# Methods for Assessing Human-Machine Performance under Fuzzy Conditions
## 4 The Center of Gravity Defuzzification Technique
### The Rectangular Fuzzy Assessment Model  
We choose, exactly as in Example 2, for discourse the set of linguistic grades \(U=\{\)A, B, C, D, F\(\}\) and we represent a student group G as a FS in \(U\) in the form G = \(\{(x,\frac{n_{x}}{n}\}\): \(x\in U\) \(\}\), where \(n\) is the total number of the students of G and \(n_{x}\) is the number of students of G whose performance is characterized by the grade \(x\) in \(U\). To be able to design the graph of the membership function \(y=m(x)\), we replace \(U\) with a set of real intervals as: F \(\rightarrow[0,1)\), D \(\rightarrow[1,2)\), C \(\rightarrow[2,3)\), B \(\rightarrow[3,4)\), A \(\rightarrow[4,5]\). Consequently, we have that \(y_{1}=m(x)=m(F)\) for all \(x\) in \([0,1)\), \(y_{2}=m(x)=m(\mathrm{D})\) for all \(x\) in \([1,2)\), \(y_{3}=m(x)=m(\mathrm{C})\) for all \(x\) in \([2,3)\), \(y_{4}=m(x)=m(\mathrm{B})\) for all \(x\) in \([3,4)\) and \(y_{5}=m(x)=m(\mathrm{A})\) for all \(x\) in \([4,5)\). Since the membership values of the elements of \(U\) in G have been defined in terms of the corresponding frequencies, we obviously have that: \(\sum\limits_{i=1}^{5}y_{i}=m(\mathrm{A})+m(\mathrm{B})+m(\mathrm{C})+m( \mathrm{D})+m(\mathrm{F})=1\). The graph of the membership function \(y=\text{m}(x)\) is shown in Figure 3, where the level's area S between the graph and the OX axis is equal to the sum of the areas of the five rectangles \(\text{Si}\), i = 1, 2, 3, 4, 5, one side of which has length 1 and lies on the OX axis. That is why this method was termed as RFAM. The COG coordinates \((x_{c},y_{c})\) are calculated by the well known from mechanics formulas:  
\[x_{c}=\frac{\iint\limits_{S}xdxdy}{\iint\limits_{S}dxdy},\;y_{c}=\frac{\iint \limits_{S}ydxdy}{\iint\limits_{S}dxdy} \tag{8}\]  
Here, we have that: \(\iint\limits_{S}dxdy=\sum\limits_{i=1}^{5}y_{i}=1\). In addition, \(\iint\limits_{S}xdxdy=\sum\limits_{i=1}^{5}\iint\limits_{F_{i}}xdxdy=\sum \limits_{i=1}^{5}\int\limits_{0}^{y_{i}}dy\int\limits_{i-1}^{i}xdx=\sum \limits_{i=1}^{5}y_{i}\int\limits_{i-1}^{i}xdx=\sum\limits_{i=1}^{5}y_{i}\int \limits_{i-1}^{i}xdx=\sum\limits_{2}^{5}\sum\limits_{i=1}^{5}(2i-1)y_{i}\) and \(\iint\limits_{S}ydxdy=\sum\limits_{i=1}^{5}\iint\limits_{F_{i}}ydxdy=\sum \limits_{i=1}^{5}\int\limits_{0}^{y_{i}}ydy\int\limits_{i-1}^{i}dx=\sum \limits_{i=1}^{n}\int\limits_{0}^{y_{i}}ydy=\frac{1}{2}\sum\limits_{i=1}^{n} y_{i}^{2}\). Substituting the above values of the double integrals into Equation (8), one finds that  
\[ \begin{array}{l}x_{\text{C}}=\frac{1}{2}(y_{1}+3y_{2}+5y_{3}+7y_{4}+9y_{5})\\ y_{\text{C}}=\frac{1}{2}({y_{1}}^{2}+{y_{2}}^{2}+{y_{3}}^{2}+{y_{4}}^{2}+{y_{5}} ^{2})\end{array}  \tag{9}\]  
However, \((y_{i}-y_{j})^{2}={y_{i}}^{2}+{y_{j}}^{2}-2\;y_{ij}\geq 0\), or \({y_{i}}^{2}+{y_{j}}^{2}\geq 2\;y_{ij}\) for \(i,j=1,2,\ldots,5\), with the equality holding if, and only if, \(y_{1}=y_{2}=y_{3}=y_{4}=y_{5}\). Therefore, \(1=(\sum\limits_{i=1}^{5}y_{i})^{2}=\sum\limits_{i=1}^{n}{y_{i}}^{2}+2(y_{1}y_ {2}+y_{1}y_{3}+\ldots+y_{4}y_{5})\geq 5\sum\limits_{i=1}^{n}{y_{i}}^{2}\), or \({y_{1}}^{2}+{y_{2}}^{2}+{y_{3}}^{2}+{y_{4}}^{2}+{y_{5}}^{2}\leq\frac{1}{5}\), with the equality holding if and only if \(y_{1}=y_{2}=y_{3}\) = \(y_{4}\) = \(y_{5}=\frac{1}{5}\). Thus, Equation (9) shows that the unique minimum \(y_{c}=\frac{1}{10}\) corresponds to the COG \(\text{F}_{\text{m}}\) (\(\frac{5}{2},\frac{1}{10}\)). The ideal case is when \(y_{1}=y_{2}=y_{3}=y_{4}=0\) and \(y_{5}=1\). Then, from Equation (9), one finds that \(\text{x}_{\text{C}}=\frac{9}{2}\) and \(\text{y}_{\text{C}}=\frac{1}{2}\). Therefore, the COG in this case is the point \(\text{F}_{\text{i}}\) (\(\frac{9}{2},\frac{1}{2}\)). On the other hand, the worst case is when \(y_{1}=1\) and \(y_{2}=y_{3}=y_{4}=y_{5}=0\). Therefore, from Equation (9), one finds that the COG is the point \(\text{F}_{\text{X}}\) (\(\frac{1}{2},\frac{1}{2}\)).