Download Review for Quiz-1 - Georgia Institute of Technology

Review for Quiz-2
ECE3600 - Fall 2015
Prof. John Copeland
3-29-2015
Computer Networking: A Top Down Approach Featuring the Internet,
5th edition.
Jim Kurose, Keith Ross
Base material copyright 1996-2006
Addison-Wesley, July 2004.
J.F Kurose and K.W. Ross, All Rights Reserved
Quiz 2 -1
ECE3600 Quiz-2 Review
• Chap. 4 – IP Addresses and Routing
• Chap. 5 – LANs, Ethernet Networks
• Chap. 6 – Wireless LANs
• Chap. 7 – Multimedia
Quiz 2 -2
Chapter 4: Network Layer
IP Addresses
• 4. 1 Introduction
• 4.2 Virtual circuit
and datagram
networks
• 4.3 What’s inside a
router
• 4.4 IP: Internet
Protocol
–
–
–
–
Datagram format
IPv4 addressing
ICMP
IPv6
Quiz 2 -3
Longest prefix matching
Prefix Match
11001000 00010111 0001 0
11001000 00010111 0001 1000
11001000 00010111 0001 1
otherwise
Link Interface
0
1
2
3
Examples
DA: 11001000 00010111 0001 1000 1010 1010
Which interface?
DA: 11001000 00010111 0001 0110 1010 0001
Which interface?
Matches links 1 & 2 - Link 1 the longest.
Only matches Link 0
DA: 11001000 00010111 0001 1100 1010 1010
Only matches Link 2
Q3-4
Quiz 2 -4
IP datagram format
IP protocol version
number
header length
(bytes)
“type” of data
max number
remaining hops
(decremented at
each router)
upper layer protocol
to deliver payload to
how much overhead
with TCP?
r 20 bytes of TCP
r 20 bytes of IP
r = 40 bytes + app
layer overhead
32 bits
type of
ver head.
len service
length
fragment
16-bit identifier flgs
offset
upper
time to
header
layer
live
checksum
total datagram
length (bytes)
for
fragmentation/
reassembly
32 bit source IP address
32 bit destination IP address
Options (if any)
data
(variable length,
typically a TCP
or UDP segment)
E.g. timestamp,
record route
taken, specify
list of routers
to visit.
Q3-5
Quiz 2 -5
IP Fragmentation and Reassembly
Example
r 4000 byte datagram
r MTU = 1500 bytes
1480 bytes in
data field
offset =
1480/8
length ID fragflag
=4000 =x =0
offset
=0
One large datagram becomes
several smaller datagrams
length
=1500
ID fragflag
=x =1
offset
=0
length
=1500
ID fragflag
=x =1
offset
=185
length
=1040
ID fragflag
=x =0
offset
=370
Quiz 2 -6
Subnets – have a
contiguous block of IP
addresses which have
the first N bits in
common (a "/N").
Recipe
• To determine the subnets,
detach each interface
from its host or router,
creating islands of isolated
networks. Each isolated
network is called a subnet.
223.1.1.0/24
223.1.2.0/24
223.1.3.0/24
Subnet mask: /24 (24 1's (32-24) 0's
11111111 11111111 11111111 00000000
Dotted decimal notation: 255.255.255.0
Q3-7
Quiz 2 -7
IP addressing: CIDR
CIDR: Classless InterDomain Routing
– subnet portion of address of arbitrary length
– address format: a.b.c.d/x, where x is # bits
in subnet portion of address
subnet
part
Addr.
Mask:
host
part
11001000 00010111 00010000 00000000
11111111 11111111 11111110 00000000
Addr.
Mask:
200.23.16.0/23
255.255.254.0 Inverted = 0.0.1.255
Network Address = Host Address (Bitwise AND) Network Mask = A & M
Minimum Host Address* = Network Address
Maximum Host Address* = Network Address (OR) [Inverted Network Mask]
* (reserved, not assigned to host)
Quiz 2 -8
IP Address Bitwise Calculations
200.23.16.0/20
1101000 00011001 0001xxxx xxxxxxxx
From this, to get Network Mask, 0 or 1 -> 1, x -> 0
11111111 11111111 11110000 00000000
Minimum Host Address: x -> 0
1101000 00011001 00010000 00000000
Maximum Host Address: x -> 1
1101000 00011001 00011111 11111111
Minimum host address is the “Network Address”
Maximum host address is the “Broadcast Addr.”
Q3-9
Quiz 2 -9
NAT: Network Address Translation
2: NAT router
changes datagram
source addr from
10.0.0.1, 3345 to
138.76.29.7, 5001,
updates table
2
NAT translation table
WAN side addr
LAN side addr
1: host 10.0.0.1
sends datagram to
128.119.40.186, 80
138.76.29.7, 5001 10.0.0.1, 3345
……
……
S: 10.0.0.1, 3345
D: 128.119.40.186, 80
S: 138.76.29.7, 5001
D: 128.119.40.186, 80
138.76.29.7
S: 128.119.40.186, 80
D: 138.76.29.7, 5001
3
1
10.0.0.4
S: 128.119.40.186, 80
D: 10.0.0.1, 3345
10.0.0.1
10.0.0.2
4
10.0.0.3
4: NAT router
3: Reply arrives
changes datagram
dest. address:
dest addr from
138.76.29.7, 5001
138.76.29.7, 5001 to 10.0.0.1, 3345
NAT Table for Internal Servers must be configured manually
(Port Forwarding).
Quiz 2 -10
ICMP: Internet Control Message Protocol
•
•
•
used by hosts & routers to
communicate network-level
information
– error reporting:
unreachable host, network,
port, protocol
– echo request/reply (used
by ping)
network-layer “above” IP:
– ICMP msgs carried in IP
datagrams
ICMP message: type, code plus
first 8 bytes of IP datagram
causing error
Type Code description
0
0
echo reply (ping)
3
3
3
3
3
3
0
1
2
3
6
7
dest. network unreachable
dest host unreachable
dest protocol unreachable
dest port unreachable
dest network unknown
dest host unknown
4
0
8
9
10
11
12
0
0
0
0
0
source quench (congestion
control - not used)
echo request (ping)
route advertisement
router discovery
TTL expired
bad IP header
Quiz 2 -11
Traceroute and ICMP
• Source sends series of
UDP segments to dest
– First has TTL =1
– Second has TTL=2, etc.
– Unlikely port number
• When nth datagram arrives
to nth router:
– Router discards datagram
– And sends to source an
ICMP message (type 11,
code 0)
– Message includes name of
router& IP address
• When ICMP message
arrives, source calculates
RTT
• Traceroute does this 3
times
Stopping criterion
• UDP segment eventually
arrives at destination host
• Destination returns ICMP
“host unreachable” packet
(type 3, code 3)
• When source gets this
ICMP, it stops.
Quiz 2 -12
IPv6 Header (Cont)
New: “flow label”,
“longer addresses”
Missing: fragmentation
(flags, ID, offset)
Priority: identify priority among datagrams in flow
Flow Label: identify datagrams in same “flow.”
(concept of“flow” not well defined).
Next header: identify upper layer protocol for data
“6to4 Translation”
4-byte IPv4 -> 16-byte IPv6
A.B.C.D -> :2002:aabb:ccdd/80
“:aa:”=“A in 2-char hex”
“:bb:”=“B in 2-char hex”
etc.
IPv4 address can become an IPv6
sub-net with 80 bits for “host”
addresses (1e24 hosts)
http://en.wikipedia.org/wiki/6to4
Quiz 2 -13
Example – 6to4 convert 130.207.17.25 to IPv6 address.
Convert the decimal byte-representations to hex:
130 = 0x82 207 = 0xCF 17 = 0x11 25 = 0x19
IPv6 addresses are written with colons separating every
16 bits (4 hex characters).
:0000: can be written ::
The first 16 bits 0x2002 are a reserved /16 block of
addresses reserved for IPv4 translations.
:2002::::::::/16
“6to4 Translation”
Next add the IPv4 32 bits:
4-byte IPv4 -> 16-byte IPv6
:2002:82CF:1119::::::/48
A.B.C.D -> :2002:aabb:ccdd/32
This is not just a single IPv6
“:aa:”=“A in 2-char hex”
“:bb:”=“B in 2-char hex”
address, but a block of 2^80
etc.
possible host addresses that
IPv4 address can become an IPv6
can replace private subnet
sub-net with 80 bits for “host”
addresses, like 192.168.0.0/16's. addresses (1e24 hosts)
http://en.wikipedia.org/wiki/6to4
Quiz 2 -14
Chapter 4 - IP Routing
• 4.5-6 Routing
– Distance Vector (RIP)
– Link state (OSPF)
– Hierarchical routing
(BGP)
• 4.7 Broadcast and
multicast routing
Some material copyright 1996-2006
J.F Kurose and K.W. Ross, All Rights Reserved
Computer
Networking: A Top
Down Approach
Featuring the
Internet,
5th edition.
Jim Kurose, Keith
Ross
Addison-Wesley
4/3/15
Quiz 2 -15
Distance Vector: link cost changes
Link cost changes:
good news travels fast
bad news travels slow “count to infinity”
problem!
44 iterations before
algorithm stabilizes: see
text
Poisoned reverse:
If Z routes through Y to
get to X :
Z tells Y its (Z’s)
distance to X is infinite
(so Y won’t route to X
via Z)
will this completely solve
count to infinity problem?
60
4
x
y
50
1
z
Y advertises X in 4 hops
Z sends datagrams for X to
Y
Z advertises "X in 5 hops".
Y-X link cost goes to 60
Y thinks Z can route in 5
hops, so Y advertises "X in
6", sends datagrams back
to Z.
Z sends datagrams back to
Y, advertises "X in 7".
Y sends datagrams back to
Z, advertises "X in 8".
Quiz 2 -16
RIP (Distance-Vector Algorithm)
B
M
128.230.0.0/16
Y
Router A Table
Prefix
Distance Port
128.230.
2
X
130.207.
6
N
209.196.
7
X
24.56.
9
X
A
X
C
Router B Table
Prefix
Distance Port
128.230.
2
X
130.207.
6
X
209.196.
5
M
24.56.
11
X
Z
Router C Table
Prefix
Distance Port
128.230.
2
X
130.207.
4
X
209.196.
7
X
24.56.
11
P
Construct the Routing Table for Router X. Use "L" for the port to the local LAN.
Router X Table
Prefix
Distance Port
128.230.
1
L
130.207.
5
C
209.196.
6
B
24.56.
10
A
Using Poison Reverse, construct the Updates sent from Router X to A, B, and C. (infinity -> 15).
Update X to A Table
Prefix
Distance
128.230.
1
130.207.
5
209.196.
6
24.56.
15
Update X to B Table
Prefix
Distance
128.230.
1
130.207.
5
209.196.
15
24.56.
10
Update X to C Table
Prefix
Distance
128.230.
1
130.207.
15
209.196.
6
24.56.
10
“Poison Reverse” prevents “ping-pong” routes.
Quiz 2 -17
Graph abstraction
5
2
u
v
2
1
x
Graph: G = (N,E)
3
w
3
1
5
z
1
y
2
N = set of routers = { u, v, w, x, y, z }
E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w,z), (y,z) }
Remark: Graph abstraction is useful in other network contexts
Example: P2P, where N is set of peers and E is set of TCP connections
Quiz 2 -18
Graphical Method
(animated - keep clicking)
5
5
3
v
2 2
w
5
4
3
u
3
2
1
2
x
8
1
3
1
5
z
4
2
y
1
Next Slide
Quiz 2 -19
Hierarchical OSPF
Only this router
seen outside Area
1, where it
represents all the
subnets in Area 1
Quiz 2 -20
Hierarchical Routing (BGP)
Our routing study thus far - idealization
all routers identical
network “flat”
… not true in practice
scale: with 200 million
destinations:
• can’t store all dest’s in
routing tables!
• routing table exchange
would swamp links!
administrative autonomy
• internet = network of networks
(Autonomous Systems)
• each network admin may want
to control routing in its own
network
Quiz 2 -21
Interconnected ASes
3c
3a
3b
AS3
2a
1c
1a
1d
1b AS1
Intra-AS
Routing
algorithm
Inter-AS
Routing
algorithm
Forwarding
table
2c
AS2
2b
• Forwarding table is
configured by both
intra- and inter-AS
routing algorithm
– Intra-AS sets entries for
internal dests
– Inter-AS & Intra-As sets
entries for external dests
Quiz 2 -22
Broadcast Routing
• Deliver packets from source to all other
nodes
• Source duplication is inefficient:
duplicate
duplicate
creation/transmission
R1
R1
duplicate
R2
R2
R3
R4
source
duplication
R3
R4
in-network
duplication
Source duplication: how does source
determine recipient addresses?
Quiz 2 -23
In-network duplication
• Flooding: when node receives brdcst pckt,
sends copy to all neighbors
– Problems: cycles & broadcast storm
• Controlled flooding: node only brdcsts pkt
if it hasn’t brdcst same packet before
– Node keeps track of pckt ids already brdcsted
– Or reverse path forwarding (RPF): only
forward pckt if it arrived on shortest path
between node and source
• Spanning tree
– No redundant packets received by any node
Quiz 2 -24
Chap. 6 - Wireless LANs
What is:
Wireless Access Point (AP)
IEEE 802.11a, b, g - WiFi
IEEE 802.15 WiMax
CDMA/CA (Collision Avoidance)
"Hidden Terminal" problem
How helped by "CA": RTS, CTS packets
Modulation Techniques
TDMA - Time slots
CDMA - Chip codes
MAC Packet Header
3rd Address needed on radio side - why
Same as Ethernet on wired side (no AP MAC
address)
Chap. 6,7,8 Review
Quiz 2 -25
802.11 frame: addressing
R1 router
H1
Internet
AP
R1 MAC addr H1 MAC addr
dest. address
AP MAC addr H1 MAC addr R1 MAC addr
address 1
address 2
source address
802.3 frame
(Ethernet)
address 3
802.11 frame
Chap. 6,7,8 Review
Quiz 2 -26
Cellular Networks
Modulation Techniques
Combined FDMA & TDMA - Freq. Div &Time slots
CDMA - Chip codes
Hexagonal Cells
Why Hex, where is antenna located (serves 3 cells)
Parade of Technolgies
AMPS (Advanced Mobile Phone Sys.): FM, 1 call/freq. 0.8
GHz
2G (2nd Generation)
IS-136 - Combined FDMA & TDMA
GSM - Combined FDMA & TDMA
IS-95 - CDMA
2.5G - Enhanced data rates: GPRS, EDGE - (144 to 380 Kbps)
3G - GSM with CDMA
Chap. 6,7,8 Review
Quiz 2 -27
Mobility
• Handoff
– used first in Cellular Telephone Systems
– Connection stretches along path through
previous cells
Chap. 6,7,8 Review
Quiz 2 -28
Chap. 7 - Multimedia
What is:
QoS - Quality of Service. Guarantees:
Packet delay less than a Maximum
Bandwidth greater than a Minimum
Defined by SLA, Service Level Agreement, w ISP
Policing
Token Bucket algorithm
Streaming Media
Out-of-Band Control Channel (RTSP)
Client Buffering - smoothes out delay variations
Interactive Media
requires short round-trip time
limits use of large buffers
less delay variation tolerance
Chap. 6,7,8 Review
Quiz 2 -29
Bandwidth Needed for Media
• Streaming Media - Not compressed
• Bandwidth = (Sampling Rate)*log2(Levels)
• Telephone - (PCM)
– 8000 samples/s, 256 levels (8 bits)=64 kb/s
– Compressed voice = 5kb/s to 12kb/s
• CD Music, Stereo (2 channels)
– 44,100 samples/s, 65k levels (16 bits)= 1.5 Mb/s
– Compressed (MP3) 96 to 160 kb/s
• Video
– Compressed -(MPEG4) < 1 Mbps)
Chap. 6,7,8 Review
Quiz 2 -30
Streaming Multimedia: UDP or TCP?
UDP
• server sends at rate appropriate for client
(oblivious to network congestion !)
– often send rate = encoding rate = constant rate
– then, fill rate = constant rate - packet loss
• short playout delay (2-5 seconds) to compensate
for network delay jitter
• error recover: time permitting
TCP
•
•
•
•
send at maximum possible rate under TCP
fill rate fluctuates due to TCP congestion control
larger playout delay: smooth TCP delivery rate
HTTP/TCP passes more easily through firewalls
Chap. 6,7,8 Review
Quiz 2 -31
Video - Recovery from Packet Loss
• TCP - too slow
• UDP - Requires (one or all)
– FEC (Forward Error Correction)
– Interleaving
• lone dropouts -> many short dropouts
– Application-Level Recovery
• Freeze picture
• Extrapolate across gap
Chap. 6,7,8 Review
Quiz 2 -32
VoIP Setting up a Connection
• SIP - Session Initiation Protocol
• Finds "callee" by IP, email address, or telephone no.
• Call Management
– add new media streams
– conference calls
– transfer and hold
Chap. 6,7,8 Review
Quiz 2 -33
Compare SIP with H.323
• H.323 is another signaling
protocol for real-time,
interactive
• H.323 is a complete, vertically
integrated suite of protocols for
multimedia conferencing:
signaling, registration, admission
control, transport and codecs.
• SIP is a single component. Works
with RTP, but does not mandate
it. Can be combined with other
protocols and services.
• H.323 comes from the
ITU (telephony).
• SIP comes from IETF:
Borrows much of its
concepts from HTTP. SIP
has a Web flavor,
whereas H.323 has a
telephony flavor.
• SIP uses the KISS
principle: Keep it simple
stupid.
Chap. 6,7,8 Review
Quiz 2 -34
Content distribution networks (CDNs)
Content replication
• Challenging to stream large
files (e.g., video) from single
origin server in real time
origin server
in North America
• Solution: replicate content at
hundreds of servers
CDN distribution node
throughout Internet
– content downloaded to CDN
servers ahead of time
– placing content “close” to
user avoids impairments
(loss, delay) of sending
content over long paths
CDN server
– CDN server typically in
CDN server
in S. America CDN server in Asia
edge/access network
in Europe
Chap. 6,7,8 Review
Quiz 2 -35
Scheduling And Policing Mechanisms
• scheduling: choose next packet to send on link
• FIFO (first in first out) scheduling: send in
order of arrival to queue
– real-world example?
– discard policy: if packet arrives to full queue: who
to discard?
• Tail drop: drop arriving packet
• priority: drop/remove on priority basis
• random: drop/remove randomly
Chap. 6,7,8 Review
Quiz 2 -36
Policing Mechanisms
Token Bucket: limit input to specified Burst Size and
Average Rate.
• bucket can hold b tokens
• tokens generated at rate r token/sec unless bucket
full
• over interval of length t: number of packets
admitted less than or equal to (r t + b).
Chap. 6,7,8 Review
Quiz 2 -37
IETF Differentiated Services
• Edge Router
– Marks packets as "in-profile" or "out-profile"
– Marks packets as to "Class of Service"
• Core Routers
– "Per Class" traffic management
– Assured Forwarding of "in-profile" packets
RSVP - Resource Reservation Protocol
• Reserves resources (bandwidth, delay
limits, …) along an end-to-end path.
Chap. 6,7,8 Review
Quiz 2 -38