Computer implemented method for dissimilarity computation between two yarns to be used for setting of a textile machine in a textile process, and computer program product

Abstract

The setting of textile machinery parameters is an important aspect that combines implicit knowledge of workers and engineers with explicit knowledge. As yarn and fabrics involved in a textile process are multicomponent artefacts, in order to automatize this process of machine configuration, a method for dissimilarity computation between two yarns is proposed including one or a combination of four algorithms to evaluate the similarity between two yarns, each composed by a list of materials. The method has proved to be successful for spinning setting and it can be applied in other steps of a textile process like weaving.

Claims

1. Computer implemented method for dissimilarity computation between two yarns to be used for setting of a textile machine in a textile process for manufacturing a textile product, wherein in the textile process a first yarn is used and said setting involving the use of at least a second yarn selected from several candidate yarns, both said first and said at least second yarns being identified by physical properties including at least count and by a list of materials, each material in turn being defined by percentage of presence, belonging to a family of materials and by some physical material properties including finesses and length, comprising: a) automatically computing material dissimilarity values of all possible combination of the materials of said list of materials of the first and second yarns; and b) automatically calculating a dissimilarity value between the first and second yarns by applying an algorithm using as inputs the list of materials of the first and second yarns and said computed material dissimilarity values, applying of said algorithm including a weighted aggregation using said material dissimilarity values computed of different combination of pairs of materials of said first and second yarns, where the weights depend on the presence and/or percentage of these materials in the yarns.

2. A computer implemented method according to claim 1 wherein said first and second yarns are different in that having a different percentage of the same materials and/or in that they include a list of different materials and/or in having a different value for some material properties.

3. A computer implemented method according to claim 1, wherein said computing of material dissimilarity values uses a textile expert knowledge that provides at least dissimilarity between each pair of materials.

4. A computer implemented method according to claim 3, wherein said textile expert knowledge further provides an optimal range of length and optimal range of fineness between each pair of materials.

5. A computer implemented method according to claim 4, wherein said algorithm performs comparisons among pairs of materials of said first and second yarns with an equivalent percentage in common and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values.

6. A computer implemented method according to claim 4, wherein said algorithm performs comparisons among pairs of materials of said first and second yarns taking into account the main material with an equivalent percentage and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values.

7. A computer implemented method according to claim 4 wherein said algorithm disregards first a common part from both first and second yarns involving a set of pairs of materials with an equivalent percentage and dissimilarity equal to 0, and if the percentages are not equal, then only the lower percentage is disregarded and then all the possible combinations among the pairs of the list of the remaining materials of both yarns are compared obtaining several material dissimilarity values and then performing a weighted aggregation of the material dissimilarity values wherein the weight for each pair of materials is the product of both material percentages divided by the percentage of the remaining uncommon part.

8. A computer implemented method according to claim 4 wherein said algorithm performs an iterative comparison selecting the possible combinations among the list of pairs of materials of said first and second yarns by percentage operating by decreasing order of percentage obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values and in case the number of materials in a list being not the same, each material in a list without pair a maximum dissimilarity value equal to 1 is added to said aggregation, wherein each material dissimilarity weight is computed as the mean value of both percentages of each pair of materials.

9. A computer implemented method according to claim 2, wherein said algorithm is an average or a combination of two or more of the following algorithms a11 to a14: a11: performing comparisons among pairs of materials of said first and second yarns with an equivalent percentage in common and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values; a12: performing comparisons among pairs of materials of said first and second yarns taking into account the main material with an equivalent percentage and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values; a13: disregarding first a common part from both first and second yarns involving a set of pairs of materials with an equivalent percentage and dissimilarity equal to 0, and if the percentages are not equal, then only the lower percentage is disregarded and then all the possible combinations among the pairs of the list of the remaining materials of both yarns are compared obtaining several material dissimilarity values and then performing a weighted aggregation of the material dissimilarity values wherein the weight for each pair of materials is the product of both material percentages divided by the percentage of the remaining uncommon part; and a14: performing an iterative comparison selecting the possible combinations among the list of pairs of materials of said first and second yarns by percentage operating by decreasing order of percentage obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values and in case the number of materials in a list being not the same, each material in a list without pair a maximum dissimilarity value equal to 1 is added to said aggregation, wherein each material dissimilarity weight is computed as the mean value of both percentages of each pair of materials.

10. A computer implemented method according to claim 9, wherein a result useful for setting of a textile machine using said at least second yarn is computed from a weighted aggregation of a dissimilarity value obtained from a method according to an average or a combination of two or more of said algorithms of claim 9 and other dissimilarities values regarding physical properties of said at least second yarn including at least count of the involved yarns obtained from a textile expert knowledge.

11. A computer implemented method according to claim 1, wherein said other dissimilarities values also comprise sector, as a non-physical property of the involved yarns to be considered in the weighted aggregation.

12. A computer implemented method according to claim 1, wherein said computed material dissimilarity values of all possible combination of the materials, of said list of materials of the first and second yarns, are comprised between 0 and 1.

13. A computer program product comprising instructions that when executed in a processor performs a method according to claim 1.

14. The computer program product according to claim 13 wherein said instructions when executed in a processor further performs a method wherein the algorithm is an average or a combination of two or more of the following algorithms a11 to a14: a11: performing comparisons among pairs of materials of said first and second yarns with an equivalent percentage in common and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values; a12: performing comparisons among pairs of materials of said first and second yarns taking into account the main material with an equivalent percentage and proceeds iteratively selecting the combinations of the pairs to be compared having a lower dissimilarity value being the corresponding weight for each pair of materials the smallest percentage and then comparing among them the rest of materials, obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values; a13: disregarding first a common part from both first and second yarns involving a set of pairs of materials with an equivalent percentage and dissimilarity equal to 0, and if the percentages are not equal, then only the lower percentage is disregarded and then all the possible combinations among the pairs of the list of the remaining materials of both yarns are compared obtaining several material dissimilarity values and then performing a weighted aggregation of the material dissimilarity values wherein the weight for each pair of materials is the product of both material percentages divided by the percentage of the remaining uncommon part; and a14: performing an iterative comparison selecting the possible combinations among the list of pairs of materials of said first and second yarns by percentage operating by decreasing order of percentage obtaining several material dissimilarity values and then performing a weighted aggregation of said dissimilarity values and in case the number of materials in a list being not the same, each material in a list without pair a maximum dissimilarity value equal to 1 is added to said aggregation, wherein each material dissimilarity weight is computed as the mean value of both percentages of each pair of materials.

15. The computer program product according to claim 14, wherein a result useful for setting of a textile machine using said at least second yarn is computed from a weighted aggregation of a dissimilarity value obtained from a method according to an average or a combination of two or more of said algorithms all -a14 and other dissimilarities values regarding physical properties of said at least second yarn including at least count of the involved yarns obtained from a textile expert knowledge.

16. The computer program product according to claim 13 wherein said other dissimilarities values also comprise sector, as a non-physical property of the involved yarns to be considered in the weighted aggregation.

Description

BRIEF DESCRIPTION OF THE DRAWINGS

[0039] FIG. 1 is a reduced Distance Material Family table, obtained from technical expert gathered information, scaled in [0,1].

[0040] FIG. 2 is a diagram showing the hierarchy of the material families and subtypes of the fibres of a yarn.

[0041] FIG. 3 is a block diagram showing the components and steps of the method according to this invention.

DETAILED DESCRIPTION OF EXEMPLARY EMBODIMENTS

[0042] Two yarns of different composition can have similar behaviour from the textile point of view and, therefore, one may be a substitute for the other and the textile machinery settings can be reused.

[0043] Given two yarns, their physical properties and the composition of their fibres are compared. The physical characteristics of the yarn that are measurable can be compared using their numeric value with existing distance metrics such as the Euclidean. Typically, these characteristics refer to different physical aspects of the yarn such as thickness, torsion, elongation or resistance. These characteristics depend on how the yarn is produced and the materials it is composed of. Other features like the sector are qualitatively modelled because they cannot be modelled numerically and what is only known is if they are equal or different. The composition of the yarn is a combination of different fibre types with a percentage of presence.

[0044] Fibres can be classified into different families depending on the material that are composed: cotton, viscose, silk, wool, etc. At the same time, each family has different types of fibres. The differences between fibre types from the same family are based on certain physical characteristics of the fibres such as the length and/or fineness. However, in general, materials from the same family with different physical characteristics are more similar than those from different families with but similar physical characteristics according to the experts' knowledge. In FIG. 2 of the drawings, a hierarchy of the families and subtypes of the fibres is shown.

[0045] Therefore, given two yarns, its composition and the physical characteristics of their fibres, a computer implemented method is proposed to calculate how similar these two yarns are. The complexity of this procedure lies in comparing different compositions since both the number of materials and their percentage are variable.

[0046] A yarn can be understood as a list (LM (H.sub.i)) of different materials (components) or fibres and each material (MAT.sub.i(H.sub.j)) has a certain percentage of presence (PERC.sub.i(H.sub.j)). A material can be a composition itself (yarn) or be composed of fibres of the same type (material). Usually, the main material (higher presence) defines the behaviour of the yarn and is therefore more important than the other materials.

[0047] The different types of fibres or materials are classified into different families of materials belonging to the (MATFAM.sub.i(H.sub.j). And the materials/fibres of the same family are differentiated by certain characteristics or physical properties. Typically these include the diameter (FINENESS.sub.i (H.sub.j)) and length (LENGTH.sub.i(H.sub.j)). Therefore, the composition of a yarn can be generalized to a list of materials (LM) where each material MAT.sub.j has: a percentage of presence, PERC.sub.j, a material family MATFAM.sub.j and some physical characteristics of the fibres which describe fibres that make up this particular material.

[0048] The different fibre types are classified into different material families (see FIG. 2). The specificity of theses materials would depend on the needs of the end user where the method is applied. Fibre types of the same family can have different physical characteristics. Typically these characteristics include the fineness FINENESS.sub.j and length LENGTH.sub.j of the fibres.

[0049] Given two yarns (H.sub.1, H.sub.2) the degree of dissimilarity among them is calculated taking into account: [0050] The characteristics of the yarn: [0051] Physical properties: count, etc. [0052] Other properties: sector, etc. [0053] The composition (materials, percentage and material properties)
and a criteria of how to compare the different materials (by pairs) and its percentages

[0054] The yarns H.sub.1, H.sub.2 can be modelled as follows:

H.sub.2=<PHYSICAL PROPERTIES (H.sub.1), OTHER PROPERTIES (H.sub.1), LM (H.sub.1)>
PHYSICAL PROPERTIES (H.sub.1)=<COUNT (H.sub.1), PROP.sub.2 (H.sub.1), . . . , PROP.sub.L (H.sub.1)>
OTHER PROPERTIES (H.sub.1)=<SECTOR (H.sub.1), OT.PROP.sub.2(H.sub.1), . . . , OT.PROP.sub.T (H.sub.1)>
LM (H.sub.1)=<MAT.sub.1(H.sub.1), . . . , MAT.sub.N (H.sub.1)>
MAT.sub.i(H1)=<PERC.sub.i(H.sub.1), MATFAM.sub.i(H.sub.1), MATERIAL PROPERTIES(H.sub.1)>
MATERIAL PROPERTIES (H.sub.1)=<FINENESS.sub.i(H.sub.1), LENGTH.sub.i(H.sub.1), . . . ,M.PROP.sub.k(H.sub.1)>
H.sub.2=<PHYSICAL PROPERTIES (H.sub.2), OTHER PROPERTIES (H.sub.2), LM (H.sub.2)>
PHYSICAL PROPERTIES (H.sub.2)=<COUNT (H.sub.2), PROP.sub.2(H.sub.2), . . . ,PROP.sub.L(H.sub.2)>
OTHER PROPERTIES (H.sub.2)=<SECTOR (H.sub.2), OT.PROP.sub.2 (H.sub.2), . . . , OT.PROP.sub.T(H.sub.2)>
LM (H.sub.2)=<MAT.sub.1(H.sub.2), . . . , MAT.sub.M (H.sub.2)>
MAT.sub.i(H.sub.2)=<PERC.sub.i(H.sub.2), MATFAM.sub.i(H.sub.2), MATERIAL PROPERTIES(H.sub.2)>
MATERIAL PROPERTIES (H.sub.2)=<FINENESS.sub.i(H.sub.2), LENGTH.sub.i(H.sub.2), . . . , M.PROP.sub.k(H.sub.2)>

[0055] COUNT (H.sub.i) (in Nm) is the number of meters of yarn per kg (smaller values indicate higher yarn diameter) and it can be numerically modelled and SECTOR (Hi) is a qualitative label that designates the area of production and it can be qualitatively modelled.

[0056] Count is an important property of the description of a yarn, but there are also other important physical properties that can be taken into account if necessary such as the tenacity and yarn twist. Likewise, there are other properties that may be important for the description of the yarn and that can vary depending on the application of the yarn and in this case the sector has been highlighted.

Calculation of the Dissimilarity Between Two Yarns

[0057] The dissimilarity between two yarns is defined as a weighted sum of the dissimilarity of their features:

[00003]$\mathrm{Dissim}\left(H_1,H_2\right)=W_{\mathrm{COUNT}}*{\mathrm{Dissim}}_{\mathrm{COUNT}}\left(H_1,H_2\right)+\underset{i=2}{\overset{L}{.Math.}}.Math.W_i*{\mathrm{Dissim}}_{{\mathrm{PROP}}_i}\left(H_1,H_2\right)+W_{\mathrm{SECTOR}}*{\mathrm{Dissim}}_{\mathrm{SECTOR}}\left(H_1,H_2\right)+\underset{i=2}{\overset{T}{.Math.}}.Math.W_i*{\mathrm{Dissim}}_{\mathrm{OT}.Math.{\mathrm{PROP}}_i}\left(H_1,H_2\right)+W_{\mathrm{LM}}*{\mathrm{Disim}}_{L.Math..Math.M}\left(H_1,H_2\right)$

wherein ΣW.sub.i=W.sub.COUNT+Σ.sub.i=2.sup.LW.sub.i+W.sub.SECTOR+Σ.sub.i=2.sup.TW.sub.i+W.sub.LM=1 and all the dissimilarities are comprised between 0 and 1.

[0058] I.e. a weighted sum where one term is the composition (list of materials) and the rest can be physical properties (e.g. count) or other properties (e.g. sector).

[0059] In this case only taking into account the COUNT, and SECTOR the formula would be:

Dissim(H.sub.1,H.sub.2)=W.sub.COUNT*Dissim.sub.COUNT(H.sub.1,H.sub.2)+W.sub.SECTOR*Dissim.sub.SECTOR(H.sub.1,H.sub.2)+W.sub.LM*Dissim.sub.LM(H.sub.1,H.sub.2)

wherein ΣW.sub.i=W.sub.COUNT+W.sub.SECTOR+W.sub.LM=1

and

Dissim(H.sub.1,H.sub.2)∈[0,1],Dissim.sub.COUNT(H.sub.1,H.sub.2)∈[0,1],Dissim.sub.SECTOR(H.sub.1,H.sub.2)∈[0,1],Dissim.sub.LM(H.sub.1,H.sub.2)∈[0,1]

Calculation of Dissimilarity According to the COUNT Attribute.

[0060] According to the expert opinion, the dissimilarity between two small values is higher than among larger values. Thus, the dissimilarity does not follow a linear growth and, therefore, a relative measure that takes into account this effect is proposed to be used:

[00004]${\mathrm{Dissim}}_{\mathrm{COUNT}}\left(H_1,H_2\right)=\frac{/\mathrm{COUNT}\left(H_1\right)-\mathrm{COUNT}\left(H_2\right)/}{\max \left(\mathrm{COUNT}\left(H_1\right),\mathrm{COUNT}\left(H_2\right)\right)}$

Calculation of Dissimilarity According to the SECTOR Attribute.

[0061] SECTOR is a qualitative feature since it cannot be measured numerically. Therefore, whether both yarns belong to the same sector or not only can be assessed according to the following formula:

[00005]${\mathrm{Dissim}}_{\mathrm{SECTOR}}\left(H_1,H_2\right)=\{\begin{array}{cc}0& \mathrm{si}.Math..Math.\mathrm{SECTOR}\left(H_1\right)=\mathrm{SECTOR}\left(H_2\right)\\ 1& \mathrm{si}.Math..Math.\mathrm{SECTOR}\left(H_1\right)\ne \mathrm{SECTOR}\left(H_2\right)\end{array}$

Calculation of Dissimilarity According to the List of Materials (LM)

[0062] For the composition of the yarn, the combination of four algorithms A1-A4 (that will be explained in detail in the following examples) is proposed. These algorithms are weighted sums of different combinations of pairs of materials, where the weights depend on the presence of these materials in the yarn. Each algorithm has a different strategy for choosing the pairs of materials to be compared and for calculation of its weights.

[0063] The dissimilarity between the two yarns may be the result of any of the presented algorithms or a combination of any of them (see FIG. 3), for example, the average of the four.

[00006]${\mathrm{Dissim}}_{\mathrm{LM}}\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)=\frac{\begin{array}{c}{A}_1\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)+A_2.Math.\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)+\\ A_3\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)+A_4\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)\end{array}}{4}$$.Math.\mathrm{wherein},.Math..Math.A_1\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)=\mathrm{MinAlg}\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)$$.Math.A_2\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)=\mathrm{MainMinAlg}\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)$$.Math.A_3\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)=\mathrm{CrossAlg}\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)$$.Math.A_4\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)=\mathrm{MainHigherAlg}\left(\mathrm{LM}\left(H_1\right),\mathrm{LM}\left(H_2\right)\right)$

EXAMPLES

[0064] The “distance” term will be used in this section as equivalent to “dissimilarity” and component of a yarn would mean here a material thereof.

[0065] 1. Example of Distance Between Two Yarns

[0066] Note: Count and Sector are avoided for this example.

[0067] 1.1 Yarns to Compare:

[0068] Yarn 1: [0.6 PC (1.5, 38), 0.3 PL (3.3, 60), 0.1 CO−t1 (1.4,20)]

[0069] Yarn 2: [0.5 CO−t1 (1.4,20), 0.25 LI (1.6, 40), 0.15 W (8.85, 50), 0.1 CO−t2(1.5,22)]

[0070] That means that yarn 1 contains 3 components: [0071] 1. A component of type PC from the family of PC with a presence of 60% (0.6 of 1). The fibres of this component have a 1.5 of fineness and 38 of length. [0072] 2. A component of type PL from the family of PL with a presence of 30% (0.3 of 1). The fibres of this component have a 3.3 of fineness and 60 of length. [0073] 3. A component of type CO−t1 from the family of CO with a presence of 10% (0.1 of 1). The fibres of this component have a 1.4 of fineness and 20 of length. [0074] Yarn 2 is represented in the same way.

[0075] The main components are PC and CO respectively. [0076] 1.2 Weight for the Weighted Aggregation:

[0077] For the Initialization of weights following values from technical expert knowledge have been estimated:

W.sub.MATFAM=0.75

W.sub.FINENESS=0.25*0.7=0.175

W.sub.LENGTH=0.25*0.3=0.075

(*) This is just an example. From the experts we know that W.sub.FINENESS>W.sub.LENGTH

[0078] 1.3 Distance for the Physical Properties of Components

[0079] For scaling purpose, we assume that fineness ∈[0,10] and length ∈[0, 50]. The distance of fineness is assessed for this example with Relative distance:

[00007]$\mathrm{dist}\left(A,B\right)=\frac{.Math.A-B.Math.}{\max \left(A,B\right)}$

[0080] The length distance is assessed with a normalized absolute distance:

[00008]$\mathrm{dist}\left(A,B\right)=\frac{.Math.A-B.Math.}{{\mathrm{LENGTH}}_{\mathrm{MAX}}-{\mathrm{LENGTH}}_{\mathrm{MIN}}}=\frac{.Math.A-B.Math.}{50-0}$

[0081] 1.4 Interpretation of the Expert's Table:

TABLE-US-00001 CO WO 0.3 0.7 5 7 5 6
0.3=distance between WO and CO (scaled in [0,1])
0.7=similarity between WO and CO (scaled in [0,1]) (it is complementary and not used) 5 7=Range of ratios of fineness. The optimal fineness ratio between WO and CO

[00009]$\left({\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{{\mathrm{CO}}_{\mathrm{fineness}}}\right)$

is in [5,7]. That means that the fineness of WO is in 5 to 7 times greater to the CO fineness to consider that the distance of WO and CO is 0.3. Otherwise the distance should be greater.

[0082] 5 6=Range of ratios of length. The optimal length ratio between WO and CO

[00010]$\left({\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{{\mathrm{WO}}_{\mathrm{length}}}{{\mathrm{CO}}_{\mathrm{length}}}\right)$

is in [5, 6]. That means that the length of WO is between 5 and 6 times greater to the CO length to consider that the distance of WO and CO is 0.3. Otherwise the distance should be greater.

[0083] The reduced Distance Material Family table scaled in [0,1] is represented in FIG. 1

[0084] 2. Assessment of the Distance of Two Yarns (Yarn 1, Yarn 2)

[0085] First all of possible combinations of components in both yarns are compared in order to save time in the four algorithms. Since all of them use a subset of these comparisons for this example, it is clearer to assess all before than to assess them when they are needed.

[0086] Then the four algorithms are assessed and finally, and for this example, the average of the four algorithms is calculated.

[0087] 2.1 Distance Between Materials

[0088] Since of all the algorithms use the distance between two components. First, we compute these distances of all of possible combinations. Regarding that this distance involves material family, fineness and fibre length.

(*) Range of length=LENGTH.sub.MAX−LENGTH.sub.MIN=50−0=50
(**) If we have WO in Yarn 1 and CO in yarn 2, la distance.sub.MATFAM=0,3. If

[00011]${\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{{\mathrm{CO}}_{\mathrm{fineness}}}$

is smaller than 5 then the expected.sub.fineness of CO is assessed

[00012]$\left(\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{MIN}}}}=\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{5}\right)$

of if is greater than 7,

[00013]$\left(\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{MAX}}}}=\frac{{\mathrm{WO}}_{\mathrm{fineness}}}{7}\right).$

[0089] This process is analogous to the length assessment.

[00014]$.Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.1\right)=0.56$$1..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.1\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.7$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.5}{1.4}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{1.5}{1}=1.5$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.4-1.5.Math.}{\max \left(1.4,1.5\right)}=0.067$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{38}{20}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{38}{1}=38$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.20-38.Math.}{{\max }_{\mathrm{length}}}=0.36$$9..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.1\right)=0.75*0.7+0.175*0.067+0.075*0.36=0.56$$.Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{LI}\right)=0.48$$10..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{LI}\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.11..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.6$$.Math.12..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.5}{1.6}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.13..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{1.5}{1}=1.5$$.Math.14..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.6-1.5.Math.}{\max \left(1.6,1.5\right)}=0.062$$.Math.15..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{38}{40}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.16..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{38}{1}=38$$.Math.17..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.40-38.Math.}{{\max }_{\mathrm{length}}}=0.24$$18..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{LI}\right)=0.75*0.6+0.175*0.062+0.075*0.24=0.48$$.Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{WO}\right)=0.63$$.Math.1..Math..Math.\mathrm{Distance}\left(\mathrm{PC},\mathrm{WO}\right)=.Math.2..Math..Math.w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.3..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.7$$4..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.5}{8.85}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{22.2},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{14.2}\right]$$.Math.5..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{1.5}{\frac{1}{14.2}}=21.3$$.Math.6..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.8.85-21.3.Math.}{\max \left(8.85,21.3\right)}=0.58$$.Math.7..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{38}{50}\in \left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{2},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=0$$9..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{WO}\right)=0.75*0.7+0.175*0.58+0.075*0=0.63$$.Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.2\right)=0.55$$1..Math..Math.\mathrm{distance}.Math.\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.2\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.7$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.5}{1.5}\in \left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=0$$.Math.5..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{38}{22}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.6..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{38}{1}=38$$.Math.7..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.22-38.Math.}{{\max }_{\mathrm{length}}}=0.32$$8..Math..Math.\mathrm{distance}\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.2\right)=0.75*0.7+0.175*0+0.075*0.32=0.55$$.Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.1\right)=0.84$$1..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.1\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.9$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{3.3}{1.4}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{3.3}{1}=3.3$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.4-3.3.Math.}{\max \left(1.4,3.3\right)}=0.58$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{60}{20}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{60}{1}=60$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.20-60.Math.}{{\max }_{\mathrm{length}}}=0.8$$9..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.1\right)=0.75*0.9+0.175*0.58+0.075*0.8=0.84$$.Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{LI}\right)=0.79$$1..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{LI}\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.9$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{3.3}{1.6}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{3.3}{1}=3.3$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.6-3.3.Math.}{\max \left(1.6,3.3\right)}=0.51$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{60}{40}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{1.05},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{1.05}\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{60}{1}=60$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.40-60.Math.}{{\max }_{\mathrm{length}}}=0.4$$9..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{LI}\right)=0.75*0.9+0.175*0.51+0.075*0.4=0.79$$.Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{WO}\right)=0.53$$1..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{WO}\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.5$$3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{3.3}{8.85}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{15},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{13.3}\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{3.3}{\frac{1}{13.3}}=43.89$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.8.85-43.89.Math.}{\max \left(1.4,3.3\right)}=0.8$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{60}{50}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{1.31},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{1.05}\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{60}{1/1.05}=63$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.50-63.Math.}{{\max }_{\mathrm{length}}}=0.26$$9..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{WO}\right)=0.75*0.5+0.175*0.8+0.075*0.26=0.53$$.Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.2\right)=0.83$$1..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.2\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.9$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{3.3}{1.5}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{3.3}{1}=3.3$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.5-3.3.Math.}{\max \left(1.5,3.3\right)}=0.54$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{60}{22}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{60}{1}=60$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.22-60.Math.}{{\max }_{\mathrm{length}}}=0.76$$9..Math..Math.\mathrm{distance}\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.2\right)=0.75*0.9+0.175*0.54+0.075*0.76=0.83$$.Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.1\right)=0$$1..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.1\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.4}{1.4}\in \left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=0$$.Math.5..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{20}{20}\in \left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.6..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=0$$.Math.7..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.1\right)=0$$.Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{LI}\right)=0.28$$1..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{LI}\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.3$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.4}{1.6}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{1.4}{1}=1.4$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.6-1.4.Math.}{\max \left(1.6,1.4\right)}=0.125$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{20}{40}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{20}{1}=20$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.40-20.Math.}{{\max }_{\mathrm{length}}}=0.4$$9..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{LI}\right)=0.75*0.3+0.175*0.125+0.075*0.4=0.28$$.Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{WO}\right)=0.3$$1..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{WO}\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0.3$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.4}{8.85}\in \left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{7},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{5}\right]$$.Math.4..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=0$$.Math.5..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{20}{50}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=\frac{1}{6},{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=\frac{1}{5}\right]$$.Math.6..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{20}{1/5}=100$$.Math.7..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.50-100.Math.}{{\max }_{\mathrm{length}}}=1$$.Math.8..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{WO}\right)=0.75*0.3+0.175*0+0.075*1=0.3$$.Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.2\right)=0.14$$1..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.2\right)=w_M*{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}+w_F*{\mathrm{distance}}_{\mathrm{fineness}}+w_L*{\mathrm{distance}}_{\mathrm{length}}$$.Math.2..Math..Math.{\mathrm{distance}}_{{\mathrm{material}}_{\mathrm{family}}}=0$$.Math.3..Math..Math.{\mathrm{ratio}}_{{\mathrm{fineness}}_{\mathrm{real}}}=\frac{1.4}{1.5}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.4..Math..Math.{\mathrm{expecte}}_{\mathrm{fineness}}=\frac{1.4}{1}=1.4$$.Math.5..Math..Math.{\mathrm{distance}}_{\mathrm{fineness}}=\frac{.Math.1.5-1.4.Math.}{\max \left(1.5,1.4\right)}=0.067$$.Math.6..Math..Math.{\mathrm{ratio}}_{{\mathrm{length}}_{\mathrm{real}}}=\frac{20}{22}.Math.\left[{\mathrm{ratio}}_{{\mathrm{ideal}}_{\min }}=1,{\mathrm{ratio}}_{{\mathrm{ideal}}_{\max }}=1\right]$$.Math.7..Math..Math.{\mathrm{expected}}_{\mathrm{length}}=\frac{20}{1}=20$$.Math.8..Math..Math.{\mathrm{distance}}_{\mathrm{length}}=\frac{.Math.22-20.Math.}{{\max }_{\mathrm{length}}}=0.04$$9..Math..Math.\mathrm{distance}\left(\mathrm{CO}-t.Math..Math.1,\mathrm{CO}-t.Math..Math.2\right)=0.75*0+0.175*0.067+0.075*0.04=0.014$

[0090] Finally, the summary of the distance between components is the following:

distance (PC,CO−t1)=0.56

distance (PC,LI)=0.48

distance (PC,WO)=0.63

distance(PC,CO−t2)=0.55

distance(PL,CO−t1)=0.84

distance(PL,LI)=0.79

distance(PL,WO)=0.53

distance(PL,CO−t2)=0.83

distance(CO−t1, CO−t1)=0

distance(CO−t1,LI)=0.28

distance(CO−t2,WO)=0.3

distance(CO−t1, CO−t2)=0.014

[0091] 2.2 Assessment of the Four Algorithms

[0092] 2.2.1 First Approach: Algorithm A1 (Min Algorithm)

[0093] This algorithm does not take into account the main components. So, it is iteratively selecting the combinations with smaller distance. So, the first step it is to know the distance of all combinations. The following list contains all the combinations ordered by the distance:

distance(CO−t1,CO−t1)=0

distance(CO−t1,CO−t2)=0.014

distance(CO−t1,LI)=0.28

distance(CO−t2,WO)=0.3

distance(PC,LI)=0.48

distance(PL,WO)=0.53

distance (PC,CO−t2)=0.55

distance(PC,CO−t1)=0.56

distance(PC,WO)=0.63

distance(PL,LI)=0.79

distance (PL,CO−t2)=0.83

distance (PL,CO−t1)=0.84

[0094] Then, for this algorithm, the order of the combinations is the following:

[0095] 1. Start:

[0096] Yarn 1 [0.6 PC (1.5, 38), 0.3 PL (3.3, 60), 0.1 CO−t1 (1.4,20)]

[0097] Yarn 2 [0.5 CO−t1 (1.4,20), 0.25 LI(1.6, 40), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0098] 2. (CO−t1, CO−t1) is the combination with minimum distance and minimum percentage min(0.1,0.5)=0.1, then 0.1 CO−t1 is disgarded from the both yarns

[0099] Remaining yarns materials to be compared:

[0100] Yarn 1 [0.6 PC (1.5, 38), 0.3 PL (3.3, 60)]

[0101] Yarn 2 [0.4 CO−t1(1.4,20), 0.25 LI (1.6, 40), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0102] 3. (PC, LI) is the combination with minimum distance and min(0.6, 0.25)=0.25 is the minimum percentage. Then 0.25 PC is extracted from yarn 1 and 0.25 Li is extracted from yarn 2.

[0103] Remaining yarns materials to be compared:

[0104] Yarn 1 [0.35 PC (1.5, 38), 0.3 PL (3.3, 60)]

[0105] Yarn 2 [0.4 CO−t1(1.4,20), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0106] 4. (PL, WO) is the combination with minimum distance with minimum percentage: 0.15

[0107] Remaining yarns materials to be compared:

[0108] Yarn 1 [0.35 PC (1.5, 38), 0.15 PL (3.3, 60)]

[0109] Yarn 2 [0.4 CO−t1(1.4,20), 0.1 CO−t2(1.5,22)]

[0110] 5. (PC, CO−t2) is the combination with minimum distance with minimum percentage: 0.1

[0111] Remaining yarns materials to be compared:

[0112] Yarn 1 [0.25 PC (1.5, 38), 0.15 PL (3.3, 60)]

[0113] Yarn 2 [0.4 CO−t1 (1.4,20)]

[0114] 6. (PC, CO−t1) is the combination with minimum distance with minimum percentage: 0.25

[0115] Remaining yarns materials to be compared:

[0116] Yarn 1 [0.15 PL (3.3, 60)]

[0117] Yarn 2 [0.15 CO−t1 (1.4,20)]

[0118] 7. PL,CO−t1)) is the combination with minimum distance and the last one, with minimum percentage: 0.15

[0119] Remaining yarns materials to be compared:

[0120] Yarn 1: [ø]

[0121] Yarn 2: [ø]

[0122] 8. distance(yarn1, yarn2)=0.1*0+0.25*0.48+0.15*0.53+0.1*0.55+0.25*0.56+0.15*0.84=0.52

Representation of how the portions of materials are compared with other portion of materials with the same percentage:

[0123] 2.2.2 Second Approach: Algorithm A2 (Main Min Algorithm)

[0124] The algorithm maps the material of the first yarn with the material of the second yarn. First, the main materials are taken into account and then the rest of materials. The distance between materials is the following:

distance(CO−t1,CO t1)=0

distance(CO−t1,CO−t2)=0.014

distance(CO−t1,LI)=0.28

distance(CO−t2,WO)=0.3

distance(PC,LI)=0.48

distance(PL,WO)=0.53

distance(PC,CO−t2)=0.55

distance(PC,CO−t1)=0.56

distance(PC,WO)=0.63

distance(PL,LI)=0.79

distance(PL,CO−t2)=0.83

distance(PL,CO−t1)=0.84

[0125] 1. We take the main materials (0.6 PC, 0.5 CO−t1), then we use the min percentage (0.5). Notice that in this example there is only 1 combination of main materials. [0126] 1. main1=max_component(yarn 1)=0.6 PC [0127] 2. main2=max_component(yarn 2)=0.5 CO−t1 [0128] 3. p=min_percentages (main1, main2)=0.5 [0129] 4. The main combination is: 0.5 (PC,CO−t1)

[0130] 2. Now, we have (without mains):

[0131] Yarn 1 [0.1PC (1.5,38),0.3 PL (3.3, 60), 0.1 CO−t1(1.4,20)]

[0132] Yarn 2 [0.25 LI (1.6, 40), 0.15 W (8.85, 50),0.1 CO−t2(1.5,22)]

[0133] 3. Select between combinations depending on the distance, as in the first algorithm: [0134] a. 0.1 (CO−t1,CO−t2) is the combination with minimum distance and the last one. [0135] Remaining yarns materials to be compared: [0136] Yarn 1 [0.1PC (1.5,38),0.3 PL (3.3, 60)] [0137] Yarn 2 [0.25 LI (1.6, 40), 0.15 W (8.85, 50)] [0138] b. 0.1 (PC,LI) is the combination with minimum distance and the last one. [0139] Remaining yarns materials to be compared: [0140] Yarn 1 [0.3 PL (3.3, 60)] [0141] Yarn 2 [0.15 LI (1.6, 40), 0.15 W (8.85, 50)] [0142] c. 0.15 (PL,WO) is the combination with minimum distance and the last one. [0143] Remaining yarns materials to be compared: [0144] Yarn 1 [0.15 PL (3.3, 60)] [0145] Yarn 2 [0.15 LI (1.6, 40)] [0146] d. 0.15 (PL, LI) is the combination with minimum distance and the last one. [0147] Remaining yarns materials to be compared: [0148] Yarn1: [ø] [0149] Yarn 2: [ø]

[0150] 4. Distance (yarn.sub.1, yarn.sub.2)=0.5*0.56+0.1*0.014+0.1*0.48+0.15*0.53+0.15*0.79=0.53

Representation of how the portions of materials are compared with other portion of materials with the same percentage but selecting first the main portions (50% PC and 50% CO−t1)

[0151] 2.2.3 Third Approach: Algorithm A3 (Cross Algorithm)

[0152] This algorithm does not take into account the main materials. First, the common part (distance=0) are removed and the all the possible combinations are compared.

[0153] Using the distance between the materials assessed in previous algorithms, we have:

distance(CO−t1,CO−t1)=0

distance(CO−t1,CO−t2)=0.014

distance(CO−t1,LI)=0.28

distance(CO−t2,WO)=0.3

distance(PC,LI)=0.48

distance(PL,WO)=0.53

distance(PC,CO−t2)=0.55

distance(PC,CO−t1)=0.56

distance(PC,WO)=0.63

distance(PL,LI)=0.79

distance(PL,CO−t2)=0.83

distance(PL,CO−t1)=0.84

[0154] 1. Removing common part from:

[0155] Yarn 1 [0.6 PC (1.5, 38), 0.3 PL (3.3, 60), 0.1 CO−t1(1.4,20)]

[0156] Yarn 2 [0.5 CO−t1(1.4,20), 0.25 LI (1.6, 40), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0157] Common part=0.1 (CO−t1, CO−t1)

[0158] 2. The common part is extracted and the rest is:

[0159] Yarn 1 [0.6 PC (1.5, 38), 0.3 PL (3.3, 60)]

[0160] Yarn 2 [0.4 CO−t1(1.4,20), 0.25 LI (1.6, 40), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0161] 3. Then the distance of comparing all the combinations is:

[00015]$4..Math..Math.\mathrm{distance}\left(\mathrm{yarn}.Math..Math.1,\mathrm{yarn}.Math..Math.2\right)=\frac{\left[\begin{array}{c}\begin{array}{c}0.6*0.4*d\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.1\right)+0.6*0.25*d.Math.\left(\mathrm{PC},\mathrm{LI}\right)+\\ 0.6*0.15*d\left(\mathrm{PC},\mathrm{WO}\right)+0.6*0.1*d\left(\mathrm{PC},\mathrm{CO}-t.Math..Math.2\right)+\end{array}\\ \begin{array}{c}0.3*0.4*d\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.1\right)+0.3*0.25*d.Math.\left(\mathrm{PL},\mathrm{LI}\right)+\\ 0.3*0.15*d\left(\mathrm{PL},\mathrm{WO}\right)+0.3*0.1*d\left(\mathrm{PL},\mathrm{CO}-t.Math..Math.2\right)\end{array}\end{array}\right]}{0.6+0.3}$$.Math.\frac{\left[\begin{array}{c}0.6*0.4*0.56+0.6*0.25*0.48+\\ 0.6*0.15*0.63+0.6*0.1*0.55+\\ 0.3*0.4*0.84+0.3*0.25*0.79+\\ 0.3*0.15*0.63+0.3*0.1*0.83\end{array}\right]}{0.9}=0.5.Math.\mathrm{.66}$

Representation of how the portions of materials that are not common to both yarns are compared all against all.

[0162] 2.2.4 Fourth Approach: Algorithm A4 (MainHigher Algorithm)

[0163] This algorithm selects the combinations in base to the percentages.

[0164] Using the distance between the materials assessed in previous algorithms, we have:

distance(CO−t1,CO−t1)=0

distance(CO−t1,CO−t2)=0.014

distance(CO−t1,LI)=0.28

distance(CO−t2,WO)=0.3

distance(PC,LI)=0.48

distance(PL,WO)=0.53

distance(PC,CO−t2)=0.55

distance(PC,CO−t1)=0.56

distance(PC,WO)=0.63

distance(PL,LI)=0.79

distance(PL,CO−t2)=0.83

distance(PL,CO−t1)=0.84

Notice that in this case there is only one combination of main materials and the number of materials in both yarns is the same. Therefore, the resulting algorithm is:

[0165] 1. Start:

[0166] Yarn 1 [0.6 PC (1.5, 38), 0.3 PL (3.3, 60), 0.1 CO−t1(1.4,20)]

[0167] Yarn 2 [0.5 CO−t1(1.4,20), 0.25 LI (1.6, 40), 0.15 W (8.85, 50)), 0.1 CO−t2(1.5,22)]

[0168] 2. We select the main component: 0.6 PC 0.5 CO−t1

[0169] 3. We select the materials depending on the higher percentage, therefore are: [0170] a. 0.3PL, 0.25 LI [0171] b. 0.1 CO−t1,0.15 WO [0172] c. −, 0.1 CO−t2

[0173] 4. distance(yarn1,yarn2)=mean(0.6,0.5)*d(PC,CO−t1)+mean (0.3,0.25)* d(PL, LI)+mean(0.1,0.15)*d(CO−t1,WO)+0.1/2*dif f.sub.max=

[0174] 5. 0.55*0.56+0.275*0.79+0.125*0.3+0.05*1=0.613

Representation of how materials are compared with other materials depending on the percentage of presence is shown. One proceeds from more percentage to less. In this example where the number of materials is different, it is shown that the last material of yarn 2 it is not compared with a material but it count as a maximum distance.

[0175] 2.3 Final Result

[0176] The result is the average of four algorithms:

Final Distance=mean(0.52,0.53,0.566,0.613)=0.56

[0177] 3. Material family codes

[0178] This is the used list, but this list could be shorter or longer, more specific or more general [0179] WO: Wool [0180] COR: Regenerated Cotton [0181] CO: Cotton [0182] PC: Acrylic [0183] AR: Aramid [0184] LI: Flax/Linen [0185] PA: Polyamide/Nylon [0186] PP: Polypropylene [0187] PL: Polyester [0188] SE: Silk [0189] VI: Viscose [0190] XX: Rest of families