Optimization by hcj

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									      Calculus-Based Optimization
                  AGEC 317
Economic Analysis for Agribusiness and Management
Readings

• Optimization Techniques
Why Consider Optimization?
• Consumers maximize utility or satisfaction.
• Producers or firms maximize profit or minimize
  costs.
• Optimization is inherent to economists.
Setting-Up Optimization Problems
•   Define the agent’s goal: objective function and
    identify the agent’s choice (control) variables
•   Identify restrictions (if any) on the agent’s
    choices (constraints). If no constraints exist,
    then we have unconstrained minimization or
    maximization problems.
    If constraints exist, what type?
    Equality Constraints (Lagrangian)
    Inequality Constraints (Linear Programming)
Mathematically,
     Optimize y = f(x1, x2, . . . ,xn)
 subject to (s.t.)
     gj (x1, x2, . . . ,xn) ≤ bj
                               or
                               = bj       j = 1, 2, . . ., m.
                               or
                               ≥ bj

  y = f(x1, x2, . . . ,xn) → objective function
      x1, x2, . . . ,xn → set of decision variables (n)
      optimize → either maximize or minimize
               gi(x1, x2, . . . ,xn) → constraints (m)
Constraints refer to
     restrictions on resources
           legal constraints
                 environmental constraints
                       behavioral constraints
Review of Derivatives
                                                         dy
•   y=f(x): First-order condition:                            f ' ( x)
                                                         dx
                                                          2
•           Second-order condition:                      d y
                                                            2
                                                                 f ' ' ( x)
                                                         dx
•   Constant function: y  f ( x)  a                      f ' ( x)  0
•   Power function: y  f ( x)  ax     b        f ' ( x)  baxb1
                                               dy
•   Sum of functions: y  f ( x)  g ( x)           f ' ( x)  g ' ( x)
                                               dx
                                    dy
•   Product rule: y  f ( x) g ( x)     f ' ( x) g ( x)  f ( x) g ' ( x)
                                    dx
•   Quotient rule: y     f ( x)    dy f ' ( x) g ( x)  f ( x) g ' ( x)
                                       
                         g ( x)     dx                g ( x)2
•   Chain rule: y  f ( g ( x)) dy
                                            f ' ( g ( x)) g ' ( x)
                                      dx
Unconstrainted Univariate
Maximization Problems: max f(x)
•   Solution:
  • Derive First Order Condition (FOC): f’(x)=0
  • Check Second Order Condition (SOC): f’’(x)<0
• Local vs. global: If more than one point satisfy both
   FOC and SOC, evaluate the objective function at
   each point to identify the maximum.
Example
   PROFIT = -40 + 140Q – 10Q2
Find Q that maximizes profit

 dPROFIT
          140 – 20Q = set 0
   dQ
           Q=7

 d 2 PROFIT - 20 < 0
         2
           
      dQ

  max profit occurs at Q = 7
     max profit = -40 + 140(7) – 10(7)2
     max profit = $450
Minimization Problems: Min f(x)
•   Solution:
  • Derive First Order Condition (FOC): f’(x)=0
  • Check Second Order Condition (SOC): f’’(x)>0
• Local vs. global: If more than one point satisfy both
   FOC and SOC, evaluate the objective function at
   each point to identify the minimum.
 Example
COST = 15 - .04Q + .00008Q2

            Find Q that minimizes cost

 dCOST
        -.04 + .00016Q = set 0
   dQ

  2
                           Q = 250
d COST
     2
        .00016 > 0
  dQ

Minimize cost at Q = 250
     min cost = $10
     Unconstrained Multivariate Optimization
•          Max                y  g ( x, z )                                     •           Min            y  g ( x, z )

                            y                                                                             y           g x ( x, z )
•          FOC:                    x
                                         g x ( x, z )                           •           FOC:                 x
                                                                                                           y           g z ( x, z )
                            y           g z ( x, z )                                                            z
                                 z
                                                                                                           2 y
•          SOC:               y
                               2
                                   g xx ( x, z )  0                            •           SOC:          x 2
                                                                                                                  ( x, z )  g xx ( x, z )  0
                             x 2

                                                                                                          2 y
                              y 2
                                   g zz ( x, z )  0                                                     z 2
                                                                                                                  ( x, z )  g zz ( x, z )  0
                             z 2


    2 y
    x 2
                   2 y
           ( x, z ) z 2 ( x, z )     2 y
                                        xz   ( x, z )   
                                                           2 y
                                                           zx
                                                                        
                                                                  ( x, z )  0   2 y
                                                                                 x 2
                                                                                                2 y
                                                                                        ( x, z ) z 2 ( x, z )    
                                                                                                                   2 y
                                                                                                                   xz   ( x, z )    2 y
                                                                                                                                        zx
                                                                                                                                                     
                                                                                                                                               ( x, z )  0
Example
PROFIT is a function of the output of two products
               (e.g.heating oil and gasoline)
                      Q1              Q2

   PROFIT  60  140 Q1  100 Q2  10 Q12  8Q2  6Q1Q2
                                               2



   dPROFIT
            140  20 Q1  6Q2  set 0
     dQ1
  dPROFIT
           100  16 Q2  6Q1  set 0
    dQ2
                        20Q1  6Q2  140
                   so, 6Q1  16Q2  100

Solve Simultaneously Q1 = 5.77 units
                     Q2 = 4.08 units
Second-Order Conditions
     2                   d 2 PROFIT
   d PROFIT
         2
             20                    6
      dQ1                  dQ1 dQ2

 d 2 PROFIT
        2
             16
      dQ2
                                        2
 d 2 PROFIT  d 2 PROFIT   d 2 PROFIT 

         2
             
                     2     dQ dQ   0
                                       
      dQ1         dQ2          1  2  

(-20)(-16) – (-6)2 > 0
    320 – 36 > 0
    we have maximized profit.
Constrained Optimization
•    Solution: Lagrangian Multiplier Method
•    Maximize y = f(x1, x2, x3, …, xn)
•       s.t. g(x1, x2, x3, …, xn) = b
•    Solution:
    • Set up Lagrangian:
    L( x1 , x 2 ,..., x n ,  )  f ( x1 , x 2 ,..., x n )   g ( x1 , x 2 ,..., x n )  b .
    • FOC:            L x1 ( x1, x 2,...,xn,  )  0
                     ...
                      L xn ( x1, x 2,...,xn,  )  0
                      L ( x1, x 2,...,xn,  )  g ( x1, x 2,...,xn)  b  0
Lagrangian Multiplier
• Interpretation of Lagrangian Multiplier λ: the
  shadow value of the constrained resource.
  • If the constrained resource increases by 1 unit,
    the objective function will change by λ units.
    Example
Maximize Profit =  60  140 Q1  100 Q2  10 Q12  8Q2  6Q1Q2
                                                      2


              subject to (s.t.) 20Q1 + 40Q2 = 200
   Could solve by direct substitution

               Note that 20Q1 = 200 – 40Q2
                        or Q1 = 10 – 2Q2

Maximize Profit =
    60  140(10  2Q2 )  100Q2  10(10  2Q2 )2  8Q2  6Q2 (10  2Q2 ).
                                                      2



               Combine like terms.

Maximize Profit = 340  160Q2  36Q22


                           Q2  2.22 units
                            Q1  5.56 units
                    How arrived?
Lagrangian Multiplier Method
Formulate Lagrangian Function
L PROFIT  60  140Q1  100Q2  10Q12  8Q2  6Q1Q2
                                           2


  (20Q1  40Q2  200)

Maximizing L profit Maximizes the profit function as long as 20Q1  40Q2  200.
Decision variables are Q1 , Q2 , .
dL profit
             140  20Q1  6Q2  20  set 0
 dQ1
dL profit
             100  16Q2  6Q1  40  set 0
 dQ2
dL profit
             (20Q  40Q2  200)  set 0
  d
(1) 20Q1  6Q2  20  140
(2) 6Q1  16Q2  40  100
(3)         20Q1  40Q2  200


Now, we have three equations in three unknowns.

Use equations (1) and (2) to eliminate  .
      20  140  20Q1  6Q2
      40  100  6Q1  16Q2


Multiply the first equation by 2
      40  280  40Q1  12Q2
      40  100  6Q1  16Q2
Therefore,280  40Q1  12Q2  100  6Q1  16Q2 .
                      or
           34Q1  4Q2  180
   also    20Q1  40Q2  200


            Q1  5.56 units
              Q2  2.22 units
          Same answer as before


     Wh Q1  5.56 units and Q2  2.22 units
      en
                    .774
 measures the change in the value of the objective function
  resulting from a one unit change in the value of the constraint.

								
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