Linear Optimization: Possibilities

[mathmari]

Hey!

To product a product $3$ square plates $A$ with measures $300\times 500$mm and $2$ square plates $B$ with measures $400\times 600$mm are needed. These plates must be cut out by two different workpieces with measres $1000\times 1000$mm (Type I) and $600\times 1500$mm (Type II). The plates, for technology reason, must be cut through and no just cut in.
Totally $500$ products must be produced. The costs for the workpieces are proportional to their area, i.e., if a workpiece of Type I has the price 1, then a workpiece of Type II has the price $0.9$ ($1000000\ mm^2$ to $900000\ mm^2$).

There are $4$ possibilities, to cut plates A and/or B from a workpiece of Type I:
(1) 4A (scrap $16\%$)
(2) 3A, 1B (scrap $13\%$)
(3) 1A, 2B (scrap $31\%$)
(4) 3B (scrap $28\%$)

There are $3$ possibilities, to cut plates A and/or B from a workpiece of Type II:
(5) 4A ($8\%$ scrap)
(6) 2A, 2B (scrap $0\%$)
(7) 3B (scrap $24\%$)

Try to find these possibilities.

Formulate the problem as a linear optimization problem. Let $x_1, x_2, \ldots , x_7$ be the number of workpieces, that are necessary for the schemes (1), (2), ... , (7). The costs for the workpieces must be minimized.

What does it mean to find the possibilities? What do we have to do? Could you give me a hint? (Wondering)

 

[I like Serena]

Hey!

To product a product $3$ square plates $A$ with measures $300\times 500$mm and $2$ square plates $B$ with measures $400\times 600$mm are needed. These plates must be cut out by two different workpieces with measres $1000\times 1000$mm (Type I) and $600\times 1500$mm (Type II). The plates, for technology reason, must be cut through and no just cut in.
Totally $500$ products must be produced. The costs for the workpieces are proportional to their area, i.e., if a workpiece of Type I has the price 1, then a workpiece of Type II has the price $0.9$ ($1000000\ mm^2$ to $900000\ mm^2$).

There are $4$ possibilities, to cut plates A and/or B from a workpiece of Type I:
(1) 4A (scrap $16\%$)
(2) 3A, 1B (scrap $13\%$)
(3) 1A, 2B (scrap $31\%$)
(4) 3B (scrap $28\%$)

There are $3$ possibilities, to cut plates A and/or B from a workpiece of Type II:
(5) 4A ($8\%$ scrap)
(6) 2A, 2B (scrap $0\%$)
(7) 3B (scrap $24\%$)

Try to find these possibilities.

Formulate the problem as a linear optimization problem. Let $x_1, x_2, \ldots , x_7$ be the number of workpieces, that are necessary for the schemes (1), (2), ... , (7). The costs for the workpieces must be minimized.

What does it mean to find the possibilities? What do we have to do? Could you give me a hint? (Wondering)

Hey mathmari! (Smile)

I think we have to figure out how to cut a workpiece of type I (with only cuts through) into 4 plates of type A with 16% scrap.
Let's see, that might be something like:
\begin{tikzpicture}[ultra thick, >=stealth', shorten >=1pt]
%preamble \usetikzlibrary{arrows,patterns}
\fill[pattern=north west lines, pattern color=gray] (0,9) -- (10,9) -- (10,10) -- (0,10) -- cycle;
\draw (0,3) -- (10,3);
\draw (0,6) -- (10,6);
\draw (0,9) -- (10,9) -- (10,10) -- (0,10) -- cycle;
\draw (5,0) -- (5,10);
\draw[blue] (0,0) -- (10,0) -- (10,10) -- (0,10) -- cycle;
\node[above] at (5,10) {Workpiece of Type I};
\node at (2.5,9.5) {Scrap};
\node at (7.5,9.5) {Scrap};
\node at (2.5,7.5) {Plate A};
\draw[xshift=-5, thin, <->] (0,0) -- node

{300 mm} (0,3);
\draw[xshift=-5, thin, <->] (0,3) -- node

{300 mm} (0,6);
\draw[xshift=-5, thin, <->] (0,6) -- node

{300 mm} (0,9);
\draw[xshift=-5, thin, <->] (0,9) -- node

{100 mm} (0,10);
\draw[yshift=-5, thin, <->] (0,0) -- node[below] {500 mm} (5,0);
\draw[yshift=-5, thin, <->] (5,0) -- node[below] {500 mm}(10,0);
\end{tikzpicture}

Hmm... that's strange... I get 6 plates of type A with 10% scrap...

Do you know what is going on? Could it perhaps be that plate A should be 300 x 700 mm? (Wondering)​

 

[mathmari]

When the plate A is 300 x 700mm we get the following:

View attachment 6408

or not? (Wondering)

Is the scrap now 16% ? (Wondering)

 

[I like Serena]

Can't we fit another 300x700 in the scrap at the right of size 300x1000? (Wondering)

 

[mathmari]

Can't we fit another 300x700 in the scrap at the right of size 300x1000? (Wondering)

Oh yes. So, it must be:

View attachment 6409

or not? (Wondering) So we $\frac{100\times 700+300\times 300}{1000\times 1000}=\frac{7\cdot 10^4+9\cdot 10^4}{10^6}=\frac{16\cdot 10^4}{10^6}=\frac{16}{100}$, i.e., $16\%$ scrap. To formulate the problem as a linear programming problem do we use the 7 schemes for the constraints? And for the objective function, since we want to minimize the cost, do we have to use also the information that the costs for the workpieces are proportional to their area? (Wondering)

 

[I like Serena]

Yes and yes.

How many plates A and B do we need? (Thinking)

 

[mathmari]

How many plates A and B do we need? (Thinking)

For the production of one product we need 3 square plates A. Since we want to produce 500 products we need (at least) $3\cdot 500=1500$ A-plates.

For the production of one product we need 2 square plates B. Since we want to produce 500 products we need (at least) $2\cdot 500=1000$ B-plates.

The A-plates are in the schemes (1), (2), (3), (5), (6). Therefore, the condition is
\begin{equation*}4x_1+3x_2+x_3+4x_5+2x_6\geq 1500\end{equation*}

The B-plates are in the schemes (2), (3), (4), (6), (7). Therefore, the condition is
\begin{equation*}x_2+2x_3+3x_4+2x_6+3x_7\geq 1000\end{equation*} Is everything correct so far? (Wondering) To minimize the cost for the workpieces is equivalent to minimize the scrap? (Wondering)

 

[I like Serena]

Yes and yes.

What would the objective function be? (Wondering)

 

[mathmari]

What would the objective function be? (Wondering)

We have that the costs for the workpieces are proportional to their area, i.e., if a workpiece of Type I has the price $1$, then a workpiece of Type II has the price $0.9$.

So, is the objective function then the following:
\begin{align*}&Z=0.16\cdot 1\cdot x_1+0.13\cdot 1\cdot x_2 +0.31\cdot 1\cdot x_3+0.28\cdot 1\cdot x_4+0.08 \cdot 0.9\cdot x_5+0 \cdot 0.9\cdot x_6+0.24 \cdot 0.9\cdot x_7 \\ &\Rightarrow Z=0.16 x_1+0.13 x_2 +0.31 x_3+0.28 x_4+0.072 x_5+0.216 x_7\end{align*}
? (Wondering)

 

[I like Serena]

Looks good. Does that mean that we're done? (Wondering)

 

[mathmari]

Looks good. Does that mean that we're done? (Wondering)

With that part I think yes. We solve that problem with Simplex, right? (Wondering) At which point did we need the possibilities of the first part of the exercise? (Wondering)

 

[I like Serena]

Simplex yes.
And no, we didn't need to figure out the configurations - they were given.
At best it helps a bit to understand the problem statement. (Nerd)