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Give IP –addresses to the networks below with
three point-to-point networks and three local area
networks. The available address space is
10.38.161.0 – 10.38.161.67 and the interface of
R3 router towards LAN-3 is 10.38.161.1.
3 hosts
7 hosts
R2
R1
LAN-1
R3
10.38.161.1
17 hosts
LAN-3
LAN-2
SOLUTION:
1. Find the point-to-points networks and the local networks.
2. Define the address space needed in each network (how many
interfaces?)
3. What is the size of the block (power of 2)?
4. Start allocating the IP –addresses from R3 (10.38.100.1).
5. What is the mask?
6. Remember the rule:
network address/number of addresses in the network = Interger
Remarks:
Address 10.38.100.12/28 (or 255.255.255.240) cannot be a
network address. Why not?
How to calculate the network address and broadcast address?
SOLUTION
1. Find the point-to-points networks and the local networks.
R1-R2
3 hosts
R2
R1
LAN-1
7 hosts
LAN-2
R1-R3
R2-R3
R3
17 hosts
LAN-3
SOLUTION
2. Define the address space needed in each network (how many
interfaces?)
3. What is the size of the block (power of 2)?
LAN-3: 17 hosts+NWA+BCA+R3-interface=20  32 addr (5 bits)
LAN-2: 7 hosts+1+1+1=10  16 addr (4 bits)
LAN-1: 3 hosts+1+1+1=6  8 addr (3 bits)
R1-R2: 2 hosts+1+1=4  4 addr (2 bits)
R1-R3: 2+1+1=4  4 addr (2 bits)
R2-R3: 2+1+1=4  4 addr (2 bits)
0
31|32
32 addresses
47|48
16 addr
8 addr
55|56 59|60 63|64 67|
4 a.
4 a.
4 a.
SOLUTION:
4. Start allocating the IP –addresses from R3 (10.38.100.1).
5. What is the mask?
LAN-3: 10.38.100.0  10.38.100.31 (mask 27 or 255.255.255.224) 32 addr
LAN-2: 10.38.100.32  10.38.100.47 (mask 28 or 255.255.255.240) 16 addr
LAN-1: 10.38.100.48  10.38.100.55 (mask 29 or 255.255.255.248) 8 addr
RI-R2: 10.38.100.56  10.38.100.59 (mask 30 or 255.255.255.252) 4 addr
RI-R3: 10.38.100.60  10.38.100.63 (mask 30 or 255.255.255.252) 4 addr
R2-R3: 10.38.100.64  10.38.100.67 (mask 30 or 255.255.255.252) 4 addr
totally 68 addr
6. Remember the rule:
network address/number of addresses in the network = Interger
LAN-3: 0/32=0Integer
R1-R2: 56/4=14Integer
LAN-2: 32/16=2Integer
R1-R3: 60/4=15Integer
LAN-1: 48/8=6Integer
R2-R3: 64:4=16Integer
Address 10.38.100.12/28 (or 255.255.255.240) cannot be a
network address. Why not?
Mask=28  28 bits for network 32-28=4 bits for hosts16 addr
NWA/addresses  12/16 NOT Integer
How to calculate the network address and broadcast address?
10.38.100.12/28 = 10.38.100.12/255.255.255.240
AND
OR
7
128
2^7
6
64
2^6
5
32
2^5
4
16
2^4
3
8
2^3
2
4
2^2
1
2
2^1
0
1
2^0
0
0
0
0
1
1
0
0
= 12
1
1
1
1
0
0
0
0
= 240 mask
0
0
0
0
0
0
0
0
= 0 NWA
0
0
0
0
1
1
1
1
= INV mask
0
0
0
0
1
1
1
1
= 15 BCA
Suppose we think that 12 is NWA and mask is 28.
There are 4 bits available for hosts =16 addr (015).
7
128
2^7
6
64
2^6
5
32
2^5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
16
2^4
0
0
0
0
1
3
8
2^3
2
4
2^2
1
1
1
1
1
1
1
1
0
0
etc. etc.
1
2
2^1
0
0
1
1
0
0
1
2^0
0
1
0
1
0
= 12
= 13
= 14
= 15
= 16  needs 5 bits