A campus network is allocated the address block 10.40.96.0/19. The block is divided into equal-size subnets, each of which must accommodate at least 500 host addresses, and the number of subnets is maximized. The number of usable host addresses in each resulting subnet is:
- (A)254 addresses
- (B)500 addresses
- (C)508 addresses
- (D)510 addresses
Show worked solution
Answer: (D)
Nine host bits form the smallest field that meets the 500-host requirement, and the resulting /23 subnet leaves 510 usable addresses once the network and broadcast values are reserved.
Two addresses in every IPv4 subnet are unusable, the all-zeros network identifier and the all-ones broadcast address, so the host field must exceed the required host count by that margin.
Eight host bits fall short at 254 usable addresses, so nine bits is the smallest qualifying field and it fixes the subnet prefix.
The /19 block supplies 13 host bits, four of which become the subnet field, so sixteen equal /23 subnets fit and each carries the same capacity.
FE Reference Handbook — Electrical and Computer Engineering: Computer Networks, IPv4 Addressing and Subnet Masks
Why the other choices appear
- (A)Uses eight host bits rather than nine, giving 2^8 - 2 = 254 usable addresses, which fails the 500-host requirement.
- (B)Takes the stated requirement itself as the subnet capacity, omitting the rounding of the host field up to the next power of two.
- (C)Subtracts four reserved addresses from the 512-address block instead of the two (network and broadcast) that are actually unusable, giving 508.