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Extra Credit Problem #2
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Taedrin
Federal Navy Academy
Gallente Federation
#1 - 2011-11-01 14:46:04 UTC  |  Edited by: Taedrin
(Extra Credit Problem #1 can be found here)

This is actually the first Extra Credit Problem which I was given by my friend's friend. This is a very simple problem requiring nothing more than your basic high school algebra.


Quote:
Bob needs to go on a quick shopping trip. Unfortunately, his hometown - city A - does not have the store he wants to shop at, so he needs to drive quite some distance. The closest store is in city B. He leaves city A and travels to city B at a speed of 30. When he arrives at city B, he quickly does his shopping (presume the amount of time spent shopping can be ignored). Excited that he is now the proud owner of the latest technological gadget, he rushes home at a speed of 60(mph or kph - it doesn't matter so long as you are consistent).

What is Bob's average speed for his entire shopping trip? Prove your answer.



Here are the important facts:
1) Bob travels at a speed of 30 from City A to City B. He travels at a speed of 60 from City B to City A.
2) Units of speed do not matter so long as you are consistent. I.e. if Bob lives in the United States (or the UK also, from what I understand?), he uses mph. If he lives anywhere else in the world, he uses kph.
3) Bob does not take a different route home. I.e. he travels on the same roads, just in the other direction for his return trip. Any deviations due to traffic laws can be presumed to be negligible and can be ignored for the purposes of this calculation. This really is just a simple word problem!

If this one gets solved right away, then I can give another one right away tomorrow.

EDIT: Fail quote.
Jark Vohaul
Goosefleet
Gooseflock Featheration
#2 - 2011-11-01 14:51:56 UTC
No, I will not do your homework.
Zagam
Caldari Provisions
Caldari State
#3 - 2011-11-01 14:55:12 UTC
My high-school age stepsister can do this one. In her sleep.

Are you having so much of an issue doing your homework that you have to come to a gaming forum to ask?
Taedrin
Federal Navy Academy
Gallente Federation
#4 - 2011-11-01 14:57:15 UTC
Jark Vohaul wrote:
No, I will not do your homework.


Unfortunately, my homework isn't as simple as this.
Jark Vohaul
Goosefleet
Gooseflock Featheration
#5 - 2011-11-01 15:21:40 UTC  |  Edited by: Jark Vohaul
From his other thread:
Taedrin wrote:
I am merely reciprocating the pain and suffering that my Calc professor has inflicted upon me by tricking me into joining a mathematics competition on November 5th.

Well, good luck with that. You shouldn't make it so easy for people to catch you in a lie.
Taedrin
Federal Navy Academy
Gallente Federation
#6 - 2011-11-01 15:27:45 UTC  |  Edited by: Taedrin
Jark Vohaul wrote:
From his other thread:
Taedrin wrote:
I am merely reciprocating the pain and suffering that my Calc professor has inflicted upon me by tricking me into joining a mathematics competition on November 5th.

Well, good luck with that. You shouldn't make it so easy for people to catch you in a lie.


???? I am somewhat confused how this implies I am lying?

EDIT:

Seriously, how does this imply that I am lying? Are you implying that because I am giving a simple math problem requiring nothing more than simple high school algebra that I can't possibly attempt to enter a mathematics competition (which has no requirements to enter, btw)?

Tell you what - you seem to think that I don't know how to do this. If you really want me to, I can give you the answer and the proof to this question and give the next problem right away.

Or you can give the answer, and I will prove you right or prove you wrong.

Or if you are trolling then whatever I say/offer really makes no difference.
Taedrin
Federal Navy Academy
Gallente Federation
#7 - 2011-11-01 17:40:56 UTC
Unless the trolls request for the solution sooner, or if there is no interest in this problem, I will post the solution tomorrow.

If there is no interest in the problem, I will post the solution in about 6-9 hours.
Karl Planck
Perkone
Caldari State
#8 - 2011-11-01 17:42:21 UTC
yay

i have been waiting for this

I has all the eve inactivity
Karl Planck
Perkone
Caldari State
#9 - 2011-11-01 17:56:06 UTC  |  Edited by: Karl Planck
awwwww sad panda. This is too easy

Weighted average for this velocity problem:

v_avg = (v1*t1 + v2*t2)/(t1+t2)

you can relate the times by inspection or by mathematically. The same distance is traveled both ways. Thus it will take twice as long to get there as it did to get back (t1 = 2t2). This is also easily derived from v1*t1 = v2*t2.

Using this v_avg becomes

v_avg = (v1*(2*t2) + v2*t2)/(3*t2) = (2 v1 + v2)/3 = 40

*edit: this makes intuitive sense because you know the answer has to be between 30 and 45 (30 being the average if the same speed the entire trip and 45 if the times were the same but the speeds were different). You know he spent longer going 30mph, so 40 is a reasonable answer.

I has all the eve inactivity
Taedrin
Federal Navy Academy
Gallente Federation
#10 - 2011-11-01 18:09:11 UTC  |  Edited by: Taedrin
Karl Planck wrote:
awwwww sad panda. This is too easy

Weighted average for this velocity problem:

v_avg = (v1*t1 + v2*t2)/(t1+t2)

you can relate the times by inspection or by mathematically. The same distance is traveled both ways. Thus it will take twice as long to get there as it did to get back (t1 = 2t2). This is also easily derived from v1*t1 = v2*t2.

Using this v_avg becomes

v_avg = (v1*(2*t2) + v2*t2)/(3*t2) = (2 v1 + v2)/3 = 40


This is the correct answer. It's not quite as complete as I would have liked. It would have been nice if you had formally shown the correctness of the weighted average formula you used, though.

EDIT:
While you are here, I could give you the next problem, which should be more interesting.

EDIT2: What I mean by formally showing the correctness:

definition of average speed: total distance / total time
d = v1 * t1
d = v2 * t2
As the problem states we travel the same distance twice, we have:
total distance = 2*d = d + d = v1*t1 + v2 * t2.
total time spent traveling = t1 + t2

Therefore:
v_avg = (v1*t1 + v2*t2)/(t1 + t2)

And then continue from there.
Karl Planck
Perkone
Caldari State
#11 - 2011-11-01 18:18:09 UTC
Taedrin wrote:
Karl Planck wrote:
awwwww sad panda. This is too easy

Weighted average for this velocity problem:

v_avg = (v1*t1 + v2*t2)/(t1+t2)

you can relate the times by inspection or by mathematically. The same distance is traveled both ways. Thus it will take twice as long to get there as it did to get back (t1 = 2t2). This is also easily derived from v1*t1 = v2*t2.

Using this v_avg becomes

v_avg = (v1*(2*t2) + v2*t2)/(3*t2) = (2 v1 + v2)/3 = 40


This is the correct answer. It's not quite as complete as I would have liked. It would have been nice if you had formally shown the correctness of the weighted average formula you used, though.

EDIT:
While you are here, I could give you the next problem, which should be more interesting.

EDIT2: What I mean by formally showing the correctness:

definition of average speed: total distance / total time
d = v1 * t1
d = v2 * t2
As the problem states we travel the same distance twice, we have:
total distance = 2*d = d + d = v1*t1 + v2 * t2.
total time spent traveling = t1 + t2

Therefore:
v_avg = (v1*t1 + v2*t2)/(t1 + t2)

And then continue from there.


lol meh, weighted average is a well known formula, no need to reinvent the wheel

I has all the eve inactivity
Taedrin
Federal Navy Academy
Gallente Federation
#12 - 2011-11-01 18:22:23 UTC  |  Edited by: Taedrin
Karl Planck wrote:


lol meh, weighted average is a well known formula, no need to reinvent the wheel


For the next problem, I have two choices to give you. I have the solution for one, but not the other.

The first one (which I have a solution to) is a logic problem, somewhat related to the Assignment Problem.

The second one (which I do NOT have a solution to) is from Game Theory.

EDIT: I can't type.
Karl Planck
Perkone
Caldari State
#13 - 2011-11-01 18:32:22 UTC
Taedrin wrote:
Karl Planck wrote:


lol meh, weighted average is a well known formula, no need to reinvent the wheel


For the next problem, I have two choices to give you. I have the solution for one, but not the other.

The first one (which I have a solution to) is a logic problem, somewhat related to the Assignment Problem.

The second one (which I do NOT have a solution to) is from Game Theory.

EDIT: I can't type.


I am more specialized in logic, but i don't really care either way, just doing it for fun ;)

I has all the eve inactivity
Taedrin
Federal Navy Academy
Gallente Federation
#14 - 2011-11-01 18:39:27 UTC  |  Edited by: Taedrin
The logic one it is then:

Consider the following matrix
11 10 25 19 16
24 17 12 15 3
13 5 14 2 18
23 4 1 8 22
6 20 7 21 9

Because the forums are terribad for mathematical figures, here is the Latex'd matrix in .gif format:
Copy and paste, don't click:
http://latex.codecogs.com/gif.latex?\left[%20\begin{array}{ccccc}%2011%20&%2010%20&%2025%20&%2019%20&%2016%20\\%2024%20&%2017%20&%2012%20&%2015%20&%203%20\\%2013%20&%205%20&%2014%20&%202%20&%2018%20\\%2023%20&%204%20&%201%20&%208%20&%2022%20\\%206%20&%2020%20&%207%20&%2021%20&%209%20\end{array}%20\right]

Pick 5 elements, such that:
1) no 2 elements share the same column or row
2) the smallest element in your choice is as large as possible.

Prove that there is no better choice.

(/me crosses fingers that links work with strange symbols in them)

EDIT: Why the hell are the forums randomly copying portions of my post when I edit it?
Karl Planck
Perkone
Caldari State
#15 - 2011-11-01 18:44:25 UTC
oh my thats a doozy

I has all the eve inactivity
Karl Planck
Perkone
Caldari State
#16 - 2011-11-01 18:53:09 UTC
If i got your matrix right

11 17 14 8 9
10 12 2 22 24
25 15 18 23 20
19 3 12 4 7
16 24 5 1 21
11 3 2 1 20
10 24 18 8 7
25 17 12 22 21
19 12 5 23 9
16 15 14 4 6

is the list of numbers that satisfy the first requirement (diagonal "strips" of the matrix)

In the above, line 3 is the line that has the highest low number of 15. So the numbers that satisfy are

25, 15, 28, 23, 20

This is a much more direct proof, a more general one is probably possible with some linear algebra tricks which i am too rusty with.

I has all the eve inactivity
Karl Planck
Perkone
Caldari State
#17 - 2011-11-01 18:54:11 UTC
lol, i remember when homework used to be fun like this. I rarely get fun stuff anymore.

I has all the eve inactivity
Taedrin
Federal Navy Academy
Gallente Federation
#18 - 2011-11-01 19:00:01 UTC  |  Edited by: Taedrin
I'm sorry to say, but that answer isn't correct Sad Let me double check to make sure I did the matrix correctly.

EDIT:
the matrix is correct. NOW it's correct.

The problem is that you incorrectly assume that if no two elements share the same row or column that they must form a 'diagonal strip'

EXAMPLE:
11, 4, 12, 2, 9

satisfies the first restraint, but is NOT a diagonal strip.

EDIT2: The matrix is slightly wrong, the 12 in the first column is supposed to be a 13.
EDIT3: The matrix is fixed now
Karl Planck
Perkone
Caldari State
#19 - 2011-11-01 19:14:27 UTC
not going to be as easy as I had thought. We'll see how this goes.

I has all the eve inactivity
Karl Planck
Perkone
Caldari State
#20 - 2011-11-01 20:31:52 UTC
Sigh, my method is a bit lame so I am not a bit fan of this, but I am pretty sure this is correct.

23, 17, 25, 21, 18

You can start from column one and work over (you could do it by rows on the same procedure). You can pretty much assume that any number less than 10 is a no go, and as I showed earlier, the largest small number is 15 (so it should be higher than 15). You start with the highest number you can, 24 and work to the right hitting the highest number you can, not doubling up on rows.

24,20,25,8,…

So no, that kills off your 24 possibility, because everything less than 25 in c3<15 and everything that you can hit in c2 <15. So, move onto 23

23,20,25,15,18 This ties the highest low number on my first attempt. Changing it up

23,17,25,21,18

This has to be it, because you cannot choose anything lower than 23 on c1, and you cannot go lower than 17 on c2. Since 17 is the limiting number, this HAS to be the correct ensemble.


I has all the eve inactivity
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