# A problem solving question dealing with logarithms :/

• Jan 12th 2011, 10:10 PM
danzig
A problem solving question dealing with logarithms :/
I seem to be stumped on this question for my homework:

"If a telephone network is designed to carry "C" telephone calls simultaneously, then the number of switches needed per call must be at least log2 C. If the network can carry 10,000 calls simultaneously, how many switches would be needed for one call and for 10,000 simultaneous calls?"

..That would be the problem, now I seem to have thought that log2 C = y would be the way to approach it, solving for "y", the number of switches but, that doesnt seem to work since there arent any whole numbers that could be used as an exponent for 2 to give me 10,000. I seem to be in my own little world with this one. Any help or insight would be great (Nod).
• Jan 12th 2011, 10:46 PM
Chris L T521
Quote:

Originally Posted by danzig
I seem to be stumped on this question for my homework:

"If a telephone network is designed to carry "C" telephone calls simultaneously, then the number of switches needed per call must be at least log2 C. If the network can carry 10,000 calls simultaneously, how many switches would be needed for one call and for 10,000 simultaneous calls?"

..That would be the problem, now I seem to have thought that log2 C = y would be the way to approach it, solving for "y", the number of switches but, that doesnt seem to work since there arent any whole numbers that could be used as an exponent for 2 to give me 10,000. I seem to be in my own little world with this one. Any help or insight would be great (Nod).

Keep in mind that problem says that the number of switches needed per call must be at least \$\displaystyle \log_2 C\$. Than means if you get a decimal value when evaluating \$\displaystyle \log_2 C\$, then you should round up to the next integer.

Does this clarify things?
• Jan 12th 2011, 11:48 PM
danzig
Yes it does clarify that I should possibly round up, but if I solve for log2 10,000 = y i get, y = 14 which 2^14 is 16,384.. seemed much of an obscure number to be right. and if i solve y for log2 1 = y than y = 0 when I get those numbers the problem doesnt seem to make sense in my brain.
• Jan 12th 2011, 11:56 PM
Unknown008
Yes, 14 is the answer for the 10 000 calls, meaning that if ever there was 10 000 simultaneous calls, the 14 switches will be able to handle them, while 13 wouldn't.

For the second part, y is not zero, because if you have no switch at all the telephone network will not work. So, the answer for part 2 is 1 switch.
• Jan 12th 2011, 11:59 PM
Chris L T521
Quote:

Originally Posted by danzig
Yes it does clarify that I should possibly round up, but if I solve for log2 10,000 = y i get, y = 14 which 2^14 is 16,384.. seemed much of an obscure number to be right. and if i solve y for log2 1 = y than y = 0 when I get those numbers the problem doesnt seem to make sense in my brain.

\$\displaystyle \log_2(10000)\approx 13.2877\implies 2^{13.2877}\approx 10000\$. But the solution we want is 14 because it doesn't make sense to have 13.2877 switches per call (if we're assuming the number of switches is integer valued).

I hope this makes sense!
• Jan 13th 2011, 12:11 AM
danzig
Quote:

Originally Posted by Unknown008
Yes, 14 is the answer for the 10 000 calls, meaning that if ever there was 10 000 simultaneous calls, the 14 switches will be able to handle them, while 13 wouldn't.

For the second part, y is not zero, because if you have no switch at all the telephone network will not work. So, the answer for part 2 is 1 switch.

Ahh! this make's perfect sense to me now! Thank alot Unknown and Chris for clarification :D!