A two-stage data envelopment analysis approach to solve extended transportation problem with non-homogenous costs

Ali Hadi; Saeid Mehrabian

doi:10.3934/naco.2022006

Article Contents

2023, Volume 13, Issue 2: 258-272. Doi: 10.3934/naco.2022006

This issue Previous Article Iterative method for solving split common fixed point problem of asymptotically demicontractive mappings in Hilbert spaces Next Article An adaptive block iterative process for a class of multiple sets split variational inequality problems and common fixed point problems in hilbert spaces

A two-stage data envelopment analysis approach to solve extended transportation problem with non-homogenous costs

Ali Hadi^1, and
Saeid Mehrabian^2, ,

1.
Department of Mathematics, Rasht Branch, Islamic Azad University, Rasht, Iran
2.
Department of Mathematics, Faculty of Mathematical Science Computer, Kharazmi University, Tehran, Iran

^*Corresponding author: Saeid_Mehrabian@khu.ac.ir
^*Corresponding author: Saeid_Mehrabian@khu.ac.ir

Received: March 2021

Revised: February 2022

Early access: March 2022

Published: January 2023

Abstract / Introduction Full Text(HTML) Figure(1) / Table(5) Related Papers Cited by

Abstract

The transportation problem is a particular type of linear programming problem in which the main objective is to minimize the cost. In marked contrast to the classical real-world transportation model, shipping supplies from one source to a destination cause several costs and benefits, each of which is incomparable to another. The extended transportation problem was first introduced in a study conducted by Amirteimoori [1]. In contrast, many important questions regarding the production possibility set, the place of costs, the benefits, and the essence of these costs were not fully addressed yet. Therefore, this paper focuses on transportation models that do not provide explicit output. This method is helpful because it is designed for a specific purpose: to send goods and supply-demand at the lowest cost and decision-maker; does not suffer from the confusion of costs and the various consequences of placing them costs and outputs. Furthermore, this model improves the contradiction between the essence of the problem and the input/output-oriented data envelopment analysis. In this paper, previous models that can not incorporate all the sources of inefficiency have been solved. We apply the slack-based measure(SBM) to calculate all identified inefficiency sources. A numerical example is considered to show the approach's applicability, as mentioned above, to actual life situations. As a result, the optimal costs achieved via the proposed method are more realistic and accurate by obtaining a more representative efficiency assessment. This example proved our proposed approach's efficiency, providing a more efficient solution by corporate all sources inefficiency and presenting efficient costs for each path.

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Mathematics Subject Classification: Primary: 90C05, 90C08; Secondary: 90B06.

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Figure 1. Extended transportation problem

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Table 1. The data for example

	I	J	K	$ S_i $
A	(531, 3500,500)	(431,380,600)	(395, 3950,400)	10
B	(394, 2850,600)	(418, 2395,700)	(512, 2590,485)	13
C	(405,310,800)	(512,409, 1000)	(412,390, 1100)	11
D	(355,290,705)	(493,385,617)	(570,419,518)	7
E	(299,415,585)	(398,512,490)	(315,255,380)	9
F	(319,512,488)	(464,215,305)	(435,355,512)	9
G	(619,612,619)	(490,510,505)	(354,550,490)	4
H	(456,299,601)	(394,512,432)	(439,499,519)	6
$ d_j $	30	25	14

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Table 2. Results for example

	I	J	K
A	$ \tilde{e}_{AI}=0.5743 $ $ \bar{e}_{AI}=0.7988 $	$ \tilde{e}_{AJ}=0.8475 $ $ \bar{e}_{AJ}=1,0000 $	$ \tilde{e}_{AK}=0.5743 $ $ \bar{e}_{AK}=1,0000 $
B	$ \tilde{e}_{BI}=0.6009 $ $ \bar{e}_{BI}=1,0000 $	$ \tilde{e}_{BJ}=0.5754 $ $ \bar{e}_{BJ}=1,0000 $	$ \tilde{e}_{BK}=0.6009 $ $ \bar{e}_{BK}=1,0000 $
C	$ \tilde{e}_{CI}=0.8978 $ $ \bar{e}_{CI}=1,0000 $	$ \tilde{e}_{CJ}=0.579 $ $ \bar{e}_{CJ}=0.7830 $	$ \tilde{e}_{CK}=0.8978 $ $ \bar{e}_{CK}=0.8551 $
D	$ \tilde{e}_{DI}=1,0000 $ $ \bar{e}_{DI}=1,0000 $	$ \tilde{e}_{DJ}=0.6646 $ $ \bar{e}_{DJ}=0.9454 $	$ \tilde{e}_{DK}=1.3844 $ $ \bar{e}_{DK}=1,0000 $
E	$ \tilde{e}_{EI}=1,0000 $ $ \bar{e}_{EI}=1,0000 $	$ \tilde{e}_{EJ}=0.9446 $ $ \bar{e}_{EJ}=0.6886 $	$ \tilde{e}_{EK}=1,0000 $ $ \bar{e}_{EK}=1,0000 $
F	$ \tilde{e}_{FI}=1,0000 $ $ \bar{e}_{FI}=1,0000 $	$ \tilde{e}_{FJ}=1,0000 $ $ \bar{e}_{FJ}=1,0000 $	$ \tilde{e}_{FK}=1,0000 $ $ \bar{e}_{FK}=0.8134 $
G	$ \tilde{e}_{GI}=0.7021 $ $ \bar{e}_{GI}=1.2044 $	$ \tilde{e}_{GJ}=1.189 $ $ \bar{e}_{GJ}=0.7541 $	$ \tilde{e}_{GK}=0.7021 $ $ \bar{e}_{GK}=1,0000 $
H	$ \tilde{e}_{HI}= 1,0000 $ $ \bar{e}_{HI}=1,0000 $	$ \tilde{e}_{HJ}=1,0000 $ $ \bar{e}_{HJ}=1,0000 $	$ \tilde{e}_{HK}=1,0000 $ $ \bar{e}_{HK}=0.9195 $

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Table 3. Composite efficiency

	I	J	K
A B	$ e_{AI}=0.6682 $ $ e_{BI}=0.7507 $	$ e_{AJ}=0.9175 $ $ e_{BJ}=0.7305 $	$ e_{AK}=0.7296 $ $ e_{BK}=0.7507 $
C	$ e_{CI}=0.9461 $	$ e_{CJ}=0.6657 $	$ e_{CK}=0.8653 $
D	$ e_{DI}=1,0000 $	$ e_{DJ}=0.7805 $	$ e_{DK}=1,0000 $
E	$ e_{EI}=1,0000 $	$ e_{EJ}=0.7963 $	$ e_{EK}=1,0000 $
F	$ e_{FI}=1,0000 $	$ e_{FJ}=1,0000 $	$ e_{FK}=0.8971 $
G	$ e_{GI}=0.7281 $	$ e_{GJ}=0.8285 $	$ e_{GK}=0.8287 $
H	$ e_{HI}=1,0000 $	$ e_{HJ}=1,0000 $	$ e_{HK}=0.9597 $

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Table 4. Results comparing

Arc	Destination aspect	Source aspect	Average	SBM-Without-Input
AI	0.5743	0.7988	0.6865	0.6682
AJ	0.8475	1	0.9237	0.9175
AK	0.5743	1	0.7871	0.7296
BI	0.6009	1	0.8004	0.7507
BJ	0.5754	1	0.7877	0.7305
BK	0.6009	1	0.8004	0.7507
CI	0.8978	1	0.9489	0.9461
CJ	0.579	0.783	0.681	0.6657
CK	0.8978	0.8351	0.8664	0.8653
DI	1	1	1	1
DJ	0.6646	0.9454	0.805	0.7805
DK	1	1	1	1
EI	1	1	1	1
EJ	0.9446	0.6883	0.8164	0.7963
EK	1	1	1	1
FI	1	1	1	1
FJ	1	1	1	1
FK	1	0.8134	0.9067	0.8971
GI	0.7021	0.7541	0.7281	0.7272
GJ	0.6575	1	0.8287	0.7934
GK	0.7021	1	0.851	0.825
HI	1	1	1	1
HJ	1	1	1	1
HK	1	0.9195	0.9597	0.9581
Mean	0.825783	0.939067	0.882404	0.866746
STD	0.182328	0.099468	0.108758	0.122179

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Table 5. Optimal costs

	I	J	K
A	0	10	0
B	10	0	3
C	11	0	0
D	0	0	7
E	9	0	0
F	0	9	0
G	0	0	4
H	0	6	0

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