In this section, some examples are provided to illustrate the potential application of the proposed method.
Example 5.1 (T1PyFN model)
Consider a transportation problem with the conditions of Table 3.
Table 3 Input data for Pythagorean transportation problem of type-I
Step 1 Calculate the score value of each PFN cost, replace all them by its score value and obtain a crisp transportation problem. This step is shown in Table 4.
Table 4 The defuzzified Pythagorean fuzzy transportation problem of Example 5.1
Step 2 To check the balance of transportation we execute:
$$\begin{aligned} \sum {a_{i} } &= 26 + 24 + 30 = 80\;{\text{and}}\\ \sum {b_{i} } &= 17 + 23 + 28 + 12 = 80.\end{aligned} $$
Therefore, it is a balanced transportation problem.
Step 3 Now to find an initial basic feasible solution, we proceed with Algorithm 1 as it mentions in the main algorithm. After maximum allotting in the cell (2, 3) we get a new table. Table 5 shows the first allotment with penalties (Tables 6, 7, 8).
Table 5 First allotment with penalties in Example 5.1
Table 6 Second allotment with penalties in Example 5.1
Table 7 Third allotment with penalties in Example 5.1
Table 8 Complete allotment in Example 5.1
Table 9 shows the initial basic feasible solution of Example 5.1.
Table 9 The initial basic feasible solution of Example 5.1
Hence, the initial basic feasible solution (IBFS) is as follows:
$$\begin{aligned} (O_{1} ,D_{1} ) &= x_{11} = 17,\;(O_{1} ,D_{2} ) = x_{12} = 9,\\ (O_{2} ,D_{3} ) &= x_{23} = 24,\;(O_{3} ,D_{2} ) = x_{32} = 14,\\ (O_{3} ,D_{3} ) &= x_{33} = 4,\;{\text{and}}\;(O_{3} ,D_{4} ) = x_{34} = 12.\end{aligned} $$
Also, the minimum cost of IBFS is obtained as follows:
$$ Min = 17 \times 0.335 + 9 \times 0.545 + 24 \times 0.26 + 14 \times 0.815 + 4 \times 0.6 + 12 \times 0.9 = 41.45. $$
Steps 4–5 Now, we test the optimality of the transportation problem. Since the m + n − 1 = 6, it is a degenerate solution, and we need to proceed to test the optimality. To obtain the optimality, we use Lingo software. Therefore, the optimal solution is as follows:
$$\begin{aligned} (O_{1} ,D_{1} ) &= x_{11} = 17,\;(O_{1} ,D_{2} ) = x_{12} = 9,\\ (O_{2} ,D_{3} ) &= x_{23} = 24,\;(O_{3} ,D_{2} ) = x_{32} = 14,\\ (O_{3} ,D_{3} ) &= x_{33} = 4,\;{\text{and}}\;(O_{3} ,D_{4} ) = x_{34} = 12.\end{aligned} $$
Step 6 Now put all \( x_{ij} \) in the above equation; so we get:
$$ {\text{Min}} = 17 \times 0.335 + 9 \times 0.545 + 24 \times 0.26 + 14 \times 0.815 + 4 \times 0.6 + 12 \times 0.9. $$
Minimum cost = 41.45.
Example 5.2 (T2PyFN model)
Consider type-II PyFN problem with the conditions of Table 10 where the costs are crisp, but the demand and supply are PFN. The supplies are denoted as Pythagorean fuzzy numbers, i.e.,
$$\begin{aligned} \tilde{s}_{1}^{P} &= \left( {\theta^{P} ,\delta^{P} } \right) \approx \left( {0.7, \, 0.1} \right),\;\tilde{s}_{2}^{P} \approx \left( {0.8, \, 0.1} \right),\\ \tilde{s}_{3}^{P} &\approx \left( {0.9, \, 0.1} \right).\end{aligned} $$
Table 10 Input data for Pythagorean transportation problem of type-II
Similarly, the demands are also denoted as Pythagorean fuzzy numbers, respectively. We note that \( \theta^{P} ,\delta^{P} \) represent the maximum degree of the membership (i.e., the degree of acceptance of quantity) and the non-membership (i.e., the degree of rejection of quantity) respectively. Moreover, they will also satisfy the inconsistence information under these conditions, i.e., \( 0 \le \theta^{P} \le 1, \) \( 0 \le \delta^{P} \le 1, \) \( 0 \le \left( {\theta^{P} } \right)^{2} + \left( {\delta^{P} } \right)^{2} \le 1 \).
Solution
After executing the steps 1–3, we get the initial basic feasible solution as follows:
$$\begin{aligned} (O_{1} ,D_{1} ) &= x_{11} = 0.3350,\;(O_{1} ,D_{2} ) = x_{12} = 0.405,\\ (O_{2} ,D_{3} ) &= x_{23} = 0.8150,\;(O_{3} ,D_{2} ) = x_{32} = 0.295,\;{\text{and}}\\ (O_{3} ,D_{4} ) &= x_{34} = 0.6050.\end{aligned} $$
Also, the minimum cost of an initial basic feasible solution is
$$ Min = 0.0335 \times 0.3350 + 0.0545 \times 0.405 + 0.295 \times 0.0815 + 0.026 \times 0.8150 + 0.07 \times 0.6050 = 0.132978. $$
Again, execute the steps 4–6, we get the optimum solution, i.e.:
$$\begin{aligned} (O_{1} ,D_{1} ) &= x_{11} = 0.3350,\;(O_{1} ,D_{2} ) = x_{12} = 0.405,\\ (O_{2} ,D_{3} ) &= x_{23} = 0.8150,\;(O_{3} ,D_{2} ) = x_{32} = 0.295,\;{\text{and}}\\ (O_{3} ,D_{4} ) &= x_{34} = 0.6050.\end{aligned} $$
$$ Min = 0.0335 \times 0.3350 + 0.0545 \times 0.405 + 0.295 \times 0.0815 + 0.026 \times 0.8150 + 0.07 \times 0.6050, $$
Minimum cost = \( 0.132978. \)
Example 5.3
Consider type-III PyFN problem with the conditions of Table 11. Here, the supplies are denoted as Pythagorean fuzzy. Similarly, the demands are also denoted as Pythagorean fuzzy numbers \( \left( {\theta^{P} ,\delta^{P} } \right) \), respectively. We note that \( \theta^{P} ,\delta^{P} \) represent the maximum degree of the membership (i.e., the degree of acceptance of quantity) and the non-membership (i.e., the degree of rejection of quantity), respectively. The cost values are also in Pythagorean fuzzy numbers where \( C^{P} \approx (C_{a}^{P} ,C_{r}^{P} ) \) represents the degree of acceptance and rejection of cost.
Table 11 Input data for Pythagorean transportation problem of type-III
Solution
After executing the steps 1–3, we get the initial basic feasible solution as follows.
After executing the step 1–3, we get the initial basic feasible solution, i.e., \( (O_{2} ,D_{1} ) = x_{21} = 0.335, \) \( (O_{1} ,D_{2} ) = x_{12} = 0.135, \) \( (O_{2} ,D_{2} ) = x_{22} = 0.48, \) \( (O_{2} ,D_{3} ) = x_{23} = 0.085, \) \( (O_{3} ,D_{3} ) = x_{33} = 0.815, \) and \( (O_{1} ,D_{4} ) = x_{14} = 0.6050; \) so the minimum IBFS cost is 0.322775.
Again, by steps 4–6, we get the optimum solution i.e., \( (O_{2} ,D_{1} ) = x_{21} = 0.335, \) \( (O_{1} ,D_{2} ) = x_{12} = 0.22, \) \( (O_{2} ,D_{2} ) = x_{22} = 0.48, \) \( (O_{3} ,D_{4} ) = x_{34} = 0.085, \) \( (O_{3} ,D_{3} ) = x_{33} = 0.815, \) and \( (O_{1} ,D_{4} ) = x_{14} = 0.52, \), that the minimum cost is 0.31895.