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SUBNETTING IN PACKET TRACER

OLYMPIACOS
Level 1
Level 1

Hello guys, I am Tasos from Greece. I have a very difficult project for my university, to define subnets in the program Packet Tracer. The project is the following:

We are given the network 192.168.224.0/22 and we have to define 8 subnets in descending order, according to the number of hosts. We are also given the following tables:

LAN

Host R1, Interface Fa0/0, 62 hosts

Host R2, Interface Fa0/0, 5 hosts

Host R2, Interface Fa0/1, 16 hosts

Host R3, Interface Fa0/0, 10 hosts

Host R4, Interface Fa0/0, 100 hosts

 

WAN

Hostname                   Address to be Assigned                          Number of Hosts

 R1-R2                        R1-First host address                                             2
R1-R3                         R1-First host address                                             2
R2-R3                         R2-First host address                                             2
R3-R4                         R3-First host address                                             2

 

We are also given the following rules:

The computers will receive the first useful IP address of the subnet and the server
penultimate beneficial respectively.
• All FastEthernet ports on a router receive the latest useful IP of the corresponding
subnet.
• R1-R2 will use the first WAN subnet, R1-R3 will use
the second WAN subnet and the R2-R3 connection will use the third WAN subnet.
• The DCE (serial) interfaces of R1 and R2 have a clock rate of 56000.
 

 

 

Any help is appreciated.

1 Reply 1

Martin L
VIP
VIP

 

this is basic of CCNA level VLSM and subnetting problem that should be in your book. (CCNA  level books).  VLSM is based on 'blocks" of subnets, aka block of 128 for /25, 64 is /26, etc till 4 is /30.  One block of /25 can give you 2x 64, so the 62 hosts per network will fit nicely into one of  /25 blocks. 

To solve this start with the largest requirement for host number, (100 hosts); then moving downwards to smallest one (2 per link). So, what is mask for getting 100 hosts?  /25 gives u 126 hosts, which is enough (you seldom will have exact number of hosts for mask)

Since you have been given 192.168.224.0/22, you must start with that prefix length, aka mask, of /22. Range of this 192.168.224.0/22 is from 192.168.224.0/22  to 192.168.227.255/22.

with this, you should be able to start building your network

We could solve this for you but then you will not learn anything; and what's purpose in that.

 

Regards, ML
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