work must be shown Enter the last usable host on the network of which the host 1
ID: 3819042 • Letter: W
Question
work must be shown
Enter the last usable host on the network of which the host 10.13.174.145/21 is a part.
What is the shorthand corresponding to a subnet mask of 255.255.128.0?
Enter the usable host range for the network of which the IP address 10.174.14.177/16 is a part.
Enter the usable host range for the network of which the IP address 10.69.121.116 255.255.128.0 is a part.
Enter the maximum number of valid subnets and usable hosts per subnet that you can get from the network 172.20.0.0 255.255.240.0.
Enter the broadcast address for the network 10.220.192.0/18.
Enter the first usable host address on the network 172.16.131.84 255.255.255.248.
Enter the network address of the subnet to which the host 192.168.38.183/27 belongs.
Enter the first usable host on the network 10.46.108.97 255.255.252.0.
What subnet mask would you use for the 172.18.0.0 network, such that you can get 40 subnets and 760 hosts per subnet?
Explanation / Answer
1) Enter the last usable host on the network of which the host 10.13.174.145/21 is a part ?
Given Class less Address is 10.13.174.145 / 21
Here 21 indicates that the 1st 21 bits of the IP address is Network address
Rest of the 11 bits is host address.
Address in Binary is 00001010.00001101.10101110.10010001 (Bold bits are network address)
Last usable host is 00001010.00001101.10101111.11111110 (Bold bits are host address)
Hence IP address of last usable host is 10.13.175.254
2) What is the shorthand corresponding to a subnet mask of 255.255.128.0?
The subnet mask 255.255.128.0 in binary is 11111111.11111111.10000000.00000000
Here 1's indicates Network address bits and 0's indicates Host address bits
Hence Network address bits is = 17
Host address bits is = 15
3) Enter the usable host range for the network of which the IP address 10.174.14.177/16 is a part?
Given Class less Address is 10.174.14.177 / 16
Here 16 indicates that the 1st 16 bits of the IP address is Network address
Rest of the 16 bits is host address.
Address in Binary is 00001010.10101110.00001110.10110001 (Bold bits are network address)
Usable host range is 00001010.10101110.00000000.00000001 (Bold bits are host address)
to
00001010.10101110.11111111.11111110
4) Enter the usable host range for the network of which the IP address 10.69.121.116 255.255.128.0 is a part?
If we make and operation on IP and network mask then we will get the network address
10.69.121.116 and 255.255.128.0
10 and 255 =10 as 00001010 and 11111111 will be 00001010
69 and 255 = 69 and 121 and 128 = 0 as 01111001 and 10000000 = 00000000
Hence the network address is 10.69.0.0
Usable host range is 00001010.0100101.00000000.00000001 (Bold bits are host address)
to
00001010.0100101.11111111.11111110
5) Enter the maximum number of valid subnets and usable hosts per subnet that you can get from the network 172.20.0.0 255.255.240.0?
Here Network address is 172.20.0.0
Subnet Mask is 255.255.240.0 which is 11111111.11111111.11110000.00000000 in binary
Here 4 1's in 3rd octet indicates the we borrowed 4 bits from host address
Hence maximum number of valid subnets is = 24 = 16.
The rest 12 bits are for host hence valid hosts is = 212 - 2 = 4094.
6) Enter the broadcast address for the network 10.220.192.0/18 ?
Here the network address is 1st 18 bits which is 10.220.11000000.00000000
Hence Network mask is 255.255.192.0
Hence the broadcast address is 10.220.11111111.11111111
7) Enter the first usable host address on the network 172.16.131.84 255.255.255.248 ?
Here by anding IP with mask address we will get the
network address as 172.16.131.01010000 (bold letters are network address)
first usable host address on the network is 172.16.131.01010001 which is 172.16.131.81
8) Enter the network address of the subnet to which the host 192.168.38.183/27 belongs ?
Here the 1st 27 bits is network address which is 192.168.38.10110111
Hence the network address is 192.168.38.10100000 which is 192.168.38.160
9) Enter the first usable host on the network 10.46.108.97 255.255.252.0 ?
by anding both we will get the network address as 10.46.01101100.00000000
first usable host is 10.46.01101100.00000001 which is 10.46.108.1
10)What subnet mask would you use for the 172.18.0.0 network, such that you can get 40 subnets and 760 hosts per subnet?
40 subnets means we have to borrow 6 bits from the host address so that
26 64 subnets possible.
Hence the subnet mask is 255.255.11111100.00000000
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