*5.1. Case Study*

Jharkhand is the eastern state of India, which has 40 percent of the mineral resources of the country, and is the second leading state in terms of mineral wealth, after Chhattisgarh state. It is also known for its vast forest resources. Jamshedpur, Bokaro, and Dhanbad, cities in Jharkhand, are famous for industries from all over the world. After that, it is known as being the state in India that has widespread poverty state, because it is primarily a rural state, as 76 percent of the population lives in villages that depend on agriculture and wages from agriculture. Only 30 percent of the villages are connected by roads, and only 55 percent of the villages have access to electricity and other facilities. But in the today's life, many are looking for ways to make changes in order to better their lives, and, accordingly, many move to the urban cities for better jobs. To stop this emigration, the Jharkhand governmen<sup>t</sup> wants to set up agricultural-based industries in the rural areas. For this, the governmen<sup>t</sup> organized the "Momentum Jharkhand" global investor summit 2017, in Ranchi, to invite companies to invest in the rural areas. The governmen<sup>t</sup> announced the various facilities that were available to be set up as five food processing plants in the rural areas, and the five attributes required for selection of the companies to set them up, namely, project cost (Q1), technical capability (Q2), financial status (Q3), company background (Q4), and other factors (Q5). The three companies that were interested in this projects, Surya Food and Agro Pvt. Ltd. (s1), Mother Dairy Fruit and Vegetable Pvt. Ltd. (s2), and Parle Products Ltd. (s3), were taken as in the form of the alternatives. Then, the main object of the governmen<sup>t</sup> was to choose the best company among them for the task. In order to fulfill this, a decision maker evaluated these and gave their preferences in the term of T-SFS, and their preference values were summarized in the form of a decision-matrix, shown in Table 1 as follows.

**Table 1.** Input information related to each alternative.


The given problem was solved using two approaches. First it was solved using new interactive operators showing their applicability. Then it was solved using geometric aggregation operators proposed in [34], showing their failure.

#### **Solution using proposed operators:**

**Step 1** With some calculations, it was found that all the values in Table 1 were T-SFNs for *t* = 3. **Step 2** By taking *ω* = (0.18, 0.22, 0.16, 0.21, 0.23)*<sup>T</sup>* we found *Pjk* and their values were summarized as below.


**Step 3** Now we had to find the score of each attribute of all alternatives, and their computed values were given as below


By comparing the score values, we had

$$\begin{array}{l} SC(P\_{12}) > SC(P\_{15}) > SC(P\_{11}) > SC(P\_{14}) > SC(P\_{13})\\ SC(P\_{22}) > SC(P\_{25}) > SC(P\_{23}) > SC(P\_{21}) > SC(P\_{24})\\ SC(P\_{34}) > SC(P\_{32}) > SC(P\_{35}) > SC(P\_{31}) > SC(P\_{33}) \end{array}$$

Based on above score analysis, we found *<sup>P</sup>σ*(*jk*) and summarized them as


**Step 4** By using the normal distribution-based method, we go<sup>t</sup> *w* = (0.1117, 0.2365, 0.3036, 0.2365, 0.1117)*<sup>T</sup>*, and by using the defined aggregation operators, we had

$$= \left( \begin{array}{c} P\_1 = \, ^\circ - SFFGIA\_{\omega,w}(P\_{11}, P\_{12}, P\_{13}, P\_{14}, P\_{15})\\ \sqrt[3]{\prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(1k)}}^3\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(1 - \left(m\_{\overline{P}\_{\sigma(1k)}}^3 + l\_{\overline{P}\_{\sigma(1k)}}^3 + n\_{\overline{P}\_{\sigma(1k)}}^3\right)\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(l\_{\overline{P}\_{\sigma(1k)}}^3\right)^{w\_j} \right) \\\ \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - l\_{\overline{P}\_{\sigma(1k)}}^3\right)^{w\_j} \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(1k)}}^3\right)^{w\_j}} \\\ = \left( 0.9380, 0.4264, 0.4928 \right) \end{array} \right)$$

$$= \left( \begin{array}{c} P\_2 = \, ^\circ - SFHIGA\_{\omega, w}(P\_{21}, P\_{22}, P\_{23}, P\_{24}, P\_{25})\\ \sqrt[3]{\prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(2k)}}^3\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(1 - \left(m\_{\overline{P}\_{\sigma(2k)}}^3 + l\_{\overline{P}\_{\sigma(2k)}}^3 + n\_{\overline{P}\_{\sigma(2k)}}^3\right)\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(l\_{\overline{P}\_{\sigma(2k)}}^3\right)^{w\_j} \right) \\\ \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - l\_{\overline{P}\_{\sigma(2k)}}^3\right)^{w\_j}, \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(2k)}}^3\right)^{w\_j}} \\\ = \ (0.9420, 0.3390, 0.5296) \end{array} \right)$$

$$= \left( \begin{array}{c} P\_3 = \, ^\circ - SFHGIA\_{\omega, w} (P\_{31} \, P\_{32} \, P\_{33} \, P\_{34} \, P\_{35})\\ \sqrt[3]{\prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(3k)}}^3\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(1 - \left(m\_{\overline{P}\_{\sigma(3k)}}^3 + t\_{\overline{P}\_{\sigma(3k)}}^3 + n\_{\overline{P}\_{\sigma(3k)}}^3\right)\right)^{w\_j} - \prod\limits\_{j=1}^5 \left(t\_{\overline{P}\_{\sigma(3k)}}^3\right)^{w\_j} \right) \\\ \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - t\_{\overline{P}\_{\sigma(3k)}}^3\right)^{w\_j}} \sqrt[3]{1 - \prod\limits\_{j=1}^5 \left(1 - n\_{\overline{P}\_{\sigma(3k)}}^3\right)^{w\_j}} \\ = (0.9779, 0.9713, 0.3906) \end{array} \right)$$


$$\text{SC}(P\_3) > \text{SC}(P\_1) > \text{SC}(P\_2)$$

The comparison of score values indicated that *P*3 had a greater score value. So, the third company was the best option. Thus, by using the new geometric interaction averaging operators a MADM problem was successfully solved.

#### **Solution using aggregation operators proposed in [34]:**

**Step 1** The input preferences related to each alternative was summarized in Table 1 for *t* = 3.

**Step 2** By using weight vector *ω* = (0.18, 0.22, 0.16, 0.21, 0.23)*<sup>T</sup>* we found *P jk*as follows

$$\begin{aligned} \mathbf{k} &= 1 \\ \mathbf{j} &= 1 \\ \mathbf{j} &= 2 \\ \mathbf{j} &= 2 \\ \mathbf{j} &= 3 \\ \mathbf{j} &= 3 \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 2 \\ \mathbf{0.5395} \\ \mathbf{0.5816} \\ \mathbf{0.5816} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 2 \\ \mathbf{0.5941} \\ \mathbf{0.5816} \\ \mathbf{0.5841} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 3 \\ \mathbf{0.4665} \\ \mathbf{0.5125} \\ \mathbf{0.1703} \\ \mathbf{0.14370} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 3 \\ \mathbf{0.4665} \\ \mathbf{0.1258} \\ \mathbf{0.1703} \\ \mathbf{0.1481} \\ \mathbf{0.1621} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 4 \\ \mathbf{k} &= 5 \\ \mathbf{0.5768} \\ \mathbf{0.1259} \\ \mathbf{0.1484} \\ \mathbf{0.1621} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 4 \\ \mathbf{0.5851} & \mathbf{0.0485} \\ \mathbf{0.1252} \\ \mathbf{0.1252} \\ \mathbf{0.1252} \\ \mathbf{0.1252} \\ \end{aligned} \quad \begin{aligned} \mathbf{k} &= 5 \\ \mathbf{0.1485} & \mathbf{0.1621} \\ \mathbf{0.1252} & \mathbf{0.1621} \\ \mathbf{0.1252} & \mathbf{0.1621} \\ \mathbf{0.1252} & \mathbf{0.1252} \\ \mathbf{0.1621} \\ \mathbf{0.1621} \\ \mathbf{0.1203} \end{$$

**Step 3** Now, we had to find the score of each attribute of all alternatives.


By comparing the score values, we had

$$\begin{array}{l} \text{SC}(P'\_{12}) > \text{SC}(P'\_{11}) > \text{SC}(P'\_{13}) > \text{SC}(P'\_{14}) > \text{SC}(P'\_{15})\\ \text{SC}(P'\_{22}) > \text{SC}(P'\_{23}) > \text{SC}(P'\_{25}) > \text{SC}(P'\_{21}) > \text{SC}(P'\_{24})\\ \text{SC}(P'\_{34}) > \text{SC}(P'\_{32}) > \text{SC}(P'\_{31}) > \text{SC}(P'\_{35}) > \text{SC}(P'\_{33}) \end{array}$$

Based on above score analysis, we found *P σ*(*jk*)

k = 1 k = 2 k = 3 k = 4 k = 5 j = 1 ⎛⎜⎝ 0.9041, 0.4665, 0.4125 ⎞⎟⎠ ⎛⎜⎝ 0.7054, 0.5359, 0.5816 ⎞⎟⎠ ⎛⎜⎝ 0.4655, 0.2759, 0.0928 ⎞⎟⎠ ⎛⎜⎝ 0.4874, 0.2825, 0.4063 ⎞⎟⎠ ⎛⎜⎝ 0.5738, 0.3486, 0.5221 ⎞⎟⎠ j = 2 ⎛⎜⎝ 0.6776, 0.1703, 0.3095 ⎞⎟⎠ ⎛⎜⎝ 0.7330, 0.3817, 0.5614 ⎞⎟⎠ ⎛⎜⎝ 0.4553, 0.1571, 0.4184 ⎞⎟⎠ ⎛⎜⎝ 0.5180, 0.4384, 0.5816 ⎞⎟⎠ ⎛⎜⎝ 0.3826, 0.0891, 0.6086 ⎞⎟⎠ j = 3 ⎛⎜⎝ 0.5863, 0.1845, 0.2033 ⎞⎟⎠ ⎛⎜⎝ 0.4811, 0.3650, 0.1032 ⎞⎟⎠ ⎛⎜⎝ 0.4370, 0.1259, 0.1931 ⎞⎟⎠ ⎛⎜⎝ 0.5563, 0.0708, 0.5221 ⎞⎟⎠ ⎛⎜⎝ 0, 0, 0.4662 ⎞⎟⎠

**Step 4** By using the normal distribution-based method, we go<sup>t</sup> *w* = (0.1117, 0.2365, 0.3036, 0.2365, 0.1117)*<sup>T</sup>*, and by using the defined aggregation operators, we had

*P* 1 = *T* − *SFHGIA<sup>ω</sup>*,*<sup>w</sup>*-*<sup>P</sup>* 11, *P* 12, *P* 13, *P* 14, *P* 15. = ⎛⎜⎝ 3K 5∏*<sup>j</sup>*=<sup>1</sup>5m3*Pσ*(1*k*) + <sup>i</sup>3*Pσ*(1*k*)6wj − 5∏*j*=1 (*i*3*Pσ*(1*k*))*wj* , 5∏*<sup>j</sup>*=<sup>1</sup>*iPσ*(1*k*)*wj* , 3K1 − 5∏*j*=151 − *<sup>n</sup>*3*Pσ*(1*k*)6*wj* ⎞⎟⎠ = (0.5750, 0.3533, 0.4473) *P* 2 = *T* − *SFHGIA<sup>ω</sup>*,*<sup>w</sup>*-*<sup>P</sup>* 21, *P* 22, *P* 23, *P* 24, *P* 25. = ⎛⎜⎝ 3K 5∏*<sup>j</sup>*=<sup>1</sup>5m3*Pσ*(2*k*) + <sup>i</sup>3*Pσ*(2*k*)6wj − 5∏*j*=1 (*i*3*Pσ*(2*k*))*wj* , 5∏*<sup>j</sup>*=<sup>1</sup>*iPσ*(2*k*)*wj* , 3K1 − 5∏*j*=151 − *<sup>n</sup>*3*Pσ*(2*k*)6*wj* ⎞⎟⎠ = (0.5384, 0.1970, 0.5721) *P* 3 = *T* − *SFHGIA<sup>ω</sup>*,*<sup>w</sup>*-*<sup>P</sup>* 31, *P* 32, *P* 33, *P* 34, *P* 35. = ⎛⎜⎝ 3K 5∏*<sup>j</sup>*=<sup>1</sup>5m3*Pσ*(3*k*) + <sup>i</sup>3*Pσ*(3*k*)6wj − 5∏*j*=1 (*i*3*Pσ*(3*k*))*wj* , 5∏*<sup>j</sup>*=<sup>1</sup>*iPσ*(3*k*)*wj* , 3K1 − 5∏*j*=151 − *<sup>n</sup>*3*Pσ*(3*k*)6*wj* ⎞⎟⎠ = (0, 0, 0.3692)

This seems meaningless because membership and abstinence of only one T-SFN is zero, but existing operators make a whole aggregated value zero.

**Step 5** This step involved the computation of score values:

$$\begin{array}{l} \text{SC}(P\_1) = 0.1006 \\ \text{SC}(P\_2) = -0.0312 \\ \text{SC}(P\_3) = -0.0503 \end{array}$$

**Step 6** By comparing score values, we go<sup>t</sup>

$$\text{SC}(P\_1) > \text{SC}(P\_2) > \text{SC}(P\_3)$$

From the above example, the applicability of the proposed operators could easily be checked by comparing the results obtained using new and existing geometric aggregation operators. It was noticed that whenever membership and abstinence of one TSFN became zero, then the aggregated value using existing aggregation operators seemed impractical. However, the aggregated value using new geometric interactive aggregation operators seemed significant and consistent.

#### *5.2. Advantages of the Proposed Work*

In this section, we prove the generalization of proposed work over the existing literature. Here we observed that under some certain conditions the proposed aggregation operators became the existing aggregation operators under different environment, which shows the superiority of our proposed work.

Consider the T-SFWGIA operator defined as

$$(T - SFWGIA\_{\overline{w}}(P\_1, P\_2, \dots, \dots, P\_k) = \begin{pmatrix} \sqrt{\prod\_{j=1}^k \left(1 - n\_j^t\right)^{w\_j} - \prod\_{j=1}^k \left(1 - \left(m\_j^t + \overleftarrow{1}\_j^t + n\_j^t\right)\right)^{w\_j}} - \prod\_{j=1}^k \left(i\_j^t\right)^{w\_j} \\\sqrt{1 - \prod\_{j=1}^k \left(1 - i\_j^t\right)^{w\_j}} \sqrt{1 - \prod\_{j=1}^k \left(1 - n\_j^t\right)^{w\_j}} \end{pmatrix} \tag{9}$$

(1) If we take *t* = 2, the Equation (9) becomes spherical fuzzy weighted geometric interaction averaging operator (SFWGIA operator) and we have

$$SFWGIA\_w(\mathbf{P}\_1, \mathbf{P}\_2, \dots, \mathbf{P}\_k) = \begin{pmatrix} \sqrt{\prod\_{j=1}^k \left(1 - n\_j^2\right)^{w\_j}} - \prod\_{j=1}^k \left(1 - \left(\mathbf{m}\_j^2 + \mathbf{i}\_j^2 + n\_j^2\right)\right)^{w\_j} - \prod\_{j=1}^k \left(\mathbf{i}\_j^2\right)^{w\_j} \\\sqrt{1 - \prod\_{j=1}^k \left(1 - \mathbf{i}\_j^2\right)^{w\_j}} \sqrt{1 - \prod\_{j=1}^k \left(1 - n\_j^2\right)^{w\_j}} \end{pmatrix}$$

(2) If we take *t* = 1, the Equation (9) becomes picture fuzzy weighted geometric interaction averaging operator (PFWGIA operator) and we have

$$PFWGIA\_w(P\_1, P\_2, \dots, \dots, P\_k) = \begin{pmatrix} \prod\_{j=1}^k \left(1 - n\_j\right)^{w\_j} - \prod\_{j=1}^k \left(1 - \left(\mathbf{m}\_{\bar{j}} + \mathbf{i}\_{\bar{j}} + n\_{\bar{j}}\right)\right)^{w\_{\bar{j}}} - \prod\_{j=1}^k \left(i\_{\bar{j}}\right)^{w\_{\bar{j}}}\\ 1 - \prod\_{j=1}^k \left(1 - i\_{\bar{j}}\right)^{w\_{\bar{j}}}, \; 1 - \prod\_{j=1}^k \left(1 - n\_j\right)^{w\_{\bar{j}}} \end{pmatrix}$$

(3) If we take *t* = 2 and *i* = 0, the Equation (9) becomes Pythagorean fuzzy weighted geometric interaction averaging operator (PyFWGIA operator) and we have

$$\Pr\_{\mathbf{Y}}\text{FWGIA}\_{\mathbf{w}}(\mathbf{P}\_{\mathbf{l}},\mathbf{P}\_{\mathbf{2}},\dots,\mathbf{Z},\mathbf{P}\_{\mathbf{k}}) = \begin{pmatrix} \sqrt{\prod\_{j=1}^{k} \left(1 - n\_{j}^{2}\right)^{w\_{j}} - \prod\_{j=1}^{k} \left(1 - \left(\mathbf{m}\_{j}^{2} + n\_{j}^{2}\right)\right)^{w\_{j}}} \\ \sqrt{1 - \prod\_{j=1}^{k} \left(1 - n\_{j}^{2}\right)^{w\_{j}}} \end{pmatrix}$$

(4) If we take *t* = 1 and *i* = 0, the Equation (9) becomes intuitionistic fuzzy weighted geometric interaction averaging operator (IFWGIA operator) and we have

$$IFWGIA\_w(P\_1, P\_2, \dots, \dots, P\_k) = \begin{pmatrix} \prod\_{j=1}^k \left(1 - n\_j\right)^{w\_j} - \prod\_{j=1}^k \left(1 - \left(\mathbf{m}\_j + n\_j\right)\right)^{w\_j} \\ 1 - \prod\_{j=1}^k \left(1 - n\_j\right)^{w\_j} \end{pmatrix}$$

Similarly, T-SFOWGIA and T-SFHGIA operators can be converted to the existing operators. All of this clearly indicated that our proposed work could be used in the problems described in existing literature, but the operators of existing literature are unable to deal with problems of T-spherical fuzzy information. For example, if we look at Example 2, it can be seen that none of the existing operators can be applied to such problems where information is in the form of T-SFNs.
