I am trying to tackle the following subnetting problem: Prepare subnetting for an organization that requires 2000 hosts to be placed on the network. The maximum number of hosts on a subnetwork is 300. The organization doesn't need more then 24 subnetworks in the foreseen future. As a solution, for the first three subnetworks indicate the first and last host address in the subnet.

So far this is what I have come with: 9 bits will be needed for the host addresses and 5 for the subnet. Thus the remaining 18 bits are reserved for the network.

I think this will require class B addressing, so I choose to assign the following IP to the network:

Subnet 1: 1st host -, last host- | and I reserved for subnet address and broadcast respectively

Is this reasoning correct, will subnet 2 start at

  • Forget class-based subnetting. That concept died with the global adoption of CIDR a long time ago. – Flup Oct 13 '14 at 7:40
  • well, if you use 9-bits for the host addresses, you get 510 addresses per network and 128 nets, so each net is exactly twice as large as a standard class C. that means your 0 network runs from .0.1 - .1.254 with the broadcast at .1.255. Remember, every bit you steal from the net mask doubles your host address space. you can use this utility to verify your answers: subnet-calculator.com . Also, remember, while it may not be exactly what your instructors want, lots of organizations use a class A even though they just have a few thousand hosts. no point in making it complicated. – Frank Thomas Oct 13 '14 at 8:37
  • @FrankThomas thanks a lot for your input. I have studied some more and used your suggestions. This is what I came up with. I need 300 hosts on each subnet, the closest I can get to this is 2^9=512. Hence my subnet mask would be or /23 and I will be able to accomodate 128 subnets. So my 1st subnet starts at .0.0 and ends at .1.45, .0.0 and .1.45 being the network and broadcast addresses. Accommodating 300 hosts. I'd appreciate your feedback? – Nic Oct 13 '14 at 21:29
  • well, 510 = 255 + 255, so the first network can't end at 1.45. you can't get exactly 300 hosts using a network mask. 510 is the nearest you can get. you could start allocating IPs from .0 up to .1.45 and then stop, but your network's broadcast is still .1.255, and the next network's ID is at .2.0 (.2.0 - .3.255). – Frank Thomas Oct 13 '14 at 22:01
  • Remember, the only things you can configure in an IP host are the IP address, the subnet mask, and the default gateway. everything else, like the calculations to derive the network ID, broadcast, and routing to other networks, must be inferred from these three bit of config. There is no way to tell all the devices in a network that the number of hosts is 300, and that the devices should use that value when calculating the address layout of the network(s). That is what the net mask is for, but as you have seen, netmasks require generalizations, and can't provide exact numbers of hosts. – Frank Thomas Oct 13 '14 at 22:29

From a real-world point of view, you've been given maximum limits, and a current situation: No more than 24 subnets ever, no subnet may exceed 300 hosts, ever, provision 2000 hosts now.

This seems to mean that the largest subnet ought to not exceed a /24, as one can exceed 300 hosts with a /23, which means that a great portion of IP addresses would be wasted if you used a /23.

And, while a /24 can address 254 hosts, you might wish to reserve a few for addressing the network equipment, which are technically, not hosts. At 250 hosts per /24, 2000 hosts can be served by 8 /24s, exactly.

In practice, one tries to avoid using all available addresses in a subnet: What happens when the organization adds its 2001st host? The problem, as stated, does not say that 2000 hosts are the maximum, and pragmatically, it wouldn't be a sound assumption in the real world.

Putting 2000 hosts (at 250 hosts per /24) into 8 /24 subnets would give 100% utilization, 0% available for growth before network reconfiguration, with 16 subnets remaining of 24 max.

Into 9 /24s: 89% address utilization, 11% available, allowing room for 243 additional hosts in total; 27 per network, before network reconfig, with 15 subnets remaining.

Into 10 /24s: 80% utilization, 20% available.room for 500 additional hosts; 50 per network. 14 subnets remaining.

11: 73% utilized, 27% available, room for 748 additional hosts, 68 per network, 13 subnets remaining.

12: 67% utilized, 33% available, room for 996 additional hosts, 83 per network, 12 subnets remaining. (My personal preference)

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