IST 220 Exam 1 Notes · 2005. 10. 7. · IST 220 Exam 1 Notes Prepared by Dan Veltri Exam 1 is scheduled for Wednesday, October 6th, in class. Exam review will be held Monday, October

IST 220 Exam 1 Notes Prepared by Dan Veltri

Exam 1 is scheduled for Wednesday, October 6th, in class. Exam review will be held Monday, October 4th, in class.

Chapter 1 & 2 The internet is expanding rapidly 1994: 3.5 million hosts 1997: 15 million hosts 2001: 109 million hosts The internet is complicated, but not hard to understand. We can easily understand the internet as a transport system.

TRANSPORT SYSTEM INTERNET

City Computer City name Computer name A guy A message A car A packet Made of metal, plastic In the form of electric current, called a signalHighway, tunnel Copper wire, fiber optic Rest, lunch, gasoline Reboost the signal A route A route

Q1. This route has __4__ computers? Q2. This route has __3__ links (connections)?

Q3. This route has __3__ hops*? Q4. A route that has 200 hops has _201_ computers? * A hop has 3 parts

Problem: When you drive, you can check a map to determine what cities are on the route. However, there is no map of the internet, so how do you know who the forwarders are? Answer: Use traceroute, a software tool. Example command: C:\> traceroute allspice.lcs.mit.edu The Result: 1. philadelphia4.t3.ans.net (140.222.27.221) 20ms 25ms 15ms 2. New-York2.t3.ans.net (140.223.37.9) 100ms 120ms 140ms 3. Cambridge1.bbnplanet.net (40.1.122) 300ms 280ms 260ms 4. Allspice.lcs.mit.edu (18.26.0.115) 339ms 279ms 279ms Notes: (1.) There are 3 forwarders (Philly, New York, Cambridge) (2.) The numbers in parenthesizes above, ex. 140.222.27.221, are IP addresses (3.) Three individual packets are sent to each destination. The three times outputted after the location and IP address represent the round trip time of each packet to the destination. Step 1: This tool sends three probing packets to the first forwarder

Step 2: This tool sends three probing packets to the 2nd forwarder which are boosted by the first forwarder

In this example, Q1. What is the average round trip time from personal.psu.edu to New York? Ans. (100+120+140) / 3 = 120ms Q2. During the whole process, how many packets are forwarded by New York? Ans. 12, 3 to Cambridge and 3 back, and 3 to Boston and 3 back. Networking Tools: Traceroute Ping Whois Finger FTP

Telnet Nslookup Snoop Netstat

Chapter 4 – Transmission Media A. Intro - In traveling, a man is carried by a car, and the car is transmitted through highways. - In computer networks, messages are carried by a parent as a singal, and the signal is

transmitted through wire, fiber optic, radio waves, etc. (transmission media) - No transmission media is perfect and every transmission media has resistance. - These limitations determine the way a protocol should be designed and a network

should work B. Copper Wires Property 1. Signal Decadence - Due to the resistance of the copper, the strength of a signal will decrease along its travel “run out of gas” property Property 2. Interference

a. we assume wire 1 and 2 are parallel and close b. signal 1 emits a small amount of electromagnetic energy, which can travel

through the air in the form of an electromagnetic wave. c. When the wave encounters wire 2, it will generate a small electric current in wire

2, denoted signal 3. d. Sometimes, signal 3 can be strong enough to make itself indistinguishable from

signal 2. Q1. When signal 3 is generated in wire 2, how many signals are in wire 1? Ans. 2, interference is two way

Property 3. Circuit loop properties

Q2. Can this message be transmitted to Computer 2? Ans. NO! A circuit loop is needed, a property of electricity.

- Therefore, in networking, 2 wires are needed to transmit a single message. - Simplest Ethernet cable has 4 wires for duplex communication because 2 signals are needed and 2 wires are needed for each signal. How to resolve interference problem? Answer 1: Coaxial Cable (TV)

One coaxial cable is enough to transmit a signal because the metal hood functions as wire 2 which can create a circuit Answer 2: use twisted pair cable

- no metal hood

(1) Prevents the two wires from radiating energy that interferes with each other (2) Makes this pair less susceptible to other pairs

Coaxial Twisted Pair

More protection Cheaper More difficult Easier to use

C. Fiber Optic

Note 1: the transmitter transforms bits to pulses of light Note 2: the receiver transforms pulses of light into bits Fiber is best used for long distance Advantages of fiber optic

1. not susceptible to electric interference 2. much faster ~10 Gbps (10x faster than fastest Ethernet)* 3. carry more info 4. much less resistance – replay distance 1-2 miles (copper is about 100 meters) 5. no need for circuits

* 1 Kbps = 1024 bits per second 1 Mbps = 1 million bits per second 1 Gbps = 1000 Mbps Modem = 56 Kbps Cable modem = 800 Kbps LAN = 100 Mbps Disadvantages of fiber optic:

1. difficult to install 2. difficult to locate where fiber breaks 3. fragile 4. difficult to repair 5. expensive – the fiber material is extremely cheap, expensive because of

disadvantages 1 - 4 above D. Radio Waves Examples: AM, FM, Satellite radio, Cell phone, PDA, Bluetooth, wireless networks Idea: electromagnetic radiation can be used to transmit data in the form of radio waves. Such transmissions are also called RF transmissions. (RF = radio frequency) How do radio waves work?

Computer 1 wants to communicate with computer 2. The transmitter of computer 1 will convert bits into radio waves and the receiver on computer 2 will convert the radio waves back into bits. Q1. Will computer 3 receive this message? Ans. Yes Property 1 – transmitted in all directions Property 2 – Radio waves never bend! Property 3 – Radio waves can penetrate walls, etc. They can hurdle obstacles that are

not too large. Property 4 – cannot penetrate thick metal Property 5 – In order to broadcast to a wider range

(1) longer antenna (2) stronger signal

E. Satellites Satellites use radio waves - Good for mountain areas - Good for long distance communication - Quite expensive

Step 1. Alice sends the message to satellite Step 2. The satellite receives Alice’s message, then reboosts the radio wave before

forwarding the message to Bob. Other transmission mediums: (do not need to know for exam)

Infrared Ex. TV remote

Laser Ex. Laser Pointer

Microwave

Chapter 5 - Local Asynchronous Communication A. Introduction A Transmission System consists of (1) NIC I (2) Cable (3) NIC II

How can we send a message from computer 1 to computer 2 in electric currents? Requirements:

(1) We need a circuit (2) Hardware to handle electric currents. Such hardware is usually called a Network

Interface Card (NIC) Assume computer 1 wants to send message “PL” to computer 2. Step 1: Computer 1 will translate the message “PL” into bits. This is done using a standard code table (ASCII) – refer to page 683, appendix 2 Translating ASCII code into binary bits: Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F Binary 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

Optimization: the first bit in every translated character is “0”, so it is removed. So to send a character, 7 bits are sent Step2: use an electric current to translate the 14 bits to computer 2

Step 3: The receiver of computer 2 will decode the electronic current to get the bits Note 1: To make this idea work, the transmitters and receivers of computers 1 and 2 must follow the same rules. Rule 1: use positive voltage to represent bit “1” Rule 2: use negative voltage to represent bit “0” Rule 3: use zero voltage to represent idle. Note 2: To send out bit “1”, the transmitter places a positive voltage on the wire for a short period of time, called the holding time, then returns to zero voltage on the wire. Note 3: The holding time cannot be too short; otherwise, the receiver cannot receive the bit Question #1: Which part of the signal is first to be received by the receiver?

Answer: The left hand side starting after time=0

Question #2: To make this idea work, that is, to ensure that the receiver will always correctly receive every bit without confusion, do we need to have the same holding time for each bit?

Answer: NO, because in this scheme, every two bits will be separated by an idle state of zero voltage

Challenge: This scheme uses 3 voltages. This can be too complex or expensive! Is there a way you could only use 2 voltages to do this job without loosing the correctness of the signal? Idea 1: use positive voltage for “1” Use negative voltage for “0” Transmitter turns off when idle Idea 2: To send bit “1”, jump from positive negative To send bit “0”, jump from negative positive (has ghost bit problem if transmitter does not turn off) Challenge continued: what if we do not allow jumps?

Best Idea: fix the holding time; this is, making the holding time the same for every bit.

Using only two voltages to send bits - Using 3 voltages is easier - To use 2 voltages we need to be careful because the receiver could be confused - If we do not allow idles, then we can do it correctly if we make the holding time constant. How to handle idle?

(a) turn off the problem is the restart time (b) use start & stop bit (c) use different holding times – good idea but won’t be discussed

Use start and stop bits following the RS-232 Standard

(1) positive – ‘1’ ‘+15’ negative – ‘0’ ‘-15’ negative – idle

(2) each character has 7 bits (3) holding times are constant (4) idles are allowed only between characters (5) During the transmission of a character, idles are NOT allowed (6) Before any character is transmitted, a start bit (an extra bit ‘1’) must be

transmitted (7) When a character ends, a stop bit (and extra bit ‘0’) is transmitted

How Quick is the Transmission? Speed of RS232 measured in bits/second Types of bits (bit’1’, bit ‘0’, start bit, stop bit) # of bits in example K – 18 bits 9 bits per character Total transmission time for example K (holding time = 10ms): 180 ms plus idle Note: The transmitter uses a holding time to transmit bits. If it uses less time, it is a quicker transmitter. If it uses more time, it is a slower transmitter. Question: If no idle used, __ bits can be transmitted is 1 second? (Holding time = 10ms) 1 second = 1000 milliseconds 1000ms/10ms = 100 bits per second In Summary:

(1) The real speed of RS-232 is 9600 bps, modem is 5600 bps, DSL is 640,000, IST 208 is 100,000,000.

a. RS-232 is too slow (2) RS-232 con only reach 50 feet (3) People found that the ceiling speed (max speed) is affected by two things

a. Noises – small amount of background interference b. Bandwidth – each wire has a finite bandwidth

Noises: Sources of noise – ex. Refrigerator, lightning, electric trains Engineers use signal-to-noise ratio (S/N) to quantify noises

Due to the refrigerator, some noises are caused inside the wire. If the strength of the signal is 100; and the strength of the noise is 10, we know S/N = 100/10 = 10 times In communication, engineers also use decibels (dB) to be the measurement units of S/N S/N in decimal S/N in dB

10 10 100 20 1000 30 10,000 40 Bandwidth:

a. Fourier, many years ago, found that every signal, such as RS-232, is a summary of a number of continuous (possibly infinite), periodic, oscillating signals, and each such periodic signal, called a harmonic, is at a unique frequency.

b. When a signal travels through a wire, all the harmonics at a frequency larger than a certain threshold will be filtered off.

c. This threshold is called the bandwidth of the wire. If you combine all the harmonics (periodic signals: H1,H2,H3,H4,etc.) you create the overall signal. In this case, the RS-232 signal. The harmonics are the components of the signal.

H1 can be split into 5 identical pieces, each such piece is called a period.A period (or cycle) = the time needed to transmit each piece. Period for H1 = 1 second Period for H2 = .5 seconds

Frequency: Equals the number of periods per second H1: within 1 second, one period can be transmitted, so the frequency is f1 = 1Hz (Hertz) H2: within 1 second, two periods can be transmitted, so the frequency is F2 = 2 Hz The meaning of bandwidth:

(1) not all harmonics can go through the wire (2) Some can, some cannot (3) There is a threshold, for example, 3 Hz, that control who can go through (4) In this example, H1 and H2 can go through, H3 is “killed”

Bandwidth, threshold, is unique for each wire In most cases, only 70% of signal is received because of bandwidth and noise restrictions. However 70% is enough to receive the bit. Never expect to receive 100% of signal. Ceiling Speed Ceiling speed is determined by noises, measured by S/N, and bandwidth, denoted B (HZ) Case 1: When there are no noises, ceiling speed D = 2*B*Log2*K D = max speed in bps K= the number of voltages used For example: RS-232, k=2 D = 2*4000*log22 = 8000 bps Why less than 9,600 bps? Because we under-estimate B; in real world, B is very large; B = 1 GHZ Case 2: There are some noises practical Shannon Theorem, 1945 Ceiling speed C = B*Log2*(1+S/N) For example: RS-232, B = 1,000,000 HZ, S/N = 1000 C = 1,000,000 * log2*(1+1000) = 10,000,000 bps Summary: In case 1, there are no noises and ceiling speed can be amazingly high because there is no limit on k. In case 2, the ceiling speed when there are noises is irrelevant to k!! That means no matter the $$$ you have, how fancy your NIC is, you could not go beyond B*Log2*(1+S/N).

Chapter 6 – Distance What is long distance? depends, more than 50 feet 30 years ago, max distance was 50 ft (RS-232) Today: IST 208 LAN: 100 Meters, WAN: 1000 miles RS-232 cannot go long distance because waveform signals become weaker too fast when they travel. Why? Because there are too many sharp angles! Therefore, the following signal can go much longer

Sin wave, called a carrier Modulation: We can use these types of signals to transmit bits long distances via a technique called modulation (every modem does this). Step 1: Before sending the bits, the transmitter will modify the carrier slightly so that the modified carrier, called m-carrier, can carry the bits. Step 2: After receiving the m-carrier, the receiver will do demodulation to extract the bits. Three representative modulation techniques:

(1) amplitude modulation (2) frequency modulation (3) phase shift modulation

Amplitude Modulation: Idea: use different amplitude to represent bits ‘1’, ‘0’, ‘idle’ Set your amplitude rules Example: 30v- ‘1’ 15v – idle 7.5v – ‘0’

Frequency Modulation: Idea: use different frequencies to represent bits ‘1’, ‘0’, ‘idle’ Set your frequency rules Example: .5 HZ = ‘1’ 1 HZ = ‘idle’ 2 HZ = ‘0’ Rule 1: Don’t change the amplitude Rule 2: Synchronized Steps to do Frequency Modulation: Step 1 – Piece by piece Step 2 – a. For each piece, figure out how many periods you should draw in 1 second b. Figure out the total time you need to handle for this piece c. Figure out how many periods total for this piece d. Draw it

Phase Shift Modulation (almost every real world modem)

On a sin wave, each point corresponds to an angle. Phase: The angle for each point is called the phase of this point.

Point A’s phase is 180° Point B’s phase is 0° Here, we shift from phase 180° to 360°!

Point A’s phase is 90° Point B’s phase is 270° The phase shift is 270° – 90° = 180°

Rule: Idle – no phase shift ‘1’ – 90° phase shift ‘0’ – 180° phase shift

A: 90° B: 270° To calculate the phase shift: Step 1: Identify the two points involved in this phase shift Step 2: The phases of A & B Step 3: B – A How to use modulation/demodulation? Case 1: In a local area > 50 feet In this case, NIC will do the modulation Case 2: Wide area Two solutions

(1) VPN – Virtual private network Will do later with IP (2) In old days before internet, people use telephone lines

- Telephone network is powerful 1980: Internet had 10 nodes 1980: Telephone network was everywhere

Phone Switches are not routers - were built to make calls (voices – analog) - later, they exploited to transmit bits (digital)

Note 1: Phone Lines are twisted pairs Two ways to do such long distance networking: Way 1: Dial-up (cheap): like making phone calls, when you dial through, you can use the circuit, when you hang up you cannot. Way 2: Leased Line ($$$): Always connected, used by businesses – between divisions

More details about this process:

(1) Computer 1 sends bits to modem 1 (2) Modem 1 will modulate the bits, then send the m-carrier to Central Office 1 (3) The switch inside Central Office 1 will demodulate the m-carrier and get the bits (4) Central Office 1 will send these bits Central Office 2 via the telephone network (5) Central Office 2 get the bits (6) Central Office 2 modulates the bits and send the m-carrier to Modem 2 (7) Modem 2 demodulates the m-carrier and sends the bits to Computer 2

Step 4 further explained: How does the AT&T network relay (reboost) signals between CO1 and CO2? Two ways:

(1) Old Way: use amplifiers a. CO1 will not do demodulation; instead, CO1 will amplify the m-carrier,

and then send to PS1. b. PS1 will just amplify the m-carrier to PS2 c. PS2 amplifies m-carrier to PS3 d. PS3 amplifies m-carrier to CO2 e. CO2 amplifies m-carrier to M2 f. M2 will demodulate

PROBLEM: signal gets distorted. Amplifier amplifies distortion. As a result, many errors.

(2) New Way: use repeaters – mini modems - can remove distortion

a. Computer 1 will send bits to modem 1 b. Modem 1 sends m-carrier to CO1 c. CO1 will demodulate the bits and then modulate the bits to create a new

m-carrier d. The process continues on and distortion is removed