[abc123__]

So, I need this homework help.

This is a question of the week --- I may use any resources available, including other teachers, students, etc. Anyone except the teacher who assigned this.

Here it is:

It's snow season here in Maine and occasionally the public schools are closed due to a big snow storm. Whenever a big storm is expected, everyone wants to know as early as possible if there is no school that day. Carlin is the son of the principal of a small school in East Fugawee, and as such he has the privilege of knowing very early in the morning when school is going to be called off due to the snow. Carlin has formed a phone tree with some friends so that they will know right after he finds out.

Here is how it works. As soon as Carlin knows that there will be no school, he calls his best friends, Frank and Hobart. Frank then calls his friends Julie and Kory and Hobart calls Laura and Marcellus. The phone tree continues with Julie calling Nihaar and Saloni, and Kory calling Tara and Vincent and so on. Since you don't know these people, it isn't really important to know the names. What you do need to know is that each person will call 2 other people until everyone has been called.

Suppose that Carlin finds out there is no school and makes his first call at 6:00 am. We are going to assume that everyone wants to get back to sleep as soon as possible so they are not going to chat. Each call takes exactly one minute. If Carlin's first call was at 6:00, his second was at 6:01 am. Ten minutes later, at 6:10 am, everyone on the list is back in bed. How many phone calls were made? Some people at the end of the tree will only need to make 1 call, but don't worry about that. Find a pattern and use it to figure out the number of calls made between 6:00 am and 6:10 am.

Now, suppose that we wanted to have a similar system at the University of Southern Maine, and there are 1000 students on the phone list. If the classes are canceled due to a snow storm and calling starts at 6:00 am, at what time will the last call be made?

_________________________________________
Show your work.


If you guys could help, I'd really appreciate it.!!!!!

[supa'Wanki]

Sorry, I have a Ti-83 Silver Edition. It's a good calculator though.

[x`ind]

^ lol too stupid to figure it out.


well heres a bump. i hate math, my worse subject. and i have a regular old TI uhhh can't find it right now.

[thachhao]

I have been trying to figure this out ever since you posted this.

[kodou]

Okay.
This is probably wrong, because I did it out by hand. (Might have miscounted err).
I believe .. 192 phone calls were made.
Basically made a tree, beginning with the first person.
His two phones calls, one labeled 6:00 the other 6:01.
Then those two people called another two people .. labeled them accordingly (two would be 6:02, two would be 6:03)
(Remember that two of them can be calling someone else at the same time .. if that makes sense.)
I'd draw it all out but blaaah.
Again, I may have miscounted haha. I'm sure there an easier way to do this.
And I'm too tired atm to figure out the second part of your problem. :x

[dazzleanddestroy]

haha, i'll do this later.. XD~~~!!!

[jojoboy4]

i'll take a stab at this...

pattern:

1st iteration) 6:00 to 6:02 -> 2 calls are made
2nd iteration) 6:02 to 6:04 -> 4 calls are made since 2 will be calling at the same time
3rd iteration) 6:04 to 6:06 -> 8 calls are made
4th iteration) 6:06 to 6:08 -> 16 calls are made
5th iteration) 6:08 to 6:10 -> 32 calls are made

so in other words...

1st iteraton -> 2 total calls
2nd iteration -> 2 + 4 = 6 total calls
3rd iteration -> 6 + 8 = 14 total calls
4th iteration -> 14 + 16 = 30 total calls
5th iteration -> 30 + 32 = 62 total calls

summation of 2 ^ i where i goes from 1 to infiniti...

if i = iteration
total calls = (2 ^ i - 2) + (2^i)

example, for the 5h iteration... [(2 ^ 5) - 2] + (2 ^ 5) = 30 + 32 = 62 total calls

so if the total calls is 1000

1000 = (2 ^ i - 2) + (2 ^ i)
then i is about 9...

during the 9th iteration, over 1000 calls would have been made..

each iteration takes 2 minutes... the 9th iteration is from 6:16 to 6:18. so at 6:17, the 1000th call would have been made...

[kodou]

^
haha guess i did it waaaay wrong XD

[dazzleanddestroy]

i'll take a stab at this...

pattern:

1st iteration) 6:00 to 6:02 -> 2 calls are made
2nd iteration) 6:02 to 6:04 -> 4 calls are made since 2 will be calling at the same time
3rd iteration) 6:04 to 6:06 -> 8 calls are made
4th iteration) 6:06 to 6:08 -> 16 calls are made
5th iteration) 6:08 to 6:10 -> 32 calls are made

so in other words...

1st iteraton -> 2 total calls
2nd iteration -> 2 + 4 = 6 total calls
3rd iteration -> 6 + 8 = 14 total calls
4th iteration -> 14 + 16 = 30 total calls
5th iteration -> 30 + 32 = 62 total calls

summation of 2 ^ i where i goes from 1 to infiniti...

if i = iteration
total calls = (2 ^ i - 2) + (2^i)

example, for the 5h iteration... [(2 ^ 5) - 2] + (2 ^ 5) = 30 + 32 = 62 total calls

so if the total calls is 1000

1000 = (2 ^ i - 2) + (2 ^ i)
then i is about 9...

during the 9th iteration, over 1000 calls would have been made..

each iteration takes 2 minutes... the 9th iteration is from 6:16 to 6:18. so at 6:17, the 1000th call would have been made...

oh wow, good job.

[thachhao]

I'm not sure that's the way, jojoboy4, because two calls are not made at the same time. We have to consider that each student takes one minute to call his first friend, and then another minute for his second friend; and then repeat for their friends too.

[abc123__]

BUMP! Please help! --------

[Voltage]

alright we can model this problem as a tree

at the top level, we have two branches with the other delayed by 1 minute

this is difficult to draw on here, so i'll get right to it

first the answer: 194 calls
if we describe this as a signal/function,
then y[n] = y[n -1] + y[n - 2], where y[1] = 1 and y[2]=2

this is to say, for any given time integer n, y[n] = the number of additional calls made
t=1, y=1
t=2, y=2
t=3, y=3
t=4, y=5
t=5, y=8
...
t=10, y=89

Summation of y[1:10] = 231

not sure if you are supposed to see a correlation with the fibonnacci sequence or not

i'll take a look at the next one as well..looks like the answer is 6:14 if u continue the sequence

[Voltage]

here's the sample tree i was talking about

         (0)
       /     \  
      1        2
     / \      / \
    2   3    3   4
   /\   /   /
  3  4 4   4 
 /   
4

count at step 1: 1
count at step 2: 2
count at step 3: 3
count at step 4: 5
this process continues to follow the fibonnaci sequence

(0) 1 2 3 5 8 13 21 34 55 89 ...

[thachhao]

Woot! Thanks Voltage! I did realized that the pattern was by Fibonacci numbers! -__-

public class CalculateMinutes {
	public static void main(String[] args) {
		int totalStudents = 1000;
		int studentsCalled = 1; // including Carlin
		int minutes = 0;
		int n_1 = 1;
		int n_2 = 0;
		int fibNum = n_1 + n_2;
		while (studentsCalled <= totalStudents) { // get up to 1000 students
			studentsCalled += fibNum; // add fibonacci number
			// get the next fibonacci number
			n_2 = n_1;
			n_1 = fibNum;
			fibNum = n_1 + n_2;
			minutes++; // increment minutes
			System.out.printf("%d minutes passed, %d students were notified!\n", minutes, studentsCalled);
		}
	}
}