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An organization with a class C IP address of $195.37.5.0$ wish to divide the network into different sub networks, the subnet mask is 255.255.255.192. Which one among these is not a broadcast address in any of the sub network?

- $195.37.5.127$
- $195.37.5.191$
- $195.37.5.31$
- $195.37.5.255$

as subnet mask is 192 which is **11 000 000** so 2 bits we take for subnet bits from HID and total 4 subnets 2^2=4 are there and first subnet 00 second subnet 01 third subnet 10 and fourth subnet 11 .

now for broadcast address all Host bits are 1 that means

**first broadcast address** is 00 111 111 which is 63

**second broadcast address** is 01 111 111 which is 127

**third broadcast address** is 10 111 111 which is 191 and

**fourth broadcast address **is 11 111 111 which is 255 .

last octet ::: 63(00) , 127(01), 191(10) , and 255(11) for 195.37.5.127, 195.37.5.191, 195.37.5.255, 195.37.5.63 broadcast addresses possible for 4 subnets .

As given option c says 31 so c) is not broadcast address.

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Best answer

the subnet mask is 192 that means 11 000000 , so 2 bits from HID taken for subnets there are 4 subnets possible like 00, 01, 10 and 11 so for 00 broadcast address is 195.37.5. 63 , for 01 broadcast address is 195.37.5. 127, for 10 broadcast address is 195.37.5. 191, for 11 broadcast address is 195.37.5. 255. but option c says 195.37.5.31 which not match with anyone hence option C is not a broadcast address in any of the sub network .

I think there is problem with the test, My wrong answer it says right. Problem with lot of question please check @Bikram

Test name : Computer Networks-12

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