The document summarizes the work of a group consulting a manufacturing firm on optimizing the arrangement of cutting discs from sheets of aluminum. The group models different layouts of arranging large and small discs on sheets, analyzing the efficiency of each. Their most efficient sheet arrangement yields an 85% efficiency. For continuous rolls, they model attaching columns of discs in a repeating pattern, determining the optimal number of columns is 332 to fully use a 1m x 100m roll, yielding 86% efficiency.

1. Optimisation of Disc Pressing
Group 6: Cameron M, Curtis C, Melissa P, Sabina A and Henry S
November 2015
1 Problem Statement
A manufacturing firm has engaged your group as consultants to give advice to
the production manager. The key part of the production process requires the
cutting of circular discs from sheets of aluminium. At present the disc-pressing
machine is set to cut out 16 discs, each of diameter 0.25m, from a 1 metre square
aluminium sheet. There is also a further need to cut discs, each of diameter
0.1m, from the same sheets. The firm needs twice the number of small discs as
they do large discs. What is the best arrangement of cutting heads to minimise
wastage? As an extension, and more generally, the manager wants to invest in
a continuous strip processing machine, that will cut discs from a 1 metre wide
sheet of aluminium, delivered on rolls of 100 metres. How should the machine
best be configured?
To tackle this problem we are going to construct models of potential layouts
for the sheets to be pressed in, supported by mathematical proof. We will
then compare the efficiency each layout gives, where this efficiency is the total
percentage of area used, and conclude as to what the best layout or combination
of layouts is.
2 Assumptions
• We are assuming that the machine possesses different tool heads, which
can each be configured to press a different layout and switched between
sheets so that a combination of layouts can be used to obtain a smaller
percentage of area wasted than if we were restricted to working on a single
sheet.
• All of our calculations will be done using exact figures, but values obtained
from calculations will be displayed to 3 significant figures.
• We have accurately (to scale) drawn the discs and the sheet to develop a
number of layouts for our model, and use these diagrams to calculate the
efficiency of each layout. Therefore while our model is somewhat based
on logic, we do incorporate mathematics in order to prove certain claims.
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2. 3 Answering the Problem
3.1 Model 1
We approached this problem using decision mathematics methods initially. How-
ever these types of methods do not take into account the shape of the disc,
meaning they give impossible solutions. For example, we found that the area
of a large disc was 0.049m2
and theoretically these methods stated we could fit
20 large discs on a single sheet. Obviously we know this is incorrect because we
were told that a maximum of 16 large discs could be cut from one sheet, and
therefore we found that these methods are invalid for this problem. Due to this
we decided that we needed to take a more logical approach to this problem and
visually see how many of the large and small discs we can fit on a single sheet.
Throughout this project, we will be constantly referring to the number of
large (0.25m Diameter) and small (0.1m Diameter) discs. These will therefore
be labelled as the variables X and Y where X is the number of large discs and Y
is the number of small discs. We will also be calculating the percentage efficiency
of each pattern. This will be calculated by taking the total area covered by the
discs and dividing it by total area of the metal sheet. Since the area of a disc is
π(radius)2
the efficiency is therefore equal to:
X(π((
0.25
2
)2
)) + Y (π((
0.1
2
)2
))
In the model provided [Fig 1], the area of the sheet is 1m2
. The layout has
16X and 0Y. Each X has an area of π
64 , which means the total area used by the
large circles is π
4 , which means the total wasted area is 0.2164m2
, or equally as
a percentage of efficiency:
0.7854
1
× 100 ⇔ 0.7854 × 100 = 78.5%
Figure 1:
2

3. Figure 2: (A)
One immediate layout that will improve efficiency would be to utilise the
space between X ’s and insert a single Y as shown in [Fig 2]. However we need
to prove that this is mathematically feasible. To accomplish this, observe [Fig
3]. This is a diagram showing how the layout would work. By forming a square
using the central points of the discs we can use simple geometrical methods to
show how much space is available in the central area.
Figure 3:
Figure 4:
The sides of the square are of length 0.25m. For the purposes of this model,
let us call the red line h.
h = (0.252 + 0.252) =
√
2
4
= 0.354m
The size of the gap between X s is equal to h - (2)(Radius of Y ): 0.354 -
- 2(0.125) = 0.104m(3s.f.). Since 0.104 > 0.1, it’s shown that a Y will fit
into the gaps between the X s. So if we look at [Fig 2], this layout has 16X and
9Y at an increased efficiency of 85.6% compared to the previous layout. We will
be referring back to this pattern in the future, so from here on this layout will
be known as (A).
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4. Figure 5: (B)
Figure 6: (C)
Since this is still not near the 1:2 ratio of X to Y we require, we will remove
an X and replace the space with Y s as shown in [Fig 5]. Since three X s in a
row have a total length of 0.75m, there will be 0.25m of space between the edge
of the metal sheet and three Xs. Since each Y is only 0.1m in diameter, 2 of
these at a length of 0.2m can easily fit into this space as 0.2 < 0.25. Since this
can be done for the vertical and horizontal components, we can press 4 more
Y s in in the space left by the removal of one X. Now we have 15X and 13Y at
an efficiency of 83.8%. This layout will be referred to as (B).
Now if we remove the entire bottom row of X s we obtain the layout shown in
[Fig 6]. With a total of 4 X s removed, there will be 1m of space between the
2 edges of the metal sheet in which we can fit 10Y. We must first check that 2
rows can fit in the space because, unlike before, there are Y s which are lower
down than X s.
4

5. Figure 7:
Figure 8:
From Pythagoras’ Theorem we know that the Unknown Distance is
(0.1752
− 0.1252
) = 0.122m
Therefore there will be a minimum of 1m−(0.625m+0.122m+0.05m) = 0.203m
space remaining. As 0.203 > 0.2, therefore 2 rows of Y can fit into the space.
This layout [Fig 6] has 12X and 29Y at an efficiency of 81.7% and will be referred
to as (C). This is the first layout where we have at least a 1:2 ratio of X to Y
as desired by the problem. We can therefore print (C) but the overall efficiency
will only be 77.8% as there are 5 unnecessary Y s being produced. In order to
increase efficiency we will attempt to use this wastage. In terms of creating a
perfect 1:2 ratio of X to Y, (C) has an excess of 5Y, (A) has a need for 23Y and
(B) has a need for 17Y. Therefore, if we make one sheet of (A), one of (B) and
8 of (C) there will be no wastage as −23Y − 17Y + 8(5Y ) = 0. The efficiency
of this combination is
(85.60 + 83.84 + 8(81.68))
10
= 82.3%
3.2 Model 2
If we were to fill an entire sheet with only Y we would obtain the layout seen in
[Fig 8]. This contains 100Y and has efficiency of 78.5%. However by changing
the grid pattern to a hexagonal offset grid [Fig 9] we can reduce the wastage.
5

6. Figure 9:
Figure 10:
Figure 11: (D)
The distance between two rows here is (0.1m2 − 0.05m2) = 0.0866m as
shown in [Fig 10[. Now consider [Fig 11]. Since the first and last lines will
need to have at least their radius in space from the edge of the metal sheet, we
can have 0.9m
0.0866 = 10 rows of the hexagonal packaging in the remaining space.
Hence there can be 6 rows containing 10Y and 5 rows containing 9Y, so the
total will be 6(10) + 5(9) = 105Y . This arrangement will be referred to as (D).
The efficiency of (D) is 82.5%.
Of all the arrangements that need more Y, the most efficient is (A) and of all
the arrangements that have an excess of Y, (D) is the most efficient. Since (A)
requires 23 and (D) has 105 excess, the lowest common multiple of 23 and 105
is 2415.
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7. Therefore, if we print 23 sheets of (D) and 105 of (A) it will be an exact 1:2
ratio of X and Y. The efficiency of this is
(23(82.47) + 105(85.60))
128
= 85.0%
This is an improvement of 2.70% on our previous combination’s efficiency of
82.3%.
4 Results
Therefore, we would recommend printing 23 sheets of (D) and 105 sheets of (A)
for the most efficient production of discs. This is a form of mass production
however and if this is not possible, then as an alternative we would recommend
printing 1 sheet of (A), 1 of (B) and 8 of (C). Furthermore, if we are indeed
restricted to cutting discs with only a single layout then (C) should be used as
although it was not the most efficient, it was the only layout that satisfied the
requirement of the 1:2 ratio for X and Y.
5 Extended Model
For the extended model, we were asked to find an optimal pressing pattern if
rather than using 1m2
sheets we were using a 1m × 100m roll. An advanced
model for a 100m strip includes columns of X with Y between them as was
proved possible in Section 3.1, then at the end columns of Y to make up the
1:2 ratio, as shown below in [Fig 12].
Figure 12:
The black dots represent a repetition of the pattern that isn’t shown. Let
each column of 4X be called n, since we know the layout must be ≤ 100m we
can come up with a formula for length in terms of n. For each n there is 0.25m
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8. of length. Now for each n there must be 2(4n) Y to make up the 1:2 ratio of X
to Y.
However due to our pattern there are 3Y for each column except the last
column where there are 3(n − 1). So after the repetition of columns of X there
are columns of 10Y so the amount of these is equal to 2(4n)−3(n−1)
10 , and the
length is 2(4n)−3(n−1)
10 x 0.1m. So the total length is
n
4
+
2(4n) − 3(n − 1)
10
× 0.1 ⇒
n
4
+
5n + 3
10
× 0.1 =: (∗)
Now to fit the problem this length must be ≤ 100, so we will equate 5n+3
10 ×
0.1 to 5n+3
100 , thus it can be used in our equation.
n
4
+
5n + 3
100
≤ 100 ⇔
120n + 12
400
≤ 100 ⇔ 120n+12 ≤ 40000 ⇔ 120n ≤ 39988 ⇔ n ≤ 333.2
Now since we have a formula for n, we will input values of n into the equation
for length (*) to see the values at which it is feasible.
When n = 333 the length = 100.25m, hence n =/333.
So we’ll try n = 332, and this gives a length = 100m, so the max integer value
of n for which n
4 + 5n+3
100 ≤ 100 is satisfied is = 332.
When n = 332 there are 1328X and 2656Y. Hence the efficiency of this
model is the total area cut out over the total area of metal, which for these
values of X and Y =
1328π(0.1252
) + 2656π(0.052
)
100(1)
× 100 = 86.0%
6 Conclusion
In conclusion, if the manufacturing firm continues to press discs out of 1m2
sheets then the optimal combination of layouts to use is 23 sheets of (D) and
then 105 sheets of (A), giving an overall efficiency of 85.0%.
However if the firm decides to start pressing discs from a roll of aluminium with
dimensions 1m × 100m, then this efficiency can be improved when adopting
the layout given in Section 5, as this gives an efficiency of 86.0%, a full 1.0%
increase, which while it may not seem like much, will give a huge benefit when
produced on an industrial scale.
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