A paper mill manufactures its paper in standard reels 6 metres wide and cuts these to the widths required by its customers. This may result in some waste, which must be discarded as there is no room to store it. It has received the following orders:

  • 2.7m wide, 30 reels ordered
  • 2.1m wide, 150 reels ordered
  • 1.65m wide, 65 reels ordered

My working out;

To begin this problem I wanted to find out how many different options there were to cut the reels in order to minimise wastage. Ways denoted by pi, Width (in metres), Wastage

p1 3*1.65=4.95, wastage = 1.05

p2 2*1.65 + 2.1=5.4 wastage= 0.6

p3 1.65 + 2*2.1=5.85 wastage= 0.15

p4 2.1 + 2.7=4.8 wastage= 1.2

p5 2*2.7=5.4 wastage= 0.6

p6 2.7 + 2*1.65=6 wastage= 0

Decision variables:

Let x1 be the number of times we use p1, x2 be the number of times we use p2, x3 be the number of times we use p3 and so on.

Constraints: I don't think my constraints are right? please can some one help me

$3x_1 + 2x_2+ x_3 + 2x_6 = 65$ (to satisfy order of 1.65m)

$x_2 + 2x_3 + x_4 = 150$ (to satisfy order of 2.1m)

$x_4 + 2x_5 + x_6 = 30$ (to satisfy order of 2.7m)

$x_i \geq 0 \ \ \forall \ i=1,..,6$

Objective function:

We want to minimise waste so;

$$Min \ \ Z= 1.05x_1 + 0.6x_2 + 0.15x_3 + 1.2x_4 + 0.6 x_5$$

To solve this problem I used the Solver function on Excel.


Use the variables of the options you´ve introduced: $p_1,p_2,p_3,p_4,p_6$

The we want to minimize the waste:

$$\texttt{Min} \ = 1.05\cdot p_1 + 0.6\cdot p_2 + 0.15\cdot p_3 + 1.2\cdot p_4 + 0.6\cdot p_5$$

I just changed the notation from $x_i$ to $p_i$.

I summerize your options in a table:

$$\begin{array}{|c|c|c|c|} \hline & 2.7 & 2.1 & 1.65 & \texttt{waste} \\ \hline p_1 &0&0&3&1.05 \\ \hline p_2&0&1&2&0.6 \\ \hline p_3 &0 &2&1 &0.15 \\ \hline p_4 &1&1&0&1.2 \\ \hline p_5&2&0&0&0.6 \\ \hline p_6& 1&0&2& 0 \\ \hline \end{array}$$

Now we have the constraint that we need at least $30$ reels with a width of $2.7 \ m$. Just read off the coefficient of the column which is lables as $2.7$

$1p_4+2p_5+p_6\geq 30$

Similar for the constraints of reels with the widths $2.1$ and $1.65$

$p_2+2p_3+1p_4\geq 150$

$ 3p_1+2p_2+1p_3+2p_6\geq 65 $

Finally the domain of the variables: $p_i\in \mathbb N_0 \ \forall i=1,2,...,6$

  • $\begingroup$ Note that $p_i$ are defined as integers. $\endgroup$ – callculus Oct 29 '18 at 22:02
  • $\begingroup$ i did the constraints as equal and got 90 reels needed but a wastage of like 30m, but if you do it with a greater than or equal to constraint you get a wastage of 11.25m but need 105 reels? obviously 90 reels is less than 105 but what is the correct solution? @callculus $\endgroup$ – Charlotte Sacks Oct 29 '18 at 22:37
  • $\begingroup$ Give me 45 min to calculate it with excel. $\endgroup$ – callculus Oct 29 '18 at 22:45
  • $\begingroup$ i also need to do the solution if the person ordering the 2.1 m reels were to increase their order by 10 and by 20 reels. also if i do it with equal constraints it doesn't work for 170, excel can't find a feasible solution but works for 160. If you use greater than it solves or parts of q but its a lot higher than 90? @callculus $\endgroup$ – Charlotte Sacks Oct 29 '18 at 22:50
  • $\begingroup$ @CharlotteSacks $p_6$ has no waste. It might be a good idea to act like it has a little waste. Since it does matter how many of $p_6$ we make. The waste is not the only criteria. I set $w_6=0.01$The solution of the original problem is $(p_1,p_2,p_3,p_4,p_5,p_6)=(0,0,75,0,0,30)$ The total waste is indeed 11.25 and the LHS of the constraints are $(30,150,135)$. $\endgroup$ – callculus Oct 29 '18 at 23:40

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