Quiz / How Many Inserts ?

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, JR

Reply to
JRWheels
Loading thread data ...

(saying that the insert has 4 edges)

150 x 4 = 600 parts per insert 10,000 / 150 = 67 sets of inserts 67 x 6 = 402 inserts total 402 x 7.5 = $3,015 in inserts total

is my guess..

Reply to
tnik

10,000/150 = change sides 67 times 67/4 = each insert location of the facemill sees 17 inserts during the job 17*6 = 102 total number of inserts 102*$7.5 = $765

can you sell 102 scrap inserts?

DanP

Reply to
DanP

oops.. thats wrong.. but I guess if we can get the customer to pay the extra, we just made some good profit..

should be 10,000 / 600 = 17 sets

17 x 6 = 102 inserts 102 x 7.5 = $765 in inserts total
Reply to
tnik

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
Reply to
BottleBob

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

Reply to
brewertr

Tom:

I love Jim's little quizzes. And while he insists they are simple story problems with nothing tricky or hidden, they are fraught with little pitfalls of interpretation/extrapolation (whether intentional or not I have no idea), but it makes them similar to the problems we all deal with in real life itself.

I think you've pretty much summed up the answers given so far.

Reply to
BottleBob

Bob, Although I really am not TRYING to sneak some tricky stuff in, The Real World Machine Shop is based around a few basic fundamentals, one being math and another being communication.

To this day I PRINT everything except my name. Why? So anyone reading what I am putting down on paper can understand it compared to a Doctors Script for medicine. Clear communication can clear up the math equation. My Dad printed everything. Why? He was a Machinist Mate in the Navy in WWII. Printing was common for clear concise communication. .

If I laid out the parameters of this quiz in the sequence the equation works, it would have been too simple. So I mixed up the facts a bit. (OK I confess to that but that's all I did on the covert side)

But this is not a hard quiz. But sometimes we read too much into simple problems.

Congrats to the guys who gave the answer as $765 Cutting Tool cost. Derived from 102 total inserts used,( but not fully consumed though).

Look at the correct answers above and see how they were calculated differently but still made sense in the end. This quiz made you realize "Grouping of Inserts" due to several inserts being in the facemill at one time.

Total amount of parts divided by parts per index gets you how many indexes took place.

10,000 / 150 =3D 67 (rounded up from 66.6667) But that is considering 6 inserts get indexed that many times. 4 edges per insert will produce 600 parts per each load of inserts. Every 4 indexes, 6 new inserts go in to make another 600 parts. HMMM 6 inserts per 600 parts. 100 inserts will make 10,000 parts at that rate. Close, but not accurate enough.

You can see the math by the guys who nailed above.

Here's another mentality to figure it out.

Explanation by part # in sequence of run. Parts 1-150 uses the 1st edge of the 1st load of inserts Parts 151-300 uses the 2nd edge of the 1st load of inserts Parts 301-450 uses the 3rd edge of the 1st load of inserts Parts 451-600 uses the 4th edge of the 1st load of inserts

6 inserts consumed

follow the pattern.....

Parts 5851-6000 uses the 4th edge of the 10th load of inserts

60 inserts consumed

Parts 8851-9000 uses the 4th edge of the 15th load of inserts

90 Inserts consumed

Parts 9451-9600 uses the 4th edge of the 16th load of Inserts

96 inserts consumed

Parts 9601-9750 uses the 1st edge of the 17th load of inserts Parts 9751-9900 uses the 2nd edge of the 17th load of inserts Parts 9901-10,150 uses the 3rd edge of the 17th load of inserts

102 Inserts would have been installed into the facemill throughout 17 loads.

102 x $7.50 =3D $765.

The point is even though you may want to use a Computer, a Calculator or an Abacus, if you know the factors and work it several ways to prove the numbers you should feel confident that your calculations are correct.

When this math figures into making money or losing money on the shop floor, we all need to know how it can effect our job, our shop and even our country, regarding the success of our decisions.

If you would have bid on the job by estimating your tooling cost was $500, it would appear you would have lost $265 just in insert cost. Plus you would have been waiting for more inserts to by delivered and the machine sitting idle. More loss even than the $265 and delayed delivery, possibly overtime hours to complete it or get caught up on the next job that was delayed. Huge money wasted now. Once time is gone, all you have is overtime.

Want to figure one on Tool life measured by Minutes of Tool Life and time in the cut ? Awww Come on, it's not that hard........More to come.

JR

Reply to
JRWheels

Bad enough to under bid tooling, no reason to compound the error by not tracking production & tooling during the run.

Tom

Reply to
brewertr

Jim:

I completely agree with you there. That's one reason solid models are often better than prints for complex parts.

In this particular case I made an assumption based upon wording that, IMO, was ambiguous.

"You get 150 parts per edge"

To remove any ambiguity, that sentence could have been worded like the following:

You get 150 parts per set of edges.

You get 150 parts per every indexing of inserts.

Sure. But if I reply this time I'll, give all possible answers. LOL

Reply to
BottleBob

Best damn selling technique there is.

Tom

Reply to
brewertr

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.