Need help with this IP address subnet question. Pretty sure I understand 1-5, bu
ID: 3817147 • Letter: N
Question
Need help with this IP address subnet question.
Pretty sure I understand 1-5, but need help with #6.
- IP address is 155.1.0.0. , needs 85 subnets.
Explanation / Answer
Question 6:
First thing is that, the IP Address is 155.1.0.0 is a Class B address i.e., the first 16 bits are used to identify the network address. To create subnets, we have to borrow bits from host part.
Now, if we borrow 1 bit, then we can create 2 subnets. If we borrow 2 bits then we can create 4 subnets. The formula is 2n subnets can be created by borrowing n bits.
So, to create 85 subnets we must borrow 7 bits which gives 128 combinations unlike 64 combinations from 6 bits.
Now, if we borrow 7 bits i.e., 7 MSBs from host part then subnet mask will become 11111111 11111111 11111110 00000000 = 255.255.254.0
Direct Broadcast Address is the last address of a subnet.
Valid subnets will be 2, 4. 6,...., 252.
The last subnet for the organization that needs 85 subnets only will be X.X.190.0
Subnet Number Subnet Address Range of hosts Direct Broadcast Address 1 X.X.2.0 X.X.2.1 - X.X.3.254 (510 hosts) X.X.3.255 2 X.X.4.0 X.X.4.1 - X.X.5.254 X.X.5.255 3 X.X.6.0 X.X.6.1 - X.X.7.254 X.X.7.255 Last X.X.252.0 X.X.252.1 - X.X.253.254 X.X.253.255Related Questions
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