Formula for efficiency improvement 3

[doinel]

I have to produce 2 kind of products in one batch(product "a" and product "b&quot
If I produce only "a", the batch will be finished in T1 hours; If only "b" is produced, the batch wil be finished in T2 hours. But I have to produce "a" anb "b" in the same time.
I'm trying to find out what in the most efficient combination "a" and "b"(how many "a" and how many "b"to place in the batch,(to get a minimal finish time)and what is the optimal "a+b"( between 50 and 100 pcs)
Have any idea if is a formula, or by graph.?
Thank you,
Doinel Blaj

 

[TheTick]

I'm not sure how applicale this is, but this is the standard "combined rate" equation.

T(final)= 1/(1/T1 + 1/T2 + ....+ 1/Tn)

 

[Iskit4iam]

Tick,
I think this equation is for combining paralell events, such as Valve A empties Tank in T1, Valve B empties Tank in T2, how long will both valves take to empty Tank. I can't see how it fits this situation, but just because I can't see it doesn't mean it doesn't.

Doinel,
I think we need to know more about what is driving this inquiry. What determines the number of a and b components you need in a given time period? Is the T1 and T2 figure the run times on a single machine or a series of machines? What size batch are the times for. Can the machine be run without any set up required when changeing from a to b?

 

[doinel]

Thank you very much both of you!(TheTick & Iskit4iam)
This is about making acrylic shells using a laser mashine.The input is variable. Let say the quantity of "a" product is 50 pcs.The machine will finish them in 3.5 hours ( if we process only "a&quot.If only "b" product is 50 pcs., the machine will finish them in 6 hours.But in reality we have to combine "a" and "b" in the same batch and in the same machine.What I have to find out is the maximum input (up to 100 pcs. "a+b&quot and minimum process time.

Doinel

 

[MintJulep]

Doinel,

It seems to me that you haven't given us nearly enough information.

What actually happens in the machine? Is there a single laser that must shine on each part in turn, or are there multiple lasers?

If there is only a single laser, I think the answer is 9.5 hours.

 

[IRstuff]

Am I missing something?

Your problem as stated is a simple linear combination. The minimum time is to run only "a", which takes 7 hrs/100pc to running all "b", which takes 12 hr/100pc and the time is linear between those two points.

TTFN

 

[Iskit4iam]

I think what IRstuff is saying, and I agree, is that there is no answer to your question. Given the information we have any combination of a and b is as efficient as any other.

 

[ivymike]

If you're constrained by the amount of time you can run this machine, then you should produce whichever part has a higher contribution margin, unless something (market conditions, etc) forces you to make the other. If you're forced to make the other, make as few as you can to meet your requirement, then make your high CM part for the rest of the time.

 

[ivymike]

sorry, that should have been "higher CM/(machine hour)"

 

[RSEngineer]

Is there a setup time that varies when you combine these parts? If so, this needs to be taken into account. Otherwise this sounds like the old economics guns vs. butter argument.

If you know this classic example, the idea is that a society can produce x amount of guns or y amount of butter, but when you produce one, you reduce the amount of the other you can produce. For the sake of simplicity, I will assume doinel's earlier statement of production time is constant, meaning the first part of one type takes the same time as the last part and every part in between. The only function left is the value function, meaning the first part is more valuable than the last part. This leaves you with a solvable set of simultaneous equations.

$ = (cost function of x)(x) + (cost function of y)
hours = x/(x parts/hr) + y/(y parts/hr)

The cost function is commonly a step function, like a customer needs 20 of x and will pay $$$ for them, but another customer needs 50 and will pay $ for them, hence the first 20 are worth $$$, 21-70 are worth $ and 71+ are worth 0. As you can see, this becomes a sum function