How about some comments on my camel solution. Can anyone find a better solution? I have no proof that mine is optimal.

Since there have been no posts following up on my hints, I will give my answer to the furniture problem. It is to make all tables at the rate of 54 per hour at an hourly gross revenue of $1080. I last took an Operations Research course 40 years ago so I hope that the rust hasn't caused me to come up with the wrong, optimal answer.

Challenge you all to find a better solution. Is Karmarkar still around ??

regards,

Bill
AKA DADDIO

Yes, there is a better solution.

Expend 1000 pounds of grain moving 9000 lbs of grain 1000/19 miles or 52.63 miles.

Expend another 1000 pounds of grain moving 8000 lbs of grain 1000/17 miles or 58.82 miles.

You have now moved 8000 pounds of grain 111.45 miles.

Expend another 1000 pounds of grain moving 7000 lbs of grain 1000/15 miles or 66.67 miles. You now have 7000 lbs of grain at 178.12 miles.

Expend another 1000 pounds of grain moving 6000 lps of grain 1000/13 miles or 76.92 miles. You now have 6000 lbs of grain at 244.04.

Expend another 1000 pounds of grain moving 5000 lbs of grain 1000/11 miles or 91.91 miles. You now have 5000 lbs of grain at 335.95 miles.

Expend another 1000 pounds of grain moving 4000 lbs of grain 1000/9 miles or 111.11 miles. You now have 4000 lbs of grain at 447.06 miles.

Expend another 1000 pounds of grain moving 3000 lbs of grain 1000/7 miles or 142.86 miles. You now have 3000 lbs of grain at 589.92 miles.

Expend another 1000 pounds of grain moving 2000 lbs of grain 1000/5 miles or 200 miles. You now have 2000 lbs of grain at 789.92 miles.

You now make 3 trips of 211.08 miles. The first and third starting with 1000 lbs, the second with 211.08 lbs. 2000 - 633.24 is 1366.76 lbs delivered to city B.

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BOINC WIKI

Sorry, that's wrong, but you were veeeeery close...

Try again...

---

- Luke.

Sorry, that's wrong, but you were veeeeery close...

Try again...

The previous post on the camel problem caused me to go back and examine my solution. The "physics" of the problem is to always give the camel a maximum load of 1000 lbs (if possible) while not leaving any grain behind. So I would have to simply readjust my mileages ala John McLeod.

Looking back over my solution I left behind 1000 lbs along the way and delivered 666.66 lbs to city B. By choosing optimal distances I should be able to deliver 1666.67 lbs to city B by not leaving any grain along the way. The hitch is that the end of the journey would not allow an optimal 1000lb load on the camel, so the backtracing here would be less effiicent.

It appears that we have an infinite series here that converges to chunks of one third. I bet that 1333.32 lbs delivered to city B is the answer. I don't know the correct mileages along the way but maybe there would be an infinite number of stops to converge on my number in the limit. This would of course require an awful lot of loading and unloading of the camel but time was not a specified burden in the problem statement so this can be ignored.

If this is true then this problem should decompose into a calculus problem with extra care about not having to backtrack on the last load along the way. Tomorrow I might see if I can formulate this correctly. Time for back to bed now here in the central US time zone.

Good problems--helps to clear the rust off of things I should have learned in school.

Regards,

Bill

6. It is your task to deliver as much grain as possible from city A to city B. The cities are 1,000 miles apart. You initially have 10,000 pounds of grain. Your camel may carry up to 1,000 pounds and eats 1 pound of grain per mile traveled. You may leave grain along the way and return to it later. How much grain can you deliver to city B?

It appears that you could deliver 333.28 pounds to city B.
My reasoning is:
10 trips for a distance of 333.33 miles, nine of those would deposit 333.33 pounds of grain and one with 666.66 pounds (since no return trip is required. This would give you a total weight of grain after the first leg of 3666.63 pounds.
The second leg would be 3 trips of another 333.33 miles carrying the full load and depositing 333.33 pounds at the end. The last trip carrys 666.63 pounds of which 333.33 pounds is eaten along the way leaving 333.30 pounds to add to the total giving 1333.29 pounds at the end of the second leg.
The third trip would be special in that if you left the full 333.33 pounds then you wouldn't have enough food to return with the rest as it takes 333.33 pounds of food to travel the distance but you would only have 333.29 pounds available. So you leave 333.29 here and go back for the remaining grain.
The final leg is only .01 mile carrying the entire load and eating .01 pounds of grain leaving 333.28 pounds to deliver at city B.

---

Jim

Some people plan their life out and look back at the wealth they've had.
Others live life day by day and look back at the wealth of experiences and enjoyment they've had.

The official answer is: 1399.77 Miles....
Can anyone now provide the correct working and show why this is the answer? The Person that does as per my last sentence gets 1 Point....

If no one can answer this in the next 24 hours (to.... April 24 20:00 UTC) John McLeod VII gets the point...


Luke.

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- Luke.

The official answer is: 1399.77 Miles....
Can anyone now provide the correct working and show why this is the answer? The Person that does as per my last sentence gets 1 Point....

If no one can answer this in the next 24 hours (to.... April 24 20:00 UTC) John McLeod VII gets the point...


Luke.

I had a minor mistake doing arithmetic in my head.

It is still 2000 - (1000 - 1000/19 - 1000/17 - 1000/15 -1000/13 - 1000/11 - 1000/9 - 1000/7 - 1000/5) * 3.
2000 - (1000 - 799.92219682622159402345160859093) * 3
2000 - 200.0778031737784059765483914091 * 3
2000 - 600.2334095213352179296451742273
1399.7665904786647820703548257727
which rounds to 1399.77 or 1399.8


[edit]

The basic idea is that you move X - 1000 pounds of grain as far as the 1000 pounds of grain will go if split into the needed trips. This gives the reducing terms. You do this until it takes less than 1000 pounds of grain to move the remainder to the destination.

It takes 19 trips to move the grain to the first cache, and 17 to move to the second cache, 15 to the third, 13 to the fourth, 11 to the fifth, 9 to the sixth, 7 to the eight, 5 to the ninth, and from there the destination can be reached with less than 1000 pounds of grain, the remainder of that travel 1000 pounds is moved to the final destination.

The basic idea of my first post holds, it was just an arithmetic error.

---

BOINC WIKI

You call that SIMPLE Maths?! Wow...

---

Account frozen...

The official answer is: 1399.77 Miles....
Can anyone now provide the correct working and show why this is the answer? The Person that does as per my last sentence gets 1 Point....

If no one can answer this in the next 24 hours (to.... April 24 20:00 UTC) John McLeod VII gets the point...


Luke.

I had a minor mistake doing arithmetic in my head.

It is still 2000 - (1000 - 1000/19 - 1000/17 - 1000/15 -1000/13 - 1000/11 - 1000/9 - 1000/7 - 1000/5) * 3.
2000 - (1000 - 799.92219682622159402345160859093) * 3
2000 - 200.0778031737784059765483914091 * 3
2000 - 600.2334095213352179296451742273
1399.7665904786647820703548257727
which rounds to 1399.77 or 1399.8


[edit]

The basic idea is that you move X - 1000 pounds of grain as far as the 1000 pounds of grain will go if split into the needed trips. This gives the reducing terms. You do this until it takes less than 1000 pounds of grain to move the remainder to the destination.

It takes 19 trips to move the grain to the first cache, and 17 to move to the second cache, 15 to the third, 13 to the fourth, 11 to the fifth, 9 to the sixth, 7 to the eight, 5 to the ninth, and from there the destination can be reached with less than 1000 pounds of grain, the remainder of that travel 1000 pounds is moved to the final destination.

The basic idea of my first post holds, it was just an arithmetic error.

Well Done, You've just earned yourself 1 Point....
Standings...

1. WinterKnight - 3 Points
2= Mr. Kevvy - 1 Point
2= John Mcleod VII - 1 Point
4. TBD....

Here are 3 More....

7. Simple Math - On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

8. Algebra required - A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?

9. Calculus required - What is the value of $1, invested for one year at 100% interest, compounded infinitely?

Were are all the other mathematicians? And everyone else have a try!!!

Best Regards,
Luke.

---

- Luke.

9. Calculus required - What is the value of $1, invested for one year at 100% interest, compounded infinitely?

e bucks! Or $2.72 (2.718 rounded up to the nearest cent) for people who like the imperfection of decimals. :^)

---

9. Calculus required - What is the value of $1, invested for one year at 100% interest, compounded infinitely?

e bucks! Or $2.72 (2.718 rounded up to the nearest cent) for people who like the imperfection of decimals. :^)

Well done again, Mr. Kevvy

Answer: $2.72 or e

Standings:
1. Winterknight - 3 Points
2. Mr. Kevvy - 2 Points
3. John Mcleod VII - 1 Point
4. TBD....

Two questions (7 & 8) remain, post your answers...

And anyone with problems of there own, post them here or PM me!!

Best Regards,
Luke.

---

- Luke.

Here are the unanswered questions so far....

7. Simple Math - On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

8. Algebra required - A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?

Have a go!!!

---

- Luke.

Here are the unanswered questions so far....

7. Simple Math - On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

8. Algebra required - A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?

Have a go!!!

It's obvious by inspection that the answer is 12495 to number 7 and .414 to number 8.

Here are the unanswered questions so far....

7. Simple Math - On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

8. Algebra required - A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?

Have a go!!!

It's obvious by inspection that the answer is 12495 to number 7 and .414 to number 8.

I am the bringer of good news, and bad news....

Bad News is that your answer to 7 is wrong...
Good News is that your answer to 8 is right!!!.......

Official answer to Question 8: 0.4142
But 0.414 is close enough....

Standings:
1. WinterKnight - 3 Points
2. Mr. Kevvy - 2 Points
3. John McLeod VII - 1 Point
4. William Rothamel - 1 Point
5. TBD......

Try again on Question 7....
Or can someone else snatch a point...

Luke.

---

- Luke.

Here are the unanswered questions so far....

7. Simple Math - On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person decides to take his share himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?

8. Algebra required - A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?

Have a go!!!

It's obvious by inspection that the answer is 12495 to number 7 and .414 to number 8.

I am the bringer of good news, and bad news....

Bad News is that your answer to 7 is wrong...
Good News is that your answer to 8 is right!!!.......

Official answer to Question 8: 0.4142
But 0.414 is close enough....

Standings:
1. WinterKnight - 3 Points
2. Mr. Kevvy - 2 Points
3. John McLeod VII - 1 Point
4. William Rothamel - 1 Point
5. TBD......

Try again on Question 7....
Or can someone else snatch a point...

Luke.

I claim I am right:

Starting with 12495:
The first man took 2499 coconuts, and the monkey took 1.
This left 12495 - 2500 = 9995 coconuts.
The second man took 1999 coconuts, and the monkey took 1.
This left 9995 - 2000 = 7995 coconuts.
The third man took 1599 coconuts, and the monkey took 1.
This left 7995 - 1600 = 6395 coconuts.
The fourth man took 1279 coconuts and the monkey took 1.
This left 6395 - 1280 = 5115 coconuts.
The fifth man took 1023 coconuts and the monkey took 1
This left 5115 - 1024 = 4091 coconuts.
In the morning, each man got 818 coconuts and the monkey 1 more.

Do you find a lower number as a solution ??

Regards,

Bill


there is another solution at : 28120 and probably others

There is a smaller minimum answer...
And I and 100% sure it is correct....
12,495 is a answer, but it isn't the smallest...

Best Regards,
Luke.

---

- Luke.

There is a smaller minimum answer...
And I and 100% sure it is correct....
12,495 is a answer, but it isn't the smallest...

Best Regards,
Luke.

Luke you are too smart for me---you are right again --how about 3121 coconuts to start with.

There is a smaller minimum answer...
And I and 100% sure it is correct....
12,495 is a answer, but it isn't the smallest...

Best Regards,
Luke.

Luke you are too smart for me---you are right again --how about 3121 coconuts to start with.

Congratz Bill, you have a quick mind to figure out another solution that fast...

Official answer for Q7: 3121 Coconuts....
Standings:
1. WinterKnight - 3 Points
2. Mr. Kevvy - 2 Points
3. William Rothamel - 2 Points
4. John McLeod VII - 1 Point
5. TBD......

Questions 10,11,12 Coming soon....

For now.....

Question:
10 Beta! (Worth 1/2 of a Point)- What digit is mysteriously missing from the number Pi for the first 30+ digits, what number is it, and at what decimal place does it first occur?

Luke.

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- Luke.

i demand one half point just for reading this thread.

---

Founder of BOINC team Objectivists. Oh the humanity! Rational people crunching data!
I did NOT authorize this belly writing!

You can give my half point to the Capitalist 0 appears first in the 32nd decimal place

There is a smaller minimum answer...
And I and 100% sure it is correct....
12,495 is a answer, but it isn't the smallest...

Best Regards,
Luke.

Luke you are too smart for me---you are right again --how about 3121 coconuts to start with.

Congratz Bill, you have a quick mind to figure out another solution that fast...

Official answer for Q7: 3121 Coconuts....
Standings:
1. WinterKnight - 3 Points
2. Mr. Kevvy - 2 Points
3. William Rothamel - 2 Points
4. John McLeod VII - 1 Point
5. TBD......

Questions 10,11,12 Coming soon....

For now.....

Question:
10 Beta! (Worth 1/2 of a Point)- What digit is mysteriously missing from the number Pi for the first 30+ digits, what number is it, and at what decimal place does it first occur?

Luke.

Back on the monkey problem : it might be better to state for sure whether or not upon the next morning they divide the coconuts that remain and whether the monkey is also to get one then as well.