Kassem Okey so I have a Networking exam tomorrow 10:00AM and I'm still not sure how I'm going to do! Everytime I think I actually got it, I realize that I need to re-vise the material again! Urgh! So anyway, while solving some sub-netting exercises I came across this one: Q. An organization is granted the block 217.22.18.0/24. The administrator wants to create 16 subnets. i. Find the subnet mask. ii. Find the number of addresses in each subnet. iii. Find the first and last addresses in subnet 0. Here's what I came to: i. 11111111.11111111.11111111.00000000 ii. Log2 of 16 = 4. So 4 bits are required to create the 16 subnets, hence only 4 bits are left to allocate the Host addresses. Which means the number of addresses in each subnet = 16 - 1 -1 = 14. ii. First I need to convert the address to binary format which I do not know how to do (help anyone?). Then the first one be of the form: address.00000001 (bold for the subnet) The last one: address.0000.1110 What do you think about my solution? Am I right?
MrClass Ok here we go: Kassem, the subnet he is asking for is not the old subnet, which is 255.255.255.0, he is asking about the new subnet after subnetting. 217.22.18.0/24 -> first 24 bits represent the subnet mask of 217.22.18.0 But since we will divide in to 16 subnets -> 2^n > or = to 16 , n = 4 then we will need 4 bits from the host part for the new subnet: 11111111.11111111.11111111.00000000 becomes 11111111.11111111.11111111.11110000 So new subnet mask is 255.255.255.240 Now we use the original address plus the 4 bits to find the rest of the address ranges: 217.22.18.0 = 11011001.00010010.00010010.00000000 so we need to vary the 4 bits from 0000 to 1111: Subnet Ranges: Subnet 0: 11011001.00010010.00010010.00000000 = 217.22.18.0 Subnet 1: 11011001.00010010.00010010.00010000 = 217.22.18.16 Subnet 2: 11011001.00010010.00010010.00100000 = 217.22.18.32 Subnet 3: 11011001.00010010.00010010.00110000 = 217.22.18.48 Subnet 4: 11011001.00010010.00010010.01000000 = 217.22.18.64 Subnet 5: 11011001.00010010.00010010.01010000 = 217.22.18.80 Subnet 6: 11011001.00010010.00010010.01100000 = 217.22.18.96 Subnet 7: 11011001.00010010.00010010.01110000 = 217.22.18.112 Subnet 8: 11011001.00010010.00010010.10000000 = 217.22.18.128 Subnet 9: 11011001.00010010.00010010.10010000 = 217.22.18.144 Subnet 10: 11011001.00010010.00010010.101000000 = 217.22.18.160 Subnet 11: 11011001.00010010.00010010.10110000 = 217.22.18.176 Subnet 12: 11011001.00010010.00010010.11000000 = 217.22.18.192 Subnet 13: 11011001.00010010.00010010.11010000 = 217.22.18.208 Subnet 14: 11011001.00010010.00010010.11100000 = 217.22.18.224 Subnet 15: 11011001.00010010.00010010.11110000 = 217.22.18.240 meaning 16 address in each subnet in Subnet 0: first address is 217.22.18.0, last address 217.22.18.15
Kassem Ok MrClass, that was very helpful. I really appreciate it. I have one more question though... How do you convert from Decimal to Binary? Do you use the method explained here or is there an easier way to do it?
MrClass Does your scientific calculator contain that feature? You dont have time to calculate that unless your instructor obliges you to do so. If you need to use the old calculation method, check this: convert 17 to binary: result remainder 17 / 2 = 8 | 1 8 / 2 = 4 | 0 4 / 2 = 2 | 0 2 / 2 = 1 | 0 1 / 2 = 1 | 1 read the remainders from bottom to top => 10001 is binary of 17
Kassem Unfortunately, I'll have to do it manually during the exam! That's going to take some time no doubt! But yeh, your method is very easy to apply thanks for the tip :)
