11. + Host IP Address Major Network Mask Major (Base) Network Address Major Network Broadcast Address Total Number of Host Bits Number of Hosts Subnet Mask Number of Subnet Bits Number of Host Bits Number of Useable Hosts per Subnet Subnet Address for this IP Address 142.20.33.17 255.255.255.0 142.20.33.0 142.20.33.255 - 8 bits 2^8 2254 host 255.255.255.252 6 bits 2 bits 2^2 2 2 hosts 142.20.33.16 IP Address for the 1st Host on this Subnet 142.20.33.17 IP Address for the Last Host on this Subnet 142.20.33.18 Broadcast Address for this Subnet 142.20.33.19
I need help understanding how could I got the IP Address for the 1st Host on this Subnet, the IP Address for the Last Host on this Subnet & the Broadcast Address for this Subnet
My Explanation:
8 bits --> 17 & 0
128 64 32 16 | 8 4 2 1
0 0 0 1 0 0 0 1 -->17
+ 0 0 0 0 0 0 0 0 --> 0
0 0 0 0 0 0 0 0 --> 0
(The Major (Base) Network Address): 142.20.33.0
To find the broadcast address, we can set all the bits in the host portion of the address to 1, which gives us 142.20.33.255. Based on the subnet mask 255.255.255.0, no bits were borrowed from the host portion, meaning 8 bits remain in the host portion. Using the formula 2^h – 2 (where h represents the number of host bits), we get 2^8 - 2 = 254 hosts.
8 bits --> 17 & 252
128 64 32 16 | 8 4 2 1
0 0 0 1 0 0 0 1 --> 17
+ 1 1 1 1 1 1 0 0 --> 252
0 0 0 1 0 0 0 0 -->16
(Subnet Address for this IP Address): 142.20.33.16
Based on the network mask 255.255.255.252, we see that 252 in bit is 1111 1100. As such, 6 bits are borrowed from the host portion, leaving us with 2 host bits. Therefore, using the formula 2^2 - 2, we get 2 usable hosts.
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