Privacy-Aware Examination Results Ranking for the Balance Between Teachers and Mothers

In this section, a case study extracted from the example in Fig. 1 is offered to demonstrate the concrete running process of our proposed PERR method. Next, we introduce PERR according to the four steps specified in Fig. 2 .

Step 1: Data normalization

According to the example in Fig. 1 , a student-course performance matrix M is as follows:

(12) $$\begin{array}{l}\begin{array}{ccccccc}& & \text{English}& \text{Sport}& \text{Nature}& & \end{array}\\ M:\begin{array}{c}\text{John}\\ \text{Alex}\\ \text{Lily}\end{array}\left[\begin{array}{ccccc}70& & 5& & 30\\ 80& & 4& & 20\\ 90& & 3& & 10\end{array}\right]\end{array}$$

In concrete terms, $M$ contains the examination scores of the three students over three courses. We normalize the score values of three columns according to Eq. ( 2 ). Afterwards, we are able to get a new normalized matrix ${\mathit{M}}^{\mathrm{\#}}$ ,

(13) $$\begin{array}{l}\begin{array}{ccccccc}& & \text{English}& & \text{Sport}& & \text{Nature}\end{array}\hfill \\ M^{\#}:\begin{array}{c}\text{John}\\ \text{Alex}\\ \text{Lily}\end{array}\left[\begin{array}{ccccc}0.50& & 0.71& & 0.80\\ 0.58& & 0.57& & 0.53\\ 0.65& & 0.42& & 0.27\end{array}\right]\hfill \end{array}$$

Step 2: Determine PIS and NIS

As normalized examination scores in matrix ${\mathit{M}}^{\mathrm{\#}}$ are all positive dimensions, we determine PIS and NIS based on Eqs. ( 3 )-( 6 ) and ( 13 ). In concrete terms, PIS and NIS are shown below:

(14) $$PIS=(0.65,0.71,0.80)$$

(15) $$NIS=(0.50,0.42,0.27)$$

Step 3: Evaluate each student based on PIS and NIS

As Eq. ( 13 ) shows, the normalized examination scores of the three students can be represented by John (0.50, 0.71, 0.80), Alex (0.58, 0.57, 0.53), and Lily (0.65, 0.42, 0.27). Next, we calculate the distance of these three students with PIS,

(16) $$D_{{\mathrm{John}}^{+}}=\mathrm{Dist}((0.50,0.71,0.80),$$ $$(0.65,0.71,0.80))=0.15,$$ $$D_{{\mathrm{Alex}}^{+}}=\mathrm{Dist}((0.58,0.57,0.53),$$ $$(0.65,0.71,0.80))=0.31,$$ $$D_{{\mathrm{Lily}}^{+}}=\mathrm{Dist}((0.65,0.42,0.27),$$ $$(0.65,0.71,0.80))=0.60$$

Moreover, we calculate the distance of these three students with NIS,

(17) $$D_{{\mathrm{John}}^{-}}=\mathrm{Dist}((0.50,0.71,0.80),$$ $$(0.50,0.42,0.27))=0.60,$$ $$D_{{\mathrm{Alex}}^{-}}=\mathrm{Dist}((0.58,0.57,0.53),$$ $$(0.50,0.42,0.27))=0.31,$$ $$D_{{\mathrm{Lily}}^{-}}=\mathrm{Dist}((0.65,0.42,0.27),$$ $$(0.50,0.42,0.27))=0.15$$

Then, according to Eq. ( 11 ), we can obtain the comprehensive score for each student,

(18) $$D_{\mathrm{John}}=D_{{\mathrm{John}}^{-}}/(D_{{\mathrm{John}}^{+}}+D_{{\mathrm{John}}^{-}})=0.80,$$ $$D_{\mathrm{Alex}}=D_{{\mathrm{Alex}}^{-}}/(D_{{\mathrm{Alex}}^{+}}+D_{{\mathrm{Alex}}^{-}})=0.50,$$ $$D_{\mathrm{Lily}}=D_{{\mathrm{Lily}}^{-}}/(D_{{\mathrm{Lily}}^{+}}+D_{{\mathrm{Lily}}^{-}})=0.20$$

Step 4: Rank all students in descending order

According to the comprehensive scores of three students, i.e., $D_{\mathrm{John}}$ , $D_{\mathrm{Alex}}$ , and $D_{\mathrm{Lily}}$ derived in Eq. ( 18 ), we can rank them in descending order, i.e., John $>$ Alex $>$ Lily. Finally, we can return the ranked list to interested users.