Question

there is a reservoir, containing 100L of water. the water consumption is 1 liter at day 1 and doubles each 10 day. In how many days the reservoir will be depleted?

Answer 1

I usually avoid these kinds of question....

Answer 2

this is the usual economic problem - continues growth and finite resources. I have stumbled once upon a great video considering the problem:
http://www.youtube.com/playlist?list=PL6A1FD147A45EF50D

Answer 3

i assume it means that 1 liter a day is used and without any use; the volume double every 10 days?

Answer 4

a{0} = 100
a{1} = 100 - 10(1) + 200
a{2} = a{1} - 10(1) + 200 .... maybe?

Answer 5

Think it's 1 for 9 days, 2 for the next (9?)..

Answer 6

reading comprhension lol ... this belongs in the english section

Answer 7

sry...
day 1-10 - consumption: 1L a day
day 11-20 - consumption: 2L a day
day 21-30 - consumption: 4L a day
day 31-40 - consumption: 8L a day
etc

Answer 8

10+20+40 = 70 in 30 days

8 * d = 30; d = 30/8 = 3.7 ...

Answer 9

thanks all. 33.7 seems to be a good answer. What I'm trying to find is the mathematical way to calculate the expiration time.
There is an article on wiki, but I cannot figure it out how the calculation there works.

so, I had imaged the 100L of water as if it would be a chess board.
chess board has 64 squares, so I thought, if I "take" the squares from the chess board, each time doubling the amount of squares taken, then it would be almost the same as the problem with 100L of water.

for the chess board if I "take" the squares and each time the number of taken squares would be doubled it would take 6 steps to "take" all the squares from the chess board.
The solution was kindly provided by amistre64:
2^x=64
ln(2^x)=ln(64)
x(ln(2))=ln(64)
x=ln(64)/ln2
x=5.95 (6 steps)
http://openstudy.com/groups/mathematics/updates/4e2b00ee0b8b3d38d3b9d955
in this case 2 represents the growth rate


in case of 100l water the growth rate needs to be calculated. what we have is an example of exponential growth: doubling time - 10 days, doubling rate - (obvious) 2. the growth rate then can be calculated as following ( http://en.wikipedia.org/wiki/Doubling_time):
10=ln2*100/x
x=ln2*100/10
x=7%

then I put the growth rate into the formula above:
1.07^x=100
ln(1.07^x)=ln(100)
x(ln(1.07))=ln(100)
x=ln(100)/ln(1.07)
x=4.605/0,068
x=60

so according to this calculation it would take 60 days untill the reservouir of 100 would be depleted.

does anyone has an idie where my calculation is incorrect?

Answer 10

since this is not a "continuous" math problem; you have to delve into discrete math techniques i believe

Answer 11

and youd have rules and conditions to meet the criteria in the problem .... maybe

Answer 12

the issue with trying to solve this with a continuous math strategy is hard for me to explain at the moment. its an issue with introducing an error from the "stairstepping" to the "smooth curving" between the 2.

if i were going to program this; id set up 2 dependant variables; one for the amount of water in the tank after each day, and another that determines the doubling every 10 days....

Answer 13

W = 100; day=0

for (m=1 to 100)
for(n=1 to 10)
day = day + 1
daily consumption = m*n
W = W - daily consumption
if (W >= 0){n=10; m=100}
next n
next m

display(day)

Answer 14

even thats messed up since I dont double it every 10 days ..... still got a bug

Answer 15

I don't know why it took me so long to understand how expiration time works. see http://en.wikipedia.org/wiki/Zero_growth

Expiration time:
ln(kR/r0 + 1)
Te= ----------------
k

k - growth rate
R - Resource reserve
r0-current or starting consumption, for this matter R/r0 gives the number of days the 100L of water last if the current rate of consumption does not change (100 days)

I have checked following configurations:
1)
R 100L
k 7% (see doubling time calulation)
r0 1L (we start with 1 L consumption per day)
Te= 29,7 (still some calculations error, see excel check)
Excel check:
consumption L left day
1,07 2,07 2
1,1449 3,2149 3
1,225043 4,439943 4
1,31079601 5,75073901 5
1,402551731 7,153290741 6
1,500730352 8,654021093 7
1,605781476 10,25980257 8
1,71818618 11,97798875 9
1,838459212 13,81644796 10
1,967151357 15,78359932 11
2,104851952 17,88845127 12
2,252191589 20,14064286 13
2,409845 22,55048786 14
2,57853415 25,12902201 15
2,759031541 27,88805355 16
2,952163749 30,8402173 17
3,158815211 33,99903251 18
3,379932276 37,37896479 19
3,616527535 40,99549232 20
3,869684462 44,86517678 21
4,140562375 49,00573916 22
4,430401741 53,4361409 23
4,740529863 58,17667076 24
5,072366953 63,24903772 25
5,42743264 68,67647036 26
5,807352925 74,48382328 27
6,21386763 80,69769091 28
6,648838364 87,34652927 29
7,114257049 94,46078632 30
7,612255043 102,0730414 31


2)
R 100L
k 7% (see doubling time calculation)
r0 10L (we start with 10 L consumption per day)
Te= 7,58 (should be correct see excel check)

Excel check:
consumption L left day
10,7 20,7 2
11,449 32,149 3
12,25043 44,39943 4
13,1079601 57,5073901 5
14,02551731 71,53290741 6
15,00730352 86,54021093 7
16,05781476 102,5980257 8