• The TUGBBS forums are completely free and open to the public and exist as the absolute best place for owners to get help and advice about their timeshares for more than 27 years!

    Join tens of thousands of other owners just like you here to get any and all Timeshare questions answered!
  • TUG has now saved timeshare owners more than $17,000,000 dollars just by finding us in time to rescind a new Timeshare purchase! A truly incredible milestone!

    Read more here: TUG saves owners more than $17 Million dollars
  • Follow the TUG Member Banner as it travels the world on vacation with Timeshare owners! Also sign up to get the banner sent to you so you can submit a photo of your vacation with the banner to share with TUG! Banner Thread
  • Sign up to get the TUG Newsletter for free! Join tens of thousands of other owners who get this every week! Latest resort reviews and the most important topics discussed by owners during the week!
  • Our official "end my sales presentation early" T-shirts are available again! Also come with the option for a free membership extension with purchase to offset the cost!

    Read more Here
  • A few of the most common links here on the forums for newbies and guests!

Any math people out there?

summervaca

TUG Member
Joined
Jun 12, 2008
Messages
239
Reaction score
0
Points
226
Location
Minneapolis
I am trying to help my son with a math project and math is not my strong suit (that is an understatement.) If anyone is interested in weighing in on this on a Sunday afternoon, I'd appreciate some help.

The problem is to figure out the least amount of time it would take using a phone chain to notify all the members in a 55 person chain.

Questions:

What are some assumptions you might need to make about phone calls?

Could you modify the procedure to shorten the amount of time involved? If so, how?

So far, we have assumed 30 seconds per phone call. Also, that after the first call, there would be two people making a call - and so on. We sort of mapped it out, but wonder if we are missing any assumptions.

If this is not appropriate in the eyes of mods, please remove. Otherwise, maybe some of you will just look at this as a puzzle!

Full Disclosure: I am a special ed reading and language teacher at my son's school so I have no plans to just do his homework for him. Tuggers are always just so clever and we are a little bamboozled so I thought I'd just toss it out there. Thanks!

Debbie
 

sstamm

TUG Member
Joined
Jun 6, 2005
Messages
431
Reaction score
2
Points
16
Location
Maryland
For one, what is the age level of the students? Obviously older students would look at the problem with greater depth.

One variable my 12 yr old son would mention: who is making the phone calls? For most, 30 seconds is probably plenty. My son thinks I cannot do anything without extra chatting, so he'd have to allot me extra time. (Sometimes he is right about that, and sometimes not!!)

Good luck!! :)
 

summervaca

TUG Member
Joined
Jun 12, 2008
Messages
239
Reaction score
0
Points
226
Location
Minneapolis
That's funny about the chatting ;) . That would be me too. I'm surprised my son didn't mention that!

The students are in 6th grade, but it is an accelerated class and is actually quite in depth. I've just laid out the basics here.

Thanks for the response. All others are appreciated:D
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
If we assume that each person calls two people, then the depth of the tree is something like 5. (1 person calls 2 people, for a total of 3 people, then those 2 people call 4 more people for a total of 7 people, until you get to 16 people calling 32 more people, for a total of more than 55 people.)

I'm not sure you can make it any faster, assuming you can only call one person at a time.

-David
 
Last edited:

dioxide45

TUG Review Crew: Expert
TUG Member
Joined
May 20, 2006
Messages
34,249
Reaction score
7,717
Points
899
Location
NE Florida
Resorts Owned
Marriott's Grande Vista
Marriott's Harbour Lake
SVV - Bella
SVV - Key West
If we assume that each person calls two people, then the depth of the tree is something like 5. (1 person calls 2 people, for a total of 3 people, then those 2 people call 4 more people for a total of 7 people, until you get to 16 people calling 32 more people, for a total of more than 55 people.)

I'm not sure you can make it any faster, assuming you can only call one person at a time.

-David
Once the first person calls two people, there are now three people to make the second set of calls.
 

Big Matt

TUG Review Crew: Veteran
TUG Member
Joined
Jun 6, 2005
Messages
5,730
Reaction score
1,185
Points
449
Location
Northern Virginia
That's the key to the puzzle.

You continue to call others once you've made your initial call in the tree. The same is true with every caller. As soon as you've called the first person, you hang up and call another.

The tree doesn't expand by doubling each time. You add one person each time. In fact the limiting variable is time per call, and you can only be on the phone with one person at a time.

You can do the tree in ten levels or five minutes at 30 seconds per call.

Once the first person calls two people, there are now three people to make the second set of calls.
 

Don40

Tug Review Crew: Rookie
TUG Member
Joined
Sep 11, 2005
Messages
291
Reaction score
13
Points
228
Location
Tampa
One big assumption that we are making is that everyone will answer the phone.

The best way to reach all 55 is by e-mail or text message, which can be done in about 30 seconds.
 

summervaca

TUG Member
Joined
Jun 12, 2008
Messages
239
Reaction score
0
Points
226
Location
Minneapolis
Ahhhh! Thank you Don, and all of you. This has actually turned out to be interesting instead of frustrating for both of us.
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
Once the first person calls two people, there are now three people to make the second set of calls.
We'll just let you make all the calls. :) A phone tree doesn't work that way.

Whatever. I shouldn't have bothered with this.
 
Last edited:

PeterS

TUG Member
Joined
Jun 6, 2005
Messages
266
Reaction score
48
Points
238
A little late but...

To evaluate this you must standardize the call time... 30 seconds is fine.

The key item is whether you require the callers to keep calling...

If so, it effectively doubles the speed... here is how....

Most peoples idea of a phone tree... call two people and they each call two people... etc... but after you place your two calls, you stop. Effective for sharing the burden (only 2 calls per person) but not very effective in getting information diseminated quickly.

This means that the number of callers doubles every 1 minute (time to place two calls) 1 caller, then 2 then 4,8,16,32,64..... but every 1 minute as each person must call 2 people.

In emergency situations, the caller doesn't stop...
If the callers continue to participate then it doubles every 30 seconds (time to place one call) 1 caller and 30 seconds later and now there are 2 callers (1 old and 1 new) the 2 callers make one call (30 seconds) and you have 4 callers (2 old and 2 new) so this follows the same doubling but at twice the speed....

Hope it helps...

Pete
 

Twinkstarr

TUG Member
Joined
Jul 12, 2007
Messages
7,269
Reaction score
0
Points
36
Location
Ohio
More interesting than the poster project for Laos for my 6th grader.

Gotta go and try to get more of the spray adhesive off my fingers.:annoyed:
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
I couldn't even begin to figure out the algorithm of who calls whom if people start calling other people after they make their first 2 calls. I'm sure that maybe somebody could, but I'm not even going to try. I don't think that was the nature of the problem, and frankly, I'm not convinced that it changes the end result.

The depth of the tree for a binary algorithm is only 5. With each person calling 2 callers, that means that everybody is notified within 5 minutes. (2 calls each at 30 seconds each).

Now, if you really want to screw it up, what happens if somebody doesn't answer their phone? It's really not worth going there for an academic problem. What if somebody is living off the grid and doesn't have a phone? There are all kinds of things you could throw at this to make it more complicated than it really is.

-David
 
Last edited:

Sandy VDH

TUG Review Crew: Elite
TUG Member
Joined
Jun 6, 2005
Messages
8,386
Reaction score
2,770
Points
498
Location
Houston, TX
Resorts Owned
Wynd VIP Plat, HGVC Elite, HICV, + few others
.... or you outsource the calling to India, and everyone gets a call but has difficulting understanding the caller.:shrug:

Only kidding.....

I am a day late and a $ short, so I had to add comic relief.

You definitely have to make assumptions about people having the right numbers to call and correct caller data, no outages in the phone systems, and avg or max time for each call, and the number of times one individual can place calls.


I have a BMath Degree from the University of Waterloo.
Yes BMath, NOT BSc in Math.:wall:
http://en.wikipedia.org/wiki/Bachelor_of_Mathematics
 

Seti

Guest
Joined
Apr 18, 2007
Messages
27
Reaction score
0
Points
1
Also late, but is there definitely a rule that you only call 2 people? What happens if you call 3 at each level, or 4?
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
Also late, but is there definitely a rule that you only call 2 people? What happens if you call 3 at each level, or 4?
It takes more time to call more people at each level.

If 1 person calls 4 people, that takes 2 minutes instead of 1. Then those 4 people call 16 people, taking another 2 minutes, for a total of 4 minutes and 21 people. Then those 16 people call the remaining people taking another 1 - 2 minutes. So, where's the gain?

-David
 
Last edited:

T_R_Oglodyte

TUG Lifetime Member
Joined
Jun 6, 2005
Messages
13,605
Reaction score
4,600
Points
648
Location
Belly-View, WA
Wouldn't it be faster to just ask your kid to do it? He or she quickly composes and sends a text message to all 55 people. Takes about 30 seconds that way.

The downside is that within two minutes every one of the people has responded to everyone else, and for those people who don't have unlimited texting, the monthly cell phone bill starts skyrocketing.
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
Steve, maybe you can figure out the formula for this.

ok, so if we say that each person continues to make calls, after they make their first call, you end up with a directed graph that starts out like this:

http://www.flex.com/~dmk/phone-tree.pdf

p respresents a person as in p1 .. p55.
tn represents time period n.
I'll leave it as an exercise to the reader to fill in the rest of the directed graph.

In each time period, each person can call 1 other person.

In the first time period, 1 person calls 1 person, in total time 1t.

In the second time period, 2 people can call 2 more people, in total time 2t.

In the third time period, 4 people can call 4 more people, in total time 3t.

In the fourth time period, 8 people can call 8 more people, in total time 4t.

In the fifth time period, 16 people can call 16 more people, in total time 5t.

In the sixth time period, 32 people can call 32 more people, in total time 6t.

Since 64 > 55, the total time is 6t or 3 minutes, if t = 30 seconds.

In the earlier example, where each person called 2 people and then didn't make any more calls, the depth of the tree is 5, but each time period is doubled because it takes 2 time periods to call 2 people. So it ends up taking 5 minutes for the last person to get called. So it isn't double or half. Yes, it is shorter if people keep making calls, but I'd hate to be the one to figure out who calls whom.

so in time period t, you can call a total of 2^t people. so if you solve for the next power of 2 greater than the number of people, you have your answer I guess. Basically p = 2^t if I did this right. I think this my third shot at the forumla.

I don't remember how to solve for t. I think it's a logarithm. Steve?

-David
 
Last edited:

Art4th

TUG Member
Joined
Sep 23, 2005
Messages
520
Reaction score
2
Points
228
The problem is to figure out the least amount of time it would take using a phone chain to notify all the members in a 55 person chain.
Debbie
A chain is different than a tree. In a phone chain, each person calls one other, and so on and so on and so on...
 

summervaca

TUG Member
Joined
Jun 12, 2008
Messages
239
Reaction score
0
Points
226
Location
Minneapolis
Well, my son handed his project in.

He thought it was great reading your responses. Everytime he checked, he would report how many more people answered! He wrote up an entire report using alot of your ideas. They really got him thinking and made for good conversations, which after all, is the point of this whole process. Oh, and by the way, he told me that the TUG people are "way better at math than I am";)

As Allen Cole would say, so it goes...

Prego!
 

Elan

TUG Member
Joined
Jun 6, 2005
Messages
4,425
Reaction score
392
Points
318
Location
Idaho
so in time period t, you can call a total of 2^t people. so if you solve for the next power of 2 greater than the number of people, you have your answer I guess. Basically p = 2^t if I did this right. I think this my third shot at the forumla.

I don't remember how to solve for t. I think it's a logarithm. Steve?

-David
I'm not Steve, and have never heard of phone trees, but to solve
p=2^t for t, one takes the logarithm of both sides:

ln(p) = t * ln(2)

then divide both sides by ln(2)

ln(p)/ln(2) = t .
 
Last edited:

MULTIZ321

TUG Member
Joined
Jun 6, 2005
Messages
25,237
Reaction score
6,675
Points
898
Location
FT. LAUDERDALE, FL
Resorts Owned
BLUEWATER BY SPINNAKER HHI
ROYAL HOLIDAY CLUB RHC (POINTS)
Well, my son handed his project in.

He thought it was great reading your responses. Everytime he checked, he would report how many more people answered! He wrote up an entire report using alot of your ideas. They really got him thinking and made for good conversations, which after all, is the point of this whole process. Oh, and by the way, he told me that the TUG people are "way better at math than I am";)

As Allen Cole would say, so it goes...

Prego!
Summervaca,

Thanks for a good laugh.

Elan, Thanks for the logarithm solution. Now to just whip out the old slide rule and compute the logarithmic answers.

Richard
 

Icarus

TUG Member
Joined
Jun 6, 2005
Messages
4,095
Reaction score
0
Points
271
I'm not Steve, and have never heard of phone trees, but to solve
p=2^t for t, one takes the logarithm of both sides:

ln(p) = t * ln(2)

then divide both sides by ln(2)

ln(p)/ln(2) = t .
yeah, that seems to work, but you have to roundup the result to the next whole number. Thanks.

-David
 
Top