255.255.255.128/what is the n/w id ,broadcast id and the
number of valid host(according to disco rules)
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Answer / sachin farde
Here subnet mask is 128 i.e. /25 is the CIDR . 1 bit is
used 4 addressing.
Class-C address format is N.N.N.H(8bit.8bit.8bit.8bit)
for subnetting we used cisco rule
1)no of subnet= 2^1=2 ( 1 is the first on bit of last
octate)
2)no. of host in each subnet=2^7=128-2=126( 7 is the off
bit of last octate)
3)valid subnet=256-128=128 i.e. 0, 128
4) broadcast address= is the no. right before the next
subnet
5) valid host= nos. between network address & broadcast
address
below table gives u detail
subnet 0 128
1st valid 1 129
host
last valid 126 253
host
broadcast
address 127 254
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / venkatrajesh
255.255.255.128
i.e,X.X.X.X/25
HOST bits(X / XXXXXXX)
1 7
2^1 = 2 Subnets
2^7-2=126 hosts/Subnet
1st subnet 0-127
2nd subnet 128-255
N/W ip-0,128.
B/C ip-127,255.
numberof valid host-126/subnet(252hosts)
For EXAMPLE
192.168.1.0/25
192.168.1.0- 1st N/W IP
192.168.1.1-First Valid Host
192.168.1.126-Last Valid Host
192.168.1.127-B/C IP
Total no of valid host in first subnet is 126
192.168.1.128- 2nd N/W IP
192.168.1.129-First Valid Host
192.168.1.254-Last Valid Host
192.168.1.255-B/C IP
Total no of valid host in second subnet is 126
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / jitendera kumar sinha
but sachin as we know that ciso says that 1 st and last
block are not used for n/w ing becose total num of n/w is
2^n-2 so how can we use it
jitendera kuamr sinha
| Is This Answer Correct ? | 3 Yes | 2 No |
Answer / don
NW ID - 223.255.255.128 / 25
Broadcast Address - 223.255.255.255
Usable Address Range - 223.255.255.129 - 223.255.255.254
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / roop&
Jitu Bhai they are talkin theory point of view not
practical .
| Is This Answer Correct ? | 0 Yes | 1 No |
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