Title: Midterm Review
1Midterm Review
2Delay
- Speed of light
- 3.0 x 108 meters/second in a vacuum
- 2.3 x 108 meters/second in a cable
- 2.0 x 108 meters/second in a fiber
- Delay x Bandwidth Product
Delay
Bandwidth
3Sequence Number
- Suppose you are designing a sliding window
protocol for a 1-Mbps point-to-point link to the
moon, which has a one-way latency of 1.25
second. How many bits do you need for the
sequence number? - 2.5 seconds 1Mbps 2.5Mbits 312Kbytes. We
need a sequence range that will cover double this
value or 624K. So 20 bits are needed.
4Seven Layer Model
Application Presentation Session Transport
Application Presentation Session Transport
Email, FTP, www integer size, big
endian Synchronization(video/audio), name
space reliability, congestion control (TCP)
Routing Address framing Errors electrical signa
ls
Network Data Link Physical
Network Data Link Physical
IP Ether
Network Data Link Physical
Network Data Link Physical
5Process Model
- Process per protocol
- Context switch at each protocol boundary
- No Semaphores
- Incoming and outgoing queue
- Slow
- Process per message
- Procedure call at each protocol boundary
- Semaphores needed
- Faster
- Error prone because of multiple threads running
in a single protocol
6Avoiding Copies
- Assume Data rate of 600Mbps73MBps
- Assume 16MHz memory bus that is 16 bits wide,
this results in 32MBps - For a MB message, one copy takes 1/32 sec
- The resulting maximum data rate is 32MBps
- Two copies take 1/32 sec 1/32 sec 1/16 sec
- The resulting maximum data rate is 16MBps
- Copies are necessary between user and system
space if no special interface
7Links
Sometimes you install your own Sometimes
leased from the phone company
STSSynchronous Transport SignalOCOptical
carrier
8Non-Return to Zero (NRZ)
0 0 1 0 1 1 1 1 0 1 0 0
0 0 1 0
Bits
NRZ
- Problem Consecutive 1s or 0s
- Low signal (0) may be interpreted as no signal
- High signal (1) leads to baseline wander
- Unable to recover clock
- NRZI eliminates consecutive 1s
94B/5B
- Problem consecutive zeros
- Idea Every 4 bits of data is encoded in a 5-bit
code, with the 5-bit codes selected to have no
more than one leading 0 and no more than two
trailing 0 (i.e., never get more than three
consecutive 0s). - Resulting 5-bit codes are then transmitted using
the NRZI encoding. Achieves 80 efficiency. - We already dealt with consecutive 1s with NRZI
10Bit stuffing Example
- Original Data
- 0011111011000011111100
- Bit Stuffed
- 001111100110000111110100
- Receiver
- 01111110 00111110110000111110100 01111110
11- Sender
- multiply M(x) by xk for our example, we get
- x10 x7 x6 x4 (10011010000)
- divide result by C(x) (1101)
- Send 10011010000 - 101 10011010101, since this
must be exactly divisible by C(x)
11111001 10011010 000 Message 1101 1001
1101 1000 1101 1011 1101
1100 1101 1 000
1 101 101 Remainder
Generator 1101
12IP Addresses
13IP Addresses
14Example
A
(3,3,0)
B
B3
(5,5,0)
(7,7,0)
(3,3,0)
(5,5,0)
C
B5
(2,2,0)
(5,5,0)
D
B7
K
B2
(7,7,0)
(2,2,0)
(7,7,0)
E
F
(1,1,0)
(1,1,0)
(1,1,0)
B1
(1,1,0)
G
H
(6,6,0)
(4,4,0)
(4,4,0)
B6
B4
(6,6,0)
(4,4,0)
I
J
15Next Phase
A
(3,2,1)
B
B3
(5,1,1)
(7,1,1)
(5,1,1)
C
B5
(2,1,1)
D
B7
K
B2
(7,1,1)
E
F
(1,1,0)
(1,1,0)
(1,1,0)
B1
(1,1,0)
G
H
(4,1,1)
B6
B4
(6,1,1)
(4,1,1)
I
J
16Next Phase
A
B
B3
(5,1,1)
(5,1,1)
C
B5
(2,1,1)
(7,1,1)
D
B7
K
B2
E
F
(1,1,0)
(1,1,0)
(1,1,0)
B1
(1,1,0)
G
H
(4,1,1)
B6
B4
(4,1,1)
I
J
17Routers? Hubs? Switches
D 128.187.174.10 D 44.fe.34.56.32.d5 S
128.187.171.2 S fe.34.56.32.d5.29
D 128.187.174.10 D 44.fe.34.56.32.d5 S
128.187.171.2 S fe.34.56.32.d5.29
173
Hub
171
H1
H2
H3
H7
H8
H9
172
174
H4
H5
H6
H10
H11
H12
D 128.187.174.10 D 44.fe.34.56.32.d5 S
128.187.171.2 S fe.34.56.32.d5.29
D 128.187.174.10 D 44.fe.34.56.32.d5 S
128.187.171.2 S fe.34.56.32.d5.29
H10 IP 128.187.174.10, Ethernet 44.fe.34.56.32.d5
18Forwarding
- H1 LANE layer receives a packet for H2
- Since H2 is not in our cache, we send query to
LES for H2 ATM address - Since the query and VC setup may take seconds, we
start sending data through the BUS - We may wait until we receive several packets
before setting up a separate virtual circuit - The LES replies with the ATM address of H2
- H1 sets up a direct VC to H2 and adds an entry in
the cache for H2
19More Questions
- What are the advantages and disadvantages of
statistical multiplexing? - What are the three kinds of failures that network
designers have to worry about? - Why is the delay x bandwidth product important?
- What is jitter? Why would it be difficult to send
video over a network with a high jitter value?
20More Questions
- Given a source routing example, determine the
destination that a message will go to. (Something
like Figure 3.7) - Given a virtual circuit routing example,
determine the destination that a message will go
to. (Something like Figure 3.5) - Compare datagrams, virtual circuits and source
routing. What are the advantages and
disadvantages of each approach?
21More Questions
- What is the spanning tree algorithm and which
links would be eliminated in a problem similar to
Figure 3.11? How do bridges learn host locations? - Why does an ATM cell have 48 bytes of payload?
- What kind of routing does ATM use?
- Given a banyan network similar to Figure 3.33,
decide if there will be a contention free route
for two packets.
22More Questions
- What is the hidden node problem (page 138) and
how does 802.11 deal with it? - Describe the process for selecting an AP with a
mobile node (pages 140-141). - What is an internetwork?
- What is the service model for IP?
- How is fragmentation performed with IPv4?
- Assume a host with an IP address of
222.123.250.3. To what class does this IP address
belong? To what network is it connected?
23More Questions
- What is an ARP packet and how is it used? How
would ARP be implemented in a network such as ATM
that doesnt have broadcast capabilities? - What does an ARP cache contain? How does the ARP
protocol work? - Is source routing practical? Explain.
- What type of IP address class would be adequate
for a corporation that needs to establish up to
100 sub-networks, each of which with up to 500
hosts? Justify your choice.
24More Questions
- Assume the following routing table
- What does the router do with the following
packets? - a) 128.96.39.10 b) 128.96.40.12 c) 128.96.40.151
d) 192.4.153.17 e) 192.4.153.90 - What are the problems with stop and wait, and how
are they solved?
25Architecture
LES
Mac
Mac
Mac
Mac
Ethernet
BUS
Relay
ATM Switch
Mac
LEC
ATM
Plug and Play Allows ATM hosts to
connect Simplified Core Switch, complex edge
switch
26Architecture
Mac
Mac
Mac
Mac
Ethernet
LEC
Relay
LES
BUS
ATM Switch
Mac
ATM
PVC static configuration Simplified Edge switch,
complex core switch
27Architecture
POS
Mac
Mac
Mac
Mac
Ethernet
Relay
Relay
Ether Switch
Mac
POS
Simplified Edge switch, point to point link, not
scalable, time division multiplexing leased
line expense