politecnico di milanohome.deib.polimi.it/redondi/fcn/e5-ip addressing.pdf · to serve an old lan...
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Politecnico di Milano Scuola di Ingegneria Industriale e dell’Informazione
E5 IP Addressing
Exercise 1
o To an IP network is assigned the following set of IP addresses: n address: 208.57.0.0 n netmask: 255.255.0.0
o It is necessary to partition the network so as to serve an old LAN with 4000 hosts
o a) What netmask is necessary to define the subnet for the 4000 hosts?
o b) What network address can be used? o c) How many subnets of the same size can be
accommodated? o d) How many subnets with 60 hosts can be
accommodated and with which netmak?
11010000.00111001.00000000.00000000 1111111.11111111.00000000.00000000
Solution 1
o For the host-ID with 4000 hosts we need 12 bits (212=4096) and then:
o A netmask with 20 consecutive 1s: 255.255.240.0 o Possible subnet addresses are those obtained with any
value of the last four bits:
o For instance: o Corresponding to: 208.57.0.0/20
o 4 bits can take any value so 16 subnets in total, 15 subnets in addition to the one with 4000 hosts just assigned
11010000.00111001.xxxx0000.00000000 11010000.00111001.00000000.00000000
Solution 1
o For an host-ID with at least 60 addresses we need 6 bits (26=64). Each of the 15 subnets of previous point c) have 12 bits of the host-ID and therefore can be subdivided using 6 bits (12-6=6) into 64 smaller subnets. In total: 64x15=960.
1101000.00111001.XXXXxxxx.xx000000
16-1=15 26=64
/16 /20
. . .
. . . /26
. . .
Exercise 2
o For a network we have a class B address: 129.174.0.0. In the network we need to create at least 15 subnets connected through a router a) Subdivide the addressing space and indicate the
network address of each subnet b) How many hosts can be accommodated in each
subnet? c) To which subnet the following addresses belong:
o 129.174.28.66 o 129.174.99.122 o 129.174.130.255 o 129.174.191.255 Are they host addresses or special addresses?
Solution 2 o The network 129.174.0.0 has a host-ID of
16 bits and a net-ID of 16 bits o Thorough the netmask we can divide the
host-ID so as to obtained a subnets o With a 4 bits of the subnet ID we can
obtained 16 subnets o Therefore, the netmask will have
16+4=20 bits set to 1 and 12 to 0. n 255.255.240.0
1 0 0 0 0 0 0 0 1281 1 0 0 0 0 0 0 1921 1 1 0 0 0 0 0 2241 1 1 1 0 0 0 0 2401 1 1 1 1 0 0 0 2481 1 1 1 1 1 0 0 2521 1 1 1 1 1 1 0 2541 1 1 1 1 1 1 1 255
Solution 2
n The remaining 12 bits for the host-ID can accommodate up to 212-2=4094 hosts
n Subnet addresses are:
0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 160 0 1 0 0 0 0 0 320 0 1 1 0 0 0 0 480 1 0 0 0 0 0 0 640 1 0 1 0 0 0 0 800 1 1 0 0 0 0 0 960 1 1 1 0 0 0 0 1121 0 0 0 0 0 0 0 1281 0 0 1 0 0 0 0 1441 0 1 0 0 0 0 0 1601 0 1 1 0 0 0 0 1761 1 0 0 0 0 0 0 1921 1 0 1 0 0 0 0 2081 1 1 0 0 0 0 0 2241 1 1 1 0 0 0 0 240
129.174. .0/20
Solution 2
Ø 129.174.28.66 129.174.16.0/20 (host) Ø 129.174.99.122 129.174.96.0/20 (host) Ø 129.174.130.255 129.174.128.0/20 (host) Ø 129.174.191.255 129.174.176.0/20 (broadcast)
0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 160 0 1 0 0 0 0 0 320 0 1 1 0 0 0 0 480 1 0 0 0 0 0 0 640 1 0 1 0 0 0 0 800 1 1 0 0 0 0 0 960 1 1 1 0 0 0 0 1121 0 0 0 0 0 0 0 1281 0 0 1 0 0 0 0 1441 0 1 0 0 0 0 0 1601 0 1 1 0 0 0 0 1761 1 0 0 0 0 0 0 1921 1 0 1 0 0 0 0 2081 1 1 0 0 0 0 0 2241 1 1 1 0 0 0 0 240
129.174. .0/20
Exercise 3
o To the network in the figure is assigned the network address:
195.56.78.0/23 1100001.00111000.01001110.00000000
Internet R2
Bridge
eth0
eth1
eth2 R1 R3
eth3 R4
pp1
pp2
pp3
Rx
190.131.99.1
eth4 R5
pp4
190.131.99.2
Exercise 3 o LANs must be dimensioned to accommodate at
least a number of hosts equal to: n eth0: 150 n eth1: 60 n eth2: 55 n eth3:57 n eth4: 61
o Point-to-point links are indicated as “pp” and are assumed to be full-duplex
a) Divide the network in subnets indicating for each of them address and netmask (both LANs and P2P links)
b) Assign to router interfaces addresses compatible with the subnets they are connected to
c) Fill in routing tables of all routers
E, F, G e H 2 hosts
A 210 hosts
B 55 hosts
C 57 hosts
D, 61 hosts
Solution 3
o Subnets:
Internet R2
Bridge
eth0
eth1
eth2 R1 R3
eth3 R4
pp1
pp2
pp3
Rx 190.131.99.1
eth4 R5
pp4
Solution 3
o For Subnet A we need a hostID with 8 bits o For Subnets B, C and D we need a hostID with 6 bits o For Subnets E, F, G and H we need a hostID with 2
bits
195.56.78.0/23
/24
/26
. . . /30
Subnet A
Subnet B
Subnet C
Subnet D Subnet E
Subnet F
Subnet G
Subnet H
Solution 3
195.56.78.0/23
/24 195.56.78.0/24
195.56.79.0/24
1100001.00111000.01001110.00000000
1100001.00111000.01001111.00000000
Subnet A
Solution 3
/26
195.56.79.0/24
1100001.00111000.01001111.00000000 195.56.79.0/26
195.56.79.64/26
195.56.79.128/26
195.56.79.192/26
1100001.00111000.01001111.01000000
1100001.00111000.01001111.10000000
1100001.00111000.01001111.11000000
Subnet B
Subnet C
Subnet D
Solution 3
195.56.79.192/26
1100001.00111000.01001111.11000000
. . .
/30 195.56.79.192/30
1100001.00111000.01001111.11000100 195.56.79.196/30
1100001.00111000.01001111.11001000 195.56.79.200/30
1100001.00111000.01001111.11001100 195.56.79.204/30
. . .
Subnet E
Subnet F
Subnet G
Subnet H
Solution 3
o Interfaces:
Internet
R2
R1 R3
R4
Rx
190.131.99.1
R5
190.131.99.2
195.56.78.0/24
195.56.79.0/26
195.56.79.64/26
195.56.79.128/26
195.56.78.254/24
195.56.79.62/26
195.56.79.126/26
195.56.79.190/26
195.56.79.196/30
195.56.79.193
195.56.79.194 195.56.79.198
195.56.79.197
195.56.79.201
195.56.79.202
195.56.79.205
195.56.79.206
Solution 3
o Routing tables:
Internet
R2
R1 R3
R4
Rx
190.131.99.1
R5
190.131.99.2
195.56.79.193
195.56.79.194 195.56.79.198
195.56.79.197
195.56.79.201
195.56.79.202
195.56.79.205
195.56.79.206
network netmask first hop0.0.0.0 0.0.0.0 195.56.79.194
network netmask first hop0.0.0.0 0.0.0.0 195.56.79.198
network netmask first hop0.0.0.0 0.0.0.0 195.56.79.201
network netmask first hop0.0.0.0 0.0.0.0 195.56.79.205
network netmask first hop195.56.78.0 255.255.254.0 190.131.99.2
network netmask first hop195.56.78.0 255.255.255.0 195.56.79.193195.56.79.0 255.255.255.192 195.56.79.197195.56.79.64 255.255.255.192 195.56.79.202195.56.79.128 255.255.255.192 195.56.79.2060.0.0.0 0.0.0.0 190.131.99.1
Exercise 4 o Consider the network in the figure o Configure interfaces and provide a
routing table for routers A and B
A
Subnet 131.175.21.0/24
Subnet 131.175.16.0/24
INTERNET
Subnet 131.175.15.0/24
Subnet 131.175.70.0/24
B
Subnet 131.175.75.0/24
Solution 4
A
Subnet 131.175.21.0/24
Subnet 131.175.16.0/24
INTERNET
Subnet 131.175.15.0/24
Subnet 131.175.70.0/24
x.x.x.254
x.x.x.52 x.x.x.254 x.x.x.33
B
x.x.x.254
Subnet 131.175.75.0/24
x.x.x.254
x.x.x.254
Solution 4
network netmask first hop131.175.15.0 255.255.255.0 131.175.16.33131.175.70.0 255.255.255.0 131.175.16.330.0.0.0 0.0.0.0 131.175.21.254
Routing table of A Interfaces of A interface address netmaskA 131.175.21.52 255.255.255.0B 131.175.16.254 255.255.255.0C 131.175.75.254 255.255.255.0
network netmask first hop131.175.21.0 255.255.255.0 131.175.16.254131.175.75.0 255.255.255.0 131.175.16.2540.0.0.0 0.0.0.0 131.175.16.254
Routing table of B Interfaces of B interface address netmaskA 131.175.16.33 255.255.255.0B 131.175.70.254 255.255.255.0C 131.175.15.254 255.255.255.0
network netmask first hop0.0.0.0 0.0.0.0 131.175.16.254
Exercise 5
o The topology of the network of the CS dept of university of california is indicated in the figure. To network the following addressing space is assigned: 131.175.96.0/22.
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
Exercise 5
o Design an addressing plan with the following requirements:
o tr0: IP subnet with at least 120 host o eth0: IP subnet with at least 500 host o eth1: IP subnet with at least 120 host o eth2: IP subnet with at least 120 host o eth3: IP subnet with at least 50 host o pp0: point to point link o pp1: point to point link
o For each subnet indicate network address, netmask, and direct broadcast address.
o Assign an address to router interfaces and provide a valid routing table for R2.
Solution 5
131.175.96.0/22
131.175.98.0/23 Da suddividere
131.175.96.0/23 eth0
131.175.98.0/23 131.175.98.128/25 eth1
131.175.98.0/25 tr0
131.175.99.0/25 eth2 131.175.99.128/25 Da suddvidere
Solution 5
131.175.99.128/25
131.175.99.192/26 Da suddividere
131.175.99.128/26 eth3
131.175.99.192/26 131.175.99.196/30 pp1
131.175.99.192/30 pp0
…
… …
Solution 5 o Eth0
o Address: 131.175.96.0/23 o Broadcast: 131.175.97.255
o Tr0 o Address: 131.175.98.0/25 o Broadcast: 131.175.98.127
o Eth1 o Address: 131.175.98.128/25 o Broadcast: 131.175.98.255
o Eth2 o Address: 131.175.99.0/25 o Broadcast: 131.175.99.127
o Eth3 o Address: 131.175.99.128/26 o Broadcast: 131.175.99.191
o pp0 o Address: 131.175.99.192/30 o Broadcast: 131.175.99.195
o pp1 o Address: 131.175.99.196/30 o Broadcast: 131.175.99.199
Solution 5
Interfaces: o R2
n If1: 131.175.98.129 n If2: 131.175.99.197 n If3: 131.175.99.193 n If4: 131.175.98.1
o R1 on tr0: n 131.175.98.2
o R4 on pp1: n 131.175.99.198
o R3 on pp0: n 131.175.99.194
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
Solution 5
o Routing table for R2:
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
R1
INTERNETINTERNET
tr0
R2 R3
R4
eth0
eth1
eth2
eth3
pp0
pp1
1 2
34
network netmask next hop 131.175.99.0 255.255.255.128 131.175.99.198
131.175.99.128 255.255.255.192 131.175.99.194
0.0.0.0 0.0.0.0 131.175.98.2
Other exercises
Exercise 6
Internet
R2
R1
LAN da 10 host
R6 R5
pp1
Switch
Switch
Switch
… …
R3
Switch
Switch
Switch
… … Switch
…
R4
Switch
…
Switch
…
Switch
…
pp2
pp3
1 60 1 60
1 25
1 60 1 60
1 50 1 50
1 50
o To the network in the figure the following addressing space is assigned: 131.175.200.0/23.
Exercise 6
o Design an addressing plan compatible with the number of hosts indicated in the figure
o Identify each subnet graphically on the figure o For each subnet indicate network address, netmask,
and direct broadcast address. o Assign an address to router interfaces and provide a
valid routing table for R1.