Hi all -I saw this question and cant figure out how to work it out-all help gratefully appreciated- The mask used is 255.255.254.0 It asks which of the following 3 addresses can be used-
Am I right in thinking the mask allows 510 usable hosts per subnet in a class C and only 1 host in a Class B? Thanks alot-I know the answers but dont understand why.
Yes there are 510 usable hosts in a /23. Classes are really meaningless once you break the /8, /16 or /24 boundry. Example 10.0.0.0/23 is technically a class "A" address, it still contains 510 hosts. 172.16.0.0/23 is technically a class "B" address, it still contains 510 hosts. 192.168.0.0/23 is technically a class "C" address, it still contains 510 hosts.
When using a /23 mask there is a very quick shortcut you can use. If the 3rd octet is an even number it immeadiately tells you it contains the network address and the rest are valid hosts. If the 3rd octet is an odd number it immeadiately tells you it contains all hosts and the broadcast address.
113.10.4.0 is a network address
186.54.3.0 is a valid host, it belongs to the 186.54.2.0/23 network.
175.33.3.255 is the broadcast address for the 175.33.2.0/23 network
26.35.2.255 is a valid host on the 26.35.2.0/23 network
152.135.7.0 is a network address
17.35.36.0 is a valid host on the 17.35.35.0/23 network
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