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Midterm Review
Lab 4: dynamic routing protocols
Typo on Page 144
• Table 4.2
– Should be router 4 instead of router 1
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How Cost is Set on a Router
10.0.1.0/24
10.0.2.0/24
eth0
.2
.1
Router A
.2
10.0.3.0/24
eth1
.1
Router B
10.0.4.0/24
.2
.1
.2
Router C
• The default cost each router adds is 1.
• When Router B announces to Router C the network 10.0.2.0/24, the cost
metric is 1.
• Offset-list 0 in value Iface (increases the metric of incoming RIP packets)
• Offset-list 0 out value Iface (increases the metric of outgoing RIP packets)
• At router B, if we run offset-list 0 out 10 eth1, then the metric for
10.0.2.0/24 is 11; if we run offset-list 0 in 10 eth1, then the metric for the
network 10.0.3.0/24 announced by C becomes 11.
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Exercise (4B): count-to-infinity
1
Router3
1
1
1
10
1 Router2
Router1
Router4
• Time consuming to reproduce, but interesting.
• Why does count-to-infinity still exist with split horizon?
• Lab report due after midterm
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Why does count-to-infinity still exist with split
horizon?
Router3
1
1
1
Router4
PC3
10.0.1.0/24
1
X
Router1
10
1 Router2
Router3’s routing table:
10.0.1.0/24 ?? 1
Suppose updates happen in the following sequence:
1. The update from PC3 arrives at Router 3
2. The update from Router 3 arrives at Router 2
3. The update from Router 4 arrives at Router 2
Router2’s routing table:
10.0.1.0/24 ?? 1
Router4’s routing table:
10.0.1.0/24 Router3 3
Router2 is not Router4’s next hop.
Router4 sends to router2 the routing update
Router2’s routing table:
10.0.1.0/24 Router 4 4
This lie will be told to Router3 and
Circulates in the system  count-to-infinity
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Midterm review
What you’ll be tested on
• Basic lab commands
– E.g., ping, traceroute, tcpdump, ethereal, ifconfig, how to
copy a file, how to list a directory
• Basic trouble shooting
– E.g., I cannot ping 128.195.1.150, why?
• Basic networking concepts
– E.g., layering principle, multiplexing, and encapsulation
• Protocols we’ve covered so far
– ARP
– ICMP
– IP
• How to design a protocol
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Address translation protocol
•
•
•
•
What is it used for?
What is an ARP cache used for?
Proxy ARP
ARP is “hop-by-hop”
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Address Translation with ARP
ARP Request:
Argon broadcasts an ARP request to all stations on the
network: “What is the hardware address of
128.143.137.1?”
Argon
128.143.137.144
00:a0:24:71:e4:44
Router137
128.143.137.1
00:e0:f9:23:a8:20
ARP Request:
What is the MAC address
of 128.143.71.1?
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Address Translation with ARP
ARP Reply:
Router 137 responds with an ARP Reply which contains the
hardware address
Argon
128.143.137.144
00:a0:24:71:e4:44
Router137
128.143.137.1
00:e0:f9:23:a8:20
ARP Reply:
The MAC address of 128.143.71.1
is 00:e0:f9:23:a8:20
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ARP Packet Format
Ethernet II header
Destination
address
Source
address
Type
0x8060
6
6
2
ARP Request or ARP Reply
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10
Hardware type (2 bytes)
Hardware address
length (1 byte)
Padding
CRC
4
Protocol type (2 bytes)
Protocol address
length (1 byte)
Operation code (2 bytes)
Source hardware address*
Source protocol address*
Target hardware address*
Target protocol address*
* Note: The length of the address fields is determined by the corresponding address length fields
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Example
• ARP Request from Argon:
Source hardware address:
Source protocol address:
Target hardware address:
Target protocol address:
00:a0:24:71:e4:44
128.143.137.144
00:00:00:00:00:00
128.143.137.1
• ARP Reply from Router137:
Source hardware address:
Source protocol address:
Target hardware address:
Target protocol address:
00:e0:f9:23:a8:20
128.143.137.1
00:a0:24:71:e4:44
128.143.137.144
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ARP Cache
• Since sending an ARP request/reply for each IP datagram is
inefficient, hosts maintain a cache (ARP Cache) of current
entries. The entries expire after a time interval.
• Contents of the ARP Cache:
(128.143.71.37) at 00:10:4B:C5:D1:15 [ether] on eth0
(128.143.71.36) at 00:B0:D0:E1:17:D5 [ether] on eth0
(128.143.71.35) at 00:B0:D0:DE:70:E6 [ether] on eth0
(128.143.136.90) at 00:05:3C:06:27:35 [ether] on eth1
(128.143.71.34) at 00:B0:D0:E1:17:DB [ether] on eth0
(128.143.71.33) at 00:B0:D0:E1:17:DF [ether] on eth0
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Proxy ARP
• Proxy ARP: Host or router responds to ARP Request that
arrives from one of its connected networks for a host that is
on another of its connected networks.
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ICMP
• What is it used for?
– E.g. error reporting, route redirect
• When will an ICMP message be triggered?
• Application programs that use ICMP
– Ping, traceroute
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IP
•
•
•
•
•
•
Network order versus host order
Structure of an IP address
CIDR addressing
Route aggregation
Longest prefix match
Fragmentation
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An IP address is often written in dotted decimal
notation
• Each byte is identified by a decimal number in the range
[0..255]:
10000000
10001111
10001001
10010000
1st Byte
2nd Byte
3rd Byte
4th Byte
= 128
= 143
= 137
= 144
128.143.137.144
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Structure of an IP address
31
0
network prefix
host number
• An IP address encodes both a network number
(network prefix) and an interface number (host
number).
– network prefix identifies a network
– the host number identifies a specific host
(actually, an interface on the network).
– All hosts on the same single segment network
have the same network prefix.
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Network prefix is variable length
128
Addr
Mask
143
137
10000000
10001111
10001001
255
255
255
11111111
11111111
1111111
144
10010000
0
00000000
• A network mask specifies the number of bits used
to identify a network in an IP address.
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CIDR notation
• CIDR notation of an IP address:
– 128.143.137.144/24
– /24 is the prefix length. It states that the first 24
bits are the network prefix of the address (and
the remaining 8 bits are available for specific
host addresses)
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Network prefix versus address prefix
• CIDR notation can nicely express blocks of addresses
– An address block
– [128.195.0.0, 128.195.255.255]
– can be represented by an address prefix 128.195.0.0/16
– All addresses whose first 16 bits are the same as those in
128.195.0.0 are in the address block or match the address
prefix 128.195.0.0/16
– How many addresses are there in a /x address block?
• 2 (32-x)
• A network prefix is the first n bits in an IP address that
identifies a single-segment network.
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How to assign network prefixes from an address
prefix
•
•
•
•
An organization obtains an address prefix 10.0.0.0/16
The organization has two LANS. LAN1 has at most 500
hosts; LAN2 has at most 100 hosts.
Assign network prefixes to each LAN.
Algorithm
1. Figure out the length of the network prefix
– 232-x1 ¸ 500  x1 = 23
– 232-x2 ¸ 100  x2 = 25
2. Allocate subdivisions of 10.0.0.0/16 to each LAN
– 10.0.0.0/23 [10.0.0.0,10.0.1.255]  LAN1
– 10.0.2.0/25 [10.0.2.0, 10.0.2.127]  LAN2
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Protocol Design
human protocols:
• “what’s the time?”
• “I have a question”
• introductions
… specific msgs sent
… specific actions taken
when msgs received, or
other events
network protocols:
• machines rather than
humans
• all communication activity in
Internet governed by
protocols
protocols define format, order of
msgs sent and received among
network entities, and actions
taken on msg transmission,
receipt
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What’s a protocol?
a human protocol and a computer network protocol:
Hi
TCP connection
req
Hi
TCP connection
response
Got the
time?
Get http://www.awl.com/kurose-ross
2:00
<file>
time
Q: Other human protocols?
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How to review
• Lecture notes
• Lab and pre-lab
• Sample midterm
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