networking - How do I determine that the subnet was going to be .11.254 and .13.254? -


i figured out subnet mask both subnets 1 , 2. problem can't grasp how subnet turns 172.20.11.254 , 172.20.13.254 respectively? assume vslm, not certain. i'm learning this. got 172.20.8.0 , 172.20.6.0 subnet , know wrong now. can provide.

to determine subnet mask work 172.20.0.0 network, first @ number of hosts required each subnet:

subnet1 (connected fastethernet0/0) has 672 hosts. support 672 hosts, subnet mask of /22 required (10 host bits in 2n-2 formula afford 1022 host addresses in subnet). subnet2 (connected fastethernet0/1) has 258 hosts. support 258 hosts, subnet mask of /23 required (9 host bits in 2n-2 formula afford 510 host addresses in subnet). network address of 172.20.0.0 , masks needed fit requirements, need configure following ip address , subnet masks:

for fastethernet0/0 connection:

172.20.8.0/22 third possible subnet. (172.20.0.0/22 first possible subnet , 172.20.4.0/22 second possible subnet.) 172.20.11.254 last possible ip address in subnet. 255.255.252.0 decimal version of 22-bit mask. fastethernet0/1 connection: 172.20.12.0/23 next available subnet not overlap. 172.20.13.254 last possible ip address in subnet. 255.255.254.0 decimal version of 23-bit mask. use following commands configure sfo interfaces:  sfo>enable sfo#configure terminal sfo(config)#interface fastethernet0/0 sfo(config-if)#ip address 172.20.11.254 255.255.252.0 sfo(config-if)#no shutdown sfo(config-if)#interface fastethernet0/1 sfo(config-if)#ip address 172.20.13.254 255.255.254.0 sfo(config-if)#no shutdown sfo(config-if)#exit sfo(config)#exit sfo#copy run start

i detect desperation, let's see if can convey , understandable explanation. :-)

172.20.0.0 seems address space destined use in exercise. class b network (255.255.0.0, or /16 netmask), since we're going subnet variably, can safely forget that. example, subnet of it in small, class c subnets (all mask of 255.255.255.0, or /24), , if did use 172.20.0.0/24 1 network, 172.20.1.0/24 another, 172.20.2.0/24 another, , on. if did that, each subnet able hold no more 254 hosts (that because leave last octet - 8 bits - host portion, , have reserve 2 - first , last - subnet address , broadcast address: 2^8-2=254).

but 254 hosts not enough needs, since have requirements 672 , 258.

if use smaller sized mask (meaning larger sized network -> more hosts) /23 (255.255.254.0) have 9 bits host portion, therefore can acommodate 2^9-2=510 hosts, big enough 258, not 672. latter need /22 network (255.255.252.0), leave 10 bits host portion allowing 2^10-2=1022.

with each bit reduce in netmask, double network size. if /24 goes 172.20.0.0 172.20.0.255 (the single '0' class c network), /23 goes 172.20.0.0 172.20.1.255 (two class c networks, '0' , '1'). , /22 goes 172.20.0.0 172.20.3.255 (four class c networks). in each case first address considered network address , not assigned device, , last 1 broadcast address, , not assigned either.

so, example, choose assign 3rd /22 network (1st being 172.20.0.0 172.20.3.255, 2nd being 172.20.4.0 172.20.7.255, , 3rd being 172.20.8.0 172.20.11.255) particular subnet. 172.20.8.0/22 is. , choose assign 7th /23 subnet possible (1st '0' , '1' class c's, 2nd '2' , '3' class c's, , on) other subnet. 172.20.12.0/23 it. remember cannot overlap!

now, why chose .254 addresses router interfaces, convention. router interfaces configured use either first usable (.1) ip address or last usable (.254) ip address in subnets, @ least on lan side. note subnets' broadcast addresses 172.20.11.255 /22 , 172.20.13.255 /23. in both cases picked router interfaces address 1 below them, i.e. last usable address. have been 1 in corresponding range.

did help?


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