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In the network 202.10.4.230/27  the last but one valid IP which can be assigned to a host from the last subnet is

___________.
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202.10.4.253
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How ?
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27 bits are there for block id. The address is of class C network so 24 bits are already reserved for the network id. Remaining 3 bits for subnet id. We have 8 bits in the last octet of which most significant 3 bits are for subnet id. For the last subnet make them all 1.

For remaining 5 bits, 11111 is the broadcast address cannot be given to any host. 11110 is given to last host. 11101 is given to second last host.

So 11111101 which is 253

btw is it correct??
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yes, 253 is the answer given , but the notation given is of CIDR, how you have assumed it to be class C ?
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@Gupta731 why you answered for 2nd last it is not mentioned in the question. @Shadan Karim is 253 correct I think it should be 254 (11111110)??

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@Peeyush Pandey it is given in question that you have to find last but one, which means second last host. You need to interpretate the question statement properly :)

@Shadan Karim The first octet is 202 which is of class C. Go through the range of IP addresses in diff classes.

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I read it, it is not a commonly used phrase hope gate authorities are not going to use such Shashi Tharoor level English. :)

And yes 253 is correct last but one means second last, last but two means 3rd last and so on.

202.10.4.230/27

every octet made up of bits

8+8+8 = 24

BUT IN THE ABOVE QUESTION IT IS GIVEN 27 AND HENCE FROM THE LAST OCTET 3 BITS ARE GIVEN

230 IN BINARY 111(00110) MAKE EVRY BIT ONE IN THE BRACKET THAT WILL BE THE BROADCAST ADDRESS

1(111111) - 254 THIS IS LAST HOST

SO 202.104.253 IS LAST BUT ONE HOST
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