband is modulated to passband. (for one channel and two orthogonal channel)

- The sender cannot send the square waveform but has to send I(t) (any signal in time domain) from [0, BHz] into air from [f, f+BHz]. One Channel: - Let us say I(t) = cos(2πf1t) - Sender sends I(t) x by carrier frequency: - I(t) x cos(2πfct) = cos(2πf1t) x cos(2πfct) = ½ * [cos(2π(f1 + fc))t) + cos(2π(f1 - fc))t)] - Essentially, 1 freq becomes 2 Two Orthogonal Channel: - Sender sends I(t)cos(2πft) + Q(t)sin(2πft) - if I(t) = cos(2πfct), Q(t) = sin(2πfct) - I(t)cos(2πfct) + Q(t)sin(2πfct) = ½ x [cos(2π(f1 - fc))t) + cos(2π(f1 + fc))t)] + ½ x [cos(2π(f1 - fc))t) - cos(2π(f1 + fc))t)]

If cars "propagate" at 100 km/hr, there are two toll booths 100 km apart and it takes 12 sec for each to service each car. How long until a caravan of 10 cars is lined up before the 2nd toll booth?

- Time to "Push" entire caravan through toll booth onto highway = 12 x 10 sec = 120 sec - Time for last car to propagate from 1st to 2nd toll booth: 100km / (100km/hr) = 1 hr - 120 sec + 1 hr = 62 minutes

A link has a latency of 10 microsecond and a bandwidth of 1Gbps, how much time to send a data of 100,000 bits over the link?

- Total time = Transmission delay + Propagation delay - Total time = (datasize/bandwidth) + latency - Total time = (100,000/1Gbps) + 10us = 110us

A circuit-switching scenario in which NCS users, each requiring a bandwidth of 20 Mbps, must share a link of capacity of 150 Mbps. What is the maximum number of users that can be supported?

- ⌊total Capacity / bandwidth per user⌋ = ⌊150 Mbps / 20 Mbps⌋ = 7 - Max is 7 users because 7 can be handled by 140 Mbps, while 8 would break the capacity due to requiring 160 Mbps.

Bit

0 or 1 There is a wire between A and B. If A wants to send a bit '1', he connects the wire to the positive end of a battery. Otherwise he disconnects it from the battery. Or A can hold a radio, if '1', he sends a frequency at f1 and if '0' he sends at frequency f2. Or there is an optical fiber between A and B and if '1' A lit up a light and if '0' A does nothing. WiFi (802.11a): 54Mbit/s (5.4*107bits/s)

TCP/IP Model

Application Layer (Telnet, FTP SMTP, DNS, NNTP, HTTP) Transport layer (TCP, UDP) Internet layer (IP) Host to Network layer (Ethernet, FDDI, X.25)

Know the issues in Framing: how does the receiver know where the start is and where the end is, and how the receiver detects errors.

At the beginning of a frame there is always a preamble which contains a special signal to alert the physical layer about the beginning of a frame. _______The length of the frame can be indicated in a special field in the header. At the sender: a short token called checksum is computed for each frame and included in the frame when it is transmitted. _______At the receiver: the checksum is recomputed. If the newly computed checksum is different from the one contained in the frame, the data link layer knows that an error has occurred and takes steps to deal with it.

TCP/IP Internet Layer

A packet switching network based on connectionless communication. Hosts send packets into the network and then the packets travel independently to their destinations. Format conversion: for different networks. Packet format and protocol: IP Interface - network dependent, Ethernet: (Eth_send, Eth_recv)

Sample

A reading of the received signal at some time instant.

Constellation Diagram

A representation of a signal A bridge between bits, symbols, and physical signals Imagine a table between bit string and value of symbol 01_________-1+j 00_________1+j 10_________-1-j 11_________1-j Symbol here is a complex value, instead of a real value like 5V or 0V. Can imagine it being composed of 2 real values, may include 5V and 0V, or two signals of 0V. Value of the point is the value of the symbol, how to build the relationship between symbol and signal? Point = (x_1 + (j_1)(y_1)) Physical signal regarding this point: f(t) = x1cos(2πfct) + y1sin(2πfct) Where fc = carrier frequency (discussed later)

Protocol

A set of coordination rules governing the communication between two peer entities, i.e. format and meaning of frames/packets. Each layer has a protocol -- layer N protocol.

Carrier

A sine wave to be multiplied with the baseband waveform

Symbol

A symbol carries one or multiple bits. By varying properties of signal, the symbols (bits) are transmitted. A symbol is a segment of signal, where the length of the segment is almost always a constant. For example, ________If we change the voltage level every 1 us, then each symbol is 1 us long. ________If we can use one of 4 voltage levels, each symbol has 2 bits; 8 levels, 3 bits

Understand the bandwidth, the baseband, the passband in the communication systems.

Bandwidth ______The bandwidth is actually the range of frequencies it passes. ______The wider this range, i.e., the larger the bandwidth, the faster the signals are allowed to change, because there are higher frequency components in the signal. Suppose the system allows frequency [f, f+B] The bandwidth is B Hz The actual signal, the baseband signal, is on [0,B], there is a way to move signal on [0, B] to passband signal frequency [f, f+B]

Concept of Bandwidth and noise

Bandwidth basically means how fast your signal can change or how fast you can send out symbols. Width of the frequency range of signal or transmission (Hz) Noise means that although you sent 1 to me, I may receive something like 1+x, where x is the noise added by the medium.

Similarities between OSI & TCP/IP

Based on the concept of the stack of independent protocols. Similar functionality.

Know what the data link layer does.

Between two machines directly communicating with each other, there may be issues due to errors, computers having diff capabilities, or non-zero delays. Data link layer helps to deal with transmission errors and regulating the flow of data to prevent fast senders from swamping slow receivers.

Classification of computer networks based on topology

Bus based Ring Star Mesh Fully connected, point-to-point, broadcast

Know how to convert from data transmission rate to baud rate and vice versa.

Capacity: Rate in bits per second Baud Rate = how many symbols per second _______High Bandwidth -> high baud rate Bit rate = number of bits/symbol * Baud rate _______How to determine the number of bits per symbol? ___________Number of bits/symbol = log_2(number of symbols) _______E.g: eight-level voltage outputs, how many bits per symbol? 3 In Computer Science, bandwidth often refers to capacity

A QPSK system has 4 constellation points. A 16-QAM system has 16 constellation points. So we should always use 16-QAM.

FALSE. When the signal is weak, you cannot.

Understand the Hamming Distance between two binary sequences.

Hamming Distance = number of bits in which two binary strings disagree HDistance(00100, 00000) = 1 HDistance(00100, 00111) = 2 HDistance(00100, 11110) = 3 HDistance(00100, 11001) = 4 If an unknown codeword is received after demodulation and detection, Hamming Distance is used to determine which codeword it most likely is.

Bandwidth

How fast can signals change. Or, the width of the frequency range that can pass the system.

Data rate

How fast can we send bits, can be measured in bps. Upper bounded by the Shannon's theorem given the bandwidth and noise.

Propagation delay

How long does it take for the symbol to reach to the destination.

Know why the sequence number is needed for each frame in Protocol 2.

If the ACK is lost on the way back, the sender may send duplicate frames with the receiver having no way of knowing it's a duplicate. If every frame is given a sequence number, the sender can say they're sending a frame with sequence number m and the receiver can acknowledge which number they've received and request the next one. Prevents duplication, since it can check to see the received frame has the next expected sequence number.

Know inefficiency of Protocol 2 (Stop-and-Wait)

If the receiver takes a while to send the ACK after receiving F m: _______The sender timeouts, and resends F m. _______The receiver will get F m twice, but will send ACK m+1 both times. _______The sender then will not do anything the second time it receives ACK m+1 because it will only respond to ACK m+2.

TCP/IP Transport Layer

Allow entities at end hosts to communicate TCP (transmission control protocol): reliable connection-oriented UDP (user datagram protocol): unreliable connectionless Interface: (IP_send, IP_rec)

Packet Switching

Message is broken down into packets Each packet is sent in a Store-and-Forward manner Pipelining and multiple paths Retransmission of erroneous packets easier More overhead, fragmentation and reassembly, sequencing Example: Internet Protocol (IP), and the User Datagram Protocol (UDP).

Assuming 100 Mbps Ethernet is used to send 30 bytes (user) data, let AH=PH=SH=TH=NH=DH*=DT*= 10 bytes, what is the maximum throughput the application can observe?

Method 2: Total: header + user = (10 x 7) + 30 = 100 The percentage of user data in each frame - 30 bytes of user data / 100 bytes total data = 30% - Max throughput/The number of user data bits/s: 30% of 100 Mbps = 30 Mbps

Know the design issues in computer network.

Reliability: Make a network that operates correctly - Error detection, correction, finding a work path through a network with routing, should be done automatically The Evolution of the Network: networks grow larger and new designs emerge that need to be connected to the existing networking - Addressing/naming, internetworking (disassembling, transmitting, reassembling), scalable (network continues to work well when the network gets large) Resource Allocation: Underlying resources are limited - Statistical multiplexing (bandwidth sharing based on stats), flow control (prevents fast sender from swamping receiver), Quality of Service (mechanisms that handle real-time delivery vs. high throughput demands) Heterogeneity in the network environment: Different machines, switches, links, interfaces from different vendors.

Protocol 2 Link Efficiency EX:Assume delay of 500 ms Bandwidth of 50 kpbs Bits per frame = 1000

Protocol 2: Transmission Delay: The time to send 1000 bits is 1000/50,000 = 20ms At time 520 ms, the receiver gets the entire frame, and send ACK back. Assume ACK is small. At time 1020ms, the sender gets the ACK, and sends the next frame. And this repeats. The efficiency is 20/1020 Efficiency = Transmission delay / Round Trip Time

Understand the layered architecture for computer network

The layered model simplifies the design. Breaks the system up into different layers where each layer provides an abstraction for the layer above it. Every layer has its own responsibility. In the example given in class The task for the sender: give the letter to the mailbox The task for the mailbox: give the letter to the post office The task for the post office: give the letter to the carrier

OSI Physical Layer

The lowest, or first, layer of the OSI model. Protocols in this layer generate and detect signals so as to transmit and receive data over a network medium. These protocols also set the data transmission rate and monitor data error rates, but do not provide error correction. --- How to transfer bits correctly across the physical hardware, such as cable, radio Conversion of bits into signals, what is 0, 1? How long does a bit last? Physical layer protocol: (1) Physically connects two devices; and (2) when the sending device sends a bitstream (e.g. 0100110100), the receiving device would receive what was sent.

OSI Network Layer

The third layer in the OSI model. Protocols in this layer translate network addresses into their physical counterparts and decide how to route data from the sender to the receiver. --- How to send a packet to the destination (hop by hop)? "Directly connected domains" cannot be very large. Larger networks are formed with routers separating directly connected domains. Forwarding, routing, congestion control, format conversion (internetworking), etc. Service interface: __________ip_send(ip_dst, data) _________ip_recv(ip_src, data) E.g. on DENEB (128.226.9.14), call ip_send(128.226.3.5, data)

Baseband waveform

The waveform that is a smoothed version of the square waveform after the low pass filter. To be multiplied with the carrier.

Message Switching

The whole message is sent in a Store-and-Forward manner One node receives the whole message before sending it to the next node No setup or disconnection delay, but queuing delay, overhead in message for routing

Noise

Will be added to the signal. Random, but some statistics can be known with which we make out detection rules.

Know the examples of the computer network.

Wireless Access Networks Shared wireless access network connected end system to router via base station, "access point" ______Wireless LANS: within building (100 ft). __________WiFi: 11, 54, 450, 1300 Mbps ______Wide-area wireless access: provided by telco operators, 10's km __________Between 1 and 10 Mbps __________3G, 4G,: LTE, 5G, 6G Enterprise access networks (Ethernet) ______Typically used in companies, universities, etc ______10 Mbps, 100 Mbps, 1Gbps, 10Gbps transmission rates ______Today, end systems typically connect into Ethernet switch The Internet ______Collection of networks and routers that span the world and use the TCP/IP protocols to form a single, cooperative virtual network. ______Includes National Service Providers (NSPs) like Verizon, Network Access Points (NAPs), and Points of Presence (POP)

Know why a timer is needed at the sender side in Protocol 2.

Without a timer, if the data frame is lost, the sender will be blocked forever, because the receiver will never get the frame and send the ACK. However, if the timer is added, if the sender does not receive the ACK before the timer expires, it knows something is wrong and resends the frame. Should be slightly longer than the round trip time.

There is a network having bandwidth of 1MBps and 3 communication links. A message size 1000 bytes has to be sent. The packet switching technique is used. Each packet contains a header of 100 bytes. On the following, how many packets the message must be divided to so that total time taken is minimum? 1, 5, 10, or 20 packets?

5 Packets - Packet size = total message size / num of packets = 1000 bytes / 5 = 200 bytes - Total packet size = packet size + header size - Total packet size = 200 bytes + 100 bytes = 300 bytes - Transmission delay = packet size / bandwidth = 300 bytes / 1 MBps = 300 bytes / 10^6 = 0.3 msec - Time taken for first packet from sender to receiver = num of communication links x Transmission delay = 3 x 0.3 msec - Time taken for remaining packets: Num of remain packets x Transmission delay = 4 x 0.3 msec = 1.2 msec - Total Time Taken: Time for first packet + time for remaining = 0.9 msec + 1.2 msec = 2.1 msec

Know what a computer network is and the elements of a computer network.

A computer network is a set of computer/switches connected by communication links - The elements of a computer network include: - Hosts, end-systems (PCs, servers, smartphones, running network apps) - Communication links (wired, point-point, wireless, fiber) - Packet switches: forward packets (chunks of data) (routers and switches)

If a system uses a frequency from 2GHz to 2.2GHz, its bandwidth is 2.2GHz. T||F?

FALSE. The bandwidth = max freq- min freq - 0.2 GHz

If the carrier is on 2.412 GHz, the baseband signal is also on 2.412 GHz. T||F?

FALSE. The baseband signal always starts from 0 height.

A QPSK system has 4 constellation points. Therefore, it has 4 bits.

FALSE. The number of bits per symbol = log_2(number of symbols), meaning this constellation system has 2 bits.

The transmitter can send out two baseband signals simultaneously, because the receiver has two independent receiving circuits on two carrier frequencies.

FALSE. There are two signals, one sin(), one cos(), on the same frequency.

A communication link should always send bits as fast as possible.

FALSE. We need to worry about errors. Reliability of the data transmission is vital.

Understand how to design the detector.

Detection - given a received signal, determine which of the possible original signals was sent. There are finite number of possible original signals (2 for the binary case - 0 or 1) An Example: Detection really depends on the noise. In the binary case when the transmitted signal is either 0 or 5V with equal probability, if we know that noise takes values -3V and 2V with probability 0.7 and 0.3, respectively. How would you design the detector? ANSWER: Prob(0V) = 0.5 Prob(5V) = 0.5 n = -3V or 2V Prob(n = -3V) = 0.7 Prob(n = 2V) = 0.3 First step, check the possible outcomes Outcomes = x + n _________If send 0V, two possible outcomes (x = 0): ____________0V + (-3V) = -3V ____________Prob(x = -3V) = 0.7 ____________0V + 2V = 2V ____________Prob(x = 2V) = 0.3 _________If send 5V, two possible outcomes: ____________5V + (-3V) = 2V ____________Prob(x = 2V) = 0.7 ____________5V + (2V) = 7V ____________Prob(x = 7V) = 0.3 _________So, a total of only 3 possible outcomes: Second step, check for each outcome, what should the output be. _________No ambiguity when received -3V and 7V. This is because you can only receive -3V if noise affects 0V, and 7V can only be received if noise affects 5V. _________What should we say when receiving 2V? ____________From 0V, Prob(x = 2V) = 0.3 ____________From 5V, Prob(x = 2V) = 0.7 ____________Since it's more likely 2V came from 5V, we'll assume that when 2V is received, 5V was sent. Therefore, when 2V is received the bit is 1. _________What is the probability that we give the wrong detection result? ____________Prob(wrong result) = P(A,B) = P(A|B) x P(B) _________________Where, A = outcome is 2V and B = 0V was sent ____________P(A,B) = P(A|B) x P(B) = P(2V|0V) x P(0V) ____________P(2V|0V) = From 0V, the Prob(x = 2V) = 0.3 ____________P(A, B) = P(2V|0V) x P(0V) = 0.3 x 0.5 = 0.15 Therefore, the probability that we give the wrong detection result is 0.15.

Encapsulation in layered architecture

Ethernet header exclude preamble and start frame of 8 bytes ________Preamble: mark a frame is coming and serve synchronization purposes. Encapsulation: ________In the source node, we have the data from application layers and upon reaching the Transport layer, add TCP header to the data. ________When receiving the TCP header and data in the Network/Internet layer, assume it's all data and add IP header. ________When receiving all data in Datalink layer, add ethernet header, tail and preamble. ________Physical layer then takes data frame from datalink and encapsulates it by converting it to appropriate data signal

The capacity of a link is 10,000,000 bps according to the Shannon's theorem, the sender can send to the receiver 10,000,000 bits per second reliably.

FALSE, it's an upper limit.

The signal is continuous, so no low pass filter can break it into parts and impose different effects on different parts. T||F?

FALSE. A low pass filter doesn't care if the signal is continuous.

Bandwidth is a metric about how large the signal strength should be.

FALSE. Bandwidth is a metric of how fast the signal can change. Signal strength is related to the transmission power.

The purpose of multiplying the baseband signal with the carrier is to make the signal change faster and propagate better.

FALSE. It is because of the FCC requirements that dictate which protocols can use which frequency bands.

A low pass filter passes only signals on low power. T||F?

FALSE. Low pass filter passes only signals on low FREQUENCY.

Noise is added to the signal and is a constant.

FALSE. Noise is added to the signal but is not a constant, it's a random number ranging from very small or very large.

Understand the latency (propagation delay) and bandwidth, the relationship between bandwidth and transmission delay. Be able to differentiate propagation delay and transmission delay.

Latency (delay) of a link (propagation delay): how much time it takes for one bit to travel from one end of the link to the other end of the link Propagation Delay/Latency = length of physical link/propagation speed in medium Typically in the units of milliseconds (ms = 0.001 second) and microsecond (us = 0.000001 second). Sometimes latency is treated as overhead and ignored. When sending large messages, this may be reasonable. Bandwidth of a link: how many bits can be injected into the link in one unit of time Transmission Delay = Packet length(bits)/link bandwidth(bps) 1Mbps: 1,000,000 bits per second 1Gbps: 1,000,000,000 bits per second

Protocol 3 Link Efficiency

Link efficiency = transmission time before receiving ACK/RTT

Understand the function of low pass filter and how it affects the physical signal.

Low Pass Filter: filters out signals that are outside of the allowed frequency band before allowing the allowed signals through to the system.

Know the problems to be resolved on the Internet.

Naming, addressing Routing Speed mismatch between sender/receiver Error control Resolve contention Fragmentation/reassembly Multiplexing/Demultiplexing

TCP/IP Application Layer

No session and presentation layers Interface - socket programming: (TCP_send, TCP_recv, UDP_send, UDP_recv)

Know the effect brought by noise to signal reception.

Noise is a random value that can have either a positive or negative effect on signal reception based on the medium, and it can it affect the designs of detectors as well as the decision on what amount of symbols to include in a constellation.

Know how to get the maximum data transmission rate (bits/s) and maximum baud rate (symbol transmission rate: symbols/s) in noiseless channel-Nyquist Theorem.

Nyquist's Theorem ______Maximum Baud rate for noiseless channel ______Max Baud Rate = 2 * Bandwidth Implication: Max Bit Rate = 2 x Bandwidth x number of bits/symbol = 2 x Bandwidth x log_2(num of symbols) = Max Baud Rate * log_2(num of symbols)

Understand how the receiver regenerates the baseband signal. (for one channel and two orthogonal channel)

One channel: - The sender sends I(t)cos(2πfct), we want I(t) - Receiver multiplies it by cos(2πfct) - I(t)cos^2(2πfct) = I(t)[½ + ½cos(4πfct)] - LPF is applied and we're left with I(t) Two Orthogonal Channel: - The sender sends I(t)cos(2πft) + Q(t)sin(2πft), we want I(t) and Q(t) - Receiver multiplies received signal by cos(2πfct) to get = [I(t)cos(2πft) + Q(t)sin(2πft)] * cos(2πfct) = I(t)cos^2(2πfct) + Q(t)sin(2πft)cos(2πfct) = I(t)[½ + ½cos(4πfct)] + Q(t)[½sin(4πft) + ½sin(0)] = I(t)[½ + ½cos(4πfct)] + Q(t)½sin(4πft)(constant dropped) - And after LPF, we're left with I(t) - At the same time, the receiver also multiplies the received signal with sin(2πft) to get = [I(t)cos(2πft) + Q(t)sin(2πft)] x sin(2πft) = I(t)½[sin(4πft) - sin(0)] + Q(t)½[1-cos(4πfct)] (constant dropped) = I(t)½sin(4πft) + Q(t)[½ - ½cos(4πfct)] - After LPF we're left with Q(t) - So the sender can send TWO, waveforms at same time.

Circuit Switching

Path is set up end-to-end before any data is sent, => setup delay Source sends over the circuit, => transmission delay Source tears down circuit when done, which incurs => some delay Example: analog telephone network, optical networks

Classification of computer networks based on physical scope

Personal Area Networks (PANs) Local Area Networks (LANs) (campus, building, room) Metropolitan Area Networks (MANs) (cover cities, cable television network) Wide Area Networks (WANs) (cover larger area beyond city or state) Internet (Giant WAN) (Network of networks)

OSI Model

Physical Data Link Network Transport Session Presentation Application Functionality of the network software are partitioned

De-encapsulation in layered architecture

Physical layer de-encapsulates signal/bits into dataframe and passes it to datalink layer. Data link layer takes data frames and de-encapsulates it to check the frame header and tail to make sure it's reached the correct destination and there are no errors. Network/Internet layer takes data packet/datagram from datalink layer and de-encapsulates it to check IP header to make sure it's been routed correctly. Transport layer takes data segments from the network layer and de-encapsulates it to check the TCP header and reassemble the data segments into data streams. Application layers de-encapsulate the data further and forwards data to the necessary applications.

Know what the sender and receiver does respectively in Protocol 1 with no error in data transmission.

Sender: _____Grab data from the network layer _____Send to physical layer. Goto 1. Receiver: _____Get data from the physical layer _____Give it to the network layer. Goto 1.

Know what the sender and receiver does respectively in Protocol 3 (Go-back-N) assuming that the window size is 1 at the receiver side.

Sender: ________Assuming window is from m - m+n-1, curr sequence number = S ________If S is within window, send a frame S to physical layer and start timer, increment S ________If you receive ACK w ____________If it's outside window, ignore ____________If it's inside window, consider all frames in window with sequence number no more than w-1 ACKed, let m=w. ________If timeout for frame m, retransmist frame m(S= m). Receiver: ________With a window size of 1 at the receiver side, there is no buffer layer. ________The receiver will receive a frame and send it directly to the network layer ________Wait to get data from physical layer. ________If data has frame sequence m, gite it to network layer, m += 1 ________When frame m is lost and no ACK, sender needs to transmit m as well as frames m+1, m+2, ... again

Know what the sender and receiver does respectively in Protocol 3 (Go-back-N) assuming that the window size is more than 1 at the receiver side

Sender: ________No change on the sender side. Receiver: ________If current receiving window size is from m - m+n-1 ________If the receiver has received every frame up to frame m-1, then it's missing frame m. ________If m = 3 and n = 6, then receiver has received 0, 1, 2 ________If the receiver then receives, 3, 4, and 5. It updates m immediately to 6, and sends the 3 received frames to the network layers.

Know the basic wireless system, the basic components in the transmitter and the receiver.

Sender: Given the bit stream, convert it to the baseband waveform, then multiply with the carrier waveform (2.4GHz if 802.11g, 1.8GHz in some cell phones), and send. Receiver: Given the signal received from the antenna, multiply it with a locally generated carrier, send it to the low pass filter, regenerate the baseband waveform. The basic components in the Transmitter and Receiver are the ____source encoder, ____channel encoder, ____modulators. Channel encoders format data into codewords which allow error correction to be performed by the receiving end.

Know what the sender and receiver does respectively in Protocol 2 (Stop and Wait) if there could be errors during data transmission after taking timer and sequence number into consideration

Sender: Ignore unexpected ACKs Receiver: Ignore unexpected frames, but still send ACK Two Rules: ______Sender: Receiving ACK m+1 means to send F m+1 next. Sending F m+1 is triggered by receiving ACK m+1. ______Receiver: Receiving F m means to send an ACK m+1. Sending ACK m+1 is triggered by receiving F m, meaning to expect to receive F m+1.

Understand different types of services in layered architecture.

Service defines what a layer can, but not how it does it. Connection-oriented service & connectionless service -- order of packets Connection-oriented -- like the telephone. Establish connection, use it, release it. Packets are in received in order they were sent Connectionless (datagram) -- like the postal system. Each message carries its destination's address and is routed to the destination independently. Packets may not be received in order they were sent Reliable and Unreliable - loss or no loss of packets Reliable: all packets are sent and received correctly, e.g. file transfer Unreliable: packets sent may not be received Example of 4 combinations of connection service types Send: 1 2 3 4 5 Receive: 1 2 3 4 5 - Reliable connection-oriented 1 3 2 5 4 - Reliable connectionless 1 2 4 - Unreliable connection-oriented 3 1 4 - Unreliable connectionless Reliable connection-oriented is too costly for most cases and many don't need it. Phones use unreliable connection-oriented while late packets are useless for video streaming

Know how to get the maximum data transmission rate (bits/s) in noisy channel-Shannon's Theorem.

Signal-to-Noise Ratio (SNR) ______Ratio of the signal power S to the noise power S/N ______Measured in dB or decibels ______10 log_10(S/N) ______S/N = 10 => 10dB, S/N = 100 => 20dB For noisy channel with bandwidth B and SNR S/N ______Max data rate (channel capacity): C = B * log_2(1+S/N) ______Channel capacity means how many bits you can send out per second reliably Therefore, Max data rate = min(Blog_2(1+S/N, 2Blog_2M) ______Where M = num of possible symbols, B = bandwidth

Understand how to calculate the window size in Protocol 3. EX:Assume delay of 500 ms Bandwidth of 50 kpbs Bits per frame = 1000

Size of sender window is determined by the bandwidth and link delay Window size = 2BD + 1 Where B = Bandwidth D = link delay / num bits per frame Window size = 2 x 50,000 x (0.5s/1000) + 1 = 51

Filter

Some device or software that can remove certain frequency components in the received waveform.

Communication is achieved by changing the medium in certain detectable patterns.

TRUE

The speed of a link is 10,000,000 bps, it means that the sender can send to the receiver 10,000,000 bits per second reliably.

TRUE

A noise is random and can take any value with non-zero probability, the reliability of any physical link can never be 100 percent.

TRUE.

If a Wi-Fi channel occupies 2.4-2.425 GHz, it means that the bandwidth of the baseband signal is no more than 25 MHz. T||F?

TRUE. The frequency is not changed from passband to baseband, so the bandwidth from 2.4-2.425 GHz = 25 MHz, is the limit.

Know the key idea of Protocol 3 (Go-back-N)

The key idea of Protocol 3, Go-Back-N, is to keep sending data even if we do not receive the acknowledgement frame. Send data up until the limit of a window size or you receive ACK. After receiving the first ACK back, sender will receive one ACK in every frame transmission time (assuming frame size is fixed and no error).

Differences between OSI & TCP/IP

The concepts of services/interfaces/protocols are clear in the OSI model, not as much in the TCP/IP model The OSI model was devised before the protocols were invented, it misses some important issues. The TCP/IP model was devised after the protocols were designed so the model may not fit other protocols Differences in the network layer: _______ TCP/IP: only connectionless (IP protocol) _______ OSI: connectionless and connection-oriented

OSI Session Layer

The fifth layer in the OSI model. This layer establishes and maintains communication between two nodes on the network. It can be considered the "traffic cop" for network communications. --- Allows users to establish, manage (including providing security) and terminate connections between applications at each end., e.g., allow remote login, provide file transfer service. ______Session management ______Authentication ______Authorization ______Synchronization, e.g. insertion of checkpoints in large data transfers

OSI Transport Layer

The fourth layer of the OSI model. In this layer protocols ensure that data are transferred from point A to point B reliably and without errors. this layer services include flow control, acknowledgment, error correction, segmentation, reassembly, and sequencing. --- End to end communication ______First layer that runs at end points but not at intermediate hops. ______Connection establishment/management/termination, segmentation/error control/flow control ______Reliability, probing data rate, deal with congestions, etc ______Service interface: __________tcp_send(ip+port, data); __________tcp_recv(ip+port, data); __________tcp_send(128.226.3.5:80, data);

Link Efficiency

The percentage of time the link is utilized for sending data.

Understand the physical signal in the frequency domain after Fourier analysis.

The physical signal can be represented as a single-value function g(t) - Any reasonably periodic function, g(t) with period T, can be constructed as the sum of a (possibly infinite) number of sines and cosines. - After Fourier analysis, it translates the signal from time domain to frequency domain. - The physical signal g(t) is made up of a_n, b_n, c. - Fundamental frequency: f = 1/T - a_n and b_n are sine and cosine amplitudes of the nth harmonics (terms) - c is a constant.

OSI Data Link Layer

The second layer in the OSI model. This layer bridges the networking media with the Network layer. Its primary function is to divide the data it receives from the Network layer into frames that can then be transmitted by the Physical layer. --- How to transfer frames correctly between source and destination addresses __________Send frames (a unit of data that has a logical meaning) Reliably transfer frames over a link, how to identify a frame, error control, speed mismatch between senders and receivers. Divided into Media Access Control (MAC) and Logical Link Control (LLC) layers Below data link layer, functionality is done in hardware Above data link layer, functionality is done in software Ethernet (data link layer) service interface would be something like this: _______Eth_send(Eth_dst, data, ...); _______Eth_recv(Eth_src, data, ...); // such routines form the service interface For two machine connected over the same Ethernet domain, software can call the service interface functions to communicate

Know why we need a sequence number in Protocol 3

The sequence number in Protocol 3 is used by the receiver to remove ambiguity and prevent duplicates. The receiver uses the sequence number attached to the frame to make sure that when a frame with seq# m is received, the sender has got all ACK for all previous frames with the same seq# m, so the sender cannot be sending a duplicate previously received frame.

OSI Application Layer

The seventh layer of the OSI model. Application layer protocols enable software programs to negotiate formatting, procedural, security, synchronization, and other requirements with the network. --- Variety of protocols that are commonly used, e.g., SMTP, FTP, Telnet _____Email, File transfer, Virtual Terminal: these are all network applications

OSI Presentation Layer

The sixth layer of the OSI model. Protocols in the Presentation layer translate between the application and the network. Here, data are formatted in a schema that the network can understand, with the format varying according to the type of network used. The Presentation layer also manages data encryption and decryption, such as the scrambling of system passwords. ---- Concerned with representation of transmitted data: characteristics/numbers -> binary numbers that machine can understand _______Binary representation, floating point convention _______Data compression: reduce the number of bits before transmission _______Encryption/decryption

Assuming that A is sending to D a message of 100,000 bits over 2 nodes. Each link has a latency of 10 us and a bandwidth of 1Gbps. What is the time to send the message using message switching?

Transmission time = Datasize / Bandwidth x Total time = ((Num of nodes - 1) x (transmission time + latency) = 3 x ((100,000/1Gbps)+ 10us) = 3 x 110us = 330us

TCP/IP Host to Network Layer

Undefined, rely on the existing technology - must be able to send IP packets over this layer. FDDI = Fiber Distributed Data Interface, used mainly for corporate and carrier X.25: Packet Switched networks allow remote devices to communicate with each other over private digital links, pop in 1970s and 80s

Classification of computer networks based on technology and features

Using Wireless (EX. Wireless LAN) Satellites Radio Public Switched Telephone Network (PSTN) The Mobile Telephone System 1G, 2G, 3G, 4G, 5G, 6G Cable Television Sensor Network Internet of Things

Virtual Circuit Switching

Virtual circuit (VC) mimics a circuit switched connection (to some extent) by using packet switching technology Virtual circuit is set up, typically by the initial packet(s) Packet size in a VC network is usually much smaller than that in a packet switched network. Subsequent packets (do not have the header) follow the virtual circuit Switches do more work to maintain VS information. Example: TCP

Entity

an active element in a layer (machine, process).

Peer entities

entities on the corresponding layers on different machines. Exchange well-defined pieces of information

Primitive

calling functions that manage communication between adjacent protocol layers within the same communications node)

Service

set of primitives (operations) provided by one layer to its upper layer. Ex. To send Data, the sender invokes a Data.Request specifying the packet to be sent. At the receiver, a Data.Indication primitive is passed up to the corresponding layer, presenting the received packet to the peer protocol entity.

Message, packet, frame

structured sequence of bits that are exchanged

Service Interface

the interface between upper layer and lower layer, it defines the services offered. Upper layer: service user. Lower layer: service provider