The grey ones were provided and we were asked to fill in the red ones with just
ID: 3731934 • Letter: T
Question
The grey ones were provided and we were asked to fill in the red ones with just that amount of information. Could you please explain to me how to get those fields from the provided info?
Netmask (dotted decimal) Network Address Host Number Total count of host addresses in IPv4 CIDR Address Broadcast Addres:s 128.193.43.35 /16 255.255.0.0 128.193.0.0128.193.255.255 11043 65536-2 65534 128.193.225.244 /20 255.255.240.0 128.193.224.0 128.193.239.255500 128.193.43.35 /23 255.255.254.0 128.193.42.0 128.193.43.255 291 35 4096-2-4094 512-2 = 510 64-2 = 62 128.193.43.35 /26 255.255.255.192 128.193.43.0128.193.43.63Explanation / Answer
For the first, third and fourth parts, the IP is provided along with the subnets. In that case,
To find the netmask for xxx.xxx.xxx.xxx/num , write bits '1' num number of times and write '0' (32- num) number of times. now divide them in groups of 8 and add decimal points in between.
For example,
16 = 11111111. 11111111. 00000000. 00000000 = 255.255.0.0
23 = 11111111. 11111111. 11111110. 00000000 = 255.255.254.0
To find the Network Address (NA), just take bitwise AND of the ip with the subnet. For example,
IP: 128.193.43.35 = 10000000.11000001.00101011.00100011
Subnet: 255.255.0.0=11111111.11111111.00000000.00000000
Therefore, the NA is : 10000000.11000001.00000000.00000000 = 128.193.0.0
To find the Broadcast Addresss(BA), convert all the host bits( the last (32-num) bits) to 1.
IP: 128.193.43.35 = 10000000.11000001.00101011.00100011
taking logical OR with 00000000.00000000.11111111.11111111
The BA comes out to be : 10000000.11000001.11111111.11111111 = 128.193.255.255
The number of host addresses is calculated mathematically as
Host = 2^(32- num) -2
For first case, num=16
Host = 2^(32-16) -2 = 65536 - 2 = 65534
The host number can be found by the following steps:
1. Subtract NA from IP
128.193.43.35
- 128.193.00.00
----------------------
000.000.43.35
2. Now we have four blocks of numbers. Let the result after subtraction be a.b.c.d . The HOST NUMBER is now found by:
a*(256*256*256)+ b*(256*256)+c*(256)+ d
For the first case,
HN = 0*(256^3)+ 0*(256^2)+ 43*256+35 = 11043
For the second case, the Subnet can be found by writing the netmask in bitwise form and calculating the number of '1' 's .
The IP can be found by writing the host number in power of 256 and adding to the NA.
Example , 500 = 0*(256^3) + 0*(256^2) + 1*(256) + (244) = 000.000.001.244
NA = 128.193.224.0
IP = NA+ HN = 128.193.225.244
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