Here we go with a similar Facemill Insert quiz:
Same Facemill with 6 Inserts.
Same Inserts with 4 edge.
Same job with 10,000 parts to be cut.

Same Insert cost of $7.50 each.
But the Operator says Tool Life is 60 Minutes per Edge
Total Cutting Time for Each Part is 30 Seconds.
What is the Total Cost of Inserts used on this job?
I can't wait to see all of the correct answers come flowing out of
this one. I have confidence in all of you.
Good Luck,
JR

Why do I think im back in arithmitic class in grade school?
2 pc. per minute or 120 pcs per hour..
insert set for tool will last 4 hours using the four sides of the insert
or 480 pcs per each set of 6 inserts. ( if you don't drop any in the
chips.... :) )
you will need 21 sets of six inserts or 126 inserts at 7.50 each for
a total price of 945 dollars.
now here comes the real question... JR comes into your shop and sees
you running the job and says " I have a new insert for that material
that will give you 20 percent more life than the one you are using. The
insert costs 15 percent more than the one you are using but this one
will put more cash in your pocket."
Considering a billing rate of 60 dollars per hour and a down time of
five minutes for each insert change or buying another face mill for 400
bucks so the downtime for insert changes will be 0, how much will you
save and is it worth buying the second face mill?
John

John,
Great question, again you must do the math to see where your doillars
are going because I wouldn't trust just any tooling guy either.
By getting 20% more tool life that would be 72 minutes of cutting
time. 144 part per edge. 576 per load of 6 inserts.
10,000 parts divided by 576 rounds up to 18 Loads, times 6 inserts 108 inserts. Total cost now is 108 * ($7.50*1.15) = 931.50 which is a
savings of $13.50.
With 3 less loads saving 15 minutes at $1 per minute that would be an
additional $15 adding up to $28.50 more in your pocket.
It would not justify buying the $400 facemill as 18 loads times 5
minutes is only 90 minutes =$90 plus the $28.50 from above totaling
$118.50 would put you in the hole $281.50.
However, instead of achieving more tool life with the new more costly
insert, lets increase the SFM & Chipload enough to get the same tool
life and reduce the cycletime by 3 seconds. That would be a savings of
500 minutes = $500 in labor, take that money and buy the additional
$400 facemill that will reduce your overall throughput time by an
additional 5 minutes per 21 load =$105 plus the $100 left over that
you didn't spend while buying the facemill so you would have a $205
cost saving, a new facemill and the next job will go into the machine
10 hours (500 minutes + 105 minutes) ahead of schedule enabling you to
quote faster deliveries to kick your competitions ass without having
to buy additional machines due to now having open capacity. Cost
avoidance of a new mill would be from $50,000 to $300,000 depending on
the mill of course. Which should show once again tooling is a small
factor compared to other costs a shop owner deals with, mainly labor.
Although these are fun, hypothetical situations, getting more parts
per edge is what a Tooling Salesman will try to do. Getting more
performance, saving you thousands of dollars week is what a good
technical partner that you buy tools from can do for you.
Thanks for the quiz sequel. That was fun.
JR.
(original amount) =126 inserts at $7.50*1.15=$1,086.75

The actual time to flip the inserts as well as changing them is 10,000
divided by 120 or 84 times (rounded off from 83.333) 84 x 5 minutes is
420 minutes which would be well worth buying another cutter as well as
the having an instant backup cutter if an "oops " happens. Hopefully
you will still have two good face mills for the next job coming down the
road. Also what real price can you put on having backup tooling. I
think of it as buying insurance.

If I were in this situation I woulr probably buy the second face mill
and load it with the new inserts and run it alternately with the other
one and compare the inserts. ( you can never completely trust what the
salesman says :) ) Also talk the salesman out of a free pack of
inserts if you buy a new face mill but don't tell him you were going to
buy the face mill anyway.
John

On Wed, 16 Jul 2008 17:49:55 -0700 (PDT), JRWheels

30 sec. cycle time = 2 PPM
60 min. per edge x 4 edges = 240 min
240 min x 2 PPM = 480 parts per insert replacement
6 inserts per replacement x $7.5 each = $45.00
$45.00 / 480 parts = $.09375 tooling cost per part
10,000 * $.09375 = $937.50
$937.50 is the total cost of inserts (wear) for this job which is a
direct answer to the question. This is what we "used" on this job, the
unused portion is returned to stock to be used on other jobs.
++++++++++++++++++++++++++++++++++++++
However if we over think the question then:
10,000 / 480 parts per insert replacement = 20.833333, round up to 21
insert changes x 6 inserts per change = 160 inserts x $7.5 = $945.00
Tom

Not unused but not worn out ether. In the previous question he asked
how many inserts we need to buy. IIRC, in this question he asked how
much we use. We did not "USE" them all up, we only used a portion of
group 21.
21 sets = 480 * 21 = 10,080 parts
Picking nits but this is a face mill, standard inserts, so you still
can run the equivalent of 80 parts which is equal to nearly 3/4 of an
hour of production before you need to change the inserts. Don't know
many shops that throw inserts out before they are used up. I am not
discounting that there is a point of diminishing returns but again it
wasn't part of the OP.
There is value still in the inserts that haven't been used up. What if
there was 200 or 400 parts wear left do you use them on the next job
or throw them out?
When talking high production shop or production job shop $0.001 can
lose you a large contract so picking nits is required.
If we were talking specially ground inserts for a non-repeating job
then the cost "IS" $945.00. However for specially ground inserts, I
would want a little insurance and spend $990.00 and buy 22 sets
because 21 sets is cutting it a little too close IMO.
Tom

Jim:
10k parts @ .5 min. each = 5,000 min. for total job.
5k min. / 60 min. for one set of 6 insert edges = 83.333 total sets of
edges. 83.333 / 4 edges = 20.833 sets of inserts. Rounded up to 21
sets of inserts. 21 X 6 = 126 Inserts 126 X $7.50 = $945.00
Now just don't tell me you meant 60 minutes for EACH single insert
edge. LOL

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