R3 R1 R2 7 hosts LAN-2 17 hosts LAN-3 R3 R1 R2 3 hosts LAN-1 Give IP –addresses to the networks below with three point-to-point networks and three local.

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Presentation transcript:

R3 R1 R2 7 hosts LAN-2 17 hosts LAN-3 R3 R1 R2 3 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 – and the interface of R3 router towards LAN-3 is

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 ( ). 5. What is the mask? 6. Remember the rule: network address/number of addresses in the network = Interger Remarks: Address /28 (or ) cannot be a network address. Why not? How to calculate the network address and broadcast address?

R3 R1 R2 3 hosts7 hosts 17 hosts SOLUTION 1. Find the point-to-points networks and the local networks. LAN-1 LAN-3 LAN-2 R1-R2 R1-R3 R2-R3

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 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 ( ). 5. What is the mask? LAN-3:  (mask 27 or ) 32 addr LAN-2:  (mask 28 or ) 16 addr LAN-1:  (mask 29 or ) 8 addr RI-R2:  (mask 30 or ) 4 addr RI-R3:  (mask 30 or ) 4 addr R2-R3:  (mask 30 or ) 4 addr totally 68 addr 6. Remember the rule: network address/number of addresses in the network = Interger LAN-3: 0/32=0  IntegerR1-R2: 56/4=14  Integer LAN-2: 32/16=2  IntegerR1-R3: 60/4=15  Integer LAN-1: 48/8=6  IntegerR2-R3: 64:4=16  Integer

Address /28 (or ) 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

/28 = / = = 240 mask = 0 NWA = INV mask How to calculate the network address and broadcast address? AND = 15 BCA OR

= = = = = 16  needs 5 bits etc. Suppose we think that 12 is NWA and mask is 28. There are 4 bits available for hosts =16 addr (0  15).