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Nightmare Math Problem.

Blogs > Sgany
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Sgany
Profile Blog Joined July 2010
United Kingdom790 Posts
Last Edited: 2011-08-21 02:07:12
August 21 2011 02:03 GMT
#1
Ok tonight while I was watching a movie with the family it started buffering as it was being streamed. This made me think what is the number of seconds(to keep it simple) I must pause the video to, until I can watch it without it buffering again. Sounds simple right, well it turns into a nightmare.

The formula I have got are different for nearly every ratio.

A ratio of 10seconds playback(We shall call this V) and 12 seconds Buffering(We will call this B) is completely different than a ratio of 3V: 7B

For 3v: 7b I found out the % ratio which was 1%:2.33333333% and then worked out the critical mass was 53%, but I am unable to work out the critical mass for 10v:12b and other different ratios.

Could anyone help me come up with a global solution for this problem?

NaDa <3, MMA <3, Bisu <3,
Probe1
Profile Blog Joined August 2010
United States17920 Posts
Last Edited: 2011-08-21 14:03:50
August 21 2011 02:04 GMT
#2
It'd help if you quit smoking cigars when you're trying to solve complex formulas.

Edit: <3 I meant no cruelness but looking back I was a bit crass. Thanks for not jumping down my throat for the joke :p
우정호 KT_VIOLET 1988 - 2012 While we are postponing, life speeds by
Sgany
Profile Blog Joined July 2010
United Kingdom790 Posts
August 21 2011 02:07 GMT
#3
On August 21 2011 11:04 Probe1 wrote:
It'd help if you quit smoking cigars when you're trying to solve complex formulas.


Fixed :D Thanks for that sir
NaDa <3, MMA <3, Bisu <3,
cz
Profile Blog Joined August 2007
United States3249 Posts
August 21 2011 02:11 GMT
#4
On August 21 2011 11:07 Sgany wrote:
Show nested quote +
On August 21 2011 11:04 Probe1 wrote:
It'd help if you quit smoking cigars when you're trying to solve complex formulas.


Fixed :D Thanks for that sir


Thanks for what?
Lixler
Profile Joined March 2010
United States265 Posts
Last Edited: 2011-08-21 02:12:24
August 21 2011 02:11 GMT
#5
I too have devoted much thought to this most interesting of problems. Basically, we assume a constant loading rate and assume that this rate is lower than the playing rate, which is always 1 second/1second. In order to pause the least, we need our loaded amount to reach the total at exactly the same moment our watched amount reaches the total. So we need to pre-load Total - Loading rate*playtime. But playtime = total, so we can simplify to total*(1-loading rate). For a video of, say, 10 seconds and a loading rate of .5 seconds/second, we get that we need to load 5 seconds in advance. During those 5 seconds, we will load 2.5 seconds, putting us at 7.5 seconds. During those 2.5 seconds, we will load 1.25 seconds, putting us at 8.75 seconds. And we will asymptotically approach having watched the video, exactly as Zeno would have predicted.
huameng
Profile Blog Joined April 2007
United States1133 Posts
August 21 2011 02:12 GMT
#6
Can you explain what that ratio means? I feel like just finding the sum of an infinite geometric series will get your answer but then again I don't really understand your post.
skating
Sgany
Profile Blog Joined July 2010
United Kingdom790 Posts
August 21 2011 02:12 GMT
#7
On August 21 2011 11:11 cz wrote:
Show nested quote +
On August 21 2011 11:07 Sgany wrote:
On August 21 2011 11:04 Probe1 wrote:
It'd help if you quit smoking cigars when you're trying to solve complex formulas.


Fixed :D Thanks for that sir


Thanks for what?


When I typed 3V: 7B it came up as 3V:7B, he pointed out the error
NaDa <3, MMA <3, Bisu <3,
cz
Profile Blog Joined August 2007
United States3249 Posts
August 21 2011 02:13 GMT
#8
On August 21 2011 11:11 Lixler wrote:
I too have devoted much thought to this most interesting of problems. Basically, we assume a constant loading rate and assume that this rate is lower than the playing rate, which is always 1 second/1second. In order to pause the least, we need our loaded amount to reach the total at exactly the same moment our watched amount reaches the total. So we need to pre-load Total - Loading rate*playtime. But playtime = total, so we can simplify to total*(1-loading rate). For a video of, say, 10 seconds and a loading rate of .5 seconds/second, we get that we need to load 5 seconds in advance. During those 5 seconds, we will load 2.5 seconds, putting us at 7.5 seconds. During those 2.5 seconds, we will load 1.25 seconds, putting us at 8.75 seconds. And we will asymptotically approach having watched the video, exactly as Zeno would have predicted.


Zeno, the guy who works at Best Buy? Never knew he was good about this stuff, even tho he sold me a sweet LG plasma a week ago...
Hypertension
Profile Joined April 2011
United States802 Posts
August 21 2011 02:19 GMT
#9
How long it takes to buffer 1 second = B
Length of video in seconds = V
Waiting time = B*V-V
Buy boots first. Boots good item.
Sgany
Profile Blog Joined July 2010
United Kingdom790 Posts
August 21 2011 02:22 GMT
#10
On August 21 2011 11:19 Hypertension wrote:
How long it takes to buffer 1 second = B
Length of video in seconds = V
Waiting time = B*V-V


That is simple to work out, what I want to know is the time it takes for the video WHILE buffering to reach a point where it is not completely buffered, but I can play it without it having to buffer at all.
NaDa <3, MMA <3, Bisu <3,
Severedevil
Profile Blog Joined April 2009
United States4839 Posts
Last Edited: 2011-08-21 02:33:51
August 21 2011 02:24 GMT
#11
Okay. You know how long the video is supposed to last; we'll call that the 'video time' and denote it by v. You don't know how long the video will take to buffer; we'll call that the 'buffer time', b.)

Now, how long do you have to pause your video and let it pre-buffer? The answer is however long it takes to buffer the full video, minus the amount of time it takes to play the full video, or b - v. If you want that as a percentage of the total buffer time, it's (b - v) / b, which is equal to 1 - (v / b).

You've already found v / b; it's the ratio you calculated. So just subtract it from 1, and you're done. (Multiply by a 100 if you want a percentage.)

So, for 10v:12b, you get 1 - (10/12) = ~17%. (This is the 'critical mass' point. You can start playing the video, and it will finish buffering right before you finish watching it. Assuming the video buffers at a constant rate, which may not be true.)

Also, I'm stealing this as a highly relatable algebra problem for students.
My strategy is to fork people.
sidr
Profile Blog Joined May 2008
United States55 Posts
Last Edited: 2011-08-21 02:31:04
August 21 2011 02:29 GMT
#12
I may be understanding this wrongly, but suppose we have L seconds left in our video and we buffer at a rate or R (R < 1, else no problem) playseconds/second. Then, buffering P (0<P<1) percent of the way and then playing the rest of the video, we want time in playseconds to buffer the rest = (L-PL) / R = L. Thus, L-PL = RL so PL = (1-R) L or P = 1-R. For your case, if R = (10 playseconds) / (12seconds), P = 1/6, i.e. if you let the video buffer for up to 1/6 of the remaining time, then play, you should have uninterrupted playback. I'm unsure exactly what critical mass means in your statement, however, so I may have completely misunderstood the problem

Edit: darn beaten
zEMPd
Profile Joined June 2011
Angola259 Posts
August 21 2011 02:30 GMT
#13
I have come up with the answer 7.
Hypertension
Profile Joined April 2011
United States802 Posts
August 21 2011 02:34 GMT
#14
On August 21 2011 11:22 Sgany wrote:
Show nested quote +
On August 21 2011 11:19 Hypertension wrote:
How long it takes to buffer 1 second = B
Length of video in seconds = V
Waiting time = B*V-V


That is simple to work out, what I want to know is the time it takes for the video WHILE buffering to reach a point where it is not completely buffered, but I can play it without it having to buffer at all.



That is the waiting time, if you wait B*V-V seconds, you can start watching safely, assuming your internet has zero hiccups.
Buy boots first. Boots good item.
Sgany
Profile Blog Joined July 2010
United Kingdom790 Posts
Last Edited: 2011-08-21 02:43:28
August 21 2011 02:35 GMT
#15
On August 21 2011 11:24 Severedevil wrote:
Okay. You know how long the video is supposed to last; we'll call that the 'video time' and denote it by v. You don't know how long the video will take to buffer; we'll call that the 'buffer time', b.)

Now, how long do you have to pause your video and let it pre-buffer? The answer is however long it takes to buffer the full video, minus the amount of time it takes to play the full video, or b - v. If you want that as a percentage of the total buffer time, it's (b - v) / b, which is equal to 1 - (v / b).

You've already found v / b; it's the ratio you calculated. So just subtract it from 1, and you're done. (Multiply by a 100 if you want a percentage.)

So, for 10v:12b, you get 1 - (10/12) = ~17%.

Also, I'm stealing this as a highly relatable algebra problem for students.


I love you and the other guys who mentioned this <3 It is a lot simplier if you do not use odd B to V ratios like 3v: 7b and ratios like 1v: 7b instead thank you sir :D
NaDa <3, MMA <3, Bisu <3,
ghrur
Profile Blog Joined May 2009
United States3786 Posts
August 21 2011 02:35 GMT
#16
On August 21 2011 11:24 Severedevil wrote:
Okay. You know how long the video is supposed to last; we'll call that the 'video time' and denote it by v. You don't know how long the video will take to buffer; we'll call that the 'buffer time', b.)

Now, how long do you have to pause your video and let it pre-buffer? The answer is however long it takes to buffer the full video, minus the amount of time it takes to play the full video, or b - v. If you want that as a percentage of the total buffer time, it's (b - v) / b, which is equal to 1 - (v / b).

You've already found v / b; it's the ratio you calculated. So just subtract it from 1, and you're done. (Multiply by a 100 if you want a percentage.)

So, for 10v:12b, you get 1 - (10/12) = ~17%. (This is the 'critical mass' point. You can start playing the video, and it will finish buffering right before you finish watching it. Assuming the video buffers at a constant rate, which may not be true.)

Also, I'm stealing this as a highly relatable algebra problem for students.


What this guy said. It's basically a geometric series.
darkness overpowering
itachisan
Profile Blog Joined June 2011
Canada109 Posts
August 21 2011 03:08 GMT
#17
nerds

User was warned for this post
SarR
Profile Joined June 2011
476 Posts
August 21 2011 05:41 GMT
#18
On August 21 2011 12:08 itachisan wrote:
nerds

User was warned for this post

Paraphrase:-
"Watching all these people talk about stuff that I cant understand makes me feel like a dumbass and left out because it reminds me that I'm intellectually inadequate which renders me unable to participate in intelligent discussions. In order to salvage some pride, I'm going to attempt to make them feel like they are the losers by labeling them using a word which traditionally carries such connotations thereby making me feel superior by virtue of being intellectually inadequate. In other words I get to be "cool" by not knowing shit"
itachisan
Profile Blog Joined June 2011
Canada109 Posts
August 21 2011 15:52 GMT
#19
On August 21 2011 14:41 SarR wrote:
Show nested quote +
On August 21 2011 12:08 itachisan wrote:
nerds

User was warned for this post

Paraphrase:-
"Watching all these people talk about stuff that I cant understand makes me feel like a dumbass and left out because it reminds me that I'm intellectually inadequate which renders me unable to participate in intelligent discussions. In order to salvage some pride, I'm going to attempt to make them feel like they are the losers by labeling them using a word which traditionally carries such connotations thereby making me feel superior by virtue of being intellectually inadequate. In other words I get to be "cool" by not knowing shit"


^ nerd.

User was temp banned for this post.
xxpack09
Profile Blog Joined September 2010
United States2160 Posts
August 21 2011 16:10 GMT
#20
On August 22 2011 00:52 itachisan wrote:
Show nested quote +
On August 21 2011 14:41 SarR wrote:
On August 21 2011 12:08 itachisan wrote:
nerds

User was warned for this post

Paraphrase:-
"Watching all these people talk about stuff that I cant understand makes me feel like a dumbass and left out because it reminds me that I'm intellectually inadequate which renders me unable to participate in intelligent discussions. In order to salvage some pride, I'm going to attempt to make them feel like they are the losers by labeling them using a word which traditionally carries such connotations thereby making me feel superior by virtue of being intellectually inadequate. In other words I get to be "cool" by not knowing shit"


^ nerd.


rofl

gets warned then immidiately makes the same post again

Is there a reason you agressively lash out at people who have different interests than you?
LordOfDabu
Profile Blog Joined December 2003
United States394 Posts
August 21 2011 18:12 GMT
#21
So let's make the problem a bit harder since I think the current one has been solved.

Let's say that a video has length T and that at time t of the video, the video/buffer ratio is given by r(t). (Above we did the case when r(t) is constant for all t). Say, for example, the video is T=10 seconds, and that the ratio is given by r(t) = (t+1)/11, 0 < t < T. What is the minimum time that you will remain paused throughout the video?

Not sure if this makes sense, but I believe a simple answer can still be given (using Calculus), provided r(t) isn't too messy. Curious to see if anyone gets an answer that agrees with mine.

My answer: + Show Spoiler +
I think it's just (int_0^T 1/r(t) dt) - T
Think fast. Click faster.
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