Re: Quiz / How Many Inserts ?

On Wed, 16 Jul 2008 10:20:27 -0700, BottleBob <bottlbob@xxxxxxxxxxxxx>
wrote:



JRWheels wrote:
Here's a nice story problem to figure out if you want to.

You want to quote a job that needs a facemilling operation done to it.
How much will the cost of the inserts be?
You have all of the material & labor costs figured out and from
running similar jobs you know what kind of tool life to expect and now
you want to estimate how much cost you will have in purchasing
facemill inserts.

Here is what you know about the job, the cutter and the tool life.
The cutter takes 6 inserts.
You get 150 parts per edge
Cost per Insert is $7.50
The job has 10,000 parts for you to run.
Each insert has 4 edges.

How much will the cost of the inserts be for the entire job?

Nothing tricky or hidden, simple story problem.

Good Luck,


Jim:

My first assumption is that when you stated the you get 150 parts
per edge, that you're meaning 150 part PER INSERT EDGE, and not 150
parts per SET of 6 INSERT EDGES. If I have made an incorrect
assumption there, all that follows after that is in error.

10,000 parts / 150 parts per edge = 66.66666 edges.

66.66666 edges / 4 edges per insert = 16.66666 inserts.

Now since the inserts can only be bought in multiples of 4 edges per
insert, you'd have to buy 17 inserts (4 X 17 = 68 edges)

17 inserts X $7.50 = $127.50

BUT, the cutter body is loaded in multiples of 6, the next higher
multiple of 6 would be 18, so you'd have to load a total of 18 inserts
to complete the job. 18 inserts X $7.50 = $135



Or from another perspective: 6 inserts per cutter X 4 edges = 24
edges per full cutter load.

24 edges X 150 edges per part = 3,600 parts.

10,000 parts / 3,600 part = 2.7777 full cutter loads. Since you
can't have a partial cutter load without overloading some of the
other inserts and probably lowering the 150 parts per edge estimate,
you would have to use the next higher full cutter load. Which would
be 3, so we have 3 X 6 insert = 18 inserts = $135


Looks like the OP can be interpreted different ways.

If 150 parts per edge means you get 150 parts before you need to index
(all) the inserts in the holder:

150 Parts Per Edge X 4 Edges = 600 parts where all 6 inserts need to
be replaced.

6 inserts * $7.5 per = $45.00

$45.00 / 600 = $.075 tooling cost per part

10000 * $.075 = $750.00

Depending upon how you read and interpret the OP you may wish to round
up to 17 holder changes * $45.00 = $765.00

+++++++++++++++++++=

Or if 150 parts per edge means 150 parts per edge x 4 edges x 6
inserts in holder, it yields 3,600 parts before all inserts in the
holder need to be changed.

6 inserts * $7.5 per = $45.00

$45.00 / 3600 = $.0125 tooling cost per part

10000 * $.0125 = $125.00

Depending upon how you read and interpret the OP you may wish to round
up to 3 holder changes, 3 * $45.00 = $135.00

Tom
.

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