<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">AJOR</journal-id><journal-title-group><journal-title>American Journal of Operations Research</journal-title></journal-title-group><issn pub-type="epub">2160-8830</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/ajor.2022.126016</article-id><article-id pub-id-type="publisher-id">AJOR-121554</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Solving Fuzzy Multi-Objective Linear Programming Problem by Applying Statistical Method
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Samsun</surname><given-names>Nahar</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Marin</surname><given-names>Akter</given-names></name><xref ref-type="aff" rid="aff2"><sup>2</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Md.</surname><given-names>Abdul Alim</given-names></name><xref ref-type="aff" rid="aff3"><sup>3</sup></xref></contrib></contrib-group><aff id="aff3"><addr-line>Department of Mathematics, Bangladesh University of Engineering and Technology, Dhaka, Bangladesh</addr-line></aff><aff id="aff1"><addr-line>Department of Basic Sciences and Humanities, University of Asia Pacific, Dhaka, Bangladesh</addr-line></aff><aff id="aff2"><addr-line>General Education Department, City University, Dhaka, Bangladesh</addr-line></aff><pub-date pub-type="epub"><day>25</day><month>10</month><year>2022</year></pub-date><volume>12</volume><issue>06</issue><fpage>293</fpage><lpage>309</lpage><history><date date-type="received"><day>5,</day>	<month>September</month>	<year>2022</year></date><date date-type="rev-recd"><day>27,</day>	<month>November</month>	<year>2022</year>	</date><date date-type="accepted"><day>30,</day>	<month>November</month>	<year>2022</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, the statistical averaging method and the new statistical averag
  ing methods have been used to solve the fuzzy multi-objective linear programming problems. These methods have been applied to form a single objective 
  function from the fuzzy multi-objective linear programming problems. At first, 
  a numerical example of solving fuzzy multi-objective linear programming prob
  lem has been provided to validate the maximum risk reduction by the proposed method. The proposed method has been applied to assess the risk of damage due to natural calamities like flood, cyclone, sidor, and storms at the coastal areas in Bangladesh. The proposed method of solving the fuzzy multi-objective linear programming problems by the statistical method has been 
  compared with the Chandra Sen’s method. The numerical results show
   
  that the proposed method maximizes the risk reduction capacity better than
   Chandra Sen’s method.
 
</p></abstract><kwd-group><kwd>Fuzzy Multi-Objective Linear Programming Problem</kwd><kwd> Fuzzy Linear Programming Problem</kwd><kwd> Chandra Sen’s Method</kwd><kwd> Statistical Averaging Method</kwd><kwd> New Statistical Averaging Method</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>Real world circumstances are not always deterministic. There exist different kinds of uncertainties in social, industrial and economic systems. Different kinds of uncertainties are defined as stochastic uncertainty and fuzziness [<xref ref-type="bibr" rid="scirp.121554-ref1">1</xref>]. A system with a stochastic uncertainty is solved by the stochastic optimization technique using the probability theory. Also, fuzzy programming technique is widely used to solve problem with uncertainty. A system with uncertainty can be optimized to reduce the risk factors in the system with fuzzy conditions. An optimization of a system with uncertainty using fuzzy conditions is called a fuzzy Optimization. Modelling under a fuzzy environment is called fuzzy modelling. Fuzzy linear programming is one of the most frequently applied fuzzy decision making techniques. We get fuzzy linear programming problem (FLPP) by interchanging the parameters of linear programming problem by fuzzy numbers. Several methods have been proposed in the literature to solve FLPP.</p><p>Firstly, Bellman and Zadeh in [<xref ref-type="bibr" rid="scirp.121554-ref2">2</xref>] proposed decision making in fuzzy condition. They explained the use of decision making in fuzzy condition by using a controlled system which is either stochastic or deterministic with multistage decision processes, Zimmermann proposed a formulation of FLPP using theory of fuzzy sets. He used FLPP to the linear vector maximum problem and found that the solutions obtained using FLPP are always efficient. Considering parameters ambiguity, Tanaka and Asai also proposed a formulation of FLPP to find a logical solution [<xref ref-type="bibr" rid="scirp.121554-ref3">3</xref>]. They highlighted that the FLPP with fuzzy numbers may be considered as a model of decision problems in a system with influential human estimations. Not only linear programming problems (LPP), but also multi-objective LPP (MOLPP) and multi-objective nonlinear programming problems were solved using fuzzy approaches which are available in the literature.</p><p>Laganathan and Lalitha [<xref ref-type="bibr" rid="scirp.121554-ref4">4</xref>] solved a multi-objective nonlinear programming problem using α-cut method in fuzzy approach and compared the solution with the solution obtained by Zimmermann [<xref ref-type="bibr" rid="scirp.121554-ref5">5</xref>] who used membership function. Beher et al. [<xref ref-type="bibr" rid="scirp.121554-ref6">6</xref>] also used α-cut method to solve multi-objective linear programming problem (MOLPP) in fuzzy approach and compared the solutions to the solution obtained by Zimmermann [<xref ref-type="bibr" rid="scirp.121554-ref5">5</xref>] who used membership function.</p><p>Thakre et al. [<xref ref-type="bibr" rid="scirp.121554-ref7">7</xref>] solved an objective function with constraint matrix and cost coefficients which are fuzzy in nature. They used MOLPP with the constraints to solve FLPP and proved that the solutions are independent of weights. Nahar and Alim [<xref ref-type="bibr" rid="scirp.121554-ref8">8</xref>] discussed Chandra Sen’s approach, statistical averaging method and new statistical averaging method to solve MOLPP. For MOLFPP, a new geometric average technique was proposed by Nahar and Alim [<xref ref-type="bibr" rid="scirp.121554-ref9">9</xref>].</p><p>Veeramani and Sumathi [<xref ref-type="bibr" rid="scirp.121554-ref10">10</xref>] solved fuzzy linear fractional programming problem (FLFPP) where the cost, resources and technological coefficients of the objective function were triangular fuzzy numbers. In the solution procedure, they converted the FLFPP into a multi-objective linear fractional programming problem.</p><p>There are a lot of methods for solving fully fuzzy linear programming problems in the literature. Ebrahim nejad and Tavana [<xref ref-type="bibr" rid="scirp.121554-ref11">11</xref>] converted the FLPP into an equivalent crisp linear programming problem and solved by simplex method. Here they proposed a new concept in which the coefficients of objective function and the values of the right hand side are represented by trapezoidal fuzzy numbers and other parts are represented by real numbers. A general form of fuzzy linear fractional programming problem with trapezoidal fuzzy numbers is proposed by Das [<xref ref-type="bibr" rid="scirp.121554-ref12">12</xref>].</p><p>Lotfi et al. [<xref ref-type="bibr" rid="scirp.121554-ref13">13</xref>] proposed a method to obtain the approximate solution of fully fuzzy linear programming problems. A method to solve fully fuzzy linear programming problems is proposed by Amit Kumar et al. [<xref ref-type="bibr" rid="scirp.121554-ref14">14</xref>] using idea of crisp linear programming and ranking function. FLPP is solved by S. Nahar et al. [<xref ref-type="bibr" rid="scirp.121554-ref15">15</xref>]. To solve FLPP, Nahar, S. et al. [<xref ref-type="bibr" rid="scirp.121554-ref16">16</xref>] used weighted sum method. Here both equal and unequal weight has been used. In this paper there is a discussion for ranking function. Here they used triangular and trapezoidal fuzzy number. Sen, C. [<xref ref-type="bibr" rid="scirp.121554-ref17">17</xref>] developed averaging technique of multi-objective optimization for rural development planning. Akter, M. et al. [<xref ref-type="bibr" rid="scirp.121554-ref18">18</xref>] proposed fuzzy synthetic evaluation method for risk assessment on the natural hazard. Nishad, A. [<xref ref-type="bibr" rid="scirp.121554-ref19">19</xref>] used alpha cut fuzzy number for solving fractional programming problem in fuzzy field. Pitam, S., [<xref ref-type="bibr" rid="scirp.121554-ref20">20</xref>] developed goal programming approach for fuzzy multi-objective linear and fuzzy multi-objective linear programming problem.</p><p>A good averaging represents maximum characteristics of the data. There are five averages, among them mean, median, mode are called simple averages. Geometric mean and harmonic mean are called special averages. In this work we consider arithmetic mean, geometric mean and harmonic mean. In this paper, we solve fuzzy multi-objective linear programming problem by applying Chandra Sen’s method, statistical averaging method and new statistical averaging method. These methods are applied for making single objective from multi-objective linear programming problem. A real life example is given to reduce risk in coastal region. From that data list a FLPP is structured. From that FLPP we convert into MOLPP.</p></sec><sec id="s2"><title>2. Concept of a Fuzzy Set</title><p>The concept of a fuzzy set is an extension of the concept of a crisp set. A crisp set on a universal set U is defined by its characteristic function from U to {0, 1}. A fuzzy set on a domain U is defined by its membership function from U to [0, 1]. Let U be a nonempty set, to be called the universal set or the universe of discourse or simply a domain. Then by a fuzzy set on U is meant a function A : U → [ 0 , 1 ] . A is called the membership function, A(x) is called the membership grade of x in (U, A). We also write A = { ( A , A ( x ) ) : x ∈ U } .</p></sec><sec id="s3"><title>3. Fuzzy Linear Programming Problem</title><p>A linear programming problem with fuzzy values is called fuzzy linear programming problem. In this paper, any fuzzy number is denoted by using ~ above fuzzy number, e.g., c ˜ j , a ˜ i j etc. Consider a fuzzy linear programming problem as in Equation (1).</p><p>( c ˜ , x ) = f i ( x j ) = f i ( x ) = max z ˜ = ∑ j = 1 n c ˜ j x j</p><p>subject   to         ∑ j = 1 n a ˜ i j x j ≤ b ˜ i ;   1 ≤ i ≤ m ,   ∃   x j &gt; 0 (1)</p><p>The membership functions of a i j and b i have been expressed as in Equation (2) and (3), respectively.</p><p>μ a i j ( x ) = { 1 ;                         x &lt; a i j a i j + d i j − x d i j ;   a i j ≤ x ≤ a i j + d i j 0 ;                 x ≥ a i j + d i j (2)</p><p>μ b i ( x ) = { 1 ;             x ≤ b i j b i + p i − x p i ;   b i ≤ x ≤ b i + p i 0 ;           b i + p i ≤ x (3)</p><p>Let a ˜ i j = ( m i j , l i j , r i j ) and b ˜ i = ( d i , e i , f i ) be fuzzy numbers. Therefore, the constraints in Equation (1) can be modified as in Equation (4).</p><p>max z ˜ = ∑ j = 1 n c ˜ j x j subject   to     ∑ ( m i j , l i j , r i j ) x i j ≤ ( d i , e i , f i )   ∀ i = 1 − m                         x j ≥ 0 , j = 1 − n (4)</p><p>Theorem: For any two triangular fuzzy numbers Thakre et al. [<xref ref-type="bibr" rid="scirp.121554-ref8">8</xref>] A = (s<sub>1</sub>, l<sub>1</sub>, r<sub>1</sub>) and B = (s<sub>2</sub>, l<sub>2</sub>, r<sub>2</sub>)</p><p>A ≤ B   iff   s 1 ≤ s 2       s 1 − l 1 ≤ s 2 − l 2       s 1 + r 1 ≤ s 2 + r 2 (5)</p><p>max z ˜ = ∑ j = 1 n c ˜ j x j subject   to     ∑ m i j x j ≤ d i     ∑ ( m i j − l i j ) x j ≤ d i − e i   ∀ i     ∑ ( m i j + r i j ) x j ≤ d i + f i         x j ≥ 0   ( ∀ j ) (6)</p><p>where membership function of c ˜ j ( x ) is</p><p>μ ˜ c j = { x − α i β i − α i ;   α i ≤ x ≤ β i γ i − x γ i − β ;       β i ≤ x ≤ γ i 0 ;         elsewhere (7)</p></sec><sec id="s4"><title>4. Numerical Examples</title><p>Consider the fuzzy multi-objective linear programming problem (FMOLPP) as in Equation (8).</p><p>    max   z ˜ 1 = ( 7 , 10 , 14 ) x 1 + ( 20 , 25 , 35 ) x 2     max   z ˜ 2 = ( 10 , 14 , 25 ) x 1 + ( 25 , 35 , 40 ) x 2 Subject   to :         ( 3 , 2 , 1 ) x 1 + ( 6 , 4 , 1 ) x 2 ≤ ( 13 , 5 , 2 )         ( 4 , 1 , 2 ) x 1 + ( 6 , 5 , 4 ) x 2 ≤ ( 7 , 4 , 2 ) (8)</p><p>where the membership function of c ˜ 1 , c ˜ 2 , c ˜ 3 , c ˜ 4 are</p><p>μ c ˜ 1 ( x ) = { x − 7 3 ;       7 &lt; x ≤ 10 14 − x 4 ;   10 &lt; x ≤ 14 0 ;       elsewhere (9)</p><p>μ c ˜ 2 ( x ) = { x − 20 5 ;   20 &lt; x ≤ 25 35 − x 10 ;   25 &lt; x ≤ 35 0 ;           elsewhere (10)</p><p>μ c ˜ 3 ( x ) = { x − 10 4 ;     10 &lt; x ≤ 14 25 − x 9 ;         14 &lt; x ≤ 25 0 ;           elsewhere (11)</p><p>μ c ˜ 4 ( x ) = { x − 25 5 ;     25 &lt; x ≤ 35 40 − x 5 ;     35 &lt; x ≤ 40 0 ;           elsewhere (12)</p><p>From Equation (8) we get,</p><p>max   z 1 = 7 x 1 + 20 x 2 max   z 2 = 10 x 1 + 25 x 2 max   z 3 = 14 x 1 + 35 x 2 max   z 4 = 25 x 1 + 40 x 2 (13)</p><p>Subject to</p><p>3 x 1 + 6 x 2 ≤ 13 x 1 + 2 x 2 ≤ 8 4 x 1 + 7 x 2 ≤ 15 4 x 1 + 6 x 2 ≤ 7 3 x 1 + x 2 ≤ 3 x 1 + 10 x 2 ≤ 9 x 1 , x 2 ≥ 0 (14)</p><p>For the first objective function in Equation (13) with constraints in Equation (14), by applying simplex algorithm we get</p><p>ϕ 1 = 20.3529 with (0.4706, 0.8529)</p><p>Similarly, for the second objective function in Equation (13) with constraints in Equation (14), we get</p><p>ϕ 2 = 26.0294 with (0.4706, 0.8529)</p><p>Similarly for third objective function in Equation (13) with constraints in Equation (14), we get</p><p>ϕ 3 = 36.4412 with (0.4706, 0.8529)</p><p>And for last objective function in Equation (13) with constraints in Equation (14), we get</p><p>ϕ 4 = 45.8824 with (0.4706, 0.8529)</p><sec id="s4_1"><title>4.1. Chandra Sen’s Method</title><p>Chandra Sen’s method is a multi-objective optimization technique which is used for making single objective from multi-objectives. In the last three decades several new multi-objective optimization techniques have been developed.</p><p>Applying Chandra Sen’s method [<xref ref-type="bibr" rid="scirp.121554-ref17">17</xref>] for making single objective function from multi objective function</p><p>max   z = z 1 ϕ 1 + z 2 ϕ 2 + z 3 ϕ 3 + z 4 ϕ 4 = 1 20.3529 ( 7 x 1 + 20 x 2 ) + 1 26.0294 ( 10 x 1 + 25 x 2 )     + 1 36.4412 ( 14 x 1 + 35 x 2 ) + 1 45.8824 ( 25 x 1 + 40 x 2 ) = x 1 ( 0.344 + 0.384 + 0.384 + 0.545 ) + x 2 ( 0.983 + 0.96 + 0.96 + 0.872 ) = 1.657 x 1 + 3.775 x 2</p><p>Thus the single objective function becomes</p><p>max z = 1.657 x 1 + 3.775 x 2 (15)</p><p>For this objective function in Equation (15) with constraints in Equation (14), we get</p><p>z = 3.9996 with (0.4706, 0.8529)</p></sec><sec id="s4_2"><title>4.2. Statistical Averaging Method</title><p>There are three mean known as arithmetic mean, geometric mean and harmonic mean. For ungrouped raw data, mean is defined as the sum of the objectives divided by the number of observations. It is easy to understand and easy to calculate. If the number of items is sufficiently large, it is more accurate and more reliable. Geometric mean of a series containing n observations is the nth root of the product of the values. Harmonic mean of a set of observations is defined as the reciprocal of the arithmetic average of the reciprocal of the given values.</p><p>Applying arithmetic mean, geometric mean and harmonic mean among ϕ 1 , ϕ 2 , ϕ 3 , ϕ 4</p><p>A . M = 20.3529 + 26.0294 + 36.412 + 45.8824 4 = 32.176 G . M = 20.3529 &#215; 26.0294 &#215; 36.412 &#215; 45.8824 4 = 30.68 H . M = 4 1 20.3529 + 1 26.0294 + 1 36.4412 + 1 45.8824       = 4 0.049 + 0.038 + 0.027 + 0.0217 = 29.477</p><p>Arithmetic averaging method:</p><p>max   z = 1 A . M ( z 1 + z 2 + z 3 + z 4 ) = 1 32.176 ( 7 x 1 + 20 x 2 + 10 x 1 + 25 x 2 + 14 x 1 + 35 x 2 + 25 x 1 + 40 x 2 ) = 1 32.176 [ 56 x 1 + 120 x 2 ] = 1.740 x 1 + 3.729 x 2</p><p>Thus the single objective function becomes</p><p>max   z = 1.740 x 1 + 3.729 x 2 (16)</p><p>For this objective function in Equation (16) with constraints in Equation (14), we get</p><p>z = 3.9994 with (0.4706, 0.8529)</p><p>Geometric averaging method:</p><p>max   z = 1 G . M ( z 1 + z 2 + z 3 + z 4 ) = 1 G . M ( 56 x 1 + 120 x 2 ) = 1 30.68 ( 56 x 1 + 120 x 2 ) = 1.825 x 1 + 3.911 x 2</p><p>Thus the single objective function becomes</p><p>max   z = 1.825 x 1 + 3.911 x 2 (17)</p><p>For this objective function in Equation (17) with constraints in Equation (14), we get</p><p>z = 4.1947 with (0.4706, 0.8529)</p><p>Harmonic averaging method:</p><p>max   z = 1 H . M ( z 1 + z 2 + z 3 + z 4 ) = 1 H . M ( 56 x 1 + 120 x 2 ) = 1 29.477 ( 56 x 1 + 120 x 2 ) = 1.899 x 1 + 4.071 x 2</p><p>Thus the single objective becomes</p><p>max   z = 1.899 x 1 + 4.071 x 2 (18)</p><p>For this objective function in Equation (18) with constraints in Equation (14), we get</p><p>z = 4.3660 with (0.4706, 0.8529)</p><p><xref ref-type="table" rid="table1">Table 1</xref> shows the comparison between the Chandra Sen’s method and the statistical averaging methods. Here statistical averaging method consists of arithmetic mean, geometric mean and harmonic mean. The statistical averaging method gives better result than the Chandra Sen’s method. These methods are used for obtaining a single objective function from multi-objective functions. We can see the comparison graphically “as shown in <xref ref-type="fig" rid="fig1">Figure 1</xref>”.</p></sec><sec id="s4_3"><title>4.3. New Statistical Averaging Method</title><p>Choosing minimum from the optimal values of maximum type in Chandra Sen’s method we get</p><p>max   z = 1 20.3529 ( z 1 + z 2 + z 3 + z 4 ) = 0.04913 ( 56 x 1 + 120 x 2 ) = 2.75128 x 1 + 5.8956 x 2</p><p>Thus the single objective becomes</p><p>max   z = 2.75128 x 1 + 5.8956 x 2 (19)</p><p>For this objective function in Equation (19) with constraints in Equation (14), we get</p><p>z = 6.3233 with (0.4706, 0.8529)</p><p>We can find a single objective function from multi objective functions by using any method among Chandra Sen’s method, statistical averaging method, and new statistical averaging method. The <xref ref-type="table" rid="table2">Table 2</xref> shows that the statistical averaging method and the new statistical averaging method gives better optimization than the Chandra Sen’s method.</p></sec></sec><sec id="s5"><title>5. Reduce Damage and Loses at the Coastal Area in Bangladesh</title><p><xref ref-type="table" rid="table3">Table 3</xref> shows secondary data of four parameters such as cropping intensity,</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Comparison between Chandra Sen’s method and statistical averaging method</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Chandra Sen’s method</th><th align="center" valign="middle" >Arithmetic mean</th><th align="center" valign="middle" >Geometric mean</th><th align="center" valign="middle" >Harmonic mean</th></tr></thead><tr><td align="center" valign="middle" >3.9996</td><td align="center" valign="middle" >3.9994</td><td align="center" valign="middle" >4.1947</td><td align="center" valign="middle" >4.3660</td></tr></tbody></table></table-wrap><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> Comparison among Chandra Sen’s method, statistical averaging method, and new statistical averaging method</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Chandra Sen’s Method</th><th align="center" valign="middle" >Statistical Averaging Method (SAM)</th><th align="center" valign="middle" >New Statistical Averaging Method (NSAM)</th></tr></thead><tr><td align="center" valign="middle" >Z<sub>max</sub> = 3.9996 with (0.4706, 0.8529)</td><td align="center" valign="middle" >Z<sub>max</sub> = 4.3660 with (0.4706, 0.8529)</td><td align="center" valign="middle" >Z<sub>max</sub> = 6.3233 with (0.4706, 0.8529)</td></tr></tbody></table></table-wrap><table-wrap id="table3" ><label><xref ref-type="table" rid="table3">Table 3</xref></label><caption><title> Data based on the data of four coastal areas</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Coastal areas</th><th align="center" valign="middle" >Cropping intensity</th><th align="center" valign="middle" >Shelter</th><th align="center" valign="middle" >Erosion</th><th align="center" valign="middle" >Population density</th></tr></thead><tr><td align="center" valign="middle" >kutubdia</td><td align="center" valign="middle" >13</td><td align="center" valign="middle" >100</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >10</td></tr><tr><td align="center" valign="middle" >Maheskl</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >37</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >18</td></tr><tr><td align="center" valign="middle" >Pekua</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >47</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >27</td></tr><tr><td align="center" valign="middle" >Manpura</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >96</td><td align="center" valign="middle" >28</td><td align="center" valign="middle" >0</td></tr><tr><td align="center" valign="middle" >sum</td><td align="center" valign="middle" >26</td><td align="center" valign="middle" >280</td><td align="center" valign="middle" >28</td><td align="center" valign="middle" >55 = 389</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >26/389 = 0.067</td><td align="center" valign="middle" >280/389 = 0.72</td><td align="center" valign="middle" >28/389 = 0.072</td><td align="center" valign="middle" >55/389 = 0.1413</td></tr></tbody></table></table-wrap><p>shelter, erosion, and population density at four coastal areas in Bangladesh. These data set have been collected from IWFM (BUET). The objective in this paper is to reduce the risk and hazard of the coastal area during natural disasters by solving FMOLPP. The targets set in this paper are to maximize cropping intensity and shelter and to minimize erosion and population density. For these purposes, those four parameters have been defined using four variables which are defined as decision variables.</p><p>Our decision variables are x 1 , x 2 , x 3 , x 4 which are risk and vulnerability indicators. x 1 for cropping intensity, x 2 for shelter, x 3 for erosion and x 4 for population density. Constant vector is terminated value for risk reduction capacity (on expert opinion). Decision variables are in different scales. For this reason, these variables have been normalized. Constraints are established based on the data of four coastal areas. The risk reduction coefficients of the objective functions are calculated using weighted method.</p><p>max   Z = 0.067 x 1 + 0.72 x 2 min   Z = 0.072 x 3 + 0.1413 x 4 (20)</p><p>Rearranging Equation (20), we get</p><p>max   Z 1 = 0.067 x 1 + 0.72 x 2 + 0 x 3 + 0 x 4 max   Z 2 = 0 x 1 + 0 x 2 − 0.072 x 3 − 0.1413 x 4 (21)</p><sec id="s5_1"><title>5.1. Finding Fuzzy Objective Function for Fuzzy Linear Programming Problem</title><p>Writing Equation (21) in matrix form, we get</p><p>max [ Z 1 Z 2 ] = [ 0.067 0.72 0 0 0 0 − 0.072 − 0.1413 ] [ x 1 x 2 x 3 x 4 ] (22)</p><p>Triangular fuzzy number for the coefficient of x<sub>1</sub> is found from <xref ref-type="table" rid="table4">Table 4</xref>.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 0.001122 + 0.001122 2 = 0.001122 = 0.0335 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( 0.0335 − 0.0335 ,   0.0335 ,   0.0335 + 0.0335 ) = ( 0 ,   0.0335 ,   0.067 ) , triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>2</sub> is found from <xref ref-type="table" rid="table5">Table 5</xref>.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 0.1296 + 0.1296 2 = 0.1296 = 0.36 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( 0.36 − 0.36 ,   0.36 ,   0.36 + 0.36 ) = ( 0 ,   0.36 ,   0.72 ) , triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>3</sub> is found from <xref ref-type="table" rid="table6">Table 6</xref>.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 0.001296 + 0.001296 2 = 0.001296 = 0.036 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( − 0.036 − 0.036 ,   − 0.036 ,   − 0.036 + 0.036 ) = ( − 0.072 ,   − 0.036 ,   0 ) , triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>4</sub> is found from <xref ref-type="table" rid="table7">Table 7</xref>.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 0.00499 + 0.00499 2 = 0.00499 = 0.07065 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( − 0.07065 − 0.07065 ,   − 0.07065 ,   − 0.07065 + 0.0706 ) 5 = ( − 0.1413 ,   − 0.07065 ,   0 ) , triangular fuzzy number.</p><p>The objective function of fuzzy linear programming problem is</p><p>max   Z = ( 0 ,   0.00335 ,   0.067 ) x 1 + ( 0 ,   0.36 , 0.72 ) x 2     + ( − 0.072 ,   − 0.036 ,   0 ) x 3 + ( − 0.1413 ,   − 0.07065 ,   0 ) x 4</p><p>Thus the multiple objective functions become as like Equation (13)</p><p>max   Z 1 = 0 x 1 + 0 x 2 − 0.072 x 3 − 0.1413 x 4 max   Z 2 = 0.00335 x 1 + 0.36 x 2 − 0.036 x 3 − 0.07065 x 4 max   Z 3 = 0.067 x 1 + 0.72 x 2 + 0 x 3 + 0 x 4 (23)</p></sec><sec id="s5_2"><title>5.2. Finding Constraints of Fuzzy Linear Programming Problem</title><p>Triangular fuzzy number for the coefficient of x<sub>1</sub> is found from <xref ref-type="table" rid="table8">Table 8</xref>.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 1586 4 = 396.5 = 19.9123 Math_63#, triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>2</sub> is found from <xref ref-type="table" rid="table9">Table 9</xref>.</p><table-wrap id="table4" ><label><xref ref-type="table" rid="table4">Table 4</xref></label><caption><title> Coefficient ofx<sub>1</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >0.067</td><td align="center" valign="middle"  rowspan="2"  >0.0335</td><td align="center" valign="middle" >0.0335</td><td align="center" valign="middle" >0.001122</td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >−0.0335</td><td align="center" valign="middle" >0.001122</td></tr></tbody></table></table-wrap><table-wrap id="table5" ><label><xref ref-type="table" rid="table5">Table 5</xref></label><caption><title> Coefficient of x<sub>2</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >0.72</td><td align="center" valign="middle"  rowspan="2"  >0.36</td><td align="center" valign="middle" >0.36</td><td align="center" valign="middle" >0.1296</td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >−0.36</td><td align="center" valign="middle" >0.1296</td></tr></tbody></table></table-wrap><table-wrap id="table6" ><label><xref ref-type="table" rid="table6">Table 6</xref></label><caption><title> Coefficient of x<sub>3</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle"  rowspan="2"  >−0.036</td><td align="center" valign="middle" >0.036</td><td align="center" valign="middle" >0.001296</td></tr><tr><td align="center" valign="middle" >−0.072</td><td align="center" valign="middle" >−0.036</td><td align="center" valign="middle" >0.001296</td></tr></tbody></table></table-wrap><table-wrap id="table7" ><label><xref ref-type="table" rid="table7">Table 7</xref></label><caption><title> Coefficient of x<sub>4</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle"  rowspan="2"  >−0.0706536</td><td align="center" valign="middle" >0.07065</td><td align="center" valign="middle" >0.00499</td></tr><tr><td align="center" valign="middle" >−0.1413</td><td align="center" valign="middle" >−0.07065</td><td align="center" valign="middle" >0.00499</td></tr></tbody></table></table-wrap><table-wrap id="table8" ><label><xref ref-type="table" rid="table8">Table 8</xref></label><caption><title> Coefficient of x<sub>1</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >13</td><td align="center" valign="middle"  rowspan="4"  >26</td><td align="center" valign="middle" >−13</td><td align="center" valign="middle" >169</td></tr><tr><td align="center" valign="middle" >2</td><td align="center" valign="middle" >−24</td><td align="center" valign="middle" >576</td></tr><tr><td align="center" valign="middle" >6</td><td align="center" valign="middle" >−20</td><td align="center" valign="middle" >400</td></tr><tr><td align="center" valign="middle" >5</td><td align="center" valign="middle" >−21</td><td align="center" valign="middle" >441</td></tr></tbody></table></table-wrap><table-wrap id="table9" ><label><xref ref-type="table" rid="table9">Table 9</xref></label><caption><title> Coefficient of x<sub>2</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >100</td><td align="center" valign="middle"  rowspan="4"  >70</td><td align="center" valign="middle" >30</td><td align="center" valign="middle" >900</td></tr><tr><td align="center" valign="middle" >37</td><td align="center" valign="middle" >−33</td><td align="center" valign="middle" >1089</td></tr><tr><td align="center" valign="middle" >47</td><td align="center" valign="middle" >−23</td><td align="center" valign="middle" >529</td></tr><tr><td align="center" valign="middle" >96</td><td align="center" valign="middle" >26</td><td align="center" valign="middle" >676</td></tr></tbody></table></table-wrap><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 3194 4 = 798.5 = 28.2577 Math_89#, triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>3</sub> is found from <xref ref-type="table" rid="table1">Table 1</xref>0.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 588 4 = 147 = 12.1243 Math_91#, triangular fuzzy number.</p><p>Triangular fuzzy number for the coefficient of x<sub>4</sub> is found from <xref ref-type="table" rid="table1">Table 1</xref>1.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 396.75 4 = 99.1875 = 9.9593 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( 13.75 − 9.9593 ,   13.75 ,   13.75 + 9.9593 ) = ( 3.7907 ,   13.75 ,   23.7093 ) , triangular fuzzy number.</p><p>Triangular fuzzy number for the constant term is found from <xref ref-type="table" rid="table1">Table 1</xref>2.</p><p>Standard deviation, σ = ∑ ( x i − x &#175; ) 2 2 = 68 4 = 17 = 4.1231 ( x &#175; − σ ,   x &#175; ,   x &#175; + σ ) = ( 95 − 4.1231 ,   95 ,   95 + 4.1231 ) = ( 90.8769 ,   95 ,   99.1231 ) , triangular fuzzy number.</p><p>Fuzzy constraints become</p><p>( 6.0877 ,   26 ,   45.9123 ) x 1 + ( 41.7423 ,   70 ,   98.2577 ) x 2 + ( − 5.1243 ,   7 ,   19.1243 ) x 3 + ( 3.7907 ,   13.75 ,   23.7093 ) x 4 ≤ ( 90.8769 ,   95 ,   99.1231 ) (24)</p><p>Equation (24) is as like (25)</p><p>( a ,   b ,   c ) x 1 + ( d ,   e ,   f ) x 2 + ( g ,   h ,   i ) x 3 + ( j ,   k ,   l ) x 4 ≤ ( m ,   n , o ) (25)</p><p>which becomes with the help of Equation (6)</p><p>a x 1 + d x 2 + g x 3 + j x 4 ≤ m ( a − b ) x 1 + ( d − e ) x 2 + ( g − h ) x 3 + ( j − k ) x 4 ≤ m − n ( a + c ) x 1 + ( d + f ) x 2 + ( g + i ) x 3 + ( j + l ) x 4 ≤ m + o (26)</p><table-wrap id="table10" ><label><xref ref-type="table" rid="table1">Table 1</xref>0</label><caption><title> Coefficient of x<sub>3</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle"  rowspan="4"  >7</td><td align="center" valign="middle" >−7</td><td align="center" valign="middle" >49</td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >−7</td><td align="center" valign="middle" >49</td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >−7</td><td align="center" valign="middle" >49</td></tr><tr><td align="center" valign="middle" >28</td><td align="center" valign="middle" >21</td><td align="center" valign="middle" >441</td></tr></tbody></table></table-wrap><table-wrap id="table11" ><label><xref ref-type="table" rid="table1">Table 1</xref>1</label><caption><title> Coefficient of x<sub>4</sub></title></caption><table><tbody><thead><tr><th align="center" valign="middle" >x i</th><th align="center" valign="middle" >mean, x &#175;</th><th align="center" valign="middle" >x i − x &#175;</th><th align="center" valign="middle" >( x i − x &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >10</td><td align="center" valign="middle"  rowspan="4"  >13.75</td><td align="center" valign="middle" >−3.75</td><td align="center" valign="middle" >14.0625</td></tr><tr><td align="center" valign="middle" >18</td><td align="center" valign="middle" >4.25</td><td align="center" valign="middle" >18.0625</td></tr><tr><td align="center" valign="middle" >27</td><td align="center" valign="middle" >13.25</td><td align="center" valign="middle" >175.5625</td></tr><tr><td align="center" valign="middle" >0</td><td align="center" valign="middle" >−13.75</td><td align="center" valign="middle" >189.0625</td></tr></tbody></table></table-wrap><table-wrap id="table12" ><label><xref ref-type="table" rid="table1">Table 1</xref>2</label><caption><title> Constant term</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >b i</th><th align="center" valign="middle" >mean, b &#175;</th><th align="center" valign="middle" >b i − b &#175;</th><th align="center" valign="middle" >( b i − b &#175; ) 2</th></tr></thead><tr><td align="center" valign="middle" >100</td><td align="center" valign="middle"  rowspan="4"  >95</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >25</td></tr><tr><td align="center" valign="middle" >90</td><td align="center" valign="middle" >−5</td><td align="center" valign="middle" >25</td></tr><tr><td align="center" valign="middle" >98</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >9</td></tr><tr><td align="center" valign="middle" >92</td><td align="center" valign="middle" >−3</td><td align="center" valign="middle" >9</td></tr></tbody></table></table-wrap><p>First equation of (26) becomes</p><p>a x 1 + d x 2 + g x 3 + j x 4 ≤ m 6.0877 x 1 + 41.7423 x 2 + ( − 5.1243 ) x 3 + 3.7907 x 4 ≤ 90.8769</p><p>Second equation of (26) becomes</p><p>( a − b ) x 1 + ( d − e ) x 2 + ( g − h ) x 3 + ( j − k ) x 4 ≤ m − n ( 6.0877 − 26 ) x 1 + ( 41.7423 − 70 ) x 2 + ( − 5.1243 − 7 ) x 3 + ( 3.7907 − 13.75 ) x 4 ≤ 90.8769 − 95 ( − 19.9123 ) x 1 + ( − 28.2577 ) x 2 + ( − 12.1243 ) x 3 + ( − 9.9593 ) x 4 ≤ − 4.1231</p><p>Third equation of (26) becomes</p><p>( a + c ) x 1 + ( d + f ) x 2 + ( g + i ) x 3 + ( j + l ) x 4 ≤ m + o ( 6.0877 + 45.9123 ) x 1 + ( 41.7423 + 98.2577 ) x 2 + ( − 5.1243 + 19.1243 ) x 3 + ( 3.7907 + 23.7093 ) x 4 ≤ 90.8769 + 99.1231 52 x 1 + 140 x 2 + 14 x 3 + 27.5 x 4 ≤ 190</p><p>The Equation (18) is written as the following Fuzzy MOLPP</p><p>max   Z 1 = 0 x 1 + 0 x 2 − 0.072 x 3 − 0.1413 x 4 max   Z 2 = 0.00335 x 1 + 0.36 x 2 − 0.036 x 3 − 0.07065 x 4 max   Z 3 = 0.067 x 1 + 0.72 x 2 + 0 x 3 + 0 x 4 (27)</p><p>subject to</p><p>6.0877 x 1 + 41.7423 x 2 + ( − 5.1243 ) x 3 + 3.7907 x 4 ≤ 90.8769 ( − 19.9123 ) x 1 + ( − 28.2577 ) x 2 + ( − 12.1243 ) x 3 + ( − 9.9593 ) x 4 ≤ − 4.1231 52 x 1 + 140 x 2 + 14 x 3 + 27.5 x 4 ≤ 190 (28)</p><p>For the first objective function in Equation (27) with same constraints in Equation (28), by applying simplex algorithm we get</p><p>ϕ 1 = 0 with (0, 0.1459, 0, 0)</p><p>For the second objective function in Equation (27) with same constraints in Equation (28), by applying simplex algorithm we get</p><p>ϕ 2 = 0.4886 with (0, 1.3571, 0, 0)</p><p>Similarly for third objective function, in Equation (27) with same constraints in Equation (28), we get</p><p>ϕ 3 = 0.9771 with (0, 1.3571, 0, 0)</p></sec><sec id="s5_3"><title>5.3. Chandra Sen’s Method</title><p>Applying Chandra Sen’s method for making single objective function from multi objective functions</p><p>max   z = z 1 ϕ 1 + z 2 ϕ 2 + z 3 ϕ 3 = 0.0754178 x 1 + 1.47366 x 2 − 0.0736812 x 3 − 0.144599 x 4</p><p>Thus the single objective function becomes</p><p>max   Z = 0.0754178 x 1 + 1.47366 x 2 − 0.0736812 x 3 − 0.144599 x 4 (29)</p><p>For this objective function in Equation (29) with same constraints in Equation (28) we get the result 2 with (0, 1.3571, 0, 0).</p></sec><sec id="s5_4"><title>5.4. Statistical Averaging Method</title><p>Applying arithmetic mean, geometric mean and harmonic mean among ϕ 1 , ϕ 2 , ϕ 3</p><p>A . M = 0.48857 G . M = 0 H . M = 0.97716</p><p>Arithmetic averaging method:</p><p>max   z = 1 A . M ( z 1 + z 2 + z 3 ) = 0.154364 x 1 + 3.016272 x 2 − 0.150809 x 3 − 0.295963 x 4</p><p>Thus the single objective function becomes</p><p>max   Z = 0.154364 x 1 + 3.016272 x 2 − 0.150809 x 3 − 0.295963 x 4 (30)</p><p>For this objective function in Equation (30) with same constraints in Equation (28) we get the result 4.0935 with (0, 1.3571, 0, 0)</p><p>Harmonic averaging method:</p><p>max   z = 1 H . M ( z 1 + z 2 + z 3 ) = 0.07718 x 1 + 1.508099 x 2 − 0.0754031 x 3 − 0.147978 x 4</p><p>Thus the single objective becomes</p><p>max   Z = 0.07718 x 1 + 1.508099 x 2 − 0.0754031 x 3 − 0.147978 x 4 (31)</p><p>For this objective function in Equation (26) with same constraints in Equation (23) we get the result 2.0467 with (0, 1.3571, 0, 0).</p><p><xref ref-type="table" rid="table1">Table 1</xref>3 shows that the statistical averaging method, the arithmetic averaging, and the harmonic averaging gives better result than the Chandra Sen’s method.</p><table-wrap id="table13" ><label><xref ref-type="table" rid="table1">Table 1</xref>3</label><caption><title> Comparison between Chandra Sen’s method and statistical averaging method</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Chandra Sen’s method</th><th align="center" valign="middle" >Arithmetic averaging method</th><th align="center" valign="middle" >Harmonic averaging method</th></tr></thead><tr><td align="center" valign="middle" >Z<sub>max</sub> = 2 with (0, 1.3571, 0, 0)</td><td align="center" valign="middle" >Z<sub>max</sub> = 4.0935 with (0, 1.3571, 0, 0)</td><td align="center" valign="middle" >Z<sub>max</sub> = 2.0467 with (0, 1.3571, 0, 0)</td></tr></tbody></table></table-wrap><table-wrap id="table14" ><label><xref ref-type="table" rid="table1">Table 1</xref>4</label><caption><title> Comparison among Chandra Sen’s method, statistical averaging method, and new statistical averaging method</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Chandra Sen’s method</th><th align="center" valign="middle" >Statistical averaging method (SAM)</th><th align="center" valign="middle" >New statistical averaging method (NSAM)</th></tr></thead><tr><td align="center" valign="middle" >Z<sub>max</sub> = 2 with (0, 1.3571, 0, 0)</td><td align="center" valign="middle" >Z<sub>max</sub> = 4.0935 with (0, 1.3571, 0, 0)</td><td align="center" valign="middle" >Z<sub>max</sub> = 4.0933 with (0, 1.3571, 0, 0)</td></tr></tbody></table></table-wrap></sec><sec id="s5_5"><title>5.5. New Statistical Averaging Method</title><p>Choosing minimum from the optimal values of maximum type in Chandra Sen’s method we get m = 0.4886</p><p>max   z = 1 0.4886 ( z 1 + z 2 + z 3 ) = 2.0467 ( 0.075417 x 1 + 1.47366 x 2 − 0.0736812 x 3 − 0.144599 x 4 ) = 0.15435 x 1 + 3.01614 x 2 − 0.1508 x 3 − 0.29595 x 4</p><p>Thus the single objective becomes</p><p>max   Z = 0.15435 x 1 + 3.01614 x 2 − 0.1508 x 3 − 0.29595 x 4 (32)</p><p>For this objective function in Equation (27) with same constraints in Equation (23) we get the result 4.0933 with (0, 1.3571, 0, 0).</p><p><xref ref-type="table" rid="table1">Table 1</xref>4 shows that the statistical averaging method and the new statistical averaging method give better result than the Chandra Sen’s method.</p></sec></sec><sec id="s6"><title>6. Conclusion</title><p>In this paper, a fuzzy multi-objective linear programming problem has been solved using the Chandra Sen’s method, the statistical averaging method, and the new statistical averaging method. The fuzzy multi-objective linear programming problem with constraints has been established from the data of cropping intensity, shelter, erosion, and population density at four coastal areas in Bangladesh. To maximize the risk reduction capacity, those data of cropping intensity, shelter, erosion, and population density have been set as fuzzy parameters which are triangular fuzzy numbers. The solutions obtained using the statistical averaging method and the new statistical averaging method are better than that of the Chandra Sen’s method.</p></sec><sec id="s7"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s8"><title>Cite this paper</title><p>Nahar, S., Akter, M. and Alim, Md.A. 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