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supernetting , (someone's gotta know this..pls help..)
17 years 11 months ago #19487
by skylimit
"...you are never too old to learn" anon
supernetting , (someone's gotta know this..pls help..) was created by skylimit
Hi all, anyone able to clarify this for me pls?
Im having problems supernetting this CIDR block given: 235.85.105.64/27 - 235.85.105.128/27
I have to assign 1500+ IP addresses to 1500+ hosts (desktops) in 15 locations.
From what i learnt in the supernetting section of the site. Supernetting means borrowing bits from the network ID, therefore to supernet 235.85.105.64/27 borrowing 3 bits I'll have:
IP: 11101011. 01010101.01101001.01000000
Netmask: 11111111. 11111111. 11111000.00000000
network ID:11101011. 01010101.01101000.00000000 (ANDing)
ie. 235.85.104.0
broadcast: 00000000.00000000.00000111.11111111
ie. 235.85.111.255
i.e: supernet range = 235.85.104.0 - 235.85.111.255 (first supernet)
no of supernets = 2^3 - 2= 6 supernets
hosts on each supernet = 254 (2046 hosts in total)
Am i correct to this point?
Question:
How would i go about calculating the remaining 5 supernets (that is the host range, supernet iD and broadcast?I will be extremely glad to get quick response to this please
Secondly, how can i summarise the supernets (i.e. route summarization) to minimise the size of routing tables in the core.
Many thanks in advance. please help me out im stuck[/b]
Im having problems supernetting this CIDR block given: 235.85.105.64/27 - 235.85.105.128/27
I have to assign 1500+ IP addresses to 1500+ hosts (desktops) in 15 locations.
From what i learnt in the supernetting section of the site. Supernetting means borrowing bits from the network ID, therefore to supernet 235.85.105.64/27 borrowing 3 bits I'll have:
IP: 11101011. 01010101.01101001.01000000
Netmask: 11111111. 11111111. 11111000.00000000
network ID:11101011. 01010101.01101000.00000000 (ANDing)
ie. 235.85.104.0
broadcast: 00000000.00000000.00000111.11111111
ie. 235.85.111.255
i.e: supernet range = 235.85.104.0 - 235.85.111.255 (first supernet)
no of supernets = 2^3 - 2= 6 supernets
hosts on each supernet = 254 (2046 hosts in total)
Am i correct to this point?
Question:
How would i go about calculating the remaining 5 supernets (that is the host range, supernet iD and broadcast?I will be extremely glad to get quick response to this please
Secondly, how can i summarise the supernets (i.e. route summarization) to minimise the size of routing tables in the core.
Many thanks in advance. please help me out im stuck[/b]
"...you are never too old to learn" anon
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