[fw-LP]
Class 11 - INTERACTIVE PROGRAMMING (continued)
An interactive Frank-Wolfe example (continued)
By formulating the airport location problem as minimizing travel and noise, we
rewrite the two city case as
min v ( f1(x), f2(x)) = v (2x1+x2, 2x1-2+x2-2)
s.t.
x1 + x2 $ 60
xi $ 0 ( i = 1, 2).
where x1 is the distance from Cincinnati and x2 is the distance from Dayton.
Figure 1 - Determination of marginal rate of substitution
Taking the gradient of value function v(f) = λ1 f1%λ2 f2 yields
Mf1
Mf2
Mx1
Mx1
Mv
Mv
Lx v (f) '
%
Mf1 Mf1
Mf2 Mf2
Mx2
Mx2
(1)
which is evaluated at x0, x1, x2, ... etc. The initial gradient at the half way point
between Cincinnati and Dayton x0 = (30, 30), for example, is
2
Lx v (f(x 0)) ' λ1 % λ2
1
&3
&4x1
&3
&2x2
x 0' (30, 30)
2λ1& 0.00015λ2
'
.
1λ1& 0.00007λ2
(2)
1
EQUIVALENT LINEAR PROGRAM
Using the tangent as an approximation for the value function at x0, the linear
program to be solved is simply
min Lx v T(f (x 0)) x
x0X
x1
(3)
' min (2λ1& 0.00015 λ2, λ1& 0.00007 λ2)
.
x
x0X
2
Suppose the decision maker decides that the marginal rate of substitution is "fiftyfifty", or &λ1/ λ2 , through local linearized indifference curves such as the one
shown in Figure 1.
At the location x0 halfway between Cincinnati and Dayton, the decision maker is
asked about the increment of travel cost ∆ f1 for which s/he is willing to trade
against a decrement of aircraft noise ∆ f2 . The slope of this indifference curve is
precisely &λ1/ λ2. Without loss of generality, let us set λ1 = 1, which means λ2 = 1
in this example. (Here λ1 + λ2 … 1.)
Now by the following LP, the optimal solution x* = (0, 60) is determined.
min 2x1 + x2
x1 + x2 $ 60
xi$0 (i = 1, 2).
Thus at this iteration, we are moving the airport toward Cincinnati from the
halfway point between the two cities according to the steepest ascent direction d0 =
x*& x0, where x0 = (30, 30) and x* = (0, 60) or d0 = (&30, 30).
STEP SIZE
The decision maker (DM) now determines the step size α to move along the
direction x0 + α0d0. The DM, assisted by tabular or graphic displays of the function
f(x0 + αd0) = ( f1 (x0 + αd0), f2 (x0 + αd0)) determines the step size α0 between 0 and
1. One possible way to obtain the best step size α is to display the values for the
two criterion functions fi (x0 + αd0) for i = 1 and 2 as a function of α over the
2
selected values of α in a tabular or graphic way.
Figure 2 - Step size determination for travel function
The DM then determines a value of α for the most preferred values of the
corresponding criterion functions. In short, the following optimization problem is
solved: min 0#α#1 v ( f (x0 + αd0)). Suppose α0 = 0.5. We are now at x1 = x0 + α0d0
= (30, 30) + 0.5 (&30, 30) = (15, 45) and the iterations continues until the
incremental ascent of the preference function v is minuscule, as with most "hillclimbing" algorithms.
Figure 3 - Graphical display to determine step size
EXERCISE
Similar to the one for travel time, please derive the analytic formula for the noise
function that determines the step size.
It can be shown the for equal weights upon the two criteria, i.e., the decisionmaker is "indifferent" between noise and travel and α = 0.5, an airport converges
at (1.587, 58.413), or about 1.6 miles outside Cincinnati.
Iteration k
Airport location xk
0
(30, 30)
1
(15, 45)
2
(7.50, 52.50)
3
(3.750, 56.250)
4
(1.875, 58.125)
5
(0.938, 59.063)
6
(30.469, 29.5319)
3
Marginal rate of substitution
[substitution&step-size]
f2(x0)
f2(x0)
x0
f1(x0)
f1(x0)
Step size determination for travel function
f1 (x0 + αd0)
= 2 (30 + α [–30]) + (30 + α[30])
= 90 – 30 α
Graphical
G
ap ca d
display
sp ay to dete
determine
e step ssize
e