*Hotel Booking Problem*

In any trip, the problem of accommodation is one of the most important issues. The best option is always selected after comparing different residences based on some parameters, such as facilities and location of hotels and the budget. In this section, we discuss the problem of hotel booking, which is about selecting the best hotel to stay regarding a list of criteria, to provide a real-life example which shows the application of our method in decision-making problems. The input data are obtained from the "www.agoda.com" website, an online hotel booking service. This website provides some online questioners that passengers (guests) can fill them up to share their experiences about the hotel that they have staid. Guests are classified into the five main groups: families with young children, families with elder children, couples, solo travelers, and group of friends. Each hotel is characterized based on several criteria including "Comfort", "Services", "Location", and "Food" according to the guests' idea by numbers between zero and ten. Note that, here, all collected data are divided to ten to be in the unit interval [0, 1].

**Example 2.** *Let us to suppose that a travel agency in Iran wants to offer a luxury group tour for Kuala Lumpur, Malaysia, to their customers. The list of ten four-star hotels in Kuala Lumpur H* = {*h*1, ··· , *h*10}*, which are compared with each other based on the following four parameters P* = {*p*1*=Comfort, p*2*=Services, p*3*=Location, p*4*=Food*}*, is chosen by this agency from the "www.agoda.com" website. The comments of five passengers, who filled on-line questionnaires, for five different categories "Families with Young Children", "Families with Elder Children", "Couples", "Solo Travelers", and "Group of Friends", whose weighting vector is ω* = (0.25, 0.25, 0.2, 0.15, 0.15)*T, are selected by the travel agency as an input data of five experts. A hotel may be selected as the best accommodation place if at least three individuals of five people are satisfied with it. Since, in general, the importance degree of all criteria for different decision makers are not the same, this company defined the weighting vector λ* = (0.3, 0.2, 0.35, 0.15)*<sup>T</sup> for the parameters based on the frequency of these parameters in the comments of the passengers. Let this travel agency also consider two different aspects for each parameters as follows: The parameter "Comfort" includes "Cleanliness" and "Staff Performance" with the weighting vector <sup>λ</sup>p***1** = (0.7, 0.3)*T. The parameter "Services" includes "Facilities" and "Free Wi-Fi connectivity" with the weighting vector <sup>λ</sup>p***2** = (0.8, 0.2)*T. The parameter "Location" includes "Close to Tourist Attractions" and "In the Green Surroundings" with the weighting vector <sup>λ</sup>p***3** = (0.7, 0.3)*T. The parameter "Food" includes "Breakfast" and "Lunch and Dinner" with the weighting vector <sup>λ</sup>p***4** = (0.5, 0.5)*T. Tables 3–7 show the evaluation of the hotels based on these five passengers' comments by using m-polar fuzzy soft sets.*


**Table 3.** Tabular representation of *MFSS* (*f*1, *<sup>P</sup>*).

**Table 4.** Tabular representation of *MFSS* (*f*2, *<sup>P</sup>*).


**Table 5.** Tabular representation of *MFSS* (*f*3, *<sup>P</sup>*).


**Table 6.** Tabular representation of *MFSS* (*f*4, *<sup>P</sup>*).



**Table 7.** Tabular representation of *MFSS* (*f*5, *<sup>P</sup>*).

*Since the required consensus degree is α* = 3*, then L* = 5! 3!(<sup>5</sup>−<sup>3</sup>)! = 10*. All 10 different combinations of three of five objects can be listed as follows:* <sup>Δ</sup>5,3(1) = {1, 2, <sup>3</sup>}*,* <sup>Δ</sup>5,3(2) = {1, 2, <sup>4</sup>}*,* <sup>Δ</sup>5,3(3) = {1, 2, <sup>5</sup>}*,* <sup>Δ</sup>5,3(4) = {1, 3, <sup>4</sup>}*,* <sup>Δ</sup>5,3(5) = {1, 3, <sup>5</sup>}*,* <sup>Δ</sup>5,3(6) = {1, 4, <sup>5</sup>}*,* <sup>Δ</sup>5,3(7) = {2, 3, <sup>4</sup>}*,* <sup>Δ</sup>5,3(8) = {2, 3, <sup>5</sup>}*,* <sup>Δ</sup>5,3(9) = {2, 4, <sup>5</sup>}*, and* <sup>Δ</sup>5,3(10) = {3, 4, <sup>5</sup>}*. (Steps 1,2).*

*Step 3. Utilize data given in Tables 3–7 with the weighting vector ω* = (0.25, 0.25, 0.2, 0.15, 0.15)*<sup>T</sup> and M-pFSMWM operator proposed in Theorem 3 to get the collective matrix C* ¯ = [*c***¯***iy*]<sup>10</sup>×<sup>4</sup> *as below:*


*Step 4. Utilize Equation* (6) *to compute matrices R* ˜(*py*)=[*r*˜*ij*(*py*)]<sup>10</sup>×10*, where y* = 1, 2, 3, 4*, as follows:*


$$
\bar{R}(p\_2) = \begin{bmatrix}
0.5 & 0.55 & 0.52 & 0.48 & 0.51 & 0.52 & 0.49 & 0.49 & 0.49 & 0.47 \\
0.44 & 0.5 & 0.46 & 0.43 & 0.46 & 0.43 & 0.44 & 0.43 & 0.41 \\
0.47 & 0.53 & 0.5 & 0.46 & 0.49 & 0.49 & 0.46 & 0.47 & 0.46 & 0.44 \\
0.51 & 0.56 & 0.53 & 0.52 & 0.53 & 0.506 & 0.506 & 0.504 & 0.48 \\
0.48 & 0.53 & 0.507 & 0.47 & 0.5 & 0.505 & 0.47 & 0.47 & 0.47 & 0.45 \\
0.47 & 0.53 & 0.501 & 0.46 & 0.49 & 0.5 & 0.47 & 0.47 & 0.46 & 0.44 \\
0.504 & 0.56 & 0.53 & 0.49 & 0.52 & 0.52 & 0.52 & 0.5 & 0.501 & 0.49 & 0.47 \\
0.503 & 0.56 & 0.52 & 0.49 & 0.52 & 0.52 & 0.49 & 0.5 & 0.49 & 0.47 \\
0.505 & 0.56 & 0.53 & 0.49 & 0.52 & 0.53 & 0.501 & 0.502 & 0.5 & 0.47 \\
0.53 & 0.58 & 0.55 & 0.51 & 0.54 & 0.55 & 0.52 & 0.52 & 0.52 & 0.52
\end{bmatrix}
$$

$$R(p\_3) = \begin{bmatrix} 0.5 & 0.52 & 0.54 & 0.505 & 0.51 & 0.5006 & 0.51 & 0.52 & 0.52 & 0.52\\ 0.47 & 0.5 & 0.51 & 0.48 & 0.49 & 0.47 & 0.49 & 0.502 & 0.501 & 0.503\\ 0.45 & 0.48 & 0.5 & 0.46 & 0.47 & 0.46 & 0.47 & 0.48 & 0.48 & 0.48\\ 0.49 & 0.51 & 0.53 & 0.5 & 0.508 & 0.49 & 0.509 & 0.51 & 0.51 & 0.52\\ 0.48 & 0.508 & 0.52 & 0.49 & 0.5 & 0.48 & 0.501 & 0.51 & 0.509 & 0.51\\ 0.49 & 0.52 & 0.53 & 0.505 & 0.51 & 0.5 & 0.51 & 0.52 & 0.52 & 0.52\\ 0.48 & 0.506 & 0.52 & 0.49 & 0.49 & 0.48 & 0.5 & 0.509 & 0.508 & 0.51\\ 0.47 & 0.49 & 0.51 & 0.48 & 0.48 & 0.47 & 0.49 & 0.5 & 0.49 & 0.501\\ 0.47 & 0.49 & 0.51 & 0.47 & 0.48 & 0.47 & 0.49 & 0.5008 & 0.5 & 0.502\\ 0.47 & 0.49 & 0.51 & 0.47 & 0.48 & 0.47 & 0.48 & 0.49 & 0.49 & 0.5 \end{bmatrix}$$

$$R(p\_4) = \begin{bmatrix} 0.5 & 0.504 & 0.52 & 0.51 & 0.48 & 0.49 & 0.49 & 0.49 & 0.53 & 0.46\\ 0.49 & 0.5 & 0.51 & 0.506 & 0.48 & 0.49 & 0.48 & 0.49 & 0.52 & 0.45\\ 0.47 & 0.48 & 0.5 & 0.48 & 0.46 & 0.47 & 0.46 & 0.47 & 0.509 & 0.43\\ 0.48 & 0.49 & 0.51 & 0.5 & 0.47 & 0.48 & 0.47 & 0.48 & 0.52 & 0.44\\ 0.51 & 0.51 & 0.53 & 0.52 & 0.5 & 0.508 & 0.502 & 0.509 & 0.54 & 0.47\\ 0.503 & 0.508 & 0.52 & 0.51 & 0.49 & 0.5 & 0.49 & 0.501 & 0.53 & 0.46\\ 0.509 & 0.51 & 0.53 & 0.52 & 0.49 & 0.505 & 0.5 & 0.506 & 0.54 & 0.46\\ 0.502 & 0.507 & 0.52 & 0.51 & 0.49 & 0.49 & 0.49 & 0.5 & 0.53 & 0.46\\ 0.46 & 0.47 & 0.49 & 0.47 & 0.45 & 0.46 & 0.45 & 0.46 & 0.5 & 0.42\\ 0.54 & 0.54 & 0.56 & 0.55 & 0.52 & 0.53 & 0.53 & 0.53 & 0.57 & 0.57 \end{bmatrix}$$

*Step 5. Now, the matrix A* ˜ = [*a***˜***ij*]<sup>10</sup>×<sup>10</sup> *is computed, using information given in matrices R* ˜(*py*) *(y* = 1, 2, 3, 4*), to obtain a collective four-polar fuzzy soft preference matrix as the following:*


⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

*Step 6. Calculate the score value of each alternatives according to the collective 4-polar fuzzy soft preference matrix A* ˜ *and Equation* (9) *as the following: S*1 = 6.95*, S*2 = 2.15*, S*3 = 1.25*, S*4 = 5.85*, S*5 = 6*, S*6 = 4.1*, S*7 = 5.8*, S*8 = 3.65*, S*9 = 3.05*, and S*10 = 6.2*.*

*Step 7. Thus, we have: h*1*h*10*h*5*h*4*h*7*h*6*h*8*h*9*h*2*h*3*. Thus, the first hotel, called h*1*, is the best option to stay while h*3 *should not be selected for accommodation.*

In Example 2, the desirable alternative is accepted by most of the decision makers where "most" is interpreted as acceptable by 60% of decision makers , i.e., at least three individuals of total five decision makers are satisfied. To check the impact of consensus degree on the final solution, we compare the obtained results in Example 2 with the result of full agreement, given by the following example in which *α* = 5 is used as the consensus degree.

**Example 3.** *(Example 2 continued) Let us reconsider the "Hotel Booking Problem" involving M-pFSSs* (*f*1, *<sup>P</sup>*),(*f*2, *<sup>P</sup>*),(*f*3, *<sup>P</sup>*),(*f*4, *<sup>P</sup>*), *and* (*f*5, *P*) *which is discussed earlier in Example 2 where H* = {*h*1, ··· , *h*10} *is the set of alternatives and P* = {*p*1, *p*2, *p*3, *p*4} *is the set of parameters.*

*Here, the proposed Algorithm 1 is applied for case α* = 5*, where L* = *C*5,5 = 1 *and* <sup>Δ</sup>5,5(1) = {1, 2, 3, 4, <sup>5</sup>}*, to get the most desirable alternative based on a unanimous consensus. After applying Algorithm 1, the scores of alternatives are computed as follows: S*1 = 6.65*, S*2 = 2.45*, S*3 = 2.15*, S*4 = 6.55*, S*5 = 7.1*, S*6 = 4.1*, S*7 = 4.95*, S*8 = 3.55*, S*9 = 0.6*, and S*10 = 6.9*. Therefore, we get: h*5 *h*10 *h*1 *h*4 *h*7 *h*6 *h*8 *h*2 *h*3 *h*9 *which shows h*5 *is the alternative accepted by all decision makers. Hotel h*1*, which is the best option for accommodation according to the most decision makers' views, is the third most desirable alternative. Figure 2 shows the effect of different consensus degrees α* = 3 *and α* = 5 *on scores of alternatives.*

**Figure 2.** The effect of consensus degree on scores of alternatives based on the M-pFSMWM aggregation method.

Now, we reconsider Examples 2 and 3 according to Algorithm 2 which is based on M-pFSIOWA operator and M-pFSIOWG operator.

**Example 4.** *(Examples 2 and 3 continued) To apply Algorithm 2 for solving MAGDM problem given in Examples 2 and 3, we first need to re-order the M-pFSSs* (*f*1, *<sup>P</sup>*), ··· ,(*f*5, *P*) *based on weighting vector ω* = (0.25, 0.25, 0.2, 0.15, 0.15)*<sup>T</sup> for the decision makers f*1, ··· , *f*<sup>5</sup>*. Subsequently, we get: Fs*1*y<sup>i</sup>* = *f s*1 (*py*)(*ui*)*, Fs*2*y<sup>i</sup>* = *f s*2 (*py*)(*ui*)*, Fs*3*y<sup>i</sup>* = *f s*3 (*py*)(*ui*)*, Fs*4*y<sup>i</sup>* = *f s*4 (*py*)(*ui*)*, and Fs*5*y<sup>i</sup>* = *f s*5 (*py*)(*ui*) *for s* = 1, 2*; i* = 1, 2, ··· , 10*; and y* = 1, 2, 3, 4*. The associated weighting vector w* = (*<sup>w</sup>*1, *w*2, ··· , *<sup>w</sup>*5) *is then generated by wk* = *Q*( *k*5 ) − *Q*( *k*−1 5 ) *for k* = 1, ··· , 5 *where Q*(.) *is a quantifier function Q* : [0, 1] → [0, 1]*.*

*First, for case "most of the decision makers" (i.e., α* = 3*), where "most" is interpreted as* 60% *or* 35 *of all data, the quantifier function may be defined by*

$$Q\_{most}(z) = \begin{cases} 0 & \text{if } z \le \frac{1}{5} \\ \frac{z - 1/9}{\frac{2/5}{5/5}} & \text{if } \frac{1}{5} < z < \frac{3}{5} \\ 1 & \text{if } z \ge \frac{3}{5} \end{cases} \tag{12}$$

*Therefore, we get w*1 = *w*4 = *w*5 = 0*, w*2 = *w*3 = 0.5 *and subsequently w* = (0, 0.5, 0.5, 0, 0)*T. Thus, by using Equations (10) and (11), we have:*

$$[M - pFSIOWA \langle f\_{1\prime}f\_{2\prime}f\_{3\prime}f\_{4\prime}f\_{5} \rangle]^s (p\_y)(u\_i) = \frac{F\_{2yi}^s + F\_{3yi}^s}{2}$$

*and*

$$[M - pFSIOWG\langle f\_{1\prime}, f\_{2\prime}, f\_{3\prime}, f\_{4\prime}, f\_{5}\rangle]^s (p\_y)(\mu\_i) = \sqrt{F\_{2yi}^s \cdot F\_{3yi}^s}$$

*for s* = 1, 2*; i* = 1, 2, ··· , 10*; and y* = 1, 2, 3, 4*.*

> *For the unanimous consensus (i.e., α* = 5*), the quantifier function may be defined by*

$$Q\_{all}(z) = \begin{cases} 0 & \text{if } z < 1\\ 1 & \text{if } z = \frac{5}{5} = 1 \end{cases} \tag{13}$$

*Thus, the weighting vector is w* = (0, 0, 0, 0, 1)*T. Subsequently, by using Equations* (10) *and* (11)*, we have*

$$[M - pFSIOWA \langle f\_1, f\_2, f\_3, f\_{4\prime}f\_5 \rangle]^s (p\_y)(\mu\_i) = F\_{5gi}^s$$

*and*

$$[M - pFSIOWG\langle f\_1, f\_2, f\_3, f\_4, f\_5 \rangle]^s (p\_y)(u\_i) = F\_{5yi}^s$$

*for s* = 1, 2*; i* = 1, 2, ··· , 10*; and y* = 1, 2, 3, 4*.*

*Thus, based on the M-pFSIOWA operator and M-pFSIOWG operator the resultant collective m-polar fuzzy soft matrix C* ¯ = [¯*ciy*]<sup>10</sup>×<sup>4</sup> *where*

$$\begin{aligned} \mathfrak{E}\_{\mathrm{ij}} &= \left( \frac{f\_2^1(p\_\mathcal{Y})(u\_i) + f\_3^1(p\_\mathcal{Y})(u\_i)}{2}, \frac{f\_2^2(p\_\mathcal{Y})(u\_i) + f\_3^2(p\_\mathcal{Y})(u\_i)}{2} \right); \\ \mathfrak{E}\_{\mathrm{ij}} &= \left( \sqrt{f\_2^1(p\_\mathcal{Y})(u\_i)}, f\_3^1(p\_\mathcal{Y})(u\_i), \sqrt{f\_2^2(p\_\mathcal{Y})(u\_i)}, f\_3^2(p\_\mathcal{Y})(u\_i) \right); \end{aligned}$$

*and*

> *c* **¯** *iy* = - *f* 15 (*py*)(*ui*), *f* 25 (*py*)(*ui*).

*is derived for different cases α* = 3 *and α* = 5*, respectively. Then, Steps 4–7 of Algorithm 1 are used to compare the scores of alternatives for both cases α* = 3 *and α* = 5*.*

The obtained results from M-pFSMWM, M-pFSIOWA, and M-pFSIOWG operators are reported in Table 8.


**Table 8.** Comparison results of Examples 2–4 for different M-pFS-based aggregation methods.

*α* refers to consensus degree.

It can be seen that, for case *α* = 5, *h*10 is the best option selected by all methods. For *α* = 3, the M-pFSMWM method selects *h*1, however the tenth hotel, *h*10, is chosen by the two other methods. Accordingly, the scores of alternatives for different consensus degrees *α* = 3 and *α* = 5 are compared in Figure 3a–c.

(**c**) M-pFSIOWG Method

**Figure 3.** The effect of consensus degree on scores of alternatives based on different M-pFS-based aggregation methods.
