Hi,

considering the following scenario:
"You are the administrator of your company's network, which uses TCP/IP as its only protocol. The network uses the private address space of 172.16.0.0/16.
Your company projects rapid growth during the next three years. To reduce the risk of network saturation, you decide to subnet the network. You will eventually need to accommodate a maximum of 17 subnets. Each subnet will include a maximum of 1,200 hosts. Which subnet mask should you configure to meet both your current needs and your projected needs?"
A. 255.255.224.0
B. 255.255.240.0
C. 255.255.248.0
D. 255.255.254.0
E. 255.255.255.0

sorry i'm asking stupid questions...my questions are:
1) doesn't "172.16.0.0/16" mean IP address of 172.16.0.0 with a subnet mask of 255.255.0.0? why do we have to configure another subnet mask?
2) 17 subnets with each subnet including 1200 hosts will be 20400 hosts. so 2^15 is 32768 which is larger than 20400. so i thought i should leave 15 bits for these hosts and therefore 17 bits is left(32-15=17). so in binary, that would be 11111111.11111111.10000000.0000000 which in decimal would be 255.255.128.0. unfortunately my answer was wrong. Could anyone show me how to do the calculation correctly?

Thanks a lot in advance,
Anthony

You required to subnet in the question so subnet you do (just as you would if the boss told you to)

You are correct, 172.16.0.0/16 means you're using 16 bits for network ID's (presently).

Since you are requiring 17 networks...what value of 'n' is required in the equation: 2(exponent)n = 17
2 to the power of 4 is 16...not enough
2 to the power of 5 is 32...this is your answer for network ID's. This leaves (16-5 = 11) 11 bits for host ID's.
2x11 = 2048
(2048 -2 leaving 2046 host ID's...which completes the required number of host ID's)

You've already got 16 bits masked off and will be subnetting in the 3'd octet...that's why you don't need to add the already existing 16 bits in the first (network ID) equation...you're only dealing with the remaining 16 bits.

I hope this makes sense....

By the way....the answer is, your subnet mask will be 255.255.248.0 because you're using 5 bits for network ID's so counting from left to right in the 3'd octet leaves you that subnet mask according to my handy dandy table:

128 64 32 16 8 4 2 1
| | | | | | | |
128 192 224 240 248 252 254 255

top row is the exponent (2x?)
middle row represents the binary bit
bottom row is subnet mask

Argh!

My little table didn't work...if you'ld like a clearer representation just email me and I'll do it up right in a text file.

The top row represents the exponent...starting from right to left
2x1 = 1
2x2 = 4
2x3 = 8 etc etc

along the bottom row you add left to right to get the subnet mask. The first bit is 128, the second bit is 128 + 64 = 192. The 3'd bit is 192 + 32 = 224 etc etc

I use this for calculating subnet masks. Once you understand how it works and have it memorized, you never have to draw it out and it comes in very handy at times having that in your head.

Curt

There is a great subnet calc that is free from Solar Winds, it creates the tables for you.

http://support.solarwinds.net/updates/SelectProgramFree.cfm

Subnet calculators are fine but if you need a subnet calculator for this problem, you don't know how to subnet. This is a very simple subnetting question and Curt's answer is correct.

I'd just add that depending on hardware, the formula may be 2^x - 2 = 17 (17 being the number of required subnets). So the formula would end being 2^x = 19 so x = 5. 5 bits being required for the subnet ID which in this case is still the same answer of 248. In many Microsoft exams the 0 subnet is not allowed so keep this in mind.