*6.1. Implementation*

**Example 1.** *Suppose that an organization wants to construct the enterprise resource planning (ERP) system [25]. After investigating the existing vendors of ERP systems on the market, five potential ERP systems are primary determined to be chosen from, i.e., Ai* (*i* = 1, 2, 3, 4, 5)*. Decision makers utilize the PHFEs to evaluate the five alternatives with respect to four criteria, namely, function and technology (C*1*), strategic fitness (C*2*), ability of vendor (C*3*), and reputation of vendor (C*4*), and the weight vector of the criteria is w* = (0.2, 0.1, 0.3, 0.4)*. Subsequently, the PHF evaluation matrix N* = -*nij*. *is obtained as shown in Table 1.*



Then, we can determine the ranking of the five potential ERP systems using the Algorithm 1, which are presented as below.

**Step 1:** Because of all the criteria are the benefit type, the standardized PHF evaluation matrix *N* is as same as the PHF evaluation matrix *N*.

**Step 2:** Use the GPHFWA (*λ* = 1) operator to aggregate the standardized PHF evaluation matrix *N*, and the collective evaluation information of each alternative is obtained as

*n* 1 = {{0.3636, 0.3877, 0.4113, 0.4336} , {0.3862, 0.4048, 0.4136, 0.4335}, {0.0444, 0.0482, 0.0502, 0.0544}};

*n* 2 = {{0.4061, 0.4401, 0.4684, 0.5023, 0.5308, 0.5545} , {0.2563, 0.2612, 0.2612, 0.2659, 0.2662, 0.2709, 0.2709, 0.2761}, {0.0891, 0.0942, 0.0964, 0.1019}};

*n* 3 = {{0.4014, 0.4789, 0.5180, 0.5230, 0.5804, 0.6159} , {0.1627, 0.1677, 0.1725, 0.1763, 0.1779, 0.1870, 0.1884, 0.1943, 0.1999, 0.2042, 0.2061, 0.2166}, {0.0448, 0.0551}};

*n* 4 = {{0.3549, 0.3738, 0.4077, 0.4251} , {0.3834, 0.3989, 0.4018, 0.4181, 0.4248, 0.4420, 0.4452, 0.5632}, {0.0576}}; *n* 5 = {{0.3468, 0.4038, 0.4756} , {0.3321, 0.3444, 0.3516, 0.3647}, {0.0612, 0.0642}}.

**Step 3:** Compute the score values of each alternative combined with Equation (14):

> *<sup>s</sup>*(*n*1) = 0.4701,*s*(*n*2) = 0.5611,*s*(*n*3) = 0.6409,*s*(*n*4) = 0.4553,*s*(*n*5) = 0.4989.

**Step 4:** According to the score values, the ranking result of the five ERP systems is determined as *A*3 *A*2 *A*5 *A*1 *A*4.

If the GPHFWG operator is utilized in the steps above to complete the information fusion, the ranking procedures are presented as follows.

#### **Step 1:** See **Step 1**.

**Step 2:** Use the GPHFWG (*λ* = 1) operator to aggregate the standardized PHF evaluation matrix *N*, and the collective evaluation information of each alternative is obtained as

*n* 1 = {{0.2307, 0.2343, 0.2405, 0.2443} , {0.5365, 0.6663, 0.5380, 0.6673}, {0.0600, 0.0797, 0.0661, 0.0856}}; *n* 2 = {{0.2017, 0.2101, 0.2151, 0.2212, 0.2304, 0.2358} , {0.4526, 0.4550, 0.4670, 0.4693, 0.4914, 0.4936, 0.5047, 0.5070},{0.0901, 0.0993, 0.0999, 0.1090}};

*n* 3 = {{0.1986, 0.2057, 0.2081, 0.2342, 0.2427, 0.2454} , {0.3334, 0.3363, 0.3602, 0.3630, 0.3766, 0.3792, 0.4016, 0.4042, 0.4830, 0.4852, 0.5038, 0.5059}, {0.0481, 0.0634}};

*n* 4 = {{0.1024, 0.1037, 0.1044, 0.1058} , {0.5949, 0.5977, 0.6709, 0.6732, 0.7038, 0.7058, 0.7594, 0.7611}, {0.0591}}; *n* 5 = {{0.1613, 0.1660, 0.1696} , {0.5850, 0.6372, 0.6503, 0.6943}, {0.0716, 0.0805}}.

**Step 3:** Compute the score values of each alternative combined with Equation (14):

> *<sup>s</sup>*(*n*1) = 0.2813,*s*(*n*2) = 0.3197,*s*(*n*3) = 0.3778,*s*(*n*4) = 0.1808,*s*(*n*5) = 0.2240.

**Step 4:** According to the score values, the ranking result of the five ERP systems is determined as *A*3 *A*2 *A*1 *A*5 *A*4.

**Example 2.** *Suppose a university wants to introduce excellent foreign professors to improve the level of teaching and scientific research [34]. There are five foreign professors who are selected by the University's human resources department. Based on the priority level, the criteria of investigation is successively morality (C*1*), research ability (C*2*), teaching capacity (C*3*), and educational experience (C*4*); a priority relationship C*1 *C*2 *C*3 *C*4 *exists between the criteria. Then, the PHF evaluation matrix N* = -*nij*. *is presented in Table 2.*


#### **Table 2.** PHF evaluation matrix in Example 2.

Subsequently, we can determine the ranking of the five foreign professors using the Algorithm 2, which are presented as follows.

**Step 1:** Because of all the criteria are the benefit type, the standardized PHF evaluation matrix *N* is as same as the PHF evaluation matrix *N*.

**Step 2:** Compute the values of *Tij* using the Equation (54)


**Step 3:** Use the GPHFPWA (*λ* = 1) operator to aggregate the standardized PHF evaluation matrix *N*, and the collective evaluation information of each alternative is obtained as

*n* 1 = {{0.5003, 0.5177, 0.5360, 0.5382, 0.5522, 0.6230, 0.6361, 0.6516, 0.6516} , {0.0495, 0.0521, 0.0562, 0.0592}, {0.1176, 0.1345, 0.1558, 0.1783}}; *n* 2 = {{0.6290, 0.6396, 0.6446, 0.6548, 0.6723, 0.6816, 0.6861, 0.6950} , {0.0460, 0.0559, 0.0594, 0.0722}, {0.1084, 0.1175, 0.1186, 0.1287, 0.1316, 0.1427, 0.1441, 0.1562}}; *n* 3 = {{0.5335, 0.5533, 0.5537, 0.5727, 0.5996, 0.6169} , {0.0494, 0.0550, 0.0617, 0.0686} , {0.1692, 0.1732, 0.1857, 0.1902, 0.2018, 0.2066, 0.2216, 0.2269}}; ={{0.5593, 0.5820, 0.5978, 0.6185, 0.6425, 0.6609} , {0.0921, 0.1015, 0.1055, 0.1162}, {0.0947, 0.1044, 0.1159, 0.1277}};*n* 5 = {{0.5888, 0.6181, 0.6429, 0.6683} , {0.0605, 0.0670, 0.0796, 0.0881}, {0.1670, 0.1742, 0.1871, 0.1951}}.

**Step 4:** Compute the score values of each alternative combined with Equation (14)

> *<sup>s</sup>*(*n*1) = 0.6888,*s*(*n*2) = 0.7368,*s*(*n*3) = 0.6580,*s*(*n*4) = 0.6978,*s*(*n*5) = 0.6874.

**Step 5:** According to the score values, the ranking result of the five foreign professors is determined as *A*2 *A*4 *A*1 *A*5 *A*3.

If the GPHFPWG operator is utilized in the steps above to complete the information fusion, the ranking procedures are presented as follows.

**Step 1:** See **Step 1**. **Step 2:** See **Step 2**.

*n*

4 *n*

4

**Step 3:** Use the GPHFPWG (*λ* = 1) operator to aggregate standardized the PHF evaluation matrix *N*, and the collective evaluation information of each alternative is obtained as

*n* 1 = {{0.4753, 0.4963, 0.5142, 0.5204, 0.5434, 0.5966, 0.6231, 0.6455, 0.6455} , {0.0527, 0.0597, 0.0609, 0.0678}, {0.1203, 0.1402, 0.1614, 0.1804}}; *n* 2 = {{0.6049, 0.6124, 0.6267, 0.6345, 0.6376, 0.6456, 0.6606, 0.6689} , {0.0668, 0.0739, 0.0809, 0.0878}, {0.1176, 0.1230, 0.1350, 0.1403, 0.1462, 0.1515, 0.1630, 0.1682}}; *n* 3 = {{0.4297, 0.4424, 0.4902, 0.5047, 0.5788, 0.5959} , {0.0526, 0.0571, 0.0703, 0.0748}, {0.2020, 0.2110, 0.2216, 0.2304, 0.2410, 0.2495, 0.2596, 0.2680}}; = {{0.5026, 0.5363, 0.5416, 0.5769, 0.5779, 0.6155} , {0.1119, 0.1178, 0.1264, 0.1322}, {0.0994, 0.1236, 0.1297, 0.1531}};*n* 5= {{0.5596, 0.5733, 0.6141, 0.6292} , {0.0640, 0.0801, 0.0838, 0.0995}, {0.1791, 0.1975, 0.1985, 0.2164}}.

**Step 4:** Compute the score values of each alternative combined with Equation (14)

$$\mathbf{s}(\boldsymbol{\tilde{n}}\_1) = 0.6757, \mathbf{s}(\boldsymbol{\tilde{n}}\_2) = 0.7080, \mathbf{s}(\boldsymbol{\tilde{n}}\_3) = 0.6040, \mathbf{s}(\boldsymbol{\tilde{n}}\_4) = 0.6550, \mathbf{s}(\boldsymbol{\tilde{n}}\_5) = 0.6572.$$

**Step 5:** According to the score values, the ranking result of the five foreign professors is obtained as *A*2 *A*1 *A*5 *A*4 *A*3.
