Transcript Welcome to COE321: Logic Design

DHCP: Dynamic Host Configuration Protocol
Goal: allow host to dynamically obtain its IP address from
network server when it joins network
Can renew its lease on address in use
Allows reuse of addresses (only hold address while connected an
“on”)
Support for mobile users who want to join network (more shortly)
DHCP overview:
 host broadcasts “DHCP discover” msg [optional]
 DHCP server responds with “DHCP offer” msg
[optional]
 host requests IP address: “DHCP request” msg
 DHCP server sends address: “DHCP ack” msg
Network Layer
4-1
DHCP client-server scenario
A
223.1.2.1
DHCP
server
223.1.1.1
223.1.1.2
223.1.1.4
223.1.2.9
B
223.1.2.2
223.1.1.3
223.1.3.1
223.1.3.27
223.1.3.2
E
arriving DHCP
client needs
address in this
network
Network Layer
4-2
DHCP client-server scenario
DHCP server: 223.1.2.5
DHCP discover
arriving
client
src : 0.0.0.0, 68
dest.: 255.255.255.255,67
yiaddr: 0.0.0.0
transaction ID: 654
DHCP offer
src: 223.1.2.5, 67
dest: 255.255.255.255, 68
yiaddrr: 223.1.2.4
transaction ID: 654
Lifetime: 3600 secs
DHCP request
time
src: 0.0.0.0, 68
dest:: 255.255.255.255, 67
yiaddrr: 223.1.2.4
transaction ID: 655
Lifetime: 3600 secs
DHCP ACK
src: 223.1.2.5, 67
dest: 255.255.255.255, 68
yiaddrr: 223.1.2.4
transaction ID: 655
Lifetime: 3600 secs
Network Layer
4-3
DHCP: more than IP address
DHCP can return more than just allocated IP
address on subnet:
address of first-hop router for client
 name and IP address of DNS sever
 network mask (indicating network versus host
portion of address)

Network Layer
4-4
DHCP: example
 connecting laptop needs its
DHCP
UDP
IP
Eth
Phy
DHCP
DHCP
DHCP
DHCP
IP address, addr of firsthop router, addr of DNS
server: use DHCP
DHCP
DHCP
DHCP
DHCP
DHCP
DHCP
UDP
IP
Eth
Phy

DHCP request encapsulated
in UDP, encapsulated in IP,
encapsulated in 802.1
Ethernet

Ethernet frame broadcast
(dest: FFFFFFFFFFFF) on LAN,
received at router running
DHCP server

Ethernet demuxed to IP
demuxed, UDP demuxed to
DHCP
168.1.1.1
router
(runs DHCP)
Network Layer
4-5
DHCP: example
 DCP server formulates
DHCP
UDP
IP
Eth
Phy
DHCP
DHCP
DHCP
DHCP
DHCP ACK containing
client’s IP address, IP
address of first-hop
router for client, name &
IP address of DNS server

DHCP
DHCP
DHCP
DHCP
DHCP
DHCP
UDP
IP
Eth
Phy
router
(runs DHCP)

encapsulation of DHCP
server, frame forwarded to
client, demuxing up to
DHCP at client
client now knows its IP
address, name and IP
address of DSN server, IP
address of its first-hop
router
Network Layer
4-6
NAT: Network Address Translation
rest of
Internet
local network
(e.g., home network)
10.0.0/24
10.0.0.4
10.0.0.1
10.0.0.2
138.76.29.7
10.0.0.3
All datagrams leaving local
network have same single source
NAT IP address: 138.76.29.7,
different source port numbers
Datagrams with source or
destination in this network
have 10.0.0/24 address for
source, destination (as usual)
Network Layer
4-7
NAT – Network Address
Translation
Placement and operation of a NAT box.
Network Layer
4-8
Internet Control Message
Protocol
The principal ICMP message types.
5-61
Network Layer
4-9
ICMP: example
 Hannah being a great network trouble
shooter

Can test basic network connectivity using
• The ping command
– uses the ICMP sending a message called echo request
– The destination should reply with an ICMP echo reply
Network Layer 4-10
Exercises
A router has the following entries in its routing
table:
Address/mask
Next hop
135.46.56.0/22
interface 0
135.46.60.0/22
interface 1
192.53.40.0/23
interface 2
default
interface 3
1. For each of the following addresses, what does
the router do if a packet with that address
arrives: a) 135.46.63.10, b) 135.46.57.14,
c)135.46.52.2, d)192.53.40.7, e) 192.53.56.7

Network Layer
4-11
Solution
 The packets are routed as follows:
a) Interface 1
 b) Interface 0
 c) Interface 3
 d) Interface 2
 e) Interface 3

Network Layer 4-12
Exercises
 A router has just received the following
new IP addresses: 57.6.96.0/21,
57.6.104.0/21, 57.6.112.0/21, and
57.6.120.0/21. If all of them use the same
outgoing line, can they be aggregated? If
so, to what? If not, why not?
Network Layer 4-13
Exercises
 You have a class C network, and you need
to design it for 7 usable subnets with
each subnet handling a minimum of 18
hosts each. Which of the following
network masks should you use?





255.255.224.0
255.255..255.230
255.255.255.224
255.255.255.240
None of the above
Network Layer 4-14
Solution
 Answer: C
Explanation:
The default subnet mask for class C network
is 255.255.255.0. If one has to create 5
subnets, then 3 bits are required. With 3
bits we can create 6 subnets. The remaining
5 bits are used for Hosts. One can create 30
hosts using 5 bits in host field. This matches
with the requirement.
Network Layer 4-15
Exercises
 If a host on a network has the address
172.16.210.0/22, what is the address of
the subnetwork to which the host
belongs?





172.16.42.0
172.16.107.0
172.16.208.0
172.16.255.208
172.16.254.0
Network Layer 4-16
Solution
 Answer: C
Explanation:
This question is much easier then it appears when
you convert it to binary and do the Boolean operation
as shown below:
IP address 172.16.210.0 =
10101100.00010000.11010010.00000000
/22 mask = 11111111.11111111.11111100.00000000
AND result = 11111111.11111111.11010000.00000000
AND in decimal= 172 . 16 . 208 . 0
Network Layer 4-17
Exercises
 How many subnetworks and hosts are
available per subnet if you apply /28 mask
to the 210.10.2.0 class C network
30 networks and 6 hosts
 6 networks and 30 hosts
 8 networks and 32 hosts
 14 networks and 6 hosts
 None of the above

Network Layer 4-18
Solution
 Answer: E
Explanation:
A 28 bit subnet mask
(11111111.11111111.11111111.11110000) applied
to a class C network uses a 4 bits for
networks, and leaves 4 bits for hosts. Using
the 2n-2 formula, we have 24-2 (or 2x2x2x2
2) which gives us 14 for both the number of
networks, and the number of hosts.
Network Layer 4-19
Exercises
 Given that you have a class B IP address
network range, which of the subnet masks
below will allow for 100 subnets with 500
usable host addresses per subnet?





255.255.0.0
255.255.224.0
255.255.254.0
255.255.255.0
255.255.255.224
Network Layer 4-20
Solution
 Answer: C
Explanation:
Using the 2n-2 formula for host addresses, 29-2 =
510 host address, so a 9-bit subnet mask will provide
the required number of host addresses. If these 9
bits are used for the hosts in a class B network, then
the remaining 7 bits are used for the number of
networks. Again using the 2n-2 formula, we have 2n2 = 126 networks that are available.
Network Layer 4-21
Exercises
 Given the following IP address and subnet
mask: 172.16.211.12/20, find the
broadcast address associated with the
subnet that this IP address resides upon.





172.16.255.255
172.16.224.224
172.16.224.255
172.16.223.255
None of the above
Network Layer 4-22
Exercises
 Your network uses the 172.12.0.0 IP
address. You need to support 459 hosts
per network, while accommodating the
maximum number of subnets. Which mask
would you use?





255.255.0.0
255.255.128.0
255.255.254.0
255.255.255.254
255.255.255.128
Network Layer 4-23
Solution
 Answer:C
Explanation:
To obtain 459 hosts the number of host bits
will be 9. This can support a maximum of 510
hosts. To keep 9 bits for hosts means the
last bit in the 3rd octet will be 0. This gives
255.255.254.0 as the subnet mask.
Network Layer 4-24
Exercises

The LAU network was assigned the class C
network address 189.66.1.0 from the ISP. If the
administrator at LAU were to subnet this class C
network using the 255.255.255.224 subnet mask,
how many hosts will they be able to support on
each subnet?





14
16
32
30
None of the above
Network Layer 4-25
Solution
 Answer: D
Explanation:
The subnet mask 255.255.255.224 is a 27 bit mask
(11111111.11111111.11111111.11100000). It uses 3 bits
from the host Id for the network ID, leaving 5 bits
for host addresses. We can calculate the number of
hosts supported by this subnet by using the 2n-2
formula where n represents the number of host bits.
In this case it will be 5. 25-2 gives us 30.
Network Layer 4-26
Exercises
 You have been the CIDR block of
115.64.4.0/22 from your ISP. Which of the
IP addresses below can you use for a host?
115.64.8.32
 115.64.7.64
 115.64.6.255
 115.64.3.255
 115.64.5.128
 115.64.12.128

Network Layer 4-27
Solution
 Answer: B, C, E
Explanation:
115.64.4.0 =01110011.01000000.00000100.00000000
Subnet mask =
11111111.11111111.11111100.00000000= 255.255.252.0
Subnet number =
01110011.01000000.00000100.00000000=
115.64.4.0
Broadcast = 01110011.01000000.00000111.11111111=
115.64.7.255
Valid address range = 115.64.4.1 - 115.64.7.254
Network Layer 4-28