7 hosts lan-2
DESCRIPTION
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. 7 hosts LAN-2. 3 hosts LAN-1. R2. R2. R1. R1. R3. R3. - PowerPoint PPT PresentationTRANSCRIPT
R3
R1 R27 hosts
LAN-2
17 hosts
LAN-3
R3
R1 R23 hosts
LAN-1
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.
10.38.161.1
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?
R3
R1 R23 hosts 7 hosts
17 hosts
SOLUTION1. Find the point-to-points networks and the local networks.
LAN-1
LAN-3
LAN-2
R1-R2
R1-R3R2-R3
SOLUTION2. 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 47|48 55|56 59|60 63|64 67|
32 addresses 16 addr 8 addr 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=0Integer R1-R2: 56/4=14Integer
LAN-2: 32/16=2Integer R1-R3: 60/4=15Integer
LAN-1: 48/8=6Integer R2-R3: 64:4=16Integer
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 hosts16 addr
NWA/addresses 12/16 NOT Integer
10.38.100.12/28 = 10.38.100.12/255.255.255.240
7 6 5 4 3 2 1 0 128 64 32 16 8 4 2 1
2^7 2^6 2^5 2^4 2^3 2^2 2^1 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
How to calculate the network address and broadcast address?
AND
0 0 0 0 1 1 1 1 = 15 BCA
OR
0 0 0 0 1 1 0 0 = 12 0 0 0 0 1 1 0 1 = 13 0 0 0 0 1 1 1 0 = 14 0 0 0 0 1 1 1 1 = 15 0 0 0 1 0 0 0 0 = 16 needs 5 bits
etc. etc.
Suppose we think that 12 is NWA and mask is 28.
There are 4 bits available for hosts =16 addr (015).
7 6 5 4 3 2 1 0 128 64 32 16 8 4 2 1
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0