1. Nonlinear Programming
Consider the following math program:
Max Z = K*L
st K + L <= 8
K,L >= 0
place Winston Fig 2 here
Where does the optimal solution fall within the feasible
region?

2. Nonlinear Programming
Consider the following math program:
Max Z = f(x)
st <= x <= 1
where f(x) is pictured at right.
place Winston Fig 3 here
Where does the optimal solution fall within the feasible
region?

3. Nonlinear Programming
Consider the following math program:
Max Z = f(x)
st <= x <= 10
where f(x) is pictured at right.
place Winston Fig 4 here
Note there are three local optima, but only one optimal solution at point C.

4. Nonlinear Programming
Formulate the following problem as an NLP.
Truckco is trying to determine where they should locate a single warehouse. The positions in the x-y plane (in miles) of their four customers and the number of shipments made annually to each customer is given below. Locate the warehouse to minimze the total distance trucks must travel annually from the ware house to the four customers.
Customer X Coord Y Coord Number of Shipments

5. Nonlinear Programming
Truckco formulation:
Let X,Y be the coordinates of the warehouse.
Min Z = 200 * D * D * D * D4
s.t. D1 = ( ( X- 5)2 + (Y-10 )2 ) .5
D2 = ( ( X- 10)2 + (Y-5 )2 ) .5
D3 = ( ( X- 0)2 + (Y-12 )2 ) .5
D4 = ( ( X- 12)2 + (Y-0 )2 ) .5
D1, D2, D3, D4, X, Y >= 0

6. Nonlinear Programming
One-Variable Unconstrained
Example: Max Z = f(x)
s.t. nothing
Or Max Z = x - (2x2 + x4)
First determine if the concave (or convex for Min problem).
f(x) is concave if the second derivative is <= 0 for all values of x.
f’(x) = 3 – 4x – 4x3
f’’(x) = – 4 – 12x2
Where does optimum occur? 3 – 4x – 4x3 = 0

7. Nonlinear Programming
One-Variable Unconstrained
An approximate graph of x and Z is as follows: Z = x - (2x2 + x4) Z x

8. Nonlinear Programming
One-Variable Unconstrained
How do you solve for x? 3 – 4x – 4x3 = 0

9. Nonlinear Programming
Multi-Variable Unconstrained
Example: Max Z = f(x)
s.t. nothing
Or Max Z = 2x1x2 + 2x2 – x12 – 2x22
For a concave function (convex when minimizing), use an approach whereby you move in a direction of steepest ascent. In other words, if f(x) is the surface area of a 3 dimensional graph then move in a direction that has the steepest angle. Move as far in the steepest direction until you are no longer ascending. Then re-evaluate the steepest ascent at that point.
This procedure is called the gradient search procedure.

10. Nonlinear Programming
Multi-Variable Unconstrained
Gradient Search procedure:
Initialization: select some initial point x’, and some e. Go to the stopping rule.
Set xj = x’j + t(df / dxj)x=x’ for all values of j
then substitute these expressions into the expression f(x).
This expression of f(x) is in terms of t.
Find the value of t* that maximizes f(x) over t >= 0.
ReSet x’ = x’ + t* (df / dxj)x=x’ . Go to stopping rule.
Stopping rule: evaluate if | (df / dxj)x=x’ | <= e for each j. If so, stop. Else, go to step 1.

11. Nonlinear Programming
Multi-Variable Unconstrained
Example Gradient Search procedure: Z = 2x1x2 + 2x2 – x12 – 2x22
df / dx1 = 2x2 – 2x1
df / dx2 = 2x1 + 2 – 4x2
Initialization: set x’1, x’2 = 0.
Set x1 = x’1 + t(df / dx1) = 0 + t(2*0 – 2*0) = 0.
Set x2 = x’2 + t(df / dx2) = 0 + t(2* – 4*0) = 2t.
f(x1 , x2) = 2*0* 2t +2*2t – 0*0 – 2*2t*2t = 4t – 8t2.
f’(x1 , x2) = 4 – 16t. Let 4 – 16t = 0 then t* = ¼.
ReSet x’1 = x’1 + t*(df / dx1) = 0 + ¼(2*0 – 2*0) = 0.
x’2 = x’2 + t*(df / dx2) = 0 + ¼(2*0 + 2 – 4*0) = ½.
Stopping rule: df / dx1 = 1, df / dx2 = 0.

12. Nonlinear Programming
Multi-Variable Unconstrained
Example Gradient Search procedure: Z = 2x1x2 + 2x2 – x12 – 2x22
df / dx1 = 2x2 – 2x1
df / dx2 = 2x1 + 2 – 4x2
Iteration 2: x’1 = 0 x’2 = ½ .
Set x = (0,1/2) + t(1,0) = (t,1/2).
f(t,1/2) = 2*t* 1/2 +2*1/2 – t*t – 2*1/2*1/2 = t – t2 + ½.
f’(t,1/2) = 1 - 2t. Let 1 – 2t = 0 then t* = ½ .
ReSet x’ = (0,1/2) + ½ (1,0) = (½, ½).
Stopping rule: df / dx1 = 0, df / dx2 = 1.

13. Nonlinear Programming
Multi-Variable Unconstrained
Example Gradient Search procedure: Z = 2x1x2 + 2x2 – x12 – 2x22
This iterative process continues until the values of the gradient are <= some e.

14. Nonlinear Programming
Multi-Variable Constrained
Category - Convex Programming
Example: Max Z = 5x1 + 8x2 – x12 – 2x { Non-linear}
s.t x1 + 2x2 <= { Convex feasible region}
x1 , x2 >= 0