nlm0402 发表于 2011-5-26 12:10 
楼主从决策的角度总结了经济学的基本理论，不知道楼主是否知道库恩塔克方法的一般公式？
KKT(Karush-Kuhn-Tucker) conditions are the necessary condition( the first order condition) for the convex programming with functional constraints,which i think is the most important type of nonlinear programming. The convex programming problem can be expressed as following:
Max Z=F(X1,X2,…,Xn) s.t. g1(X1,…,Xn)<=b1,…,gm(X1,…,Xn)<=bm (24.1)
X1,…,Xn
where X1,X2,…,Xn are the decision variables, and gj<bj (j=1 to m) are the functional constraint,and it was assumed that the multivariate objective function F(X) is a concave downward function and the constraint functions gj (j=1 to m) are convex downward functions, and F(X) and gj(X) are all differentiable.
The whole first order condition of the convex programming above are the so-called KKT conditions, which can be described as following:
(24.2)
The conditions (1)and(2) can be combined as
(24.3)
and the conditions (3) and (4) can be combined as
(24.4)
In order to understand KKT conditions,firstly let us consider the following one-dimensional nonlinear maximization problem:
Max Z=f(x) , s.t. x>=0 (24.5)
x
where f(x) is a one-variable differentiable function, x is the decision variable.
Obviously, when the optimal point x* is greater than 0,the first order is that the derivative of f(x) at x* is equal to 0,which is called the internal solution; and if the optimal point x* coincides with 0, then the derivative of f(x) at x*=0 is less than or equal to 0,which is so-called the corner-solution.This is the one-dimensional KKT condition with variable constraint( without functional constraints),i.e.
df/dx=0,if x>0; df/dx<=0 if x=0 (24.6)
I think anyone can understand it by imagination of the function graph.
We can generalize the preceding conclusion about one-dimensional problem to a multivariate function maximization problem only with the constraints with respect to the variables Xi>=0.
Max Z=F(X)=F(X1,…,Xn) , s.t. Xi>=0 , for i=1 to n (24.7)
X1,…,Xn
Let us denote the optimal vector X*=(X1*,…,Xn*), for any variable( or the vector component) Xi*, if Xi*>0, then the first order condition should be that the partial derivative with respect to Xi of the function F(X) at X* is equal to 0; and ,if Xi*=0, then the first order condition should be that the partial derivative with respect Xi of the function F(X) at X* is less than or equal to 0. i.e.
DF/DXi=0,if Xi>0; DF/DXi<=0 if Xi=0 (24.8)
Only if some one variable Xj*=0, then the optimal vector is called a corner solution; as long as all the variables Xi>0 the optimal vector X* is called internal solution.
As for the complete KKT conditions (24.2), they are only the first order conditions on the following Lagrange function , through which the functional constrained problem is transformed into a nonnegative constrained problem as following:
Max L(X,lamda=V)=F(X)+v1(b1－g1)+…+vm(bm－gm) s.t. X>=0, V>=0 (24.9)
X,V
Apply the conclusion in the formulation (24.8) to the nonnegative constrained problem and we can obtain the complete KKT conditions (24.2). To get this result, what you only need to do is to regard the variable X and the Lagrange multiplier V as variables in (24.9).
II. The equivalence between KKT conditions and the inframarginal analyis
I think that the KKT conditions are equivalent to the inframarginal analysis,in order to understand it,let us suppose a two-dimensional problem with only one constraint:
Max Z=F(X1,X2) s.t. g(X1,X2)-b<=0 ,X1,X2>=0 (24.10)
X1,X2
where X1,X2 are the decision variables. The functional constrained maximization can be transformed into only nonnegative constrained maximization problem as following:
Max L(X1,X2,v)=F(X1,X2)+v[b-g(X1,X2)] X1,X2,v>=0 (24.11)
X1,X2,v
The KKT conditions of preceding problem are(the nonnegative constraints are omitted):
DL(X1,X2,v)/DX1=0 if X1*>0; DL(X1,X2,v)/DX1<=0 if X1*=0 (24.12.1)
DL(X1,X2,v)/DX2=0 if X2*>0; DL(X1,X2,v)/DX2<=0 if X2*=0 (24.12.2)
DL(X1,X2,v)/Dv=0 if v*>0; DL(X1,X2,v)/Dv<=0 if v*=0 (24.12.3)
There are eight combinations of the variables X1,X2,v that can be divided into two situations,">0" and "=0":(formula 24.13)
(1) X1>0,X2>0,v>0 , obviously ,this is the internal solution, in order to solve this situation, marginal analysis only is needed
(2) X1>0,X2>0,v=0,
(3) X1>0,X2=0,v>0
(4) X1>0,X2=0,v=0
(5) X1=0,X2>0,v>0
(6) X1=0,X2>0,v=0
(7) X1=0,X2=0,v>0
(8) X1=0,X2=0,v=0 , obviously, this is the trivial situation, in which what need to do is only to verify the inequality derivative condition.
Every situation from (2) to (8) corresponds to a corner solution, and every corner solution except (8) can be solved through usual marginal analysis. For example, the situation (5) can be reduced into the marginal analysis:
Max L5(X2,v)= L(0,X2,v)=F(0,X2)+v[b-g(0,X2)] X2,v>=0 (24.14)
X2,v
Because of X2* ,v*>0 , the preceding problem (24.14) has an internal solution with respect to (or within) the two-dimensional space (X2,v), therefore the marginal analysis suffices to find the optimal solution, which is so-called marginal analysis in a corner solution by Yang Xiaokai.
If we have found all the solutions from (1) to (8), then we can compare the objective values which get from every situation and choose the greatest one as the global maximum, which is so-called inframarginal analysis. Of course,strictly speaking, the situation (8) is not a marginal analysis,however, this situation will not appear in economics.
So, each situation from (1) to (7) corresponds to a marginal analysis, and comparing all the objective values from (1) to( 7) is the inframarginal analysis. Thus ,we ended the explication of the equivalence between KKT conditions and inframarginal analysis.
）。 
(24.2)
(24.3)
(24.4)
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