Optimal breakdown cost for a fixed weight

[DonAlex]

Dear All,

First of all many thanks for reading this post, and to whoever can help, many thanks for taking the time for responding.

Today I came across a problem at work, which I believe could have a mathematical solution, however I am not smart enough to find it.

Let me explain my problem:

1. I have a product whose price is determined only by the final weight. Let's say that product weights 1000 grams (=1kg).

2. The product is made out of different components whose price is determined by weight ranges. For example:
2.1. Component A: Range 100 to 200 Grams: Cost per kg is 500USD
2.2 Component A: Range 250 to 450 Grams: Cost per kg is 600USD (same component as A, but different cost for different range of weight)
2.2 Component B: Range 300 to 800 Grams: Cost per kg is 700USD

3. I sell my product at a constant price for the TOTAL weight of 1000 grams. Let's say I sell my product for 1.500 USD at 1000 grams.

4. One of the ways in which I can earn money is by smart buying of the components. I can freely chose the mixture of components.

If you have read until here, let me give you an intermediate thank you.

5. The only thing I cannot change is the final weight which is 1000 grams and I always need to use a minimum of 30% of product B and a minimum of 20% of product A of the second range.

5. Option A:
I can buy 150 grams of component A = 0.150 kg x 500 USD = 75 USD
I can buy 350 grams of component A = 0.350 kg x 600 USD = 210 USD
I can buy 500 grams of component B = 0.500 kg x 700 USD = 350 USD
The 3 components together make 1000kg and my cost is 635 USD.

6. Option B:
I can buy 0 grams of component A = 0.000 kg x 500 USD = 0 USD
I can buy 460 grams of component A = 0.460 kg x 600 USD = 276 USD
I can buy 540 grams of component B = 0.540 kg x 700 USD = 378 USD
The 3 components together make 1000kg and my cost is 654 USD.


With option A I save 19 USD. So how can I create a system where I can know what is the optimal mix of components to achieve the cheapest price.

I hope I explained it well enough. Your help is very much appreciated and your time put into this valued.

Thank you so much for reading and helping.

Regards,
Jose

 

[JeffM]

Dear All,

First of all many thanks for reading this post, and to whoever can help, many thanks for taking the time for responding.

Today I came across a problem at work, which I believe could have a mathematical solution, however I am not smart enough to find it.

Let me explain my problem:

1. I have a product whose price is determined only by the final weight. Let's say that product weights 1000 grams (=1kg).

2. The product is made out of different components whose price is determined by weight ranges. For example:
2.1. Component A: Range 100 to 200 Grams: Cost per kg is 500USD
2.2 Component A: Range 250 to 450 Grams: Cost per kg is 600USD (same component as A, but different cost for different range of weight)
2.2 Component B: Range 300 to 800 Grams: Cost per kg is 700USD

3. I sell my product at a constant price for the TOTAL weight of 1000 grams. Let's say I sell my product for 1.500 USD at 1000 grams.

4. One of the ways in which I can earn money is by smart buying of the components. I can freely chose the mixture of components.

If you have read until here, let me give you an intermediate thank you.

5. The only thing I cannot change is the final weight which is 1000 grams and I always need to use a minimum of 30% of product B and a minimum of 20% of product A of the second range.

5. Option A:
I can buy 150 grams of component A = 0.150 kg x 500 USD = 75 USD
I can buy 350 grams of component A = 0.350 kg x 600 USD = 210 USD
I can buy 500 grams of component B = 0.500 kg x 700 USD = 350 USD
The 3 components together make 1000kg and my cost is 635 USD.

6. Option B:
I can buy 0 grams of component A = 0.000 kg x 500 USD = 0 USD
I can buy 460 grams of component A = 0.460 kg x 600 USD = 276 USD
I can buy 540 grams of component B = 0.540 kg x 700 USD = 378 USD
The 3 components together make 1000kg and my cost is 654 USD.


With option A I save 19 USD. So how can I create a system where I can know what is the optimal mix of components to achieve the cheapest price.

I hope I explained it well enough. Your help is very much appreciated and your time put into this valued.

Thank you so much for reading and helping.

Regards,
Jose

This general type of problem can be solved by a mathematical technique called linear programming. Linear programming problems that are very simple can be solved relatively easily by hand, but for a business, it is probably best to buy a software package and let it solve such problems for you.

However, this problem as formulated does not need linear programming. Component A can be bought in two different price ranges. Simply buy at the cheaper price, using multiple dealers if necessary. Your minimum cost is:

700 gm of A @ $500/kg = $350
300 gm of B @ $700/kg = $210
Total cost of 1 kilogram = $560, cheaper than either of your options.

That is the lowest cost you can achieve. You cannot use less of B, and every additional gram of B that you substitute for A adds net cost.

I'd also suggest switching to a new supplier: few will demand a higher unit price for a larger quantity sold.

Moreover, if you really are stuck paying a higher unit price for larger quantities of A, it is still cheaper than B so minimize B and maximize A at the cheaper price. Basically, this requires no advanced math, just common sense and arithmetic.

200 gm of A @ $500/kg = $100
500 gm of A @ $600/kg = $300
300 gm of B @ $700/kg = $210
Total cost of 1 kilogram = $610, still cheaper than either of your options.

EDIT: Notice that the constraint of at least 20% of the weight coming from the higher priced version of A is ignored in my first solution, but the 30% minimum for B is addressed. In the second solution, both the minimum for B and the maximum order quantity for cheaper A are explicitly addressed. The 20% minimum for expensive A is not explictly addressed: it is fulfilled automatically. Thanks denis for pointing out the need for clarification.