**1.5 A Rigorous Formulation of Agents' Decision**

Having explained the relation between indifference curves and utility functions and the meaning of a unity function we are now in a position to pose the problem solved by the agents. Assume that the price of good 1 is and that of good 2 is and that the agent has a budget of and a utility function . The reader may have already guessed that the agent needs to solve a maximization problem which is formulated as below:

s.t.

Since is a monotone increasing function, the inequality can be replaced with an equality . Thus, one way of solving this problem is by substituting in terms of and solving an optimization problem with only the non-negativity constraints .

We will, however, utilize this opportunity to review the optimality conditions as they play a central role in the portfolio choice problem. Furthermore, we will explain or provide a heuristic proof for the optimality conditions. This proof highlights a relation (duality) between two systems of inequalities that is used later in the Arbitrage Pricing Theory (**APT**).

In order to state the optimality conditions in a generic way we should decide about a canonical form of an optimization problem. We will define a canonical form as a minimization problem with constraints of the type . That is, a problem of the form

s.t. ,

Where is a vector in and and , are functions from to and , .

It is easy to see that there is no loss of generality assuming such a canonical form of an optimization problem. The solution of a maximization problem is the same for a minimization problem in which the negative of the objective function is minimized. Inequality constraints could be reversed by multiplying the inequality by and an equality can be written as a pair of two inequalities. After we develop the optimality condition for the above problem we will apply it to our maximization of utility problem.

Let us first start with the intuition behind the optimality conditions. If is a minimum point of , subject to the constraint stipulated above, it must be that there is no feasible direction “along which one can move away” from and reach a point where the value of is smaller than the value of at . Being a feasible point, satisfies the constraint, that is , . The direction is feasible if there exists an ", mathvariant = "normal", fence = "false", separator = "false", stretchy..." align="center" border="0"> such that , . Moving away from along the direction , will lead to a point for some positive .

Our next step requires characterization of feasible directions and directions along which the function decreases. To accomplish this task we now define a function on such that

, for a given and in .

thus equals , and if is negative it means that moving along the direction , decreases. __Footnote 6__** **A visualization of direction of decent relative to the gradient and the value of the function along these directions is offered below for the utility function of the former section. The graph below demonstrates the direction for at a point (5,5).

The directions of the negative of the gradient is drawn as a black line in the axis that points to the origion, with two other directions that are perpendicular to it. All the direction between the two prepndcualr directions are directions of decent. The movement along these directions is also drawn on the graph of the function so one can realize that these are indeed directions of decent.

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NGFradNfun:=plot3d([[5-x,5-x,(5-x)*(5-x)],[5-x,5-x,0],[5-x,5+x,(5-x)*(5+x)],[5+x,5-x,(5+x)*(5-x)],[5-x,5+x,0],[5+x,5-x,0]],x=0..5,y=0..5,axes=box,thickness=4): |

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Util:=plot3d(x*y,x=0..10,y=0..10,axes=box,style=patchcontour,contours = 20,filled=true,orientation=[-82,56],title="Gradient, steepest decent, and directions of decent ",transparency=.2): |

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plots[display](Util,NGFradNfun); |

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Similarly if is a point on the boundary of the feasible set, i.e., it satisfies one of the constraints say as equality rather than inequality, and if the function is such that

is not a feasible direction. This is because as we move along , increases and since at , there exists an ", mathvariant = "normal", fence = "false", separator = "false", stretchy = "false", symmetric =..." align="center" border="0">such that violates the constraint since for every ε in

This is illustrated below where the vector is the red line. Moving along with .5 from the point (2,3) is demonstrated by the green line which leads to the point . Moving along from the point (2,2) with is demonstrated by the blue line which ends at . The feasible set is demonstrated by the black circle and one can visualize that the direction from the point (2,2) is a feasible direction as it is possible to move along it and still stay within the feasible set. On the other hand the direction is not a feasible direction from the point (2,3) since moving along it, even with a very small positive, will result in a point outside the feasible set. The feasible set in this illustration is given by the function

.

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V11:=plots[arrow]([<1,1>], width=[0.02, relative], head_length=[0.04, relative], color=red): |

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V1122:=plots[arrow](<2,2>,<1,1>, width=[0.02, relative], head_length=[0.04, relative], color=blue): |

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V1123:=plots[arrow](<2,3>,<1.5,1.5>, width=[0.02, relative], head_length=[0.04, relative], color=green): |

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FeasC:=plot([[2+cos(t),2+sin(t), t=0..2*Pi]], color=[black],thickness=2,scaling=constrained): |

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plots[display](V11,V1122,V1123,FeasC); |

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The derivative of with respect to is given by where ▽ denotes the gradient of (the vector of partial derivatives) and denotes the transpose of (a row vector). At the point , is positive if At the point (2,3) we can calculate as below

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Student:-MultivariateCalculus:-Gradient((x-2)^2+(y-2)^2,[x,y]=[2,3]); |

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In the graph below the gradient of at the point (2,3) is illustrated by the blue line, the direction from the point by the green line, and the direction from the point by the red line.

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V0123:=plots[arrow](<2,3>,<0,1>, width=[0.02, relative], head_length=[0.04, relative], color=blue): |

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V1Neg123:=plots[arrow](<2,3>,<1,-1>, width=[0.02, relative], head_length=[0.04, relative], color=red): |

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plots[display](V1123,V0123,V1Neg123,FeasC); |

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You may recall that if than the angle between the vector and is acute, while if the angle between the vector and is obtuse . In the case illustrated above by the green line since and hence the angle between the vector and is acute and the green line is not a feasible direction.

On the other hand the direction is a feasible direction from the point . It is represented by the red line which points inside the circle. On the circle the value of is 1 and inside the circle it is less than 1. Hence moving along this direction does not violate the constraint that requires that Feasible directions are therefore those that generate an obtuse angle with the gradients of the binding constraints. A feasible direction at a point , is characterized by the fact that for each binding constraint at , i.e., for each for which , the product of and the gradient of is negative, that is

If the point is a minimum point, it must be that there exists no feasible direction along which the objective function decreases. __Footnote 7__ Since if one existed there would be a feasible point say = for some γ>0 such that , but that means that could not be the minimum point. Thus, if is a minimum point it must be that

for every such that then ,

i.e., the function increases along each feasible direction from .

This condition can also be stated as:

Every such that for such that , must satisfy .

Put into words, every feasible direction from point , must be a direction along which does not decrease. This condition can also be geometrically visualized.

Assume that there are two binding constraints at a point , hence we have vectors representing the gradient of , and . A feasible direction must make an acute angle with each of these vectors. We start by plotting a vector representing and the vectors that are making an acute angle with it. Assume that is the vector . Two vectors that make a right angle with it are given by and , hence vectors that are making an acute angle with are between these two vectors. (Note the vectors that make a right angle with must have a product of zero with it, e.g., . This is illustrated below:

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Lg1:=plots[textplot]([1,2,`-grad(g1)`],align=ABOVE): |

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Neggradg1:=plots[arrow]({<1,2>,<-2,1>,<2,-1>}, width=[0.02, relative], head_length=[0.04, relative], color=red): |

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plots[display](Lg1,Neggradg1); |

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The vectors that are making an acute angle with are those that are between the two vectors making a right angle with it. These are the vectors with an end point above the red line (that passes via the origin).

Assume that and let us now graph and all the vectors making an acute angle with it. This can still be displayed in the figure below.

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Neggradg2:=plots[arrow]({<2,0>,<0,-2>,<0,2>}, width=[0.02, relative], head_length=[0.04, relative],color=blue): |

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Lg2:=plots[textplot]([1.6,0.1,`-grad(g2)`],align=ABOVE): |

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plots[display](Lg2,Neggradg2); |

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Let us now visualize the vectors that are making acute angles with bothandThis will be the intersection of the two sets of vectors described in the above two graphs. That is the vectors with an end point above the red line that passes via the origin, and to the left of the blue line that passes through the origin.

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plots[display](Lg1,Neggradg1,Lg2,Neggradg2); |

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Recall that if is a minimum point, the angle between each feasible andmust also be acute, that is for every which is a feasible direction. Observing the set of feasible directions in the graph above, the reader should realize that for to make an acute angle with each feasible , must be between and An example of a vector that satisfies this condition is illustrated below. It is labeled as grad(f) and it is the green vector in the graph below:

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gradf:=plots[arrow](<3,2>, width=[0.02, relative], head_length=[0.04, relative],color=green): |

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Lf:=plots[textplot]([2.7,2,`grad(f)`],align=ABOVE): |

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plots[display](Lg1,Neggradg1,Lg2,Neggradg2,gradf,Lf); |

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However algebraically for to be between and it means that is a positive linear combination of and i.e., = for some positive and In general therefore, this condition is satisfied if is a positive linear combination of , for such that . That is where the summation is over such that and the are all nonnegative. __Footnote 8__

We are now ready to stipulate the optimality conditions for the canonical minimization problem. It is common to stipulate the optimality conditions in terms of a function called Lagrangian, associated with the canonical optimization problem. The Lagrangian of our canonical minimization problem is defined to be a function of , a vector in ( is the number of constraints) and .

The optimality conditions are stated in terms of the partial derivative of the Lagrangian below:

i=1,...,n

j=1,...,m

j=1,...,m

The reader is asked to verify that the first condition above states that , the third condition ensures that if the constraint is not binding =0 and hence the sum in is only over the binding constraints. Finally the second constraint simply guarantees that is a feasible point and satisfies the constraints.

In our utility maximization problem, is , ( and are and ), the non-negativity constraints on and can be written in terms of the , and we also have one inequality constraint that can be written as an equality constraint because of the monotonic of . The equality constraint can be written as two inequality constraints.

The reader is asked to show that this will alter the optimal conditions only by allowing the associated with this constraint to be positive or negative instead of the nonnegative constraint that is imposed on λ that is associated with an inequality constraint of the type less than or equal. Our problem is a maximization problem. We can transform the maximization of to a minimization of . Therefore we have

s.t.

Writing the Lagrangian of this problem and the optimality conditions where, and are associated with

the non-negativity constraints on and respectively, yields:

If we assume that the non-negativity constraints on and are not binding, that is that the optimal solution is an interior point of the positive orthant we can try to satisfy the equations:

which implies that

or that

Note that is the slope of the budget line and is the slope of the indifference curve. Since we have the second equation:

Solving these two equations will thus produce the same solution as produced by the graphical procedure.

We conclude this chapter with an example. Consider the optimization problem

s.t.

Where , , and . The KT optimality conditions of this problem are:

This system of equalities and inequalities can be solved by Maple as demonstrated below.

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solve({c1+c2/(1.14)=80,-diff(c2+1.07*c1,c1)+lambda+eta||1=0,\
-diff(c2+1.07*c1,c2)+lambda*(1/1.14)+eta||2=0,\ c1>=0,c2>=0,eta||1<=0,eta||2<=0,eta||1*c1=0,eta||2*c2=0},\ {c1,c2,lambda,eta||1,eta||2}); |

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One of the exercises at the end of the chapter asks the reader to solve this problem using the graphing approach (which can also be done using our procedure **BudnIndif(c1+c2/1.14, c2+1.07c2))** and to explain and compare the results of the graphical approach with that of the formal approach using KT conditions.

**Footnote**

__To next Section____Back to table of contents.__