Topic 5: Constrained optimisation Flashcards
What is an example of constrained optimisation (1)
-Max π (Q1, Q2) subject to Q1 + Q2 = 1000
What is the general method to solving constrained optimization problems (3)
-The general method is the lagrange multiplier method
-L = F(x, y) - λ(g(x, y) - c)
-F(x, y) = the maximising function, g(x, y) = the cost function, c = the budget
How do we use the first order condition in solving the lagrange equation (2,1)
-With our equation, we take the first order partial derivatives of L with respect to x, y and λ, set them to 0, and then use simultaneous equations
-The FOC’s are thus necessary conditions for the max of the constrained optimisation
-F’1(c, y) = marginal utility, λg’1(x, y) = marginal cost, and optimisation is where MU - MC = 0
How would you solve f(x,y) = x^(2/3)y^(1/3) subject to 100x + 100y = 400,000 (1,3,2,1,3)
-L = x^(2/3)y^(1/3) - λ[100x + 100y - 400,000]
The FOC’s are:
-Lx = (2/3)x^(-1/3) y^(1/3) - 100λ = 0
-Ly = (1/3)x^(2/3)y^(-2/3) - 100λ = 0
-Lλ = -[100x + 100y - 400,000] = 0
Equal Lx and Ly to eachother
-(2/3)x^(-1/3) y^(1/3) = (1/3)x^(2/3)y^(-2/3)
-y = 1/2x
-Plug this value into Lλ and solve for 0
-x = 2666.66666
-y = 1333.3333
-Max production thus equal 2117 units
What is the value function for the problem (4)
-Suppose x* and y* solve max f(x, y) subject to g(x, y) = c
-assume x* and y* can be written as x(c) and y(c)
-The function can thus be written as f(c) =f(x(c), y*(c))
-This is the value function for the problem (the value of the objective function using the max values)
What does the lagrange multiplier represent (3)
-If f(C) is differentiable, then df(c) = λdc = df*(c)/dc = λ
-λ represents the rate at which the value function changes when the constraint constant C changes
-λ is sometimes called the shadow price/marginal constant
What is the tangency condition of optimisation in 3d (2)
-If g(x, y) = c passes through a level curve of f it can’t be the highest/lowest value of the function along the constraint
-Extreme points of the constrained optimisation problem can only occur when g(x, y) = c is tangent to a level curve
(g(x, y) = c is the constraint), f(x, y) = function to maximise)
What is the slope of the level curve (1)
-y’ = dy/dx = -(f1‘(x, y))/(f2‘(x, y)
What is a necessary solution for (x0, y0) to be a solution of the constrained optimisation problem (3)
-(f1‘(x, y)/(g1‘(x, y)) = (f2‘(x, y)/(g2‘(x, y))
-This represents tangency between the constraint (g) and level curve (f)
-Setting the latter ratio to λ and rearranging gives both FOC of the lagrangian
(g(x, y) = c is the constraint), f(x, y) = function to maximise)
How do we write a general problem with a multivariate function with n variables and m constraints (3)
-A general problem involves a multivariate function with n variables f(x1, …, xn) and m constraints g1(x1, xn) = c1, …, gn (x1, xn) = cn
-Formally, the problem is max f(x1, …, xn) s.t g1(x1, xn) = c1, …, gn (x1, xn) = cn
-We then have one lagrange multiplier for each constraint, so we get L = f(x1, …, xn) - Σλj[gj(x1, xn) - cj]
What does it mean if the constraint is slack/non-binding (1)
-If the optimum g(x, y) < c
When are the non-negativity constraints violated, and then what do you do (2)
-Solve the problem as max f(x, y) s.t. g(c, y) = c
-If x* or y* < 0, then the constraints are violated and you look for a maximum/minimum on the bounds (corner solution)
