Ya...really then, all it is a "2 walker" solution, with 3rd walker made to fit in:

L / (time-for-2-walkers) .

No.Quote:

>EDIT: Did your program search for cases where M may carry a person backwards?

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- June 16th 2011, 11:16 AMWilmerRe: Motorcycle and Walking Rates and Times
- June 16th 2011, 05:58 PMgodelproofRe: Motorcycle and Walking Rates and Times
see the example in #9. there you must carry B backwards. I suspect even in this case my formular gives correct answer... are you able to modify your program to include this cases, too? If so, can you still "see" my fomular "fits" perfectly what your

**modified**program does? - June 17th 2011, 10:36 PMgodelproofRe: Motorcycle and Walking Rates and Times
Sorry for my claim. It is wrong...

Consider L=1000, A=1, B=999, C=1000, M=1001. Here M>C but unfortunely there's no feasible plan.

Me, too! I didn't expect the general problem to turn out to be so complicated. Why nobody responses? I guess most people just browse the first few posts of the thread and think it's just a matter of computation of a specific problem? Anyway this thread is getting too long... Perhaps it's better to post the generalised problem to some other subforum? - June 18th 2011, 03:01 AMbjhopperRe: Motorcycle and Walking Rates and Times
Hello.

I have followed this post from the beginning.I made calculations to see if I could estimate the time for the walkers to reach the other end.My inital estimate was 4 hours I started by sending M to pick up the slowest walker and to drop him at a point where the time to do the same for the others would keep him from crossing the finish line.Rate of walkers .1 K per minute,.133, .166. Rate of M 2K per minute.It takes M to meet a 57 minutes.They approach each other at 2.1 K per minute. The trip back to the 90 mile mark takes 47 minute. The pick up point was at mile 5.7. Picking up the others took a total of 247 minutes. I can see that the drop offs should have been closer to the end.Final back and forth movements ofM to get walkers in correct positions would take little time.

bjh - June 18th 2011, 04:23 AMgodelproofRe: Motorcycle and Walking Rates and Times
Now to the first question, the answer is NO! See #21 and #22.

To the second question: Go here http://www.mathhelpforum.com/math-he...tml#post660726 - June 18th 2011, 05:28 AMWilmerRe: Motorcycle and Walking Rates and Times
Integer example (2 walkers) where 1 walker is driven backward.

L = 1080 (X to Y), M = 100, A = 20, B = 80

M and B meet after 6 hours.

M drives B backward for 2 hours.

B then walks to Y : 10 hours ; total = 18 hours

After the 1st 8 hours, A has walked 160.

M is now 120 from A: takes 1 hour to meet A.

Distance from Y now 900: takes 9 hours ; total = 18 hours. - June 18th 2011, 06:25 AMgodelproofRe: Motorcycle and Walking Rates and Times
- June 18th 2011, 06:37 AMgodelproofRe: Motorcycle and Walking Rates and Times
- June 18th 2011, 06:56 AMbjhopperRe: Motorcycle and Walking Rates and Times
godelproof you didn't answer my question.

bjh - June 18th 2011, 07:12 AMgodelproofRe: Motorcycle and Walking Rates and Times
- June 18th 2011, 09:45 AMWilmerRe: Motorcycle and Walking Rates and Times
Huh? If Motorcycle speed < fastest walker speed, then ONLY possible case is

to have ALL walkers at fastest walker speed, and the Motorcycle is useless.

That's the only way to reach destination at same time. Agree?

Remark on case where a walker is driven backward:

it is possible to get a solution where a walker is driven back BEHIND (West of) starting point X.

The wording of your problem does not prevent that. - June 18th 2011, 03:26 PMbjhopperRe: Motorcycle and Walking Rates and Times
My question. How many hours have you estimated are required to complete the task. Just give your numbers without lengthy comments.Thanks.

bjh - June 18th 2011, 05:11 PMgodelproofRe: Motorcycle and Walking Rates and Times
a1=.1 a2=.133 a3=.166 m=2

What is L? L=2.1*57=1197/10? - June 18th 2011, 06:44 PMgodelproofRe: Motorcycle and Walking Rates and Times
I don't agree(Smirk)

Say a=100, b=99, c=98 and m=99.9, L=100

what do you say? m first meets a and drive him back to somewhere between b and c, then pick up c to make everybody arrive the same time at destination. Calculations should be similar to that done here http://www.mathhelpforum.com/math-he...tml#post660891 and a feasible plan exists. - June 18th 2011, 09:03 PMWilmerRe: Motorcycle and Walking Rates and Times