You are an ISP. For the Address Block 195.200.0.0/16 answer the following questions

Question

a. If you have 320 Customers that need 128 addresses/customer - will there be enough addresses to satisfy them? and explain why? 

b. Say the first block of 64 customers want 128 addresses/customer - what would the address space look like? 

Answer 1

Hi Benny B.

The answer to Question A: Yes. 195.200.0.0/16 (or 195.200.0.0 - 195.200.255.255) allows for 65,536 Addresses (65,534 usable addresses to assign to hosts). So, if you have 320 customers that each need 128 Addresses, then that would give you (320 x 128) = 40,960 IP Addresses, which should be plenty with a good amount (24,576) left over.

The answer to Question B: If the goal is to assign one subnet that could accommodate 128 addresses for each of the 64 customers, then each of those subnets would need to have a 25-bit Subnet Mask; 255.255.255.128 or in "CIDR" (Classless Interdomain Routing) Notation: /25.

The Networks for the first block of 64 customers would look something like what you will see below. I will omit most of the subnets in between, in order to avoid taking up too much space on this page and to avoid a lot of copying and pasting :)

You would start out with Subnet Zero and work your way down:

1. 195.200.0.0/25
2. 195.200.0.128/25
3. 195.200.1.0/25
4. 195.200.1.128/25
5. 195.200.2.0/25
6. 195.200.2.128/25
7. 195.200.3.0/25
8. 195.200.3.128/25
9. 195.200.4.0/25
10. 195.200.4.128/25

.

.

(This pattern would continue until you get all the way down to the 64th Subnet/Customer)

.

.

63. 195.200.31.0/25

64. 195.200.31.128/25

Each Subnet range would appear like this:

1. 195.200.x.0 - 195.200.x.127 or
2. 195.200.x.128 - 195.200.x.255

The above would be the subnet assignments for the first block of 64 customers. Each one of those Subnets listed above counts as One Independent Network and can accommodate up to 128 Addresses, but remember, the First IP Address in the range is reserved for the "Network ID" and the Last IP Address in the range is always reserved for the "Broadcast Address". So in actuality, each subnet would have 126 Addresses that could be used for the purpose of assigning to End Hosts (Laptops, Servers, Printers, Phones, TVs, Cameras etc.).

You could even Summarize all 64 of those Customer Subnets listed above into One large Block, which would be:

195.200.0.0/19 - This would include all 64 of the customer networks ranging from:

195.200.0.0 - 195.200.31.255

That entire collection of Networks that we just assigned to each of our 64 customers totals out to 8,192 IP Addresses (128 Addresses x 64 Subnets = 8,192 Addresses).

If we had 320 Customers and we divided them up into to 5 "blocks" of 64 (64 x 5 = 320) and each "block" has 8,192 IP Addresses,

Then that means that we would have exactly 40,960 IP Addresses (8,192 x 5 = 40,960) as I had mentioned earlier. Which means we have way more than enough, as answered in Question A ;-)

Hopefully this was not confusing and you learned something from this. If you or anyone has any further questions or would like to learn more about Subnetting, please reach out to me and you will learn how to master Subnetting and many more Networking techniques! :)