Daffodil International University Library Digital Institutional Repository DIU Journal of Science and Technology Volume 5,Issue 1,January 2010 2010-01-01 EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM Das, Bimal Chandra Daffodil International University http://hdl.handle.net/20.500.11948/504 Downloaded from http://dspace.library.daffodilvarsity.edu.bd, Copyright Daffodil International University Library DAFFODIL INTERNATIONAL UNIVERSITY JOURNAL OF SCIENCE AND TECHNOLOGY, VOLUME 5, ISSUE 1, JANUARY 2010 29 EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM Bimal Chandra Das Department of Textile Engineering, Daffodil International University, Dhaka E-mail: bcdas@daffodilvarsity.edu.bd Abstract: In this paper, a computer implementation on the effect of graphical method for solving mathematical programming problem using MATLAB programming has been developed. To take any decision, for programming problems we use most modern scientific method based on computer implementation. Here it has been shown that by graphical method using MATLAB programming from all kinds of programming problem, we can determine a particular plan of action from amongst several alternatives in very short time. Keywords: Mathematical programming, objective function, feasible-region, constraints, optimal solution. 1 Introduction Mathematical programming problem deals with the optimization (maximization/ minimization) of a function of several variables subject to a set of constraints (inequalities or equations) imposed on the values of variables. For decision making optimization plays the central role. Optimization is the synonym of the word maximization/minimization. It means choosing the best. In our time to take any decision, we use most modern scientific and methods based on computer implementations. Modern optimization theory based on computing and we can select the best alternative value of the objective function. [1].But the modern game theory, dynamic programming problem, integer programming problem also part of the optimization theory having wide range of application in modern science, economics and management. In the present work I tried to compare the solution of Mathematical programming problem by Graphical solution method and others rather than its theoretic descriptions. As we know that not like linear programming problem where multidimensional problems have a great deal of applications, non-linear programming problem mostly considered only in two variables. Therefore for nonlinear programming problems we have a opportunity to plot the graph in two dimension and get a concrete graph of the solution space which will be a step ahead in its solutions. We have arranged the materials of the paper in the following way: First I discuss about Mathematical Programming (MP) problem. In second step we discuss graphical method for solving mathematical programming problem and taking different kinds of numerical examples, we try to solve them by graphical method. Finally we compare the solutions by graphical method and others. For problem so consider we use MATLAB programming to graph the constraints for obtaining feasible region. Also we plot the objective functions for determining optimum points and compare the solution thus obtained with exact solutions. 2 Mathematical Programming Problems The general Mathematical programming (MP) problems in n-dimensional Euclidean space Rn can be stated as follows: Maximize f(x) subject to gi ( x) ≤ 0 , i = 1, 2, .....,m (1) h j (x ) = 0 , j = 1, 2, ....., p (2) x ∈s (3) Where x = (x1 , x 2 ,....., x n )T is the vector of unknown decision variables and f ( x), g ( x) (i = 1, 2, 3, .......m ) h j ( x ), ( j = 1, 2, .... p ) are the real valued functions. The function f(x) is known as objective function, and inequalities 30 DAS: EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM (1) equation (2) and the restriction (3) are referred to as the constraints. We have started the MP as maximization one. This has been done without any loss of generality, since a minimization problem can always be converted in to a maximization problem using the identity min f(x) = -max (f(x)) (4) i.e, the minimization of f(x) is equivalent to the maximization of (-f(x)). The set S is normally taken as a connected subset of Rn. Here the set S is taken as the entire space Rn. The set X= {x ∈ s, g i (x)=0, i=1,2, …..,m, j=1,2, …..,p} is known to as the feasible reason, feasible set or constraint set of the program MP and any point x ∈ x is a feasible solution or feasible point of the program MP which satisfies all the constrains of MP. If the constraint set x is empty (i.e. x= φ ), then there is no feasible solution; in this case the program MP is inconsistent and it was developed by [2]. A feasible point x ∈ x is known as a global optimal solution to the program MP if f ( x ) ≤ f ( x ), x ∈ x . By [3]. 3 Graphical Solution Method The graphical (or geometrical) method for solving Mathematical Programming problem is based on a well define set of logical steps. Following this systematic procedure, the given Programming problem can be easily solved with a minimum amount of computational effort and which has been introduced by [4]. We know that simplex method is the well-studied and widely useful method for solving linear programming problem. while for the class of non-linear programming such type of universal method does not exist. Programming problems involving only two variables can easily solved graphically. As we will observe that from the characteristics of the curve we can achieve more information. We shall now several such graphical examples to illustrate more vividly the differences between linear and non-linear programming problems. Consider the following linear programming problems Maximize z = 0.5 x1 + 2 x2 Subject to x1 + x2 ≤ 6 x1 − x 2 ≤ 1 2 x1 + x 2 ≥ 6 0 .5 x1 − x 2 ≥ − 4 x 1 ≥ 1, x 2 ≥ 0 . Fig.1 Optimal solution by graphical method The graphical solution is show in Fig.1. The region of feasible solution is shaded. Note that the optimal does occur at an extreme point. In this case, the values of the variables that yield the maximum value of the objective function are unique, and are the point of intersection of the lines x1 + x2 = 6, 0.5x1 − x2 = −4 so that the optimal values of the variables 4 14 The ∗ ∗ ∗ ∗ x 1 and x 2 are x 1 = , x 2 = . 3 3 maximum value of the objective function is 4 14 z = 0 .5 × + 2 × = 10 , which was by [5]. 3 3 Now consider a non-linear programming problem, which differs from the linear programming problem only in that the objective function: 2 2 z = 10 ( x 1 − 3 . 5 ) + 20 ( x 2 − 4 ) . (5) Imagine that it is desired to minimize the objective function. Observe that here we have a separable objective function. The graphical solution of this problem is given in Fig.2 Fig. 2 Optimal solution by graphical method The region representing the feasible solution is, of course, precisely the same as that for DAFFODIL INTERNATIONAL UNIVERSITY JOURNAL OF SCIENCE AND TECHNOLOGY, VOLUME 5, ISSUE 1, JANUARY 2010 the linear programming problem of Fig1. Here, however, the curves of constant z are ellipse with centers at the point (3.5, 4). The optimal solution is that point at which an ellipse is tangent to one side of the convex set. If the optimal values of the variables are ∗ ∗ x1 and x2 , and the minimum value of the objective function is z*, then from Fig 1-2, ∗ ∗ x1 +x2 =6, ( ∗ ) ( 2 ∗ ∗ and ) 2 z = 10 x1 − 3.5 + 20 x 2 − 4 . Furthermore the slope of the z = 10(x1 − 3.5) + 20( x2 − 4 ) ∗ 2 ( ∗ 1 at x , x2 slope of ∗ 2 curve evaluated ) must be –1 since this is the x1 + x2 = 6. Thus we have the ∗ ( ∗ ) additional equation x2 − 4 = 0.5 x1 − 3.5 . We have obtained three equations involving ∗ x1 , x2 ∗ ∗ and z*. The unique solution is ∗ 31 the same as that considered above. This case is illustrated graphically in Fig.3. The optimal values of x1, x2, and z are x1* = 2, x2* = 3, and z* = 0.Thus it is not even necessary that the optimizing point lie on the boundaries. Note that in this case, the minimum of the objective function in the presence of the constraints and non-negativity restrictions is the same as the minimum in the absence of any constraints or non-negativity restrictions. In such situations we say that the constraints and non-negativity restrictions are inactive, since the same optimum is obtained whether or not the constraints and non-negativity restrictions are included. Each of the examples presented thus far the property that a local optimum was a global optimum and was introduced by [5]. As a final example, I shall examine an integer linear programming problem. Let us solve the problem x1 = 2.50 , x2 = 3.50 and z*=15. Now the 0.5 x1 + x2 ≤ 1.75 point which yields the optimal value of the objective function lies on the boundary of the convex set of feasible solutions, but it is not an extreme point of this set. Consequently, any computational procedure for solving problems of this type cannot be one which examines only the extreme points of the convex set of feasible solutions. By a slight modification of the objective function studied above the minimum value of the objective function can be made to occur at an interior point of the convex set of feasible solutions. Suppose, for example, that the objective function is 2 2 z = 10 ( x1 − 2 ) + 20 ( x 2 − 3 ) x1 + 0.30 x2 ≤ 1.50 x1 , x2 ≥ 0, x1 , x 2 intigers Max z = 0.25 x1 + x2 . The situation is illustrated geometrically in Fig.3.4.The shaded region would be the convex set of feasible solutions in the absence of the integrality requirements. When the xj are required to be integers, there only four feasible solutions which are represented by circles in Fig.4. If we solve the problem as a linear programming problem, ignoring the integrality C B O A Fig. 4 Optimal solution by graphical method Fig. 3 Optimal solution by graphical method and that the convex set of feasible solutions is requirements, the optimal solutions is x ∗1 = 0 , x ∗ 2 = 1.75, and z*= 1.75 . However it is clear that when it is required that the xj be integers, the optimal solution is x1*= 1 , x2*= 1 , DAS: EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM 32 and z*= 1.25 . Note that this is not the solution that would be obtained by the solving the linear programming problem and rounding the results to the nearest integers, which satisfy the constraints (this would give x1 = 0, x 2 = 1 ). and z = 0 .However, in the case of a NLP problem the optimal solution may or may not occur at one of the extreme points of the solution space, generated by the constraints and the objective function of the given problem. Graphical solution algorithm: The solution NLP problem by graphical method, in general, involves the following steps: Step 1: construct the graph of the given NLP problem. Step 2: Identify the convex region (solution space) generated by the objective function and constraints of the given problem. Step 3: Determine the point in the convex region at which the objective function is optimum maximum or minimum). Step 4: Interpret the optimum solution so obtained. Which has been introduced by [2]. 4 Solution of Various Kinds of Problems by Graphical Solution Method 4.1 Problem with objective function linear constraints non-Linear Maximize Z = 2 x1 + 3 x 2 Subject to the constraints x1 + x 2 2 2 ≤ 20 The point which maximizes the value z = 2 x1 + 3 x2 and lies in the convex region OABCD have to find. The desired point is obtained by moving parallel to 2 x1 + 3 x2 =k for some k, so long as 2 x1 + 3 x2 =k touches the extreme boundary point of the convex region. According to this rule, we see that the point C (2, 4) gives the maximum value of Z. Hence we can find the optimal solution at this point by [6] z Max = 2.2 + 3.4 = 16 at x1 = 2, x2 = 4. 4.2 Problem with objective function linear constraints non-linear+linear Z = x1 + 2 x 2 Maximize slt x1 + x 2 ≤ 1 2 2 2 x1 + x 2 ≤ 2 x1 , x 2 ≥ 0 . Let us solve the above problem by graphical method: For this we see that our objective function is linear and constraints are non-linear and linear. Constraints one is a circle of radius 1 with center (0, 0) and constraints two is a straight line. In this case tracing the graph of the constraints of the problem in the first quadrant, we get the following shaded region as opportunity set. x1 x 2 ≤ 8 x1 ≥ 0 , x2 ≥ 0. Let us solve the problem by graphical method: For this, first we are tracing the graph of the constraints of the problem considering inequalities as equations in the first quadrant (since x1 ≥ 0, x2 ≥ 0 ). We get the following shaded region as opportunity set OABCD. D C B A O Fig. 6 Optimal solution by graphical method Considering the inequalities to equalities C x1 + x 2 = 1 (6 ) 2 x1 + x 2 = 2 (7) 2 B Solving O A ( 6 ) and ( 7 ) 3 4 ( x1 , x 2 ) = (1, 0 ),  ,  5 5 The extreme points of the convex region are O(0, 0), A (1,0) B (3 / 5, 4 / 5) and C(0,1). We get Fig. 5 Optimal solution by graphical method 2 DAFFODIL INTERNATIONAL UNIVERSITY JOURNAL OF SCIENCE AND TECHNOLOGY, VOLUME 5, ISSUE 1, JANUARY 2010 By moving according to the above rule we see that the line x1 + 2 x2 = k touches (3 / 5, 4 / 5) the extreme point of the convex region. Hence the required solution of the given problem is 3 4 3 8 11 Z Max = + 2 ⋅ = + = = 2.2 5 5 5 5 5 3 4 at x1 = , x2 = . 5 5 4.3 Problem with objective function nonlinear constraints linear Minimize Z = x1 + x 2 2 2 subject to the constraint s: x1 + x2 ≥ 4 2x1 + x2 ≥ 5 x1 , x2 ≥ 0 Our objective function is non-linear which is a circle with origin as center and constraints are linear. The problem of minimizing Z = x1 + x2 is equivalent to minimizing the radius of a circle with origin as centre such that it touches the convex region bounded by the given constraints. First we contracts the graph of the constraints by MATLAB programming [9] in the 1st A quadrant since x1 ≥ 0, x2 ≥ 0 . 2 2 33 2 x1 + x2 = 5 and x1 + x2 = 4 Differentiating, we get 2dx1 + dx2 = 0 and dx1 + dx2 = 0 ⇒ dx 2 = −2 dx1 dx2 = −1 dx1 (9) Now, from (8) and (9) we get − x1 = − 2 ⇒ x1 = 2 x 2 x2 and − x1 = − 1 ⇒ x1 = x 2 . x2 This shows that the circle has a tangent to it(i) the line x1 + x2 = 4 at the point (2,2) (ii) the line 2 x1 + x2 = 5 at the point (2,1). But from the graph we see that the point (2,1) does not lie in the convex region and hence is to be discarded. Thus our require point is (2,2). ∴ Minimum Z = 22 + 22 = 8 at thepoint(2,2). 5 Comparison of Solution by Graphical Method and Others Let us consider the problem Maximize Z = 2 x1 + 3 x2 − x1 Subject to the constraints: 2 x1 + 2 x2 ≤ 4 x1 , x2 ≥ 0 First I want to solve above problem by graphical solution method. The given problem can be rewriting as: B   1 3 Maximize Z = −(x1 − 1) + 3 x2 +  2 Subject to the constraints x1 + 2 x2 ≤ 4 C x1 , x2 ≥ 0 Fig. 7 Optimal solution by graphical method x1 + x2 ≥ 4 and 2 x1 + x2 ≥ 5, the Since desire point must be some where in the unbounded convex region ABC. The desire point will be that point of the region at which a side of the convex region is tangent to the circle. Differentiating the equation of the circle 2 x 1 dx 1 + 2 x 2 dx 2 = 0 ⇒ dx 2 x = − 1 dx 1 x2 Considering the inequalities to equalities (8 ) We observe that our objective function is a parabola with vertex at (1, -1/3) and constraints are linear. To solve the problem graphically, first we construct the graph of the constraint in the first quadrant since x1 ≥ 0 and x2 ≥ 0 by considering the inequation to equation. Here we contract the graph of our problem by MATLAB programming [9] According to our previous graphical method our desire point is at (1/4, 15/8) DAS: EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM 34 If we consider 2 x1 = Now putting the value of x1 in (10), we get x = 15 B 2 1 3 4 −1− 3 ∂F = 0 satisfied ≡ 2− 2⋅ − = 4 2 2 ∂ x1 0 3 ∂F ≡ 3 − 2/ ⋅ = 0 satisfied 2/ ∂x2 Fig. 8 Optimum solution by graphical method Hence we get the maximum value of the objective function at this point. Therefore, Zmax = 2 x1 + 3 x2 − x1 = 97 16 1 ,x 4 2 = 15 . 8 Let us solve the above problem by using [7] Kuhn-Tucker Conditions. The Lagrangian function of the given problem is F (x 1 , x 2 , λ ) ≡ 2 x 1 + 3 x 2 − x 1 + λ (4 − x 1 − 2 x 2 ). 2 By Kuhn-Tucker conditions, we obtain ∂F ≡ 2 − 2 x1 − λ ≤ 0 , ∂x1 15 16 − 1 − 15 1 ∂F ≡ 4− − 2/ ⋅ = = 0 satisfied 8/ 4 4 4 ∂λ x1 2 at x 1 = 8  1 15 3  for this solution ∴ (x 1 , x 2 , λ ) =  , , . 4 8 2 A (a) 1 1 then x1 = . 2 4 ∂F ≡ 3 − 2λ ≤ 0 ∂x 2 λ 15 1 ∂F ∂F ≡ ×0+ × 0 = 0 satisfied + x2 ∂ x1 8 4 ∂x2 ∂F 3 ≡ ⋅ 0 = 0 satisfied 2 ∂λ Thus all the Kuhn-Tucker necessary conditions are satisfied at the point (1/4, 15/8) Hence the optimum (maximum) solution to the given NLP problem is 2 Z max = 2 x1 + 3 x 2 − x1 = 97 16 at x 1 = 1 15 . , x2 = 4 8 ∂F ≡ 4 − x1 − 2 x 2 ≥ 0 Let us solve the problem by Beale’s method. ∂λ 2 Maximize f ( x ) = 2 x1 + 3 x2 − x1 ∂F ∂F ≡ x1 (2 − 2 x1 − λ ) + x 2 (3 − 2λ ) = 0 Subject to the constraints: (c ) x1 + x2 ∂x1 ∂x 2 (b) (d ) λ ∂F ≡ λ (4 − x1 − 2 x2 ) = 0 ∂λ x1 + 2 x2 ≤ 4 with λ ≥ 0. x1 , x2 ≥ 0 Now there arise the following cases: Case (i) : Let λ = 0 , in this case we get from ∂F ∂F ≡ 2 − 2 x1 ≤ 0 and ≡ 3− 2⋅0 ≤ 0 ∂x1 ∂x 2 ⇒ 3 ≤ 0 which is impossible and this solution is to be discarded and it has been introduced by [12].Case (ii): Let λ ≠ 0 . In this case we get from λ ( 4 − x1 − 2 x2 ) = 0 4 − x1 − 2x2 = 0 ⇒ x1 + 2x2 = 4 Also from ∂ F ≡ 2 − 2 x1 − λ ≤ 0 ∂ x1 ∂F ≡ 3 − 2 λ ≤ 0 ∴ 2 x1 + λ − 2 ≥ 0 ∂x 2 and 2λ − 3 ≥ 0 ⇒ λ ≥ 3 2 3 1 If we take λ = , then 2 x1 ≥ 2 2 (10) Introducing a slack variables s, the constraint becomes x1 + 2 x2 + s = 4 x1 , x2 ≥ 0 since there is only one constraint, let s be a basic variable. Thus we have by [13] xB = (s ), xNB = (x1,x2 ) with s = 4 Expressing the basic xB and the objective function in terms of non-basic xNB, we have and f =2x1+3x2-x12. We s=4-x1-2x2 evaluated the partial derivatives of f w.r.to non-basic variables at xNB=0, we get  ∂f    = (2 − 2 x1 )x = 2 − 2⋅0 = 2  ∂ x1  x NB = 0 NB = 0  ∂f   ∂x 2   =3  x NB = 0 since both the partial derivatives are positive, the current solution can be improved. As DAFFODIL INTERNATIONAL UNIVERSITY JOURNAL OF SCIENCE AND TECHNOLOGY, VOLUME 5, ISSUE 1, JANUARY 2010 ∂f gives the most positive value, x2 will ∂x 2 enter the basis. Now, to determine the leaving basic variable, we compute the ratios:  α  α γ  γ  min  ho , ko  = min  30 , 20   α 32 α 22   α hk γ kk   4 3  , =2 = min   − 2 0  α since the minimum occurs for 30 , s will α 30 leave the basis and it was introduced by [8]. Thus expressing the new basic variable, x2 as well as the objective function f in terms of the new non-basic variables (x1 and s) we have: x1 s − 2 2 x s  2 and f = 2 x1 + 3 2 − 1 −  − x1 2 2   x1 3 2 =6+ − s − x1 2 2 Expressing the basic xB in terms of non-basic xNB , we have, x1 = and x 2 =  ∂f     ∂s  x NB =0   1 =  − 2 x1    2 = − x NB =0 3  ∂F  ; =−   2  ∂ s  x NB = 0 Now, u =0  δf   δu1 20 21 , γ 10 γ 11   = − 2u1 = 0  x NB = 0 u =0 since = x1 = 0 1 2 3 . 2  1/ 2 2  ,  =  − 2   − 1 / 2  3 .  = 4  since the minimum of these ratios correspond , non-basic variables can be to γ 10 γ 11 removed. Thus we introduce a free variable, u1 as an additional non-basic variable, defined by 1 ∂f 1 1 1  =  − 2 x1  = u1 = − x1 2 ∂ x1 22 4  Note that now the basis has two basic variables x2 and x1 (just entered). That is, we have x NB = (s , u 1 ) and x B = ( x 1 , x 2 ) . ∂f ∂f ≤ 0 for all xj in xNB and = 0, ∂u1 ∂u the current solution is optimal. Hence the optimal basic feasible solution to the given problem is: x since the partial derivatives are not all negative, the current solution is not optimal, clearly, x1 will enter the basis. For the next Criterion, we compute the ratios  α min   α 1 (4 − x 1 − x 3 ) = 15 + 1 u 1 − 1 s . 2 8 2 2 1  1 1   15 1  f = 2 − u1  + 3 + u1 − s  −  − u1  8 2 2 4 4       97 3 2 = − s − u1 . 16 2 we, again, evaluate the partial derivates of f w. r. to the non-basic variables:    1 − u1 4 The objective function, expressing in terms of xNB is, x2 = 2 −  ∂f   ∂ x1 35 1 1 , 4 = x 2 15 8 , Z * = 97 16 Similarly we can find that by Wolfe’s algorithm the optimal point is at (1/4, 15/8). which was introduced by [14]. Thus for the optimal solution for the given QP problem is M a x Z = 2 x1 + 3 x 2 − x12 = 2 ⋅ = 97 16 1 15  1  + 3 ⋅ −   4 8  4  at (x ∗ 1 , x2∗ 2 )=  1 15  ,   8   4 Therefore the solution obtained by graphical solution method, Kuhn-Tucker conditions, Beale’s method and Wolf’s algorithm are same. The computational cost is that by the graphical solution method using MATLAB Programming it will take very short time to determine the plan of action and the solution obtained by graphical method is more effective than any other methods we considered. 2 36 DAS: EFFECT OF GRAPHICAL METHOD FOR SOLVING MATHEMATICAL PROGRAMMING PROBLEM 6 Conclusion This paper has been presented a direct, fast and accurate way for determining an optimum schedule (such as maximizing profit or minimizing cost) The graphical method gives a physical picture of certain geometrical characteristics of programming problems. By using MATLAB programming graphical solution can help us to take any decision or determining a particular plan of action from amongst several alternatives in very short moment. All kinds of programming problem can be solved by graphical method. The limitation is that programming involving more than two variables i.e for 3-D problems can not be solved by this method. Non-linear programming problem mostly considered only in two variables. Therefore, from the above discussion, we can say that graphical method is the best to take any decision for modern game theory, dynamic programming problem science, economics, and management from amongst several alternatives. References [1] Greig, D. M: “Optimization”. Lougman- Group United, New York (1980). [2] Keak, N. K: “Mathematical Programming with Business Application”. Mc Graw-Hill Book Company. New York. [3] G. R. Walsh: “Methods of optimization” Jon Wiley and sons ltd, 1975, Rev. 1985. [4] Gupta, P. K. Man Mohan: “Linear programming and theory of Games” Sultan Chand & sons, New Delhi, [5] M. S. Bazaraa & C. M. Shetty: “Non-linear programming theory and algorithms”. [6] G. Hadley: “Non-linear and dynamic programming”. [7] Adadie. J: “On the Kuhn-Tucker Theory,” in NonLinear Programming, J. Adadie. (Ed) 1967b. [8] Kanti Sawrup, Gupta, P. K. Man Monhan “Operation Research” Sultan Chandra & Sons. New Delhi, India (1990) [9] “Applied Optimization with MATLAB Programming,” Venkataraman (Chepter-4). [10] Yeol Je Cho, Daya Ram Sahu, Jong Soo Jung. “Approximation of Fixed Points of Asymptotically Pseudocontractive Mapping in Branch Spaces” Southwest Journal of Pure and Applied Mathematics. Vol. 4, Issue 2, Pub. July 2003 [11] Mittal Sethi, “Linear Programming” Pragati Prakashan, Meerut, India 1997. [12] D. Geion Evans “On the K-theory of higher rank graph C*-algebras” New York Journal of Mathematics . Vol. 14, Pub. January 2008 [13] Hildreth. C: “A Quadratic programming procedure” Naval Research Logistics Quarterly. [14] Bimal Chandra Das: “A Comparative Study of the Methods of Solving Non-linear Programming Problems ”.Daffodil Int. Univ. Jour. of Sc. and Tech. Vol. 4, Issue 1, January 2009. Bimal Chandra Das has completed M. Sc in pure Mathematics and B. Sc (Hons) in Mathematics from Chittagong University. Now he is working as a Lecturer under the Department of Textile Engineering in Daffodil International University. His area of research is Operation Research.