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I have an upcoming exam in networks communication. One of the available tasks for training on subnetting is the following:

Suppose you are an ISP with a 132.1.32.0/24 address block. If 2 customers require addresses for 127 and 60 computers respectively. Do the address assignment.

Don't i need 1 bit for subnets and then will have only 126 hosts per subnet since 2^7-2 = 126 ,therefor not being able to satisfy the customer demanding 127 hosts.

Or is there some other way to solve this that have passed me by ?

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marked as duplicate by Chris S May 14 '13 at 13:30

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There are no classes and haven't been since 1993 (20 years - seriously, you might not have been born when they abolished those things). Nobody should even mention them anymore outside of a history class. –  Chris S May 14 '13 at 13:29
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Also, answer depends on how shady your teacher is. Since he's/she's apparently taught you about "classes" I'll assume he/she fits into the "those who can't, teach" category. The network ID can be used as an IP address too. So in a /25 you normally have .0, .127 and 1 IP for a router (itself a "host" by the strict definition); and 125 "usable" IPs. If you assign .0 to a host it should work, and this might be what you're teacher is after. It's bad practice to use that IP however, and not guaranteed to work on all IP Stacks. –  Chris S May 14 '13 at 13:36

1 Answer 1

i'm not sure if that's the right answer for your exam but i would give 2 subnets to customer1:

subnet one: 132.1.32.0/25
subnet two: 132.1.32.128/30

and for customer2 i would start at the end of the /24 block

subnet: 132.1.32.192/26
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