Fractional Programming

Transcription

1 Fractional Programming Natnael Nigussie Goshu Addis Ababa Science and Technology University P.o.box: 647, Addis Ababa, Ethiopia August Abstract Fractional Programming, one of the various application of non linear programming, is applicable in various fields such as Finance and Economics, of which production planning, financial and corporate planning, health care, hospital planning are some of the application examples in Fractional Programming. In general, minimization or maximization of objective functions such as productivity, return on investment, return/risk, time /cost or output/input under limited constraint are some other examples of the applications of Fractional Programming. This paper focused on how to solve Fractional Programming Problem in particular on Linear Fractional Programming. Fractional Programming Introduction In various application of non linear programming a ratio of two functions is to be maximized or minimized. In other applications the objective function involves more than one ratio of function. Ratio optimization problems are commonly called Fractional Programming. Some of the best examples of fractional programming are given in the form as, Efficiency = Technical Terms Economical Terms Growth Rate of an Economy = Max x S [Min i m out put i input i ] where S denote a feasible production plane of the economy Fractional programming deals with the optimization of one or several ratios of extended real-valued functions to constraints S given by, f(x) Min, x S (P ) Where f(x) is a ratio of one or several ratios of extended real-valued functions like f(x) = g(x) h(x), f(x) = m g i(x) g(x) i= h i(x) or f(x) = ( h,, gm(x) (x) h m(x) ). Here both functions on the numerator and denominator are extended real-valued functions and finite valued on a feasible set S of (P ). Minimizations of cost-time ratio, maximization of out put-input ratio and maximization of profit-capital ratio or profit-revenue ratio are some other examples of fractional programming problem.

2 Fractional Programming by Natnael Nigussie 2 Now a day s, different researchers in their study shows different methods of solving fractional programming problems. In 96, Gilmore and Gomory show how linear fractional programs can be solved with an adjacent vertex by following procedures just like linear programs with simplex method. Separately, in 962 Charnes and Cooper show how a linear fractional program can be reduced to a linear program by using a non-linear variable transformation. Fractional programs with one or more ratios have often been studied in the broader context of generalized convex programming. Ratios of convex and concave functions as well as composites of such ratios are not convex in general, even in the case of linear ratios, but often they are generalized as a convex. Fractional programming also overlaps with global optimization. Now we are going to see different classification of fractional programming based on the nature of the function on the numerator, denominator and the constraint. The purpose of the following overview is to demonstrate the diversity of the problem. Here after on this paper we use minimization of the problem, since maximization problem is the same as the minimization of the negative of the same problem. 2 Classification of Fractional Programming. Single-Ratio Linear Fractional Programming This kind of problem is given by g(x) Min, x S (P ) h(x) Where S = {x : l k (x) 0, k = : n}, h(x) > 0 for all x S and l k : x R, where k = : n and g, h and l k are affine functions (linear plus a constant) 2. Single-Ratio Fractional Programming This kind of problem is given by g(x) Min, x S (P ) h(x) Where g and h are extended real-valued functions which are finite valued on S, and h(x) > 0 for all x S and S is non empty closed feasible region in X.. Single-Ratio Quadratic Fractional Programming This kind of problem is given by g(x) Min, x S (P ) h(x) Where g and h are quadratic, h(x) > 0 for all x S and l k : x R where k = : n is affine functions. 4. Generalized Fractional Programming This kind of problem is given by g i (x) Sup i m Min, x S (P ) h i (x) With extended real-valued functions g i, h i h i (x) > 0 for every i = : m, and x S. : X [, ], which are finite valued on S with 5. Min-Max Fractional Programming This kind of problem is given by Min x S Max y W g(y, x) h(y, x) (P )

3 Fractional Programming by Natnael Nigussie Where S R m and W R n are non empty closed set and g : R m+n [, ] is a finite-valued function on S W. In case h : R m+n [, ] is a finite- valued positive function on S W. 6. Sum-of-Ratio Fractional Programming This kind of problem is given by m i= With h i (x) > 0 for every i = : m and x S. 7. Multi-Objective Fractional Programming This kind of problem is given by With h i (x) > 0 for every i = : m and x S. g i (x) Min,x S (P ) h i (x) ( g (x) h (x),, g m(x) ) Min,x S (P ) h m (x) The remaining paper is focus on linear fractional programming and methods of solving linear fractional programming since it is one of a particular type of fractional programming. Linear Fractional Programming Problems in which the objective function is the ratio of two affine functions and the constraints are affine inequality are called linear fractional programming problems. Such programming problems have recently been a subject to wide interest in non linear programming. One of a particular example of linear fractional programming is stack cutting problem, Gilmore and Gomory discuss a stack cutting problems in the paper industry for which under the given circumstance it is more appropriate to minimize the ratio of wasted and used amount of raw material rather than just minimizing the amount of wasted materials. This stack cutting problem is formulating as a linear fractional program, since production is naturally experienced as a linear function. Linear fractional programming can be stated precisely as l j= f(x) = c jx j + α l j= d () jx j + β l Subject to a ij x j = b i j= x j 0 where i = : m and j = : l It is assumed that the constraint set S given by () is regular, i.e the set of feasible solution is not empty and bounded, and the denominator of the objective function is strictly positive for all feasible solutions. In order to solve this linear fractional programming, we must first convert it in to which is known as standard form as Subject to f(x) = ct x + α d t (P ) x + β a i x + a i2 x a in x n = b i x j 0 where i = : m and j = : l

4 Fractional Programming by Natnael Nigussie 4 and the number of variables n may or may not be same as before. The process of conversation may require several steps. Step : If the linear fractional program originally formulated as the maximization of f(x), we can instead substitute the equivalent objective to minimization of f(x). Step 2: If any variable x j is not restricted to non negative values it can be eliminated by transformation x j = x j x j,where x j x j 0. Every real valued of x j can be expressed by non negativity values of x j and x j. Step : Finally, any inequality constraints in the original formulation can be converted to equations by the addition of non negativity slack or surplice variables. Thus, the constraints a x + a 2 x a n x n b a 2 x + a x a 2n x n b 2 Would become a x + a 2 x a n x n + x l+ = b a 2 x + a x a 2n x n x l+2 = b 2 with x l+, x l+2 0 Here x l+ and x l+2 are slack and surplus variables, respectively. In matrix or vector notation, the standard form of the linear fractional programming problems is written as Subject to f(x) = ct x + α Ax = b x 0 (P ) Where c and d are n vectors, b and x are columns of vectors having m and n components, respectively. In addition to this, A is an m n matrix, α and β are real scalars. Example. Activity Analysis to Maximize Rate of Return. There are l activities x, x 2,, x l a company may employ using the available supply of m resources R, R 2,, R m. Let b i be the available supply of R i and let a ij be the amount of R i used in operating x j at unit intensity. Let c j be the net return to the company for operating x j at unit intensity, and let d j be the time consumed in operating x j at unit intensity. Certain other activities not involving R, R 2,, R m are required of the company and yield net return α at time consumption of β. The problem is to maximize subject to the restrictions Ax = b and x 0. We note that the constraints set is non empty if b 0, that it is generally bounded (for example if a ij > 0 for all i and j) and that is positive on the constraint set if d 0 and β > 0. the rate of return ct x+α d t x+β Now let s see some important nature of the objective functions of linear fractional programming. Definition.2 Let S be a subset of a real linear space.. The set S is called convex if for all x, y S, then λx + ( λ)y S for all λ (0, ) 2. Let the set S be a non-empty and convex. A functional f : S R is called convex if for all x, y S, then f(λx + ( λ)y) λf(x) + ( λ)f(y) for all λ (0, )

5 Fractional Programming by Natnael Nigussie 5. Let the set S be a non-empty and convex, and a functional f : S R is called concave. If the function f is convex. Definition. Let f : S R, where S is a non-empty convex subset of a real linear space. The function f is said to be quasiconvex. If for each x, x 2 S, the following inequality is true. f(λx + ( λ)x 2 ) maximum{f(x 2 ), f(x )} for all λ (0, ) The function f is said to be quasiconcave if f is quasiconvex. The optimal solution of a function having quasiconcave property occurs at the extreme points of the polyhedral set. Definition.4 Let f : S R, where S is a non-empty convex subset of a real linear space. The function f is said to be strict quasiconvex. If for each x, x 2 S with f(x 2 ) f(x ), we have f(λx + ( λ)x 2 ) < maximum{f(x 2 ), f(x )} for all λ (0, ) The function f is said to be strictly quasiconcave if f is strictly quasiconvex. Definition.5 Let the set S be a non empty convex subset of a real linear space and let f : S R be a given functional which has a directional derivative at some x S in every direction x x with arbitrary x S. The functional f is called pseudoconvex at x if for all x S, Or equivalently if (x x ) f(x ) 0 f(x) f(x ) 0 f(x) f(x ) 0 (x x ) f(x ) 0 The function f is said to be pseudoconcave if f is pseudoconvex. Pseudoconvex, strictly quasiconvex, quasiconvex and convex functions are various types of function that share some desirable properties such every local minimal point also a global minimal point. Theorem.6 Let S be a non-empty open subset of real linear space and f : S R be differentiable pseudoconvex function on S, then f is both strictly quasiconvex and quasiconvex. Proof: We first show that f(x) is strictly quasiconvex. We can prove this by contradiction. Suppose that there exist x, x 2 S such that f(x ) f(x 2 ) and f(x λ ) max{f(x ), f(x 2 )} where x λ = λx + ( λ)x 2 for all λ (0, ) with out loss of generality, assume that f(x ) < f(x 2 ), so that f(x λ ) f(x 2 ) > f(x ) () Because of pseudoconvexity of f(x), f(x λ ) (x x λ ) < 0. Now, since f(x λ ) (x x λ ) < 0 and x x λ = ( λ)(x2 x) λ, then f(x λ ) (x 2 x λ ) > 0 By pseudocovexity of f(x), we must have f(x 2 ) f(x λ ) (2) By () and (2), we get f(x 2 ) = f(x λ ). Also, since f(x λ ) (x 2 x λ ) > 0, there exist a point x η = ηx + ( η)x 2 with η (0, ), such that f(x η ) > f(x λ ) = f(x 2 ). Again, by pseudoconvexity of f(x), we have f(x η ) (x 2 x η ) < 0. Similarly f(x η ) (x λ x η ) < 0. Summarizing, we must have f(x η ) (x 2 x η ) < 0 () f(x η ) (x λ x η ) < 0 (4)

6 Fractional Programming by Natnael Nigussie 6 But x 2 x η = η(xη xλ) ( η) and, hence inequality () and (4) are not compatible. This contradicts that f(x) is strictly quasiconvex. Now we have to show f(x) is also quasiconvex. Let x, x 2 S and f(x) is a lower semi-continuous. If f(x ) f(x 2 ), then, by the strict quasiconvexity of f(x), we must have f(x λ ) < max{f(x ), f(x 2 )} where x λ = λx + ( λ)x 2 for some λ (0, ) Now suppose that f(x ) = f(x 2 ), to show that f(x) is quasiconvex, we need to show that f(x λ ) f(x ) for each λ (0, ). By contradiction, suppose that f(x η ) > f(x ) where x η = ηx + ( η)x 2 for some η (0, ), since f(x) is lower semi-continuous, there exist λ (0, ) such that f(x η ) > f(x λ ) > f(x ) = f(x 2 ) (5) Note that x η can be represented as a convex combination of x λ and x 2. Hence, by the strict quasiconvexity of f(x) and since f(x λ ) > f(x 2 ), f(x η ) < f(x λ ), This contradicting to (5). This completes the proof. Strictly quasiconvex is important in non-linear programming, which assures that a local minimum over a convex set is also global minimum and quasiconvex is also important in non-linear programming, which assures the existence of optimal solution at the extreme point if the solution is exist. Lemma.7 Let f(x) = ct x+α d t x+β, and let S be a convex set such that dt x + β 0 over S, then f(x) is both pseudoconvex and pseudoconcave. Proof: Since 0, either > 0 for all x S or < 0 for all x S. Otherwise, there exist x, x 2 S such that d t x + β > 0 and d t x 2 + β < 0. Hence, for some convex combination x of x and x 2, = 0, this contradiction our assumption. First we have to show f is pseudoconvex Suppose that x, x 2 S with (x 2 x ) t f(x ) 0, we need to show that f(x ) f(x 2 ). We have f(x ) = (dt x + β)c (c t x + α)d (d t x + β) 2 Since (x 2 x ) t f(x ) 0 and () 2 > 0, it follows Therefore, (c t x 2 + α)(d t x + β) (d t x 2 + β)(c t x + α) 0 (x 2 x ) t ((d t x + β)c (c t x + α)d) 0 (c t x 2 + α)(d t x + β) (d t x 2 + β)(c t x + α) But, since d t x +β and d t x 2 +β are both either positive or negative, dividing by (d t x +β)(d t x 2 +β) > 0, We get c t x 2 + α d t x 2 + β ct x + α d t x + β, i.e f(x 2) f(x ) Therefore f is pseudoconvex. Now we have to show f is pseudoconcave. Suppose that x, x 2 S with (x 2 x ) t f(x ) 0, we need to show that f(x 2 ) f(x ) We have f(x ) = (dt x + β)c (c t x + α)d (d t x + β) 2 Since (x 2 x ) t f(x ) 0 and () 2 > 0, it follows (x 2 x ) t ((d t x + β)c (c t x + α)d) 0 (c t x 2 + α)(d t x + β) (d t x 2 + β)(c t x + α) 0 Therefore, (c t x 2 + α)(d t x + β) (d t x 2 + β)(c t x + α)

7 Fractional Programming by Natnael Nigussie 7 But, since d t x +β and d t x 2 +β are both either positive or negative, dividing by (d t x +β)(d t x 2 +β) > 0, We get c t x 2 + α d t x 2 + β ct x + α d t x + β i.e f(x 2) f(x ) Therefore f is pseudoconcave. Therefore linear fractional programming is both pseudoconvex and pseudoconcave property. This completes the proof. By definition of pseudoconvexity and pseudoconcavity, every local optimal solution is also a global optimal solution. In other word by Theorem 5, because of strictly quasiconvex and strictly quasiconcave we can ensure that a local optimal solution is also a global optimal solution. Theorem.8 Consider the problem: Subject to f(x) = ct x + α Ax = b x 0 Where c and d are n vectors, b and x are columns of vectors having m and n components, respectively, A is an m n matrix, α and β are real scalars, then the optimal solution occur at extreme point. Proof: Suppose the minimum of f(x) is attain at x 0 S = {x : Ax = b, x 0}. If there is an extreme point whose objective is equal to f(x 0 ), then the result is at hand. Otherwise, let x, x 2,, x k be the extreme points of S, and assume that f(x 0 ) < f(x j ) for j = : k. So, x 0 can be represented as Since f(x 0 ) < f(x j ) for each j, then x 0 = k λ j x j, j= k λ j =, j = : k j= f(x 0 ) < minf(x j ) = α () Now consider the upper level set S α = {x : f(x) α}. From () x j S α for j = : k and by quasicocavity of f(x), S α is convex. Hence x 0 = k j= λ jx j belongs to S α. This implies that, f(x 0 ) α. Which contradicts ().This contradiction shows that f(x 0 ) = f(x j ) for some extreme point x j, and the proof is complete. Now we can summarize fractional programming as follows, the optimal solution for a linear fractional program occurs at extreme points of the feasible region. Furthermore, every local minimum is also a global minimum because of convexity property. 4 Methods of Solving Linear Fractional Programming We have different kinds of methods for solving linear fractional programming. The first method is convexsimplex methods; this method is a minor modification of simplex method of linear programming. The other method is a procedure credited by Gilmore and Gomory for solving a linear fractional program. In addition to this, Charnes and Cooper describe another procedure of solving linear fractional programming by reducing the problem in to a linear programming by using non-linear variable transformation. Now we have to see each of the methods one by one. 4. Convex-Simplex Method Convex simplex method (Zangwill 967) is used to solve the optimal value of convex objective function subject to linear inequality constraints. However, the method is applicable to problems where the objective function is a general function having continuous partial derivatives. Because of the special structure

8 Fractional Programming by Natnael Nigussie 8 of the objective function of linear fractional programming, convex simplex method is a true generalization of linear simplex method both in sprite and in the fact that the same tableau and variable selection techniques are used. If the objective function is linear, then the convex simplex method reduced to the linear simplex method. Moreover, the convex simplex method is actually behaves like the linear simplex method whenever it encounters a linear portion of a convex objective function. Convex simplex method coincides with the linear simplex method when applied to linear fractional programming problems in the case when we initially start with a basic feasible solution to the problem. But the Convex simplex method will differ from linear simplex method in linear fractional programming if we start with any not basic feasible solution to the problem at least for the steps. Once we reach the basic feasible solution, to start with the steps are just the same as linear simplex method for the next iteration. Now our first task is to pose conditions under which a given point x* is optimal by restating kuhn-tuckre condition Consider problem Karush-Kuhn-Tucker (KKT) conditions f(x) (P ) Subject to Ax = b x 0 Where f(x) is a convex function with continuous first partial derivatives, A is an m n matrix of rank m, b and x are column vectors of m and n components, respectively. First we change the problem in to unconstraint problem by using lagrange multiplier and the problem become L(x, λ, µ) = f(x) λ t (Ax b) µ t x, where λ R m, µ R n From () and (5) we have L x (x, λ, µ) = f(x) λ t A µ t = 0 f(x) λ t A = µ t () L λ (x, λ, µ) = Ax b = 0 Ax = b (2) L µ (x, λ, µ) = x 0 x 0 () µ t x = 0 and (4) µ 0 (5) (6) f(x) t λ t A = µ t 0 f(x) t λ t A 0 ( ) From () and (4) we have From (7) ( f(x) t λ t A)x = 0 ( ) ( ) ( B f(x) t λ t B, N f(x) t λ t xb N) = 0 x N ( B f(x) t λ t B)x B + ( N f(x) t λ t N)x N = 0 ( N f(x) t λ t N)x N = 0 since x N = 0 From the above ( B f(x) t λ t B)x B = 0 and x B > 0 B f(x) t λ t B = 0 λ t = B f(x) t B ( )

9 Fractional Programming by Natnael Nigussie 9 Substituting (***) in both equation (*) and (**), then we have ( f(x) t B f(x) t B A) 0 dual feasibility ( f(x) t B f(x) t B N)x = 0 complementary slackness condition We can write the above condition component wise as follows ( f(x) x j ( f(x) x j B f(x) t B a j ) 0 j = : n dual feasibility B f(x) t B a j )x j = 0 j = : n complementary slackness condition A point satisfying dual feasibility and complimentary slackness condition is a KKT point. Because of the convexity of the objective function f(x), a point satisfying the KKT condition for a minimization problem is also a global minimum over the feasible region. Now our first task is to pose conditions under which a given point x is optimal. Lemma. formulate the optimality conditions by restating kuhn-tuckre condition. Lemma 4. Let A be any linear programming tableau with b the corresponding right hand side so that Ax = b, x 0 if and only if x is feasible. Let x be a particular feasible point and the corresponding relative cost vector is C(x ) t = ( f(x ) t B f(x )B A). Let α = min{ f x j β = min{ f x j B f(x ) t B a j, j A} and B f(x ) t B a j )x j, j A} If α = β = 0, then x is optimal. Proof: Observe that the problem subject to f(x) Ax = b x 0 Since x is feasible, x must satisfy Ax = b x 0

10 Fractional Programming by Natnael Nigussie 0 If α = β = 0, x satisfies If α = 0 If β = 0 max { f x B f(x ) t B a j } = 0 j B f(x ) t B a j 0 () x j max {( f x B f(x ) t B a j )x j } = 0 j ( f x B f(x ) t B a j )x j 0 j but by () f x B f(x ) t B a j 0 and x j 0 j ( f x j B f(x ) t B a j )x j = 0 Hence, ( f(x ) x j ( f(x ) x j B f(x ) t B a j ) 0 j = : n dual feasibility B f(x ) t B a j )x j = 0 j = : n complementary slackness condition A point satisfying dual feasibility and complimentary slackness condition are simply the karusa-kuhn- Tuckre point and because of convexity property of the objective function, a point indeed an optimal point. This completes the proof. Now consider the problem subject to f(x) Ax = b x 0 Where f(x) is a convex function with continuous first partial derivative, A is m n matrix of rank m, b and x are columns of vectors having m and n components, respectively. We can write the constraint set as Ax = x a + x 2 a x j a j + + x n a n, where x, x 2,, x n 0 and a j is the j th column of A, and assume that the rows of A are linearly independent. Suppose that m linearly independent columns a, a 2,, a m have selected from A. If we set the n m variables not associated with these columns equal to zero,then the unique solution to the resulting system of m equations in the m unknowns x, x 2,, x m is called a basic solution. The columns a, a 2,, a m are the basic columns, where x, x 2,, x m are the basic variables. If any one or more of basic variables has the values zero, then the basic solution is said to be degenerate. So we can be decompose A in to [B, N], B is m m matrix of full rank and N is an m (n m) matrix. i.e B = [a, a 2,, a m] and N is the remaining m (n m) matrix.a is an mxn matrix, b and x are columns of vectors having m and n components respectively. So, A decompose in to B and N and x

11 Fractional Programming by Natnael Nigussie decompose in to x B and x N, then we have Bx B + Nx N = b Bx B = b Nx N = b j N a j x j x B = B b j N B a j x j x B = b j N y j x j, where b = B b and y j = B a j We can write this component wise as follows x Bi = b i j N y ij x j, for all i = : m () Let z = f(x B, x N ). The partial derivative of z with respect to the nonbasic variables x j, l N, can be calculated by the chain rule as follows z = f + x j x j z x j = f x j m i= m i= f x Bi x Bi x j f x Bi y ij since x B i x j z = f B f(x)y j (2) x j x j = y ij Where x B vectors corresponding to B. i.e x B = (x, x 2,, x m) t x N vectors corresponding to N. A is the set of subscripts of the columns of matrix A, i.e j A if and only if x j is the j th component of x. B is the set of subscripts of the basic variables,i.e j B if and only if x j is in x B. N is the set of subscripts of the non basic variables,i.e j N if and only if x j is in x N. B i the subscript i indicates the i th component of the basic variables. B f(x) is the gradient of f(x) with respect to the basic vector x B. N f(x) is the gradient of f(x) with respect to the non basic vector x N. ( f(x)) t the superscript t indicates the transpose of f(x). Equation (2) is called relative cost vector C(x) or reduced gradient vector. If the objective function is linear i.e f(x) = c t x equation (2) become z j = c j c B y j is called relative cost or reduced cost. In the convex simplex method,the partial derivative changes as x j increases and may at some point fall to zero. When this happens it is no longer desirable to increase x j even though it may still be feasible to do so. Accordingly, displacement ends either when. The value of some basic variable x Br falls to zero or 2. The partial derivative z x j vanishes.

12 Fractional Programming by Natnael Nigussie 2 The point at which one of these two events first occurs is taken as the starting point of the next iteration. Thus the convex simplex method is a member of the feasible directions family. Here, it will be better to start with a basic feasible solution instead of any feasible solution to the problem because it is known that optimum of such problem occurs at the basic feasible solution to the problem. It would lead to lesser computational work. Also for such problems local optimum is global. The Algorithmic Procedure Initialization step Find a starting basic feasible solution x to the system of Ax = b, x 0 from simplex and, go to step of iteration k with k =. Iteration k The feasible point x k and tableau A k are given, where A k is the value of a matrix A at k iteration which is also equivalent to matrix A. Step : Calculates the relative cost vector(c). Where C(x ) t = ( f(x ) t f(x ) B Y ) Y = B A k f(x k ) = ( f(xk ) x k f(x k ) B = ( f(x) x k B, f(xk ) x k 2,, f(xk ) x k ) t n, f(x) x k,, f(x) B 2 x k ) t B n Let α k = min{ f x B f(x ) t B a j, j A} and j β k = min{( x B f(x ) t B a j )x j, j A} j If α k = β k = 0, terminate x k is optimal. Otherwise, go to step 2. Step 2: Determine the non-basic variable to change. Let p is an index such that α k = f x k B f(x k )y p and () p q is an index such that β k = ( x k B f(x k )y q )x k q (2) q If α k β k, increase x p from () adjusting only basic variables. If α k β k, decrease x q from (2) adjusting only basic variables. Step : Calculate x x+ and A k+ Case : x k p is to be increased and for some i, y k ip > 0. Increase x k p will drive a basic variable to zero. Let z k = (z k i )t be the x valued when that occurs. Specifically z k i = x k i, i N p z k p = x k p + p () z k B i = x k B i y k ip k, i B

13 Fractional Programming by Natnael Nigussie Where N p is the set of subscripts of the non basic variables except p and B is the set of subscripts of the basic variables, and k = xk B r y k rp = min { xk B i yip k ; yip k > 0} (4) Find x k+, where f(x k+ ) = min{f(x)/x = λx k + ( λ)z k, 0 λ } If x k+ z k, set A k+ = A k, go to iteration k with k + replacing k. Do not change basis. If x k+ = z k, pivot on yrp k and forming A k+, go to iteration k with k + replacing k, and replace x k B r with x k p in the basis. Case 2: x k p is to be increased and yip k < 0 for all i. Define z k the same as in equation () except let k =. Then attempt to determine x k+ such that f(x k+ ) = min{f(x)/x = x k + λ(z k x k ), λ 0} If no x k+ exists, terminate. The optimal solution is unbounded. If x k+ does exist, set A k+ = A k, go to iteration k with k + replacing k and the same basis. Case : x k q is decrease determining z k using equation () except defining k as k = max [ k, k 2] Where k = xk br y k rq = max { xk b i y k iq / y k iq < 0} and k 2 = x k q If y k iq 0, i = : m, let k = Find x k+, where f(x k+ ) = min{f(x)/x = λx k + ( λ)z k, 0 λ } If x k+ z k, set A k+ = A k, go to iteration k with k + replacing k. Do not change basis. If x k+ = z k, pivot on yrq k and forming A k+, go to iteration k with k + replacing k, and replace x k b r with x k q in the basis. We continue this process with the new feasible solution x k+ B till we obtain that α n = β n = 0 is satisfied at the n th iteration which is the condition of optimality and hence,x n will be the optimal solution and the corresponding optimal value of f(x) will be f(x n ). Remark 4.2 In order to calculate x k+, the next feasible solution to start with, we solve f(x k+ ) = min {f(x)/x = λx k + ( λ)y k, 0 λ } But in the case of linear fractional programming problem, by theorem 2 the optimal solution is obtained at a basic feasible solution S and one of the basic feasible solutions to the problem is optimal. Therefore, it will always be true that x k+ = z k. Theorem 4. If the objective functions is linear fractional programming, then β k = 0 for any iteration, where β k = max {( x k B f(x)y j )x k j, j A}. j Proof: First we have to show a complimentary slackness condition for non basic variables. Since x k f j = 0, for all j N we have ( x k B f(x)y j )x k j, for all j N. Therefore, a complimentary j slackness condition is hold for non basic variables x N. Now we have to proof a complimentary slackness condition for a basic variables x B i.e j B. If x j, j B,

14 Fractional Programming by Natnael Nigussie 4 then y j must be the unit vector e r. z z z z z = f = f = f = f = f B f(x)y j, for all j B where y j = B a j f x Bi y ij ( f x B y j + + f x Bi y ij + + f x Bj y jj, since y j is the unit vector f x j = 0, for all j B z x j x j = 0 for all j B f x Bm y mj ) Therefore a complimentary slackness condition is hold for basic variables. slackness condition is hold i.e ( f B f(x k )y j )x k j = 0 for all j A and Therefore complimentary This is the proof. β k = max{( f B f(x k )y j )x k j for all j A} = 0. Corollary 4.4 Consider the problem Subject to f(x) = ct x + α Ax = b x 0 Let A k be any linear programming tableau with b k the corresponding right hand side so that A k x k = b k, x 0, if and only if x is feasible. Let x k be a particular feasible point and the corresponding relative cost vector is C(x k ) t = ( f(x k ) t f(x k ) B Y )Where Y = B A k Let α k = min { f x k B f(x k )y j, j A} j If α k = 0, then x k is optimal. Proof: If α k = β k = 0, then x k is optimal by lemma and from above theorem we have β k = 0 for any iteration k, therefore α k = 0 is the optimal condition for linear fractional programming problem. Remark 4.5 Suppose that we are given an extreme point of the feasible region with basis B such that x B = B b > 0 and x N = 0. To get a lower objective function value needs to increase or decrease one of the non basic variables accordingly. Since the current point is an extreme point with x N = 0 decreasing a non basic variable is not permitted as it would violate the nonnegative restriction.

15 Fractional Programming by Natnael Nigussie 5 Gilmore and Gomory Procedure Now we have to see below a procedure credited to Gilmore and Gomory (96) for solving a linear fractional programming. Initialization step: Find a starting basic feasible solution x to the system of Ax = b, x 0 from simplex and, go to step of iteration k with k =. Iteration k The feasible point x k and tableau A k are given. Step : Compute the vector α k. If α k = 0, terminate x k is an optimal. Otherwise, go to step 2. Step 2: Let p is an index such that α k = f x k B f(x k )yp k for which f p x k B f(x k )yp k < 0. p This gives x k p is increasing from zero to some positive number. Now we can determine a basic variable which falls to zero. x k B i = b k x k py k ip > 0 () Case : x k p is increasing and for some i, y k ip > 0. k = min { xk B i yip k, for some yip k > 0} = xk B r y k rp (2) x k B r is the new non basic variables which falls to zero. Up date the table corresponding pivoting at yrp k and by substituting (2) in to () we have z k = x k i, i N p zp k = x k p+ k () zb k i = x k B i yip k k, i B We have got a new basic feasible solution z k and x k+, go to iteration k with k+ replacing k, go to step. Case 2: x k p is increasing and y k ip < 0 for all i. Here we can increase xk p as we need with out changing a basic variable to zero. The optimal solution is unbounded. Remark 4.6 We continue this process with the new feasible solution x k+ till we obtain that α k is satisfied at the n th iteration which is the condition of optimality and hence, x n will be the optimal solution and the corresponding optimal value of f(x) will be f(x n ). We now assume that x B > 0 for each extreme point. This algorithm moves from one extreme point to another. By nondegeneracy assumption, objective function strictly decreases at each iteration, so that the extreme point generated is distinct. There are only a finite number of these points, and, hence, the algorithm stops in a finite number of steps. At the end, the relative cost vector is non negative i.e α k = 0 resulting in a Karush-Kuhn-Tucker (KKT) point and because of pseudoconvexity and pseudoconcavitiy. This point indeed an optimal point. Now we can solve numerical example of linear fractional programming problem by using convex simplex algorithm.

16 Fractional Programming by Natnael Nigussie 6 Example 4.7 Consider the following problem 4x + x x + 4x Subject to 2x + x 2 6 x 2 8 2x + x 2 6 x, x 2 0 Solution: Before solving by using convex-simplex method, let we see what looks like the figure of the constraints Figure shows the feasible region with the extreme points (0, 0),(0, 6),(, 8),(4, 8) and (8, 0).The value objective function at those points is 0.8, 0.766, 0.625, and.7 respectively. Hence the optimal solution is (8, 0) and the corresponding optimal value is.7. Now we have to solve this problem by using convex-simplex method, by introducing additional slack variables x, x 4 and x 5 the problem (P) is changed to standard linear fractional programming. The problem is given by 4x + x x + 4x Subject to 2x + x 2 + x = 6 x 2 + x 4 = 8 2x + x 2 + x 5 = 6 x, x 2, x, x 4, x 5 0 Iteration : Let x = (0, 0, 8, 6, 6) since x B = B b > 0. Here x B = (x, x 4, x 5 ) = (6, 8, 6). Step : f(x ) = ( 4 25, 2 25, 0, 0, 0) and then C(x ) = ( 4 25, 2 25, 0, 0, 0). The corresponding tableau A is given by f(x 4 2 ) f(x B ) x B x x x 2 x x 4 x 5 b 0 x x x C(x )

17 Fractional Programming by Natnael Nigussie 7 α = min{ 4 25, 2 25, 0, 0, 0} = 4 25 Since α 0, we go to the next step Step 2: Let p = since α = c where c is the st component of C. We can increase x since α = 4 25 Step : Calculate x 2 and A 2 Increase x from 0 to and one of a basic variable in x has become zero. Let z = (z k i ) be the x valued when that occurs specifically z = x +, where = min{ 6 2, 2 > 0} z = = 8 z 2 = 0 z = 6 ( 2)8 = z 4 = 8 (0)8 = 8 z 5 = 6 ( 2)8 = 0 So, z = (8, 0,, 8, 0) Now we have find x 2, where f(x 2 ) = min {f(x)/λx + ( λ)y, 0 λ }. Since the solution of linear fractional programming is occurs at the extreme point, the new iteration point is x 2 = y = (8, 0,, 8, 0). We can pivot on y and the corresponding tableau A 2 is given by f(x ) f(x 2 B ) x2 B x2 x 2 x 2 2 x 2 x 2 4 x 2 5 b x x x C(x ) Since α 2 = 0. This is the condition of optimality. α 2 = min{0, 9 7, 0, 0, } = 0 Therefore the optimal solution is x 2 = (8, 0,, 8, 0) t and the corresponding optimal value of the objective function is f(x 2 ) = Methods of Charnes and Cooper Charnes and Cooper (962) solved linear fractional programming by using one additional constraint and one additional non-linear variable. Definition 4.8 The problem f(x) = ct x + α Subject to S = {x : Ax = b, x 0} is called regular if S is non empty, f(x) is not constant and if there exist M > 0 such that 0 < d t x+β < M for all x S, where S E and E = {x R n : x 0, > 0}.

18 Fractional Programming by Natnael Nigussie 8 Now, consider the problem of finding x = (x, x 2,, x n ) t Subject to f(x) = ct x + α Ax = b x 0 (P ) Where x, c and d are column of vectors all having n components, A is matrix of rank m, b is a columns of vectors having m components, α and β are real scalar constants.to avoid technical difficulty we assume that the constraints set S is regular i.e the set S of the feasible solution set is non-empty and bounded, and the denominator is strictly positive throughout the constraint set. So we can change linear fractional programming in to a linear programming by the following manner, subject to f(x) = ct x + α Ax = b x 0 (c t x + α ) Subject to Ax = b x 0 Let w = d t x+β and z = xw. Then the original problem becomes (c t z + αw) Subject to Az bw = 0 z, w 0 Now we have to minimize c t z + αw with the change of variables z = xw, this becomes embedded in the following linear program of finding z = (z, z 2,, z n ) and w in the following problem c t z + αw = g(z, w) (P 2 ) Subject to = Az bw = 0 z, w 0 Lemma 4.9 Every z, w satisfying the constraints = Az bw = 0 z, w 0, has w > 0

19 Fractional Programming by Natnael Nigussie 9 Proof: Suppose w = 0, that is (z, 0) is feasible and z must be different from 0, this means x + z would be feasible for the original problem for all > 0 and any feasible x A(x + z) = b Ax + Az = b Ax = b This contradicts S bounded. Therefore w > 0. This completes the proof. The following property extends to the case when (P ) is regular a similar result obtained by charnes and cooper under the supposition that S is a bounded non empty set. Theorem 4.0 If the problem P is regular, then for any feasible solution (z, w) of the problem P 2 there exist a feasible solution x S of the problem P such that x = z w and f(x) = g(z, w), and the converse is also true. Proof: Let w = d t x+β and z = xw or x = z w f(x) = ct x + α c t x + α c t xw + αw c t z + αw g(z, w) x is any feasible solution to P Ax = b, x 0 A( z ) = b since z = xw w Az = bw Az bw = 0 (z, w) is also a feasible solution to P 2. This completes the proof. Next we need an auxiliary result which establish the relationship between problem P and P 2 that generalizes for regular linear fractional programming a result obtained by charnes-cooper in the case when the feasible set is bounded and non empty. Theorem 4. If the problem is regular, then the following statement holds The problem P and P 2 have both optimal solution and its optimal value are equal and finite. Moreover, if (z, w ) is an optimal solution of P 2, then x = z w is an optimal solution for P and conversely if x = z w is an optimal solution of P, then there exist an optimal solution (z, w ) of P 2 such that z = x w. Proof:( :) Given (z, w ) is an optimal solution to P 2 by theorem 6, there exist is a feasible solution x such that Ax = b and x 0. We have to show x = z w is an optimal solution. Let x is a feasible solution such that Ax = b and x 0. Since > 0 by assumption and the vector (z, w) is a feasible solution to P 2. Where z = Since (z, w ) is an optimal solution to P 2. x and t = c t z +αw c t z + αw (2) ()

20 Fractional Programming by Natnael Nigussie 20 By substitution for z, z, w in (2) the equation become w (c t x + α) ct x + α The result immediately follows by dividing the left hand side by = d t z + βw c t x + α d t x + β ct x + α Therefore x is an optimal solution for P. ( :) Given x = z w is an optimal solution to P and also a feasible solution. By theorem 6, there exists a feasible solution (z, w ) such that Az bw = 0 and z 0, w > 0. We have to show (z, w ) is an optimal solution. Let (z, w) is a feasible solution such that Az bw = 0 and z 0, w > 0, Where z = x d t x+β and w = d t x+β Since x is an optimal solution to a linear fractional program c t x + α d t x + β c t x d t x + β + α d t x + β ct x + α c t x + α c t d t x + β ( z w ) + α d t x + β (w w ) x ct + α c t z (d t x + β)w + αw (d t x + β)w c t x + α c t z + αw c t z + αw Since (d t x + β)w = d t x w + βw =. Therefore, (z, w ) is an optimal solution to a linear programming.this completes the proof. To summarize, we have shown that a fractional linear program could be solved by a linear programming problem with one additional non-linear variable and one additional constraint. Now we have to see numerical example of linear fractional programming problem solved by using method of Charnes and Cooper. Example 4.2 Consider the following problem 4x + x x + 4x subject to 2x + x 2 6 x 2 8 2x + x 2 6 x, x 2 0 (P ) Solution: Since (0, 0) is a feasible point in the above problem the denominator of the objective function is strictly positive over the entire feasible region. Now we have to minimize the following problems. Let z = x w and z 2 = x 2 w () 4z + z 2 + 5w (P 2 ) Subject to 2z + z 2 6w 0 z 2 8w 0 2z + z 2 6w 0 2z + 4z 2 + 6w = z, z 2 0, w > 0

21 Fractional Programming by Natnael Nigussie 2 Graphs of constraints of the above problem is as The extreme points are (0, 0), (0, ), ( 8, 0.28) and ( 8, 0). The objective function value at those points is 0.8, 0.76, and.2777, respectively. Hence the optimal solution is ( 8, 0) and the corresponding optimal value is -.7. Now first we have to add slack variables of y,y 2 and y, artificial variable y 4 and M is big number to change in to a standard linear optimization problem P 2 problem became 4z + z 2 + 5w + My 4 subject to 2z + z 2 6w + y 0 cmz 2 8w + y 2 0 2z + z 2 6w + y 0 2z + 4z 2 + 6w + y 4 = z, z 2, y, y 2, y, y 4 0, and w > 0 Then the tableau is given by d N -4 5 d B y B y N z z 2 w b 0 y y y y z -4-2M -4M 5-6M Since min z = 5 6M < 0, increase w and decreasing y 4 to zero. Then the new tableau is given by d N -4 M d B y B y N z z 2 y 4 b 0 y y y 5 w z

22 Fractional Programming by Natnael Nigussie Since min z = 7 5 < 0, increase z and decreasing y to zero. Then the new tableau is given by d N 0 0 d B y B y N y z 2 y 4 b 0 y y y 5 w 7 z All z 0. This is the condition of optimality M Therefore the optimal feasible solution is ( 8, 0,, 8, 0, 0). Therefore z = 8, z 2 = 0, and w = from (). So x = z w = 8 and x 2 = z2 w = 0, i.e (8, 0) is the extreme optimal solution to a linear fractional problem and the corresponding optimal value is.7. 5 Conclusion This paper focused on Fractional Programming in particular Linear Fractional Programming. Such kind of programming problems has recently been a subject of wide interest in non linear programming. Linear Fractional Programming problem have a global solution because of convexity property and can be solved by different methods. Convex simplex method, Gilmore and Gomory procedure, and Charnes and Cooper method are some of the methods that solve Linear Fractional Programming. Each of the method mention above is its own property. Convex simplex method is almost identical to reduced gradient method, the difference is only it changes one non basic variables and modifying basic variables, while reduced gradient method is by generating feasible direction and convex simplex method is reduced to simplex method whenever the objective function is linear or a linear portion of the objective function is encounter, while reduced gradient method may not so reduced. Convex simplex method is identical to method of Gilmore and Gomory in a selection of basic feasible solution and pivot points when convex simplex method is applying to Linear Fractional Programming. On the other hand, the advantage of Charnes and Cooper algorithm is reducing Linear Fractional Programming in to Linear Programming to facilitating finding of solution by using simplex method but the disadvantage is it needs one more additional variables and one constraints. Lastly convex simplex method is behaves generality like the linear simplex method whenever any linear portion of the objective function are encounter. The method explained here will be useful in the solution of economic problems in which the different economic activities utilize fixed resources in proportion to the level of there values. Reference. M. Avrie, W.E. Diewert, S.Schaible and I. Zang, Generalized Concavity, plenun press, New York, A.I. Barros, Discret and Fractional Programming Techniques for Location Models, kluwer Academic publishers, Dordrecht-Bosten-Londen, A. Charnes and W.W. Cooper, Programming with linear fractional functionals, Navel Research Logistic Quarterly 9(962), W. Dinkelbach, On non linear fractional programming, Management Science (967), S. Schaible and T. Ibaraki, Fractional Programming, Europian Journal of Operational Research 2(98), no., H. Ishii, T.Ibaraki and H. Mine, Fractional Knapsack Problems, Mathematical Programming (976), no.,

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