For the story below, has there been a mathematical analysis performed
on this to find out the relationships between all the variables, to
find out when is the right time to sharpen the axe?

A man was walking in a forest one day, and he met a woodcutter. It was
a hot day, and he sat down for a smoke and engaged in friendly banter
with the woodcutter - things about the weather and such. 5 Marlboros
later, he asked, "Mr. Woodcutter, you've been making little progress
on chopping down the tree the last half hour. Perhaps your axe is too
blunt, why don't you sharpen it?"

"That's the truth! The axe has not been sharpened for a long while
now, I could make far better progress with a sharp axe!" said the
woodcutter.

"Then why not you take some time to sharpen it now? You'd make good
progress!" said the man.

"I don't really have time for that, you know. I gotta chop down all
these trees, I don't have the luxury of sharpening my axe," replied
the woodcutter.

If s is the time to sharpen the axe, t is the difference in time to
cut down a tree with a dull axe versus a sharp axe and n is the number
of trees cut down, then sharpen the axe whenever n*t>s.

the axe gets progressively duller with cutting trees - it degrades.

Oh boy - a continuous function!

The woodcutter story can be likened to an electrical cicruit.  

You basically have a capacitor here (the blade) that progressively
dulls (discharges).  Your blade can also only be so sharp (analogous
to your capacitor can only be so big).

A sharp blade will dull faster than a dull blade, which is also
exactly how a capacitor works, in the sense that a fully charged
capacitor will discharge faster than a 1/2 full capacitor.

There is one potential area where the analogy does not work, and is
where I need some clarification.  How long does it take the woodcutter
to sharpen his axe?  Assuming this is time S, as berkely used, then my
next question is whether this time is constant.  For instance, if the
woodcutter only chops down one tree, does it still take him "S"
minutes to sharpen his axe?  What about after thirty trees?  Common
sense would tell us that an almost sharp blade can be fully sharpened
faster than a extremely dull blade, but that may not be the case for
the purposes of this story.

Let's assume for now that the sharpening time is constant regardless
of blade condition.

Next, if we're going to continue with the electical analogy, then we
need to establish some groundrules.

First, I want to portray that a given amount of "axe-power" or charge
is required to cut down the tree.  This number is always constant.  In
real life, this implies that all the trees are the same size.  In the
electrical analogy, this implies you're using something like a
rudimentary camera flash, something that uses the same amount of
charge each time it fires.

Secondly, I want to portray that "axe-power", as I am defining it, is
force of blow times a sharpness value.  Our lumberjack, being the
brawny guy that he is, clearly does not get tired between blows, and
all his blows have the same force.  So, the only thing changing is the
sharpness of the axe.  Ultimately (at t = infinity), the axe reaches a
minimum sharpness, which means that no matter how hard you hit the
tree, it will take forever to chop it down.  Again, this is akin to
capacitor discharge, where at t=inf, charge remaining in the capacitor
equals zero.

Thirdly, for the purposes of this problem, I'm going to assume dulling
is a continuous function.  I know that in the real world, dulling only
happens when the axe actually hits the tree.  But in this case, I'd
like imagine the axe as being more like a chainsaw, so there are no
"hacks".  Instead, the axe degrades continuously, not in quantized
chunks.

So, if that's all the case, then blade sharpness is equal to...

A*e^-t where A is an arbitrary constant and t is in units of time.

"axe power" at time t = F*A*e^-t where F is the force exerted by lumberjack/axe.

Q = time to cut a tree = B * [1-e^-t] where B is an constant that
takes axe material, tree size, tree type, lumberjack force, etc. into
account.  You don't really have to worry about what B equals; we'll
treat it as an arbitrary constant for now.  B is a very large number,
because as t approaches infinity, time to cut a tree approaches B.

So, as long as Q <= S, it takes longer to sharpen the axe as it does
to cut down the next tree.  However, this is not really an accurate
account of when he should sharpen the axe.  After all, if he could
sharpen the axe and cut down three trees in the time it takes to not
sharpen the axe and only cut two trees, then he should obviously
sharpen the axe.

So, let's look at it this way we'll quantize it by tree and calculate
the time to cut down tree number R.

TIME = T = C*e^R where C is another arbitraty constant that takes all
the same stuff into account that B does.

Tree number 1...TIME = C*2.7
Tree number 2...TIME = C*7.4
Tree number 3...TIME = C*20.1...and so on

The final outcome suggests that if  S + T(1) < T(R) then the
lumberjack should definitely sharpen his axe.


So, let's assume S = 15 and C = 1.


T(1) = 2.7
S+T(1) = 17.7
T(2) = 7.4
T(3) = 20.1

This means that the lumberjack should sharpen his axe after tree number 2!

Does this make sense?  Let's see...assume he doesn't sharpen his axe at all...

T(1) = 2.7 -- total = 2.7
T(2) = 7.4 -- total = 10.1
T(3) = 20.1 -- total = 30.2
T(4) = 54.6 -- total = 84.8
T(5) = 148.4 -- total = 233.2
T(6) = 403.4 -- total = 636.6
T(7) = 1096.6 -- total = 1733.2

Now assume he sharpens his axe after tree number 2

T(1) = 2.7 -- total = 2.7
T(2) = 7.4 -- total = 10.5
SHARP = 15 -- total = 25.5
T(3) = 2.7 -- total = 28.2
T(4) = 7.4 -- total = 35.6
SHARP = 15 -- total = 50.6
T(5) = 2.7 -- total = 53.3
T(6) = 7.4 -- total = 60.7
SHARP = 15 -- total = 75.7
T(7) = 2.7 -- total = 78.4

As a sanity check, assume he sharpens it after tree 1.

T(1) = 2.7 -- total = 2.7
SHARP = 15 -- total = 17.7
T(2) = 2.7 -- total = 20.4
SHARP = 15 -- total = 35.4
...and so on...
T(7) = 2.7 -- total = 108.9

Another sanity check -- assume he sharpens after tree number 3.

T(1) = 2.7 -- total = 2.7
T(2) = 7.4 -- total = 10.1
T(3) = 20.1 -- total = 30.2
SHARP = 15 -- total = 45.2
T(4) = 2.7 -- total = 47.9
T(5) = 7.4 -- total = 55.3
T(6) = 20.1 -- total = 75.4
SHARP = 15 -- total = 90.4
T(7) = 2.7 -- total = 93.1

Clearly, sharpening after tree number 2, using the conditions
provided, is the best choice.

So there you have it.

Just to complicate the problem, in case Toufaroo hasn't complicated
the answer enough:  There are woodcutters and woodcutters.  A bad one
wastes a lot of chops that are ineffectual, and dulls his axe, while a
really good one makes each blow count.  And then there are trees and
trees:  nice 20 inch diameter poplars (soft wood), and 40 inch oaks.
I don't think the answer can be narrowed down to sharpening after the second tree.

I like Berkeley's formula, hoping the lumberjack is good enough   - or
his trees small enough -  that he doesn't have to deal with fractions.

It's not a simple problem folks!  I know, I complicated it WAY more
than needs be, but it's pretty hard to solve this one (correctly)
without using a continuous function.

Having said that...I may have made a slight mistake.  As I was driving
into work this morning, I passed a patch of trees and thought to
myself...do'h!  Yesterday, as I was typing up the response, I got
called away to a meeting.  I returned and continued typing, but
post-meeting, I used the wrong function.  It seems I used the function
for continuous interest, like what you'd have in an ideal savings
account instead of continuous discharge, like what you have in a
capacitor.  Whoops; I guess my fingers got ahead of my mind there...

OK, so go ahead and disregard everything after...

"So, let's look at it this way we'll quantize it by tree and calculate
the time to cut down tree number R."

I'm gonna start right before that statement and hopefully clear things up!

OK, so if we quantize by tree, then 

Q = time to cut a tree = B * [1-e^-(x*R)] where R = tree number and x
is some number, less than 1, which we'll use just to make the math a
bit cleaner.  You don't really need this x value, but if you don't
have it, rounding errors will be more likely.  I'll explain more on
this later.

As I said before, B needs to be a sufficiently large number such that
when the part in brackets approaches 1, meaning once the woodcutter is
better off using the back of the axe because the blade is so dull, it
will take a very long time to cut down the tree.

So, just for argument's sake, let's say B = 100 and x = 0.1.  The
actual value of B makes no difference, as it is factored through.  x
defines how fast the blade dulls.  It takes myoarin's comments into
account regarding there being pine trees vs maple; A cold-tempered
steel axe vs. a lead axe...

Q = 100 [1 - e^-(x*R)]

Q(tree 1) = 9.52
Q(2) = 18.1
Q(3) = 25.9
Q(4) = 32.9
Q(5) = 39.3
Q(6) = 45.1
Q(7) = 50.3
Q(8) = 55.1

you'll notice, of course, that the difference between trees 7 and 8 is
smaller than the difference between trees 1 and 2.  This is because of
what I discussed in my original post regarding how a sharp blade
degrades faster than a dull blade.

As to when it is optimal to sharpen the blade, that does depend on your x value.

Here's a spreadsheet that calculates the optimal time to sharpen the
blade based on an x factor (will have to be solved from initial
conditions; that is to say, we have to know how long it takes the
lumberjack to cut down the first tree with a brand new axe) and based
on the time required to sharpen the blade (assumed as a constant
value)

Plug these values into the BLUE boxes, and the optimal sharpening
point will be in the YELLOW box.  It's that simple!

http://www.4shared.com/dir/743663/4fcf225f/sharing.html

and the file is called lumberjack.xls

Be warned that the spreadsheet only allows you to sharpen the blade up
to after tree number 10, as I actually do have to get back to work. 
Also, there is a known bug in it which is a result of Excel limiting
me to 8 IF statements...if the optimal tree shows up as "0", it is
probably tree 9 or 10.


After figuring this whole thing out, here's my recommendation:  go buy a chainsaw.

-toufaroo

Toufaroo,
You are indefatigable, and maybe right.  If you are, I can understand
the woodcutter's logic.  ;-)

By the way, chain saws also dull  - and then are more dangerous than a
dull ax.  But one can replace the chain, if one has an extra one, and
very fast, as I saw last night on a TV show about a Finnish woodsmen
competition  - 15 seconds.
But I guess that is cheating.  The woodcutter could have a second axe,
and the oldtimers had a double-bitted one, but I have forgotten
advantage of that, not just to have a second blade, they were
different.

In a separate calculation, we need to know the optimum sharpening. 
Razor sharp obviously takes more time, but probably isn't that much
more advantageous, since the razor edge will quickly be dulled and the
time used to achieve it wasted.

I have great respect for your calculations, not that I can follow them.
I wonder if in the practical end, the answer gets close to Berkeley's?

Cheers, Myoarin

The traditional solution is for the woodcutter to take on an
aprentice, whos job it is to...
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sharpen the ax! ;-)