Subnetting a 3 router networking scheme Please help

novellof

Hey all,

The question is: Use the 192.168.156.0/22 address to create an addressing scheme to accommodate all hosts on the network. Begin the address assignments with 192.168.157.0 address.

*** Why does it say begin with 192.168.157 address???

here's the topology:

I have 3 routers in the network called HQ, Remote1 , and Remote 2.

HQ has 90 hosts on fa0/0, 60 on fa0/1, 2 on S0/0/0 and s0/0/1.
90+60+4= 154 hosts for HQ router

Remote1 has 30 hosts on fa0/0, 60 on fa0/1.

30+60 = 90 hosts for Remote 1 router

Remote 2 has 128 hosts on fa0/0, 60 on fa0/1

128+60 = 188 hosts on Remote 2 router.

so I will start with Remote 2 because it has the most hosts right? 2^n -2 = 188?

2^8-2 = 254

128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1

im stuck...

i dont know what i am doing...please help lol

fadhil

novellof wrote: »

Hey all,

The question is: Use the 192.168.156.0/22 address to create an addressing scheme to accommodate all hosts on the network. Begin the address assignments with 192.168.157.0 address.

*** Why does it say begin with 192.168.157 address???

here's the topology:

I have 3 routers in the network called HQ, Remote1 , and Remote 2.

HQ has 90 hosts on fa0/0, 60 on fa0/1, 2 on S0/0/0 and s0/0/1.
90+60+4= 154 hosts for HQ router

Remote1 has 30 hosts on fa0/0, 60 on fa0/1.

30+60 = 90 hosts for Remote 1 router

Remote 2 has 128 hosts on fa0/0, 60 on fa0/1

128+60 = 188 hosts on Remote 2 router.

so I will start with Remote 2 because it has the most hosts right? 2^n -2 = 188?

2^8-2 = 254

128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 1

im stuck...

i dont know what i am doing...please help lol

first remember that each port in a router is separate network.
hence f0/0 and f0/1 are two different networks.
also remember that S0/0/0 and s0/0/1 need only 2 host VLSM must be used so as to save bunch of ip addresses.
to subnet first find the largest number of hosts for each network(HQ has 90 hosts on fa0/0, 60 on fa0/1,Remote1 has 30 hosts on fa0/0, 60 on fa0/1,Remote 2 has 128 hosts on fa0/0, 60 on fa0/1,).
for this case Remote 2 has high number of hosts than the other interface.
we should find subnet mask through this interface. since it help to meet the requirement for the other interfaces because they low number of hosts.
2^n -2>= 128 where n must be 8 since 7 will not meet the requirement.
hence only last 8 zeros will present number host from a subnet mask
to get block size (256 -255=1).the block size is 1.
networks will be

192.168.156.0/24
192.168.157.0/24
192.168.158.0/24
192.168.159.0/24 and so on
you supposed to begin with 192.168.157.0 because there is range of networks.
hence for Remote 2 with fa0/0 and hosts 128 will assign this ip address 192.168.157.0.
the remaining ip address will be assigning to other networks(interface)
note:
variable length subnet mask must:
be used to assign for serial interface and
also for other networks with low number of hosts(optional).

novellof

fadhil wrote: »

first remember that each port in a router is separate network.
hence f0/0 and f0/1 are two different networks.
also remember that S0/0/0 and s0/0/1 need only 2 host VLSM must be used so as to save bunch of ip addresses.
to subnet first find the largest number of hosts for each network(HQ has 90 hosts on fa0/0, 60 on fa0/1,Remote1 has 30 hosts on fa0/0, 60 on fa0/1,Remote 2 has 128 hosts on fa0/0, 60 on fa0/1,).
for this case Remote 2 has high number of hosts than the other interface.
we should find subnet mask through this interface. since it help to meet the requirement for the other interfaces because they low number of hosts.
2^n -2>= 128 where n must be 8 since 7 will not meet the requirement.
hence only last 8 zeros will present number host from a subnet mask
to get block size (256 -255=1).the block size is 1.
networks will be

192.168.156.0/24
192.168.157.0/24
192.168.158.0/24
192.168.159.0/24 and so on
you supposed to begin with 192.168.157.0 because there is range of networks.
hence for Remote 2 with fa0/0 and hosts 128 will assign this ip address 192.168.157.0.
the remaining ip address will be assigning to other networks(interface)
note:
variable length subnet mask must:
be used to assign for serial interface and
also for other networks with low number of hosts(optional).

hey thanks for your response...it was really helpful but i have another question regarding remote2 network

Remote2
Fa0/0 128 /24 192.168.157.0 255.255.255.0
Fa0/1 60 /26 192.168.159.0 255.255.255.192
S0/0/0 2 /30 192.168.159.100 255.255.255.252

why do they jump from 157 to 159? whey dont they use a .158....

im looking at the answer sheet i found online...
http://haxia.org/itk/2/cisco/CCNA-2-EIGRP-Skills-Based-Assessment-Exam-Answered.pdf

novellof

fadhil wrote: »

first remember that each port in a router is separate network.
hence f0/0 and f0/1 are two different networks.
also remember that S0/0/0 and s0/0/1 need only 2 host VLSM must be used so as to save bunch of ip addresses.
to subnet first find the largest number of hosts for each network(HQ has 90 hosts on fa0/0, 60 on fa0/1,Remote1 has 30 hosts on fa0/0, 60 on fa0/1,Remote 2 has 128 hosts on fa0/0, 60 on fa0/1,).
for this case Remote 2 has high number of hosts than the other interface.
we should find subnet mask through this interface. since it help to meet the requirement for the other interfaces because they low number of hosts.
2^n -2>= 128 where n must be 8 since 7 will not meet the requirement.
hence only last 8 zeros will present number host from a subnet mask
to get block size (256 -255=1).the block size is 1.
networks will be

192.168.156.0/24
192.168.157.0/24
192.168.158.0/24
192.168.159.0/24 and so on
you supposed to begin with 192.168.157.0 because there is range of networks.
hence for Remote 2 with fa0/0 and hosts 128 will assign this ip address 192.168.157.0.
the remaining ip address will be assigning to other networks(interface)
note:
variable length subnet mask must:
be used to assign for serial interface and
also for other networks with low number of hosts(optional).

hey thanks for your reply it was really helpful . but i got another question
Remote2
Fa0/0 192.168.157.1 255.255.255.0 0.0.0.255
Fa0/1 192.168.159.1 255.255.255.192 0.0.0.63
S0/0/0 192.168.159.102 255.255.255.252 0.0.0.3
how come they have fa0/0 .157 network
then they have fa0/1 .159 network?????

so confused!!! how come it cant be a .158 network?

fadhil

novellof wrote: »

hey thanks for your reply it was really helpful . but i got another question
Remote2
Fa0/0 192.168.157.1 255.255.255.0 0.0.0.255
Fa0/1 192.168.159.1 255.255.255.192 0.0.0.63
S0/0/0 192.168.159.102 255.255.255.252 0.0.0.3
how come they have fa0/0 .157 network
then they have fa0/1 .159 network?????

so confused!!! how come it cant be a .158 network?

as I shown you earlier that there are so many networks,just an administrator choose one and assign to a certain interface. so no problem to assign .158 and .159.
for S0/0/0 the network 1192.168.159.0 is subneted more than once until reach to /30. so as to save bunch of ip addresses from being lost.
note: variable length subnet mask is used

remember that the ip address of Fa0/1 is subnetted to block size of 64 than assigned to since it satisfy the requirement.
variable length subnet mask is used

novellof

fadhil wrote: »

as I shown you earlier that there are so many networks,just an administrator choose one and assign to a certain interface. so no problem to assign .158 and .159.
for S0/0/0 the network 1192.168.159.0 is subneted more than once until reach to /30. so as to save bunch of ip addresses from being lost.
note: variable length subnet mask is used

remember that the ip address of Fa0/1 is subnetted to block size of 64 than assigned to since it satisfy the requirement.
variable length subnet mask is used

thank you