Multi-objective optimization for multi-type transportation problem in intuitionistic fuzzy environment

Abstract

Multi-objective optimization is an emerging field concerning optimization problems associated with more than one objective function, each of them has to be optimized simultaneously. Multi-objective optimization is widely used in logistics and supply chains to reduce the cost and time involved in transportation. With the increase in Global Supply Chains, many organizations are facing the challenges of delivering products to their customers at a fast pace, low cost, and high reliability. There are numerous factors that may affect the goal of an organization to optimize the cost, time, and effort during the transportation of their products to the end customers. For instance, in the existing transportation problems, the type of vehicles used for the movement of the products is not focused. Transportation of the goods is considered to utilize any type of vehicle irrespective of the nature of the goods. However, in real-life scenarios, there are certain constraints in the vehicle used to transport the finished goods or raw materials from a source to a destination. Vehicles such as tanker trucks, top open trucks, closed trucks, etc. need to be booked based on the nature of goods to be transported. Also, the cost and time of transportation are uncertain in nature. In this paper, we formulate the Multi-Objective Solid Transportation Problem (MOSTP) by considering the above issue. The uncertain parameters of the problem are considered as Pentagonal Intuitionistic Fuzzy Numbers (PIFN). Magnitude method is used for defuzzification. An algorithm to find the solution of formulated Intuitionistic Fuzzy Multi-Objective Solid Transportation problem (IFMOSTP) is provided. The proposed model is illustrated by a numerical example which is solved with the help of LINGO software.

Keywords

Intuitionistic fuzzy sets intuitionistic fuzzy numbers magnitude ranking multi-objective solid transportation problems fuzzy programming

Abbreviations and notations

Abbreviation = Description

PIFN

pentagonal intuitionistic fuzzy numbers

MOSTP

multi-objective solid transportation problems

IFMOSTP

intuitionistic fuzzy multi-objective solid transportation problems

Z _r

solution of r^th objective function

denotes the r^th penalty criterion

denotes the number of objective considered

denotes the item of the goods to be transported

denotes the destination to which the goods to be transported

denotes type of vehicle used for transportation

c_{ijk}^{r}

is unit transportation penalty for i^th item to j^th destination through k^th type of vehicle and is considered as PIFN

x_{ijk}^{r}

feasible solution for r^th objective function

A _jk

is the goods to be transported through k^th means of transport to j^th destination

B _ki

is the demand in i^th product to be transported through k^th means of transport

E _jj

is the i^th product to be transported to the j^th destination.

X ^r

set of feasible solution for r^th constraint

Z_{r}^{#}

optimal solution for r^th constraint

1 Introduction

One of the earliest applications of linear programming problems is the transportation problem. The transportation problem’s goal is to reduce the total cost of delivering a product from available sources or origin to the required destination. In 1941, Alfred Hitchcock [1] proposed the basic transportation problem. Appa [2] discussed variants of transportation problems. In 1963, Dantzig [3] developed an algorithm to solve transportation problems. Bridgen [4] presented an algorithm for solving the transportation problem, which includes equality and inequality constraints as well as availability constraints.

Shell [5] extended two-dimensional transportation problems to three-dimensional transportation problems, allowing for the consideration of three constraints. As an extension of the MODI method, Haley [6] created a solution procedure for solving solid transportation problems. He also extended solid transportation to multi-objective problems and illustrated the solution [7, 8]. In solid transportation problems, Patel and Tripathi [9] used mixed constraints. Lee and Moore [10] investigated the optimization of multi-objective transportation problems. Charnes and Cooper [11] proposed the goal programming method to solve multi-objective transportation problems. Ijiri [12], Lee [13], and others refined this technique. Bit et al, [14] proposed a fuzzy programming approach to multi-objective transportation problems in 1992. The aforementioned works were completed with precise parameters.

In real life situation due to various factors, the estimation of transportation cost, time, etc. are not precise in nature. To handle such situation Zadeh [15] introduced fuzzy sets. The idea of fuzzy sets was developed to handle uncertainty and vagueness which a simple crisp set cannot do. Attanassov [16] extended fuzzy sets to intuitionistic fuzzy sets which incorporated the value of hesitation to handle uncertainty efficiently. Intuitionistic fuzzy numbers [17] are special kind of intuitionistic fuzzy sets which are significant in solving linear programming problems. Intuitionistic fuzzy numbers can be expressed in the form of triangular, trapezoidal, pentagonal etc. Different form of uncertain parameter is discussed inTable 1. Because the parameters of MOSTP are uncertain in nature, many others have expressed an interest in solving such problems with uncertain parameters. Table 2 provides a literature review of various research studies conducted on MOSTP with uncertain parameters.

Table 1
Different forms of parameters

Uncertain parameter Existing form Information

Crisp numbers Single value Can be used only when the exact value is known.

Interval numbers Maximum and minimum value Can be used when the value lies between intervals.

Fuzzy numbers Membership function Can be used when the value is uncertain.

Intuitionistic fuzzy numbers Membership function and non-membership function Can be used when the value is uncertain or Vague. It measures the uncertainty in more efficiently since it takes into account the degree of membership function and non-membership function.

Uncertain parameter	Existing form	Information
Crisp numbers	Single value	Can be used only when the exact value is known.
Interval numbers	Maximum and minimum value	Can be used when the value lies between intervals.
Fuzzy numbers	Membership function	Can be used when the value is uncertain.
Intuitionistic fuzzy numbers	Membership function and non-membership function	Can be used when the value is uncertain or Vague. It measures the uncertainty in more efficiently since it takes into account the degree of membership function and non-membership function.

Table 2

Literature review of multi-objective

S.No	Authors name	Parameter	Problem type	Method applied	Main contribution
1	Ali Ebrahimnejed [18]	Triangular fuzzy numbers	Transportation Problem	Two step method	Optimal solution to transportation problems in uncertain nature
2	Bhatia [19]	LR flat fuzzy number	Transportation problem	A new method, based on tabular representation of fuzzy transportation problems	Sensitivity analysis of fuzzy transportation problems
3	Bhabani Mallia et al, [20]	Crisp Numbers	Multi-objective transportation problem	Lexicographical goal programming based Two Phase Approach	Solution to Multi-objective transportation problem
4	Sharif Uddin et al, [21]	Uncertain parameters in the form of normal distribution	Uncertain multi-objective transportation problem	Fuzzy goal programming	Compromise solution of a multi-objective transportation problem
5	Gowthami [22]	Triangular fuzzy numbers	Multi-objective Transportation problem	Algorithm based on Parametric form	Solution of multi objective transportation problem
6	Harish Garg [23]	Intuitionistic fuzzy set	Multi-objective transportation problem	Cuckoo search algorithm	Optimal solution to multi-objective transportation problem
7	Maqbool [24]	Intuitionistic fuzzy number	Multi-objective Transportation problem	Fuzzy Programming Algorithm	Solution of the multi-objective transportation problem
8	Laurent Lemarchand et al, [25]	Crisp Numbers	Multiobjective Optimization for Multimode Transportation Problem	Exact Algorithm and Heuristic algorithms such as PAES and NSGA2	Optimize Multimode Transportation mode
9	Shashi Aggarwal [26]	Trapezoidal intuitionistic fuzzy number	Solid transportation	New method	Compromise solution to solid transportation problems
10	Jimenez [27]	Interval and Fuzzy parameter	Solid transportation	Chanas method	Solution to solid transportation problems
11	Shiang-Tai Liu [28]	Trapezoidal fuzzy number	Solid transportation	New method based on Zadeh extension principle	Solution to solid transportation problems
12	Senthil Kumar [29]	Triangular intuitionistic fuzzy number	Solid transportation problem	PSK method	Solution to solid transportation problems
13	Amrit Das [30]	Crisp Numbers	Multi-Objective Stochastic Solid Transportation Problem	Global criterion method and fuzzy goal programming approach	Solution to Multi-Objective Stochastic Solid Transportation Problem under probabilistic constraints
14	Li [31]	Fuzzy numbers	Multi-objective Solid transportation	Genetic algorithm	Non dominated solution to multi-objective transportation problems
15	Pradip Kundu et al, [32]	Trapezoidal fuzzy	Multi-objective Solid transportation	Expected value method and minimum of fuzzy numbers	Compromise optimal solution to multi-objective transportation problems
16	Chakraborty et al, [33]	Fuzzy inequalities	Multi item multi-objective Solid transportation	New method	Solution to multi-objective multi item solid transportation problem
17	Nithishkumar [34]	Triangular and Trapezoidal fuzzy number	Multi-objective Solid transportation	Goal programming and expected method	Solution to multi-objective solid transportation problems
18	Mouhya.B et al [35]	Type-I and type-II fuzzy variables	Multi item multi-objective Solid transportation	Global criterion method	Compromise solution to multi-objective transportation problems
19	Hassan Dalman [36]	Interval number	Multi-objective solid transportation problem	Global criteria method	Solution multi-objective solid transportation problem
20	Pradip Kundu et al, [37]	Type-2 fuzzy parameters	Multi-item solid transportation problem		Solution multi-objective solid transportation problem

1.1 Drawbacks of the existing methods

Hassan Dalman, Sivri. M [36] in 2017 handled MOSTP by considering the parameters as interval numbers. The closed and bounded interval is converted into a deterministic one by using weighted factor that lies between [0, 1].The user may confuse while choosing the weighted factor. Also always it is not possible to say that cost lies within a single interval. Due to many factors there may be partition also.

Li [31] solved MOSTP with fuzzy parameters using genetic algorithm. The availability is considered to be transported by all means of transportation. But this is not possible due to factor of such as availability of space in the vehicle, availability of routes, and nature of the goods to be transported.

Mouhya [35] developed solution to multi-objective solid transportation problems in a fuzzy environment. The method followed to obtain crisp form is complicated and tedious. Trapezoidal fuzzy numbers used in this method can handle fewer factors than PIFN.

From the Table 2, it has been noticed that many works have been carried out to solve solid transportation problem and MOSTP in fuzzy environment. Only a few works have been carried out to solve solid transportation with intuitionistic fuzzy parameters. So, for in the literature, Multi-objective solid transportation with intuitionistic fuzzy parameters is not yet studied as a research paper. Hence the main purpose of this work is to study Multi-objective Solid transportation problem involving intuitionistic fuzzy numbers.

1.2 Novelty of the proposed work

In this work PIFN is used as parameters to handle uncertainty for the following factors:

This number is created primarily to capture variations in the curve at different levels that a triangular and trapezoidal fuzzy number do not.

The shape of pentagonal intuitionistic fuzzy numbers is simpler and more consistent, allowing for more natural interpretation and easier calculation.

It is difficult to represent each parameter with triangular or trapezoidal intuitionistic fuzzy numbers if it has five variables. As a result, pentagonal intuitionistic fuzzy numbers can be used to solve the problem.

1.3 Motivation of the research work

During the actual transportation, the estimation done is not accurate, due to error in the measuring technique. Let us consider the transportation time of goods from the source to a destination to be ‘h’ hours approximately. (i.e.) it is not accurately ‘h’ hours.it is either less than ‘h’ hours or greater than ‘h’ hours. This concept of variation leads to a new type of IFN called PIFN. Generally, PIFN has five parameters for each membership and a non-membership function, which is a subset of real number $R$ .

An IFN takes the form $A = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) (a_{1}^{'}, a_{2}^{'}, a_{3}^{'}, a_{4}^{'}, a_{5}^{'}) : t, u, v, w}$ with parameters $a_{1}^{'} ⩽ a_{1} ⩽ a_{2}^{'} ⩽ a_{2} ⩽ a_{3} ⩽ a_{3}^{'} ⩽ a_{4} ⩽ a_{4}^{'} ⩽ a_{5} ⩽ a_{5}^{'}$ and 0 ⩽ t ⩽ 1, 0 ⩽ u ⩽ 1, 0 ⩽ v ⩽ 1, 0 ⩽ w ⩽ 1

where, the middle point a₃ has the grade of membership 1. t and u are the grades of a₂and a₄ respectively. $a_{3}^{'}$ has the non-membership grade 0. v and w are the grades of $a_{2}^{'} and a_{4}^{'}$ respectively. Every PIFN is associated with four weights t, u, v, w, where t, u the weights associated with the membership function are and v, w are the weights associated with the non-membership function.

For an instance, the time taken for transportation from Tirupur to Chennai is approximately 9 hours. It may be less than 9 hours or greater than 9 hours. This uncertainty may be due to the factors such as traffic, road condition, weather, vehicle efficiency etc. Considering the above factors time parameter can be represented as PIFN.

Transportation time is approximately 9 hours, hence a₃ is 9 with membership grade 1. If the traffic is less than the time will reduce and is represented by giving weighted value ^′t′ to a₂. The time deviation due to less traffic is given by the interval [a₂, a₃]. If the traffic is heavy, then the time taken will increase. This is represented by giving a weighted value to ^′u′ to a₄. The time deviation due to more traffic is given by the interval [a₃, a₄]. The time delay due to weather condition is given in the interval [a₄, a₅]. The time deviation due road condition is given by the interval [a₁, a₂].

1.4 Structure of the paper

Section 2 contains preliminary definitions that is required to carry out the current research. Section 3 elaborates on the mathematical formation of intuitionistic fuzzy multi-objective solid transportation and the algorithm for obtaining the optimal solution. The proposed algorithm is validated in the following section by solving real-time problems in which the transportation parameters were modeled as pentagonal intuitionistic fuzzy numbers. Section 5 discusses the significance of the proposed research work in comparison to previous research. This paper concludes with a precise conclusion.

2 Preliminaries

Definition 2.1. An intuitionistic fuzzy set [16] A in X is given by A ={ (x, μ_A (x) , ν_A (x) , x ∈ X) }, where the function μ_A (x) : X → [0, 1] and ν_A (x) : X → [0, 1] are the degree of membership and degree of non-membership of the element x ∈ X to the set A, which is a subset of X, and for every x ∈ X, 0 ⩽ μ_A (x) + ν_A (x) ⩽ 1.

Definition 2.2 A generalized non- symmetric pentagonal intuitionistic fuzzy number Fig. 1, [38] $A = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) (a_{1}^{'}, a_{2}^{'}, a_{3}^{'}, a_{4}^{'}, a_{5}^{'}) : t, u, v, w}$ is a sub set of intuitionistic fuzzy set on a set of real number R, with parameters $a_{1}^{'} ⩽ a_{1} ⩽ a_{2}^{'} ⩽ a_{2} ⩽ a_{3} ⩽ a_{3}^{'} ⩽ a_{4} ⩽ a_{4}^{'} ⩽ a_{5} ⩽ a_{5}^{'}$ and 0 ⩽ t ⩽ 1, 0 ⩽ u ⩽ 1, 0 ⩽ v ⩽ 1, 0 ⩽ w ⩽ 1, whose membership function and non -membership function are defined as follows:

Fig. 1

Generalized non- symmetric pentagonal intuitionistic fuzzy number.

$μ_{A} (x) = {\begin{matrix} 0, x ⩽ a_{1} \\ t (\frac{x - a_{1}}{a_{2} - a_{1}}), a_{1} ⩽ x ⩽ a_{2} \\ t + \frac{(1 - t) (x - a_{2})}{(a_{3} - a_{2})}, a_{2} ⩽ x ⩽ a_{3} \\ 1, x = a_{3} \\ 1 + \frac{(u - 1) (x - a_{3})}{a_{4} - a_{3}}, a_{3} ⩽ x ⩽ a_{4} \\ u (\frac{a_{5} - x}{a_{5} - a_{4}}), a_{4} ⩽ x ⩽ a_{5} \\ 0, x ⩾ a_{5}) \end{matrix}$ $ϑ_{A} (x) = {\begin{matrix} 1, x ⩽ a_{1}^{'} \\ 1 + \frac{(v - 1) (x - a_{1}^{'})}{a_{2}^{'} - a_{1}^{'}}, a_{1}^{'} ⩽ x ⩽ a_{2}^{'} \\ v (\frac{a_{3}^{'} - x}{a_{3}^{'} - a_{2}^{'}}), a_{2}^{'} ⩽ x ⩽ a_{3}^{'} \\ w (\frac{x - a_{3}^{'}}{a_{4}^{'} - a_{3}^{'}}), a_{3}^{'} ⩽ x ⩽ a_{4}^{'} \\ w + \frac{(1 - w) (x - a_{4}^{'})}{a_{5}^{'} - a_{4}^{'}}, a_{4}^{'} ⩽ x ⩽ a_{5}^{'} \\ 1, x ⩾ a_{5}^{'}) \end{matrix}$ $A = {μ_{A}^{L 1} (x), μ_{A}^{L 2} (x), μ_{A}^{R 1} (x), μ_{A}^{R 2} (x), ν_{A}^{L 1} (x), ν_{A}^{L 2} (x), J_{A}^{R 1} (x), J_{A}^{R 2} (x)},$ $μ_{A} (x) = {\begin{matrix} 0, x ⩽ a_{1} \\ p (\frac{x - a_{1}}{a_{2} - a_{1}}), a_{1} ⩽ x ⩽ a_{2} \\ p + \frac{(1 - p) (x - a_{2})}{a_{3} - a_{2}}, a_{2} ⩽ x ⩽ a_{3} \\ 1, x = a_{3} \\ 1 + \frac{(q - 1) (x - a_{3})}{a_{4} - a_{3}}, a_{3} ⩽ x ⩽ a_{4} \\ q (\frac{a_{5} - x}{a_{5} - a_{4}}), a_{4} ⩽ x ⩽ a_{5} \\ 0, x ⩾ a_{5} \end{matrix}$

where $μ_{A}^{L 1} (x) = t (\frac{x - a_{1}}{a_{2} - a_{1}}), μ_{A}^{L 2} (x) = t + \frac{(1 - t) (x - a_{2})}{a_{3} - a_{2}}$ $μ_{A}^{R 1} (x) = 1 + \frac{(u - 1) (x - a_{3})}{a_{4} - a_{3}}, μ_{A}^{R 2} (x) = u (\frac{a_{5} - x}{a_{5} - a_{4}})$ $ϑ_{A}^{L 1} (x) = 1 + \frac{(v - 1) (x - a_{1}^{'})}{a_{2}^{'} - a_{1}^{'}}, ϑ_{A}^{L 2} (x) = v (\frac{a_{3}^{'} - x}{a_{3}^{'} - a_{2}^{'}}),$ $ϑ_{A}^{R 1} (x) = w (\frac{x - a_{3}^{'}}{a_{4}^{'} - a_{3}^{'}}), ϑ_{A}^{R 2} (x) = w + \frac{(1 - w) (x - a_{4}^{'})}{a_{5}^{'} - a_{4}^{'}}$

2.3 Magnitude of pentagonal intuitionistic fuzzy numbers

For an arbitrary Pentagonal intuitionistic fuzzy number $A = {μ_{A}^{L 1} (x), μ_{A}^{L 2} (x), μ_{A}^{R 1} (x), μ_{A}^{R 2} (x), ν_{A}^{L 1} (x), ν_{A}^{L 2} (x), ν_{A}^{R 1} (x), ν_{A}^{R 2} (x)},$

Magnitude of A [38] is given by $\begin{matrix} Mag (A) = \frac{1}{4} \int_{0}^{1} {(μ_{A}^{L 1})^{- 1} (y) + (μ_{A}^{L 2})^{- 1} (y) + (μ_{A}^{L})^{- 1} (1) + (μ_{A}^{R 1})^{- 1} \\ (y) + (μ_{A}^{R 2})^{- 1} (y) + (μ_{A}^{R})^{- 1} (1) + (ϑ_{A}^{L 1})^{- 1} (y) + (ϑ_{A}^{L 2})^{- 1} (y) + (ϑ_{A}^{L})^{- 1} (0) \\ + (ϑ_{A}^{R 1})^{- 1}) +, (ϑ_{A}^{R 2})^{- 1} (y) + (ϑ_{A}^{R})^{- 1} (0)} r (y) dy . \end{matrix}$ (1)

Where, the function r (y) is a non-negative and increasing function on [0,1] with r (0) = 0, r (1) = 1 and $\int_{0}^{1} r (y) dy = \frac{1}{2}$ . Obviously r (y) can be considered as a weighting function and can be chosen according to the situation. Here we choose r (y) = y.

The magnitude of non-symmetric Pentagonal intuitionistic fuzzy number is given by $\begin{matrix} Mag (A) = \frac{1}{24} [a_{1} t^{2} + a_{2} (1 + t) - a_{3} (t^{2} + t + u^{2} + u - 10) \\ + a_{4} (1 + u) + a_{5} u^{2} - b_{1} (v^{2} + v - 2) + b_{2} (1 + v) \\ + b_{3} (v^{2} + w^{2} + 6) + b_{4} (1 + w) - b_{5} (w^{2} + w - 2) \end{matrix}$ (2)

2.4 Ranking procedure

For any two arbitrary Pentagonal intuitionistic fuzzy numbers

A ={ (a₁, a₂, a₃, a₄, a₅) (b₁, b₂, b₃, b₄, b₅) ; p, q, s, t } and

B ={ (c₁, c₂, c₃, c₄, c₅) (d₁, d₂, d₃, d₄, d₅) ; u, v, w, z }, the ranking based on magnitude [38] is done as follows:

If Mag (A) < Mag (B) , then A ≺ B

If Mag (A) > Mag (B) , then A ≻ B

If Mag (A) = Mag (B) , then A ≈ B.

3 Solution of intuitionistic fuzzy multi-objective solid transportation problems

The mathematical formulation of the multi-objective solid transportation problem in an intuitionistic fuzzy environment is elaborated with appropriate notations in this section. An algorithm for solving the proposed mathematical model has also been described in detail.

3.1 Mathematical formulation of intuitionistic fuzzy multi-objective solid transportation problems (IFMOSTP)

Let i (i = 1, 2, 3 … m) items are to be transported to j (j = 1, 2, … . . n) destination through k (k = 1, 2 . . p) mode of transport,A_jk be the goods to be transported through k^th means of transport to j^thdestination,B_kiis the demand in i^th product to be transported through k^th means of transport,E_ij is the i^th product to be transported to the j^thdestination, Fig. 2 is representation of constraints of proposed model. $c_{ijk}^{r}$ is unit transportation penalty for i^th item to j^th destination through k^th type of vehicle and is considered PIFN.x_ijk amount of i^th item transported to j^th destination through k^th type of vehicle. Then the IFMOSTP is formulated as follows, $\begin{matrix} Minimize Z_{r} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} \sum_{k = 1}^{p} x_{ijk} c_{ijk}^{r}, \\ r = 1, 2, 3 \dots \dots . . q \end{matrix}$ (3) $Subject to \sum_{i = 1}^{n} x_{ijk} = A_{jk},$ (4) $\sum_{j = 1}^{m} x_{ijk} = B_{ki},$ (5) $\sum_{k = 1}^{p} x_{ijk} = E_{ij},$ (6)

where $\sum_{j = 1}^{n} A_{jk} = \sum_{i = 1}^{m} B_{ki}, \sum_{k = 1}^{p} B_{ki} = \sum_{j = 1}^{n} E_{ij}, \sum_{i = 1}^{m} E_{jk} = \sum_{k = 1}^{p} A_{jk},$ $\sum_{j = 1}^{n} \sum_{k = 1}^{p} A_{jk} = \sum_{k = 1}^{p} \sum_{i = 1}^{m} B_{ki} = \sum_{i = 1}^{m} \sum_{j = 1}^{n} E_{ij,} and x_{ijk} ⩾ 0 .$ (7)

Fig. 2

Representation of constraints of proposed model.

3.2 Algorithm

The steps for finding an optimal solution to an intuitionistic fuzzy multi-objective solid transportation problem are as follows.

Step: 1 Convert the transportation parameter $c_{ijk}^{r}$ , which is modeled as PIFN to crisp quantity by using magnitude given by equation (2)

Step: 2 Solve the multi-objective solid transportation by considering each objective function separately. Represent the different values of $c_{ijk}^{1}$ as in shown in the Table 3.

Table 3
Representation of the IFMOSTP

$c_{111}^{1}$ $c_{112}^{1}$ ... $c_{11 p}^{1}$ E ₁₁ $c_{121}^{1}$ $c_{122}^{1}$ ... $c_{12 p}^{1}$ E ₁₂ ... $c_{1 n 1}^{1}$ $c_{1 n 2}^{1}$ ... $c_{1 np}^{1}$ E _1n B ₁₁ B ₁₂ ... B _1p

$c_{211}^{1}$ $c_{212}^{1}$ ... $c_{21 p}^{1}$ E ₂₁ $c_{221}^{1}$ $c_{222}^{1}$ ... $c_{22 p}^{1}$ E ₂₂ ... $c_{2 n 1}^{1}$ $c_{2 n 2}^{1}$ ... $c_{2 np}^{1}$ E _2n B ₂₁ B ₂₂ ... B _2p

... ... ... ... ... ... ... ... ...

... .. .. ... ... .. ... .. . ...

$c_{n 11}^{1}$ $c_{n 12}^{1}$ ... $c_{n 1 p}^{1}$ E _n1 $c_{n 21}^{1}$ $c_{n 22}^{1}$ ... $c_{n 2 p}^{1}$ E _n2 ... $c_{mn 1}^{1}$ $c_{mn 2}^{1}$ ... $c_{mnp}^{1}$ E _mn B _m1 B _m2 ... B _np

A ₁₁ A ₁₂ A _1p A ₂₁ A ₂₂ A _2p .. A _n1 A _n2 A _np

Step: 3 The Solid transportation problem reduces to typical transportation problem whenj = 1. Apply Least Cost Method to find the feasible solution for the reduced transportation problem as represented in the Table 4.

Table 4

Representation of the reduced solid transportation problem

$c_{111}^{1}$	$c_{112}^{1}$	$c_{113}^{1}$	... ..	... ..	$c_{11 p}^{1}$	E ₁₁	B ₁₁	B ₁₂	...	B _1p
$c_{211}^{1}$	$c_{212}^{1}$	$c_{213}^{1}$	... ..	... .	$c_{21 p}^{1}$	E ₂₁	B ₂₁	B ₂₂	...	B _2p
...	...		... ..	...
				... ..
$c_{n 11}^{1}$	$c_{n 12}^{1}$	$c_{n 13}^{1}$	... ..	... .	$c_{n 1 p}^{1}$	E _n1	B _m1	B _m2	...	B _np
A ₁₁	A ₁₂	A ₁₃			A _1p

Step: 4 Check the allocations of $c_{i 1 k}^{1} .$ If all the allocations are less than B_ik,∀i, k, then move to Step 6, otherwise proceed with next step.

Step: 5 If any one allocation of $c_{i 1 k}^{1},$ exceeds B_ik form a closed path consisting of horizontal and vertical lines starting and ending with the cell $c_{i 1 k}^{1}$ having corner cells occupied. Assign the sign (+) & (–) alternatively starting with (–) sign in the corresponding cell. Let θ represents the difference between $c_{i 1 k}^{1} and B_{ik}$ . Subtract θ from cells with (–) sign and add θ to the cells with (+) sign. All the cells with (–) sign must be allocated with at least θ units.

Step: 6 Form the transportation problem for j = 2. New value of $B_{ik}^{'}$ is calculated from $B_{ik} - c_{i 1 k}^{1}$ . Use Least Cost Method to find feasible solution of the resulting transportation problem. Check each allocation of $c_{i 2 k}^{1}$ is less than $B_{ik}^{'}$ ,∀i, k. If any allocation of $c_{i 2 k}^{1}$ exceeds $B_{ik}^{'}$ , let $θ^{'} = B_{ik}^{'} - c_{i 2 k}^{1},$ then proceed as in step 5. If it is not possible to make a closed loop, then carry θ′units for $c_{i 2 k}^{1}$ from the previous Table 4 as follows.

Make a closed loop starting with (–) sign at $c_{i 1 k}^{1}$ and assign alternate (+) and (–) sign in the corner of the loop. Subtract θ′ to the allotted value of $c_{i 1 k}^{1}$ with (–) sign and add θ′ to the corresponding $B_{ik}^{'}$ . Add θ′ to the allotted value of $c_{i 1 k}^{1}$ with (+) sign and subtract θ′to the corresponding $B_{ik}^{'}$ . Now the allocations of the occupied cells $c_{i 2 k}^{1}$ is less than or equal to $B_{ik}^{'}$

Step: 7 Form transportation problem for j = 3. New value of $B_{ik}^{''}$ is calculated from $B_{ik}^{'} - c_{i 2 k}^{1}$ . Use Least Cost Method to find feasible solution of the transportation problem. Repeat the process as mentioned for j = 2. Continue the process for j = n - 1. Allocate the remaining b_ik to $c_{ink}^{1}$ . Feasible solution is obtained for r = 1as a single objective.

Step: 8 Repeat steps 2 to 7 for $c_{ijk}^{r} with r = 2, 3, \dots . . q .$ Let $X^{1} = {x_{ijk}^{1}}$ be the feasible solution set obtained for $c_{ijk}^{r}$ for r = 1.

Step: 9 Calculate the values of all ‘q’ objective functions at all these ‘q’ optimal solutions X^rfor (r = 1, 23 …… … q) . Then the payoff matrix is formed as follows:

$\begin{matrix} X^{1} & X^{2} & \dots . . & X^{q} \end{matrix}$ $\begin{matrix} Z_{1} \\ Z_{2} \\ \dots \\ Z_{q} \end{matrix} [\begin{matrix} Z_{1} (X^{1}) & Z_{1} (X^{2}) & \dots & Z_{1} (X^{q}) \\ Z_{2} (X^{1}) & Z_{2} (X^{2}) & . . & Z_{2} (X^{q}) \\ . . & . . & . . & . . \\ Z_{q} (X^{1}) & Z_{q} (X^{2}) & \dots & Z_{q} (X^{q}) \end{matrix}]$

Find the lower and upper bounds for each objective from the above matrix given by $\begin{matrix} L_{r} = Z_{r} (X^{r}), r = 1, 2, \dots \dots . q and U_{r} \\ = Max {Z_{r} (X^{1}), Z_{r} (X^{2}), Z_{r} (X^{3}), \dots Z_{r} (X^{r})} \end{matrix}$

Step: 10 Construct the function σ_r (Z_r) corresponding to r^th objective as follows: $σ_{r} (Z_{r}) = {\begin{matrix} 1 if Z_{r} ⩽ L_{r} \\ \frac{U_{r} - Z_{r}}{U_{r} - L_{r}} if L_{r} < Z_{r} < U_{r} \\ 0 if Z_{r} > U_{r} \end{matrix}$ (8)

Step: 11 Formulate the linear programming problem equivalent to the intuitionistic fuzzy multi-objective solid transportation problem as follows:

Max λ

subject to $\sum_{i = 1}^{m} \sum_{j = 1}^{n} \sum_{k = 1}^{p} x_{ijk} c_{ijk}^{r}, + λ (U_{r} - L_{r}) ⩽ U_{r} for all r$ (9)

along with Equation (4)–(7) $and λ ⩾ 0 and λ = \min_{r} {σ_{r} (Z_{r})}$ (10)

Solving the above problem, the optimal compromise solution is obtained. LINGO software was applied to find the solution.

4 Numerical example

To illustrate the IFMOSTP model (3.1), we consider a transportation scenario in which three types of food products (p₁, p₂, p₃), are to be transported to three different destinations using three different types of vehicle. The problem is to minimize the transportation cost of the vehicle, time taken, handling cost required during transportation. This factor depends on fuel rate, weather, vehicle efficiency, driver hours, etc, which is uncertain. The company is uncertain about the cost and time taken for delivering the product. Therefore, an approximate cost and time taken for transporting a product from each production unit to different places through type of vehicle is represented as PIFN.A_jk is the goods to be transported through k^th means of transport to j^thdestination.B_ikis the demand in i^th product to be transported through k^th means of transport. E_ij is the i^th product to be transported to the j^thdestination. $c_{ijk}^{1}$ is unit transportation cost for i^th item to j^th destination through k^th type of vehicle and is considered as PIFN. $c_{ijk}^{2}$ is the labor cost of i^th item to j^th destination through k^th type of vehicle and is considered as PIFN. $c_{ijk}^{3}$ is the time taken of transporting one unit of the for i^th item to j^th destination through k^th type of vehicle and is considered as PIFN.

The above data can be modeled as IFMOSTP as follows $Minimize Z_{r} = \sum_{i = 1}^{3} \sum_{j = 1}^{3} \sum_{k = 1}^{3} x_{ijk} c_{ijk}^{r}, r = 1, 2, 3$ (11) $Subject to \sum_{i = 1}^{3} x_{ijk} = A_{jk},$ (12) $\sum_{j = 1}^{3} x_{ijk} = B_{ki},$ (13) $\sum_{k = 1}^{3} x_{ijk} = E_{ij},$ (14)

where, $\begin{matrix} \sum_{j = 1}^{3} A_{jk} = \sum_{i = 1}^{3} B_{ki}, \sum_{k = 1}^{3} B_{ki} = \sum_{j = 1}^{3} E_{ij}, \sum_{i = 1}^{3} E_{jk} = \sum_{k = 1}^{3} A_{jk}, \\ \sum_{j = 1}^{3} \sum_{k = 1}^{3} A_{jk} = \sum_{k = 1}^{3} \sum_{i = 1}^{3} B_{ki} = \sum_{i = 1}^{3} \sum_{j = 1}^{3} E_{ij,} and x_{ijk} ⩾ 0 . \end{matrix}$ (15)

With $\sum_{i = 1}^{3} x_{i 11} = 4$ , $\sum_{i = 1}^{3} x_{i 12} = 3$ , $\sum_{i = 1}^{3} x_{i 13} = 3$ , $\sum_{i = 1}^{3} x_{i 21} = 6$ , $\sum_{i = 1}^{3} x_{i 22} = 8$ , $\sum_{i = 1}^{3} x_{i 23} = 4$ $\sum_{i = 1}^{3} x_{i 31} = 5$ , $\sum_{i = 1}^{3} x_{i 32} = 4$ , $\sum_{i = 1}^{3} x_{i 33} = 6$ , $\sum_{j = 1}^{3} x_{1 j 1} = 6$ , $\sum_{j = 1}^{3} x_{1 j 2} = 5$ , $\sum_{j = 1}^{3} x_{1 j 3} = 4$ , $\sum_{j = 1}^{3} x_{2 j 1} = 5$ , $\sum_{i = 1}^{3} x_{2 j 2} = 4$ , $\sum_{j = 1}^{3} x_{2 j 3} = 4$ , $\sum_{j = 1}^{3} x_{3 j 1} = 4$ , $\sum_{j = 1}^{3} x_{3 j 2} = 6$ , $\sum_{j = 1}^{3} x_{3 j 3} = 5$ , $\sum_{k = 1}^{3} x_{11 k} = 5$ , $\sum_{k = 1}^{3} x_{12 k} = 5$ , $\sum_{k = 1}^{3} x_{13 k} = 5$ , $\sum_{k = 1}^{3} x_{2 k 1} = 3$ , $\sum_{k = 1}^{3} x_{22 k} = 6$ , $\sum_{k = 1}^{3} x_{23 k} = 4$ , $\sum_{k = 1}^{3} x_{3 k 1} = 2$ , $\sum_{k = 1}^{3} x_{32 k} = 7$ , $\sum_{k = 1}^{3} x_{33 k} = 6$ .

The unit transportation of penalties are given in Tables 5, 6, and 7.

A₁₁ = 4, A₁₂ = 3, A₁₃ = 3, A₂₁ = 6, A₂₂ = 8, A₂₃ = 4, A₃₁ = 5, A₃₂ = 4, A₃₃ = 6,

B₁₁ = 6, B₁₂ = 5, B₁₃ = 4, B₂₁ = 5, B₂₂ = 4, B₂₃ = 4, B₃₁ = 4, B₃₂ = 6, B₃₃ = 5,

E₁₁ = 5, E₁₂ = 5, E₁₃ = 5, E₂₁ = 3, E₂₂ = 6, E₂₃ = 4, E₃₁ = 2, E₃₂ = 7, E₃₃ = 6,

Table 5

Penalties/unit transportation cost $c_{ijk}^{1}$

$c_{111}^{1} = {(4, 6, 9, 13, 15) (2, 5, 11, 14, 17); . 6, . 5, . 3, . 4}$	$c_{211}^{1} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 18); . 8.5, . 3, . 4}$
$c_{112}^{1} = {(7, 10, 12, 15, 18) (5, 8, 14, 17, 20); . 7, . 8, . 3, . 2}$	$c_{212}^{1} = {(3, 5, 7, 12, 14) (1, 4, 10, 13, 17); . 7, . 7, . 4, . 3}$
$c_{113}^{1} = {(8, 11, 13, 16, 19) (6, 9, 15, 18, 21); . 6, . 8, . 3, . 2}$	$c_{213}^{1} = {(7, 9, 11, 16, 18) (5, 8, 14, 17, 21); . 8, . 6, . 4, . 5}$
$c_{121}^{1} = {(3, 5, 8, 12, 14) (1, 4, 10, 13, 16); . 5, . 4, . 2, . 2}$	$c_{221}^{1} = {(2, 4, 6, 11, 13) (0, 3, 9, 12, 16); . 6, . 6, . 3, . 2}$
$c_{122}^{1} = {(9, 11, 14, 18, 20) (7, 10, 16, 19, 22); . 7, . 5, . 4, . 3}$	$c_{222}^{1} = {(5, 7, 8, 14, 16) (3, 6, 10, 15, 19); . 5, . 8, . 3, . 4}$
$c_{123}^{1} = {(7, 9, 12, 16, 18) (5, 8, 14, 17, 20); . 6, . 8, . 4, . 5}$	$c_{223}^{1} = {(10, 12, 13, 19, 21) (8, 11, 15, 20, 24); . 6, . 8, . 5, . 4}$
$c_{131}^{1} = {(6, 8, 11, 15, 17,) (4, 7, 13, 16, 19); . 8, . 6, . 3, . 3}$	$c_{231}^{1} = {(20, 22, 23, 29, 31) (18, 21, 25, 30, 34); . 8, . 7, . 2, . 2}$
$c_{132}^{1} = {(7, 9, 12, 16, 18) (5, 8, 14, 17, 20); . 5, . 5, . 3, . 2}$	$c_{232}^{1} = {(5, 7, 9, 14, 18) (3, 6, 12, 15, 20); . 5, . 5, . 2, . 2}$
$c_{133}^{1} = {(4, 6, 7, 13, 15) (2, 5, 11, 14, 17); . 5, . 7, . 4, . 4}$	$c_{233}^{1} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 7, . 7, . 2, . 3}$
$c_{311}^{1} = {(10, 12, 13, 18, 20) (8, 11, 15, 19, 21); . 6, . 6, . 3, . 3}$	$c_{322}^{1} = {(7, 9, 12, 16, 18) (5, 8, 14, 17, 20); . 7, . 7, . 3, . 3}$
$c_{312}^{1} = {(2, 4, 6, 11, 14) (1, 3, 9, 12, 16); . 7, . 7, . 5, . 3}$	$c_{323}^{1} = {(5, 7, 8, 14, 18) (3, 6, 10, 15, 20); . 5, . 3, . 2, . 2}$
$c_{313}^{1} = {(5, 7, 8, 16, 18) (3, 6, 10, 15, 19); . 5, . 7, . 3, . 2}$	$c_{331}^{1} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 18); . 5, . 7, . 3, . 4}$
$c_{321}^{1} = {(5, 7, 914, 16) (3, 6, 12, 15, 20); . 7, . 8, . 1, . 2}$	$c_{332}^{1} = {(6, 8, 11, 9, 14, 16) (4, 7, 13, 16, 20); . 5, . 5, . 4, . 4}$
$c_{333}^{1} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 20); . 8, . 6, . 3, . 3}$

Table 6

Penalties/labor cost $c_{ijk}^{2}$

$c_{111}^{2} = {(3, 5, 8, 12, 15) (1, 4, 10, 13, 16); . 6.5, . 3, . 4}$	$c_{211}^{2} = {(3, 5, 7, 12, 14) (1, 4, 10, 13, 17); . 5.7, . 4, . 4}$
$c_{112}^{2} = {(8, 11, 13, 16, 19) (6, 9, 15, 18, 21); . 6, . 8, . 3, . 2}$	$c_{212}^{2} = {(10, 12, 13, 19, 21) (8, 11, 15, 20, 24); . 6, . 8, . 5, . 4}$
$c_{113}^{2} = {(7, 10, 12, 15, 18) (5, 8, 14, 17, 20); . 7, . 8, . 3, . 2}$	$c_{213}^{2} = {(2, 4, 6, 11, 14) (1, 3, 9, 12, 16); . 7.7, . 5, . 3}$
$c_{121}^{2} = {(9, 11, 14, 18, 20) (7, 10, 16, 19, 22); . 7, . 5, . 4, . 3}$	$c_{221}^{2} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 20); . 7, . 7, . 2, . 2}$
$c_{122}^{2} = {(7, 9, 12, 16, 18) (5, 8, 14, 17, 20); . 6, . 8, . 4, . 2}$	$c_{222}^{2} = {(6, 8, 11, 15, 17) (4, 7, 13, 16, 20); . 8, . 6, . 3, . 3}$
$c_{123}^{2} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 18); . 8, . 6, . 3, . 4}$	$c_{223}^{2} = {(13, 16, 18, 22, 28) (12, 14, 21, 25, 30); . 7, . 6, . 2, . 2}$
$c_{131}^{2} = {(20, 22, 23, 29, 31) (18, 21, 25, 30, 34); . 8, . 7, . 2, . 2}$	$c_{231}^{2} = {(8, 10, 12, 17, 20) (7, 9, 15, 18, 22); . 6, . 6, . 2, . 2}$
$c_{132}^{2} = {(6, 8, 11, 15, 17) 4, 7, 13, 16, 19; . 8, . 6, . 3, . 3}$	$c_{232}^{2} = {(7, 9, 11, 16, 19) (6, 8, 14, 17, 21); . 5, . 5, . 3.2}$
$c_{133}^{2} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 7, . 7, . 2, . 3}$	$c_{233}^{2} = {(7, 9, 12, 16, 18) (5, 8, 14, 17, 20); . 4, . 4, . 2, . 4}$
$c_{311}^{2} = {(13, 15, 17, 22, 24) (11, 14, 20, 23, 27); . 5.7.4, . 4}$	$c_{322}^{2} = {(8, 10, 12, 17, 20) (7, 9, 15, 18, 24); . 7, . 6, . 2, . 2}$
$c_{312}^{2} = {(16, 18, 21, 25, 27) (14, 17, 23, 26, 29); . 7, . 7, . 2, . 2}$	$c_{323}^{2} = {(3, 5, 8, 12, 15) (1, 4, 10, 13, 19); . 5, . 5, . 3, . 3}$
$c_{313}^{2} = {(17, 20, 22, 25, 28,) (15, 18, 24, 27, 30); . 6, . 8, . 2, . 2}$	$c_{331}^{2} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 6, . 6, . 3, . 3}$
$c_{321}^{2} = {(10, 12, 13, 19, 24) (8.11.15, 20, 24); . 6, . 7.5, . 3}$	$c_{332}^{2} = {(6, 8, 11, 15, 17) (4, 7, 13, 16, 20); . 7, . 6, . 2, . 3}$
$c_{333}^{2} = {(13, 15, 17, 22, 26) (11, 14, 20, 25, 30); . 5, . 5, . 4, . 2}$

Table 7

Penalties/time taken to transport one unit item $c_{ijk}^{3}$

$c_{111}^{3} = {(7, 10, 12, 15, 18) (5, 8, 14, 17, 21); . 6.7, . 2, . 2}$	$c_{211}^{3} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 6, . 6, . 2, . 2}$
$c_{112}^{3} = {(3, 5, 8, 12, 14) (1, 4, 10, 13, 16); . 6, . 5, . 3, . 4}$	$c_{212}^{3} = {(20, 22, 23, 29, 31) (18, 21, 25, 30, 34); . 6, . 7, . 2, . 3}$
$c_{113}^{3} = {(6, 8, 11, 15, 17) (4, 7, 13, 16, 21); . 6, . 7, . 3, . 3}$	$c_{213}^{3} = {(5, 7, 9, 14, 16) (3, 6, 12, 15, 18); . 5, . 5, . 3, . 3}$
$c_{121}^{3} = {(2, 4, 6, 11, 13) (0, 3, 9, 12, 15); . 5, . 5, . 2, . 2}$	$c_{221}^{3} = {(8, 10, 12, 17, 20) (7, 9, 15, 18, 24); . 8, . 7, . 2, . 4}$
$c_{122}^{3} = {(3, 7, 8, 16, 18) (1, 5, 9, 12, 16); . 7, . 7, . 2, . 2}$	$c_{222}^{3} = {(3, 5, 8, 15, 17) (1, 4, 10, 16, 20); . 4, . 5, . 2, . 2}$
$c_{123}^{3} = {(10, 12, 13, 18, 20) (8, 11, 15, 19, 21); . 5, . 6, . 3, . 2}$	$c_{223}^{3} = {(8, 11, 13, 16, 19) (6, 9, 15, 18, 21); . 6, . 6, . 3, . 3}$
$c_{131}^{3} = {(6, 8, 11, 15, 17) (4, 7, 13, 16, 20); . 7, . 6, . 2, . 3}$	$c_{231}^{3} = {(5, 7, 9, 16, 18) (3, 6, 12, 15, 20); 6.6, . 2, . 2}$
$c_{132}^{3} = {(10, 12, 13, 18, 23) (8, 11, 15, 20, 24); . 5, . 7, . 4, . 3}$	$c_{232}^{3} = {(13, 16, 18, 22, 26) (12, 14, 20, 25, 29); . 7, . 6, . 3, . 2}$
$c_{133}^{3} = {(7, 9, 11, 16, 19) (6, 8, 14, 17, 20); . 8, . 6, . 1, . 3}$	$c_{233}^{3} = {8, 11, 13, 16, 19) (6, 9, 15, 18, 21); . 6, . 7, . 3, . 2}$
$c_{311}^{3} = {(16, 18, 21, 25, 27) (14, 17, 23, 26, 29); . 5.7.4, . 3}$	$c_{322}^{3} = {(8, 11, 13, 16, 20) (6, 9, 15, 18, 21); . 7, . 7, . 2, . 2}$
$c_{312}^{3} = {(2, 4, 6, 11, 14) (1, 3, 9, 12, 16); . 6, . 6, . 5, . 4}$	$c_{323}^{3} = {(5, 7, 9, 15, 18) (3, 6, 12, 16, 20); . 5, . 4, . 2, . 2}$
$c_{313}^{3} = {(10, 12, 13, 19, 22,) 8, 11, 15, 20, 24); . 7, . 7, . 2, . 2}$	$c_{331}^{3} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 6, . 6, . 3, . 3}$
$c_{321}^{3} = {(11, 13, 16, 20, 22) (9, 12, 18, 21, 24); . 7, . 7.2, . 2}$	$c_{332}^{3} = {(9, 11, 14, 18, 20) (7, 10, 16, 19, 22); . 6, . 7, . 3, . 3}$
$c_{333}^{3} = {(3, 5, 8, 12, 15) (1, 4, 10, 13, 17); . 5, . 6, . 3, . 4}$

Using Equation (2) $c_{ijk}^{r}$ is converted to crisp value and is given below. $c_{ijk}^{1} = [\begin{matrix} 9.6 & 12.8 & 13.8 & 8.6 & 14.6 & 12.7 & 11.6 & 12.7 & 9.1 \\ 10.3 & 8.5 & 12.3 & 7.4 & 9.7 & 14.7 & 24.6 & 10.5 & 17.3 \\ 14.6 & 7.5 & 9.9 & 10.5 & 12.7 & 9.5 & 10.5 & 11.7 & 10.4 \end{matrix}]$ $c_{ijk}^{2} = [\begin{matrix} 8.6 & 13.8 & 12.8 & 14.6 & 12.8 & 11.7 & 24.6 & 11.6 & 16.6 \\ 8.5 & 14.6 & 22.8 & 10.5 & 10.2 & 19 & 13.5 & 12.5 & 12.6 \\ 18.5 & 21.7 & 7.6 & 14.7 & 13.7 & 8.9 & 16.7 & 11.7 & 18.8 \end{matrix}]$ $c_{ijk}^{3} = [\begin{matrix} 12.8 & 8.6 & 11.9 & 24.6 & 10.3 & 13.5 & 21.8 & 7.5 & 14.6 \\ 7.3 & 8.9 & 14.3 & 13.5 & 9.3 & 13.7 & 16.7 & 13.7 & 10.5 \\ 11.7 & 14.6 & 12.3 & 10.6 & 18.1 & 13.7 & 13.4 & 14.7 & 8.7 \end{matrix}]$

First consider single objective (ie) solve $c_{ijk}^{1}, c_{ijk}^{2}, c_{ijk}^{3}$ separately to get feasible solution, $c_{ijk}^{1}$ for j = 1 solid transportation problem reduces to transportation problem $\begin{matrix} c_{i 1 k}^{1} = [\begin{matrix} 9.6 & 12.8 & 13.8 \\ 10.3 & 8.5 & 12.3 \\ 14.6 & 7.5 & 9.9 \end{matrix}] \begin{matrix} 5 \\ 3 \\ 2 \end{matrix} \\ \begin{matrix} 4 & 3 & 3 \end{matrix} \end{matrix}$

Applying least cost method the feasible solution of the transportation problem is $\begin{matrix} c_{i 1 k}^{1} = [\begin{matrix} 9.6 (4) & 12.8 & 13.8 (1) \\ 10.3 & 8.5 (1) & 12.3 (2) \\ 14.6 & 7.5 (2) & 9.9 \end{matrix}] \begin{matrix} 5 \\ 3 \\ 2 \end{matrix} \\ \begin{matrix} 4 & 3 & 3 \end{matrix} \end{matrix}$

B₁₁ = 6,B₁₂ = 5,B₁₃ = 4,B₂₁ = 5,B₂₂ = 4,B₂₃ = 4,B₃₁ = 4,B₃₂ = 6,B₃₃ = 5.

In above allocated values of $c_{i 1 k}^{1} < B_{ik} \forall i, k$ hence $x_{111}^{1} = 4, x_{113}^{1} = 1 x_{212}^{1} = 1, x_{213}^{1} = 2, x_{312}^{1} = 2$

Consider $c_{ijk}^{1}$ for j = 2 $\begin{matrix} c_{i 2 k}^{1} = [\begin{matrix} 8.6 & 14.6 & 12.7 \\ 7.4 & 9.7 & 14.7 \\ 10.5 & 12.7 & 9.5 \end{matrix}] \begin{matrix} 5 \\ 6 \\ 7 \end{matrix} \\ 6 8 4 \end{matrix}$

Revised B_ikvalues are $B_{ik}^{'} = B_{ik} -$ allocated values of $c_{ijk}^{1}$

$B_{11}^{'} = 2$ , $B_{12}^{'} = 5$ , $B_{13}^{'} = 3$ , $B_{21}^{'} = 5$ , $B_{22}^{'} = 3$ , $B_{23}^{'} = 2$ , $B_{31}^{'} = 4$ , $B_{32}^{'} = 4$ , $B_{33}^{'} = 5$

Applying least cost method the feasible solution of the transportation problem is $\begin{matrix} c_{i 2 k}^{1} = [\begin{matrix} 8.6 & 14.6 (5) & 12.7 \\ 7.4 (6) & 9.7 & 14.7 \\ 10.5 & 12.7 (3) & 9.5 (4) \end{matrix}] \begin{matrix} 5 \\ 6 \\ 7 \end{matrix} \\ \begin{matrix} 6 & 8 & 4 \end{matrix} \end{matrix}$

In the above check allocated values of $c_{i 2 k}^{1} < B_{ik}^{'} \forall i, k$ . It is observed that $c_{221}^{1} > B_{21}^{'}$ by one unit hence we find an alternative solution by forming a closed loop following the steps given in step: 5. $\begin{matrix} c_{i 2 k}^{1} = [\begin{matrix} 8.6 (+) & 14.6 (5) (-) & 12.7 \\ 7.4 (6) (-) & 9.7 (+) & 14.7 \\ 10.5 & 12.7 (3) & 9.5 (4) \end{matrix}] \begin{matrix} 5 \\ 6 \\ 7 \end{matrix} \\ \begin{matrix} 6 & 8 & 4 \end{matrix} \end{matrix}$ $\begin{matrix} c_{i 2 k}^{1} = [\begin{matrix} 8.6 (1) & 14.6 (4) & 12.7 \\ 7.4 (5) & 9.7 (1) & 14.7 \\ 10.5 & 12.7 (3) & 9.5 (4) \end{matrix}] \begin{matrix} 5 \\ 6 \\ 7 \end{matrix} \\ \begin{matrix} 6 & 8 & 4 \end{matrix} \end{matrix}$

Now $c_{i 2 k}^{1} < B_{ik}^{'} \forall i, k .$ Hence, $x_{121}^{1} = 1, x_{122}^{1} = 4$ , $x_{221}^{1} = 5$ , $x_{222}^{1} = 1$ , $x_{322}^{1} = 3$ , $x_{323}^{1} = 4$ .

Consider $c_{ijk}^{1}$ for j = 3 $\begin{matrix} c_{i 3 k}^{1} = [\begin{matrix} 11.6 & 12.7 & 9.1 \\ 24.6 & 10.5 & 17.3 \\ 10.4 & 11.7 & 10.4 \end{matrix}] \begin{matrix} 5 \\ 4 \\ 6 \end{matrix} \\ \begin{matrix} 5 & 4 & 6 \end{matrix} \end{matrix}$

Revised B_ik values are $B_{ik}^{''} = B_{ik}^{'} -$ allocated values of $c_{i 2 k}^{1}$

$B_{11}^{''} = 1$ , $B_{12}^{''} = 1$ , $B_{13}^{''} = 3$ , $B_{21}^{''} = 0$ , $B_{22}^{''} = 2$ , $B_{23}^{''} = 2$ , $B_{31}^{''} = 4$ , $B_{32}^{''} = 1$ , $B_{33}^{''} = 1$

Now allocation of every occupied cell $c_{i 2 k}^{1} (i = 1, 2, 3; k = 1, 2, 3)$ is less than or equal to the available units B_ik so the remaining units are allocated to $c_{i 3 k}^{1} (i = 1, 2, 3; k = 1, 2, 3)$ , hence basic feasible solution is $x_{131}^{1} = 1, x_{132}^{1} = 1$ , $x_{133}^{1} = 3$ , $x_{232}^{1} = 2$ , $x_{233}^{1} = 2$ , $x_{331}^{1} = 4$ , $x_{332}^{1} = 1, x_{333}^{1} = 3$ .

$X^{1} = {x_{111}^{1} = 4, x_{113}^{1} = 1 x_{212}^{1} = 1$ , $x_{213}^{1} = 2$ , $x_{312}^{1} = 2, x_{121}^{1} = 1, x_{122}^{1} = 4$ , $x_{221}^{1} = 5$ , $x_{222}^{1} = 1$ , $x_{322}^{1} = 3$ , $x_{323}^{1} = 4, x_{131}^{1} = 1, x_{132}^{1} = 1$ , $x_{133}^{1} = 3$ , $x_{232}^{1} = 2$ , $x_{233}^{1} = 2$ , $x_{331}^{1} = 4$ , $x_{332}^{1} = 1, x_{333}^{1} = 1}$

Similarly solving for $c_{ijk}^{2}$ , $c_{ijk}^{3}$ the basic feasible are obtained as

$X^{2} = {x_{111}^{2} = 1, x_{112}^{2} = 3, x_{113}^{2} = 1$ , $x_{211}^{2} = 3$ , $x_{313}^{2} = 2, x_{121}^{2} = 3, x_{122}^{2} = 2$ , $x_{222}^{2} = 4, x_{223}^{2} = 2$ , $x_{321}^{2} = 3, x_{322}^{2} = 2, x_{323}^{2} = 2$ , $x_{131}^{2} = 2$ , $x_{133}^{2} = 3$ , $x_{231}^{2} = 2$ , $x_{233}^{2} = 2$ , $x_{331}^{2} = 1, x_{332}^{2} = 4, x_{333}^{2} = 1}$

$X^{3} = {x_{111}^{3} = 1, x_{112}^{3} = 1, x_{113}^{3} = 3$ , $x_{211}^{3} = 3, x_{312}^{3} = 2, x_{121}^{3} = 5$ , $x_{222}^{3} = 4$ , $x_{223}^{3} = 2$ , $x_{321}^{3} = 1$ , $x_{322}^{3} = 4, x_{323}^{3} = 2, x_{312}^{3} = 5$ , $x_{231}^{3} = 2$ , $x_{233}^{3} = 2$ , $x_{331}^{3} = 3$ , $x_{333}^{3} = 3}$

Pay of matrix is formed as given in step 9 and is given below $\begin{matrix} \begin{matrix} X^{1} & X^{2} & X^{3} \end{matrix} \\ \begin{matrix} Z_{1} \\ Z_{2} \\ Z_{3} \end{matrix} [\begin{matrix} 461.2 & 513 & 511 \\ 589.5 & 575 & 586 \\ 524 & 546 & 527.5 \end{matrix}] \end{matrix} ∥$ $U_{1} = 513.5, L_{1} = 461.2, U_{2} = 575, L_{2} = 573,$ $U_{3} = 586.1, L_{3} = 527.5$

Linear programming problem is formed as given in equation

Max λ

subject to $\sum_{i = 1}^{3} \sum_{j = 1}^{3} \sum_{k = 1}^{3} x_{ijk} c_{ijk}^{r}, + λ (U_{r} - L_{r}) ⩽ U_{r} for all r = 1, 2, 3$ (16)

along with Equation (4)–(7) $and λ ⩾ 0 and λ = \min_{r} {σ_{r} (Z_{r})}$ (17)

Using Lingo optimizing software

We get λ = . 7655

x₁₁₁ = 4, x₁₁₂ = 1, x₁₂₁ = 1.87, x₁₂₂ = 3.12, x₁₃₁ = .12, x₁₃₂ = .89, x₁₃₃ = 4,

x₂₁₂ = 1.41, x₂₁₃ = 1.58, x₂₂₁ = 4.1, x₂₂₂ = .68, x₂₂₃ = 1.2, x₂₃₁ = .89,

x₂₃₂ = 1.8, x₂₃₃ = 1.2, x₃₁₂ = .58, x₃₁₃ = 1.41, x₃₂₂ = 4.2, x₃₂₃ = 2.79, x₃₃₁ = 4,

x₃₃₂ = 1.2, x₃₃₃ = .79.68.

Optimal objective values of $Z_{1}^{#}, Z_{2}^{#}, Z_{3}^{#}$ are 472.7, 577.4, 531.18 respective

5 Results and discussion

In this section the drawbacks of the existing methods, comparison with an existing method, advantages and limitations of our proposed research work has been outlined. Which is given in Table 8.

Table 8
Restrictions of existing transportation models along with the underlying features

Author Objective function and parameters considered Constraints

One function More than one function Uncertain parameters IFPN Item Destination Mode of transportation

Aggarwal [26] ✓ × ✓ × ✓ ✓ ✓

Mallia [20] ✓ ✓ × × ✓ ✓ ×

M.C.Uddin [21] ✓ ✓ ✓ × ✓ ✓ ×

R.Gowthami [22] ✓ ✓ ✓ × ✓ ✓ ×

A.Das [30] ✓ ✓ ✓ × ✓ ✓ ✓

Proposed model ✓ ✓ ✓ ✓ ✓ ✓ ✓

Author	Objective function and parameters considered	Constraints
Aggarwal [26]	✓	×	✓	×	✓	✓	✓
Mallia [20]	✓	✓	×	×	✓	✓	×
M.C.Uddin [21]	✓	✓	✓	×	✓	✓	×
R.Gowthami [22]	✓	✓	✓	×	✓	✓	×
A.Das [30]	✓	✓	✓	×	✓	✓	✓
Proposed model	✓	✓	✓	✓	✓	✓	✓

As seen from the Table 9 proposed method minimizes all the objective function as compared with the algorithm provided in [39 , 40–54]. Figure 3 is shows that comparison of proposed with existing methods optimal objective values.

Table 9

Comparison of proposed method with existing method

Method applied	Optimal objective value of Z₁	Optimal objective value of Z₂	Optimal objective value of Z₃	Values of Decision variable
Existing method [39]	495.8	587.2	533.9	x₁₁₁ = 2.75, x₁₁₂ = 2.25, x₂₁₂ = .71, x₂₁₃ = 2.24
				x₃₁₁ = 1.24, x₃₁₃ = .75, x₁₂₁ = 3.25, x₁₂₂ = 1.75
				x₂₂₁ = 2.75, x₂₂₂ = 2.37, x₂₂₃ = .87, x₃₂₂ = 3.87
				x₃₂₃ = 3.12, x₁₃₃ = 4, x₂₃₁ = 2.24, x₂₃₂ = 0.87,
				x₂₃₃ = .87, x₃₃₁ = 2.75, x₃₃₂ = 2.12, x₃₃₃ = 1.1
Proposed method	472.7	577.4	531.18	x₁₁₁ = 4, x₁₁₂ = 1, x₁₂₁ = 1.87, x₁₂₂ = 3.12,
				x₁₃₁ = .12, x₁₃₂ = .89, x₁₃₃ = 4, x₂₁₂ = 1.41
				x₂₁₃ = 1.58, x₂₂₁ = 4.1, x₂₂₂ = .68,, x₂₂₃ = 1.2
				x₂₃₁ = 0.89, x₂₃₂ = 1.8, x₂₃₃ = 1.2, x₃₁₂ = 0.58
				x₃₁₃ = 1.41, x₃₂₂ = 4.2, x₃₂₃ = 2.79, x₃₃₁ = 4
				x₃₃₂ = 1.2, x₃₃₃ = .79

Fig. 3

Comparison between existing methods.

The MOSTP in intuitionistic fuzzy environment has been studied for the first time, which handle uncertainty in a better way. In this work PIFN are considered as parameter for MOSTP which is the generalized form of triangular intuitionistic fuzzy numbers (ie), in $A = {(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) (a_{1}^{'}, a_{2}^{'}, a_{3}^{'}, a_{4}^{'}, a_{5}^{'}) : t, u, v, w}$ when t = u = 0, andv = w = 1 the PIFN takes the form of triangular intuitionistic fuzzy numbers. The Magnitude based ranking is used to solve IFMOSTP which is better than many other ordering methods, since it satisfies Kerre’s properties for ordering fuzzy numbers. The vehicle type used to transport is selected according to the nature of the goods to be transported. Allocation of the items to the vehicle is done by considering the volume capacity of the vehicle. The limitations of the present research work cannot be applied if the objective function is non-linear.

6 Conclusion

The necessity of multi-objective transportation problems arises, when the objective is not only to minimize the cost of transportation but also to minimize transportation time, labor effort, etc. The proposed method yields an optimal solution for the given multi-objective transportation problem when the cost and time is uncertain. In this method, Lingo software is used to find the optimal solution. Future scope is may be extended in non-linear system by using intuitionistic fuzzy numbers. This method can help decision makers with the logistics related issues of real life problems in decision making process and provide an optimal solution in a simple and effective manner.

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