Question

A large number of consecutive IP addresses are available starting at 198.16.0.0. Suppose that four organizations, A, B, C, and D, request 512, 2000, 1000, and 4000 addresses, respectively, and in that order. For each of these, give the first IP address assigned the last IP address assigned, and the mask in the w.xy.z/s notation

Answer 1

First of all we need to assign an IP address to organization A. Organization A require 4000 IP addresses. So the number of host bits can be calculated by applying 2ⁿ  = 2¹² = 4096(nearest value to 4000). Thus the number of host bits are 12. The subnet mask for the organization A will be 198.16.0.0/20. Here we write 20 because we subtract the host bits from the total bits i.e. 32 - 12 = 20. The first IP address is already given 198.16.0.0 . Now to determine the last IP Address, convert the bytes into binary. 198 --> 11000110, 16 --> 00010000, and the modified third and fourth bit will be 00001111 and 11111111 (replacing 12 bits by 1). Now converting these bits into decimal number we get the last IP Address of organization A i.e. 198.16.15.255 .

Now for organization B, it requires 2000 IP Addresses. So the number of host bits can be calculated by 2¹¹ = 2048(nearest value to 2000). The number of host bits are 11. So 11 bits will be replaced by 1 in the given IP Address. The modified IP Address will be 11000110 00010000 00010111 11111111 (adding an extra 1 because the IP Address of the next organization will be greater by 1 than the IP Address of the previous organization i.e. 11000110 00010000 00010111 11111111). Now converting this into decimal number, we get our first IP Address --> 198.16.16.0 and the last IP Address --> 198.16.23.255. The subnet mask will be 198.16.16.0/21 (32-11=21).

Organization C requires 4000 IP Addresses so just like before we will calculate the host bits = 12. So replacing 12 bits in the IP Address. The modified IP Address will be 11000110 00010000 00101111 11111111(here we incremented network part only to avoid clash of IP Addresses i.e. 11000110 00010000 0001 + 1 = 11000110 00010000 0010). Converting this into decimal number we get, our first IP Address --> 198.16.32.0 and the last IP Address --> 198.16.47.255 . The subnet mask will be 198.16.32.0/20 .

Organization D requires 8000 IP Addresses so 2¹³ = 8192. Number of host bits = 13. The modified IP Address after replacing 13 bits will be 11000110 00010000 01011111 11111111. Converting this into decimal number we get, our first IP Address --> 198.16.64.0 and the last IP Address --> 198.16.95.255. The subnet mask will be 198.16.64.0/19 (32-13 = 19).