nonlinear programming solution a
TRANSCRIPT
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Dr. A.K. Bardhan Facult of Mana ement Studies Universit of Delhi 1
Nonlinear Programming
Solutions
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Dr. A.K. Bardhan Facult of Mana ement Studies Universit of Delhi 2
Difficulties of NLP Models
Nonlinear Programs:
LinearProgram:
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Dr. A.K. Bardhan Facult of Mana ement Studies Universit of Delhi 3
Graphical Analysis of Non-linear programs
in two dimensions: An example
2 214 15( ) ( )x y + Minimize
subject to (x - 8)
2
+ (y - 9)
2
49x 2x 13x + y 24
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Where is the optimal solution?
0 2 4 6 8 10 12 14 16 18
0
2
4
6
8
10
12
14
16
18y
x
Note: the optimalsolution is not at acorner point.It is where theisocontour first hits
the feasible region.
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Local vs. Global Optima
There may be several locally optimal solutions.
x
z
10
z = f(x) max f(x)s.t. 0 x 1A
B
C
Defn: Let x be a feasible solution, then
x is a global maxif f(x) f(y) for every feasible y. x is a local maxif f(x) f(y) for every feasible y sufficiently close to x
(i.e., xj-yjxj+ for allj and some small ).
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When is a locally optimal solution also globallyoptimal?
For minimization problems The objective function is convex.
The feasible region is convex.
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W
P
2 4 6 8 10 12 14
2
4
6
8
10
12
14
Convexity and Extreme Points
We say that a set S is convex, if for every
two points x and y in S, and for every realnumber l in [0,1], lx + (1-l)y S.
The feasible region of a
linear program is convex.
x
y
We say that an element w S is anextreme point(vertex,corner point), if wis not the midpoint of any line segment
contained in S.
8
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On convex feasible regions
If all constraints are linear, then the feasible region isconvex
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On Convex Feasible Regions
The intersection of convex regions is convex
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S
Recognizing convex sets
Rule of thumb: suppose for all x, y S the midpoint of xand y is in S. Then S is convex.
xy
It is convex if
the entire line
segment is
always in S.
(x+y)/2
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Which are convex?
CB
B C B CB C
DA
12
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Joining two points on a curve
The line segment joining two points on a curve. Let f( ) be a
function, and let g(ly+(1-l)z)) = l f(y) + (1-l)f(z) for 0l.
f(y)
f(z)
(y+z)/2
f(y)/2 +f(z)/2
g(x)
g(y) = f(y)
g(z) = f(z)
g(y/2 + z/2) = f(y)/2 + f(z)/2
y z
f(x)
x
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Convex Functions
Convex Functions: f(l y + (1-l)z) l f(y) + (1-l)f(z)for every y and z and for 0l.e.g., l = 1/2 f(y/2 + z/2) f(y)/2 + f(z)/2
Line joining any pointsis above the curvef(x)
xy z
f(y)
f(z)
(y+z)/2
f(y)/2 +f(z)/2
We say strict
convexityif signis
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Concave Functions
Concave Functions: f(l y + (1-l)z) l f(y) + (1-l)f(z)for every y and z and for 0 l.e.g., l = 1/2 f(y/2 + z/2) f(y)/2 + f(z)/2
Line joining any pointsis below the curve
xy z
f(y)
f(z)
(y+z)/2
f(y)/2 +f(z)/2
We say strict
concavityif signis
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Classify as convex or concave or both or neither.
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More on convex functions
x
f(x)
-x
f(-x)
If f(x) is convex,then f(-x) isconvex.
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More on convex functions
x
y
If f(x) is convex,then K - f(x) isconcave.
x
y
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More on convex functions
If f(x) is a twice differentiable function of one
variable, and if f(x) > 0 for all x, then f(x) isconvex.
f(x) = x2.
f(x) = 2x, f(x) = 2
0
1
2
3
4
5
6
7
8
9
10
-4 -3 -2 -1 0 1 2 3 4
f(x) = - ln(x) for x > 0
f(x) = -1/x, f(x) = 1/x2
-2
-1.5
-1
-0.5
0
0.5
1
1.52
2.5
0 0.5 1 1.5 2 2.5 3 3.5 4
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0
1
2
3
4
5
6
7
8
-3 -2 -1 0 1 2 3
f(x) g(x)
Even more on convex functions
If f(x) is convex and g(x) is convex,
then so is f(x) + g(x)
f(x)+g(x)
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0
1
2
3
4
5
6
7
8
-3 -2 -1 0 1 2 3
f(x) g(x)
Even more on convex functions
If f(x) is convex and g(x) is convex,
then so is max [f(x), g(x)]
max[f(x), g(x)]
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What functions are convex?
f(x) = 4x + 7 all linear functions
f(x) = 4x2 13 some quadratic functions f(x) = ex
f(x) = 1/x for x > 0
f(x) = |x|
f(x) = - ln(x) for x > 0
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Convex functions vs. convex sets
If y = f(x) is convex, then{(x,y) : f(x) y} is a convex set
y
x
f(x)
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Local Minimum Property
A local min of a convex function on a convex feasible region is also aglobal min.
Strict convexity implies that the global minimum is unique.
The following NLPs can be solved
Minimization Problems with a convex objective function and linearconstraints
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Local minimum property
There is a unique local minimum for the function below.
y
x
f(x)
The local
minimum is
a global
minimum
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Local Maximum Property
A local max of a concave function on a convex feasible region is also aglobal max.
Strict concavity implies that the global optimum is unique.
Given this, the following NLPs can be solved
Maximization Problems with a concave objective function and linearconstraints
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Local maximum property
y
x
There is a unique local maximum for the function below.
The local
maximum isa global
minimum
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Finding a local optimal for a single variable NLP
Solving NLP's with One Variable:max f(q)
s.t. a q b
Optimal solution is either
a boundary point or
satisfies f
(q*
) = 0 and f
(q*
) < 0.
q
f(q)
a
q *
b
q
f(q)
a q * b
q
f(q)
a q * b
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Unimodal Functions
A single variable function f is unimodalif there is at most
one local maximum (or at most one local minimum) .
H fi d if f i f l i bl
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How to find if a function of several variablesis Convex or Concave?
Use of Hessian Matrix
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The Hessian MatrixH(X) is the nn matrix whose ith row are
j
f
x
2 2 2
2
1 1 2 1
2 2 2
2
1 2 2 2
2 2 2
2
1 2
. .
. .
..
.
. . .
n
n
n n n
f f f x x x x x
f f f
x x x x x
f f f
x x x x x
i.e. H(X) =
with respect toxi.
the partial derivates of (j =1,2,..n)
(i =1,2,..n)
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++
22
26),(Then
2)(If
1
21
2221
31
xxxH
xxxxxf
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Principal Minor
The ith principal minor of an nx nmatrix is the determinant of the ix imatrix
obtained by deleting (ni) rows and corresponding (ni) columns of thematrix.
Example
4122.22.6
isminorprincipalleadingsecondThe2.and6areminorsprincipalfirstThe
22
26),(Then
2)(If
11
1
1
21
2
221
3
1
++
xx
x
xxxH
xxxxxf
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Leading Principal Minor
The kth leading principal minor of an nx nmatrix is the determinant of the k
x kmatrix obtained by deleting the last (nk) rows and columns of thematrix.
Example
4122.22.6
isminorprincipalleadingsecondThe6isminorprincipalleadingfirstThe
22
26),(Then
2)(If
11
1
1
21
2
221
3
1
++
xx
x
xxxH
xxxxxf
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e.nonnegativareofminorsprincipalall,
eachforifonlyandifSonfunctionconvexais),...,,(
),...,,(ThenS.),...,,(pointeachforderivative
partialordersecondcontinuoushas),...,,(Suppose
21
2121
21
HSx
xxxf
xxxfxxxx
xxxf
n
nn
n
convex.is),(theorem,abovethefromTherefore
0.02(2)-2(2)minorprincipalsecondThe0).2equal(both
entriesdiagonaltheareHessiantheofminorsprincipalfirstThye
22
22
2),(
:Example
21
2
121
2
121
xxf
H
xxxxxxf
++
Convex Function
C F ti
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.)1(assignsame
thehaveofminorsprincipalzero-nonall;,...,2,1and
eachforifonlyandifSonfunctionconcaveais
),...,,(ThenS.),...,,(pointeachforderivative
partialordersecondcontinuoushas),...,,(Suppose
2121
21
k
nn
n
HnkSx
xxxfxxxx
xxxf
concave.is),(theorem,abovethefromTherefore
0.7(-1)(-1)-2(-4)-
minorprincipalsecondThee.nonpositivboth0)(-2,-4
entriesdiagonaltheareHessiantheofminorsprincipalfirstThye
41
12
22),(
:Example
21
2
121
2
121
xxf
H
xxxxxxf
Concave Function
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Example
A monopolist producing a single product has two types of
customers. If q1 units are produced for customer-1, thencustomer-1 is willing to pay 70 4q1 rupees. If q2 units areproduced for customer-2, then customer-2 is willing to pay 150 15q2 rupees. For q > 0, the cost of manufacturing q units is
100 - 15q rupees. To maximize profit, how much should themonopolist sell to each customer?
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Solution
f(q1, q2)= q1 (70 4q1) + q2 (100 - 15q2) 100 15q1 - 15q2
-8 0
H(q1, q2 )=
0 -30
Objective function is concave and hence (55/8, 9/2) maximize
it.
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Exercise-2 A DVD costs Rs. 55 to produce. We are considering
charging a price of between Rs. 110 and Rs. 150 for thisDVD. For a price of Rs. 110, Rs. 130 and Rs. 150, themarketing department estimates the demand for the DVDin the three regions where the DVD would be sold.
What price would maximize profit?
Price in Rs. Demand (in Thousands)Region 1 Region 2 Region 3
110 35 32 24
130 32 27 17
150 22 16 9
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Solution
The demands curves for Region 1, Region-2, Region-3
can be estimated using nonlinear regression
The curves for Regions 2 and 3 look very much similar
Region-1
0
5
10
15
20
2530
35
40
0 50 100 150 200
Price
Demand
Demand
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Region-1: Demand1= -0.87 (price)2+1.95*(price)-73.625
Region-2: Demand2= -0.75 (price)2+1.55*(price)-47.75
Region-3: Demand3= -0.125 (price)2+0.005*(price)-44.625
Total demand (TD) = Demand1+Demand2+Demand3
Profit = (price-55)*TD
Profit is a Non-linear function of one variable. The functionis concave and the optimal price is Rs. 129
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Lagrange multipliers
Lagrange multipliers can be used to solve NLPs
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Lagrange multipliers-Examples
A company is planning to spend $10,000 on advertising. It
costs $ 3,000 per minute to advertise on television and $ 1,000per minute to advertise on radio. If the firm buys x minutes oftelevision advertising and y minutes of radio advertising, thenits revenue in thousands of dollars is given by
f(x, y) = -2x2 y2 + xy + 8x + 3y
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Kuhn-Tucker Conditions
KT Conditions, sometimes known as Karush-Kuhn-Tucker
(KKT) conditions Gives necessary and sufficient conditions for
to be an optimal solution for NLP
),...,,( 21 nxxxx
mi
bxxxg
xxxf
ini
n
,...,2,1
),...,,(S.t.
),...,,(min)(ormax
21
21
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Kuhn-Tucker Conditions [contd.]
For maximization problems. If
is an optimal solution to the problem, then
must satisfy the mconstraints of the NLP, and there must existmultipliers satisfying
)
)
)m21i0
m21i0bxg
n21j0x
xg
x
xf
iii
mi
1i j
ii
j
,...,,;
,...,,;)(
,...,,;)()(
l
l
l
),...,,( 21 nxxxx
),...,,( 21 nxxxx
mlll ,...,, 21
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Kuhn-Tucker Conditions [contd.]
For minimization problems. If
is an optimal solution to the problem, then
must satisfy the mconstraints of the NLP, and there must existmultipliers satisfying
)
)
)m21i0
m21i0bxg
n21j0x
xg
x
xf
iii
mi
1i j
ii
j
,...,,;
,...,,;)(
,...,,;)()(
l
l
l+
),...,,( 21 nxxxx
),...,,( 21 nxxxx
mlll ,...,, 21
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Sufficiency of the KKT conditions:
Sense of
optimization
maximization
minimization
Required conditions
Objective
function
Solution
space
concave
convex
Convex set
Convex set
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It is simpler to verify whether a function is
concave or convex than to prove that thesolution space is a convex set.
We thus give a set of sufficient conditions
that are easier to check that the solution
space is a convex set in terms of the
convexity or concavity of the constraint
functions.
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Consider the general non-linear problem:
Maximize or minimize z =f(X)
Subject to gi(X) 0 i = 1,2,..,p
gi(X) 0 i =p+1,p+2,.., q
gi(X) = 0 i = q+1, q+2,.., r
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Sufficiency of the KKT conditions:
Sense ofoptimization
maximization
minimization
Required conditions
0
0
URS
convex
concave
linear
convex
1 ip
p+1 i q
q+1 i r
f(X) gi(X) i
0
0
URS
convex
concave
linear
1 ip
p+1 i q
q+1 i r
concave
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The conditions in the above table
represent only a subsetof the conditionsgiven in the earlier table.
The reason is that a solution space may be
convex without satisfying the conditions
of the constraint functions given in the
second table.
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Problem Use the KKT conditions to derive
an optimal solution for the following
problem:
maximize f(x1, x2) = x1+ 2x2x23
subject to x1 + x21
x1 0
x20
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Solution: Here there are three constraints
namely,
g1(x1,x2) = x1+x2- 10
g2(x1,x2) = - x1 0
g3(x1,x2) = - x2 0
Hence the KKT conditions become
01
3
3
1
2
2
1
1
1
1
x
g
x
g
x
g
x
flll
10,20, 30
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1g1(x1,x2) = 0
2g2
(x1
,x2
) = 0
3g3(x1,x2) = 0
g1(x
1,x
2)0
g2(x1,x2)0
g3(x
1,x
2)
0
Note:fis concave
gi are convex,
maximizationproblem
these KKT
conditions are
sufficient at the
optimum point
02
3
3
2
2
2
2
1
1
2
x
g
x
g
x
g
x
flll
i 1 + 0 (1)
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i.e. 11 +2 = 0 (1)
23x221 +3 = 0 (2)
1(x1 + x21) = 0 (3)
2x1 = 0 (4)
3x2 = 0 (5)
x1 + x21 0 (6)
x1 0 (7)
x2 0 (8) 1 0 (9)
2 0 (10) and 3 0 (11)
(1) gives 1 + 1 >0 (using 10)
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(1) gives1 = 1 +2 1 >0 (using 10)
Hence (3)givesx1 + x2 = 1 (12)
Thus both x1, x2cannot be zero.
So letx1>0 (4)gives2 = 0.therefore1 = 1
if nowx2 = 0, then (2) gives 201 +3= 0or3 < 0
not possible
Therefore x2 > 0
hence (5) gives3 = 0 and then (2) givesx22 = 1/3so x2 =1/3
And sox1 = 1- 1/3
Max f = 1 - 1/3 + 2/3 1/33 =1 + (2 / 33 )
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Maximizef(x) = 20x1 + 10 x2
Subject tox12 + x2
21
x1 + 2x22
x10, x20
KKT conditions become
20 - 21x12+3= 0
10 - 21x222 +4= 0
1 (x12+ x2
21) =0
1 (x1+ 2x22) =0
3x1 =0
(0,1)
(4/5,3/5)
x 0
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4x2 =0
x12 + x2
2 1
x1 + 2x2 2
x1 0
x2 0
1 0
2 0
3 0
4 0
F h fi i i l h f ( ) h
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From the figure it is clear that max f occurs at (x1, x2)where
x1, x2>0.
3 = 0,4 = 0supposex1 + 2x22 0
2 = 0 ,therefore we get 20 - 21x1=0
10 - 21x2=0
1x1=10 and1x2=5, squaring and adding we get
12
= 125 1 = 55therefore x1 = 2/5, x2 = 1/5, f= 50/5 >22
0 x + x 2 = 0
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2 0 x1 + x22 = 0
Thereforex1=0, x2=1, f =10
Or x1= 4/5, x2= 3/5, f=22
Therefore max f occurs at x1 = 2/5, x2 = 1/5
P bl U h KKT di i d i
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Problem Use the KKT conditions to derive
an optimal solution for the following
problem:
minimizef(x1, x2) = x12+ x2
subject to x12
+ x22
9x1+x2 1
Solution:Here there are two constraints,
namely, g1(x1,x2) = x12+x2
2- 9 0
g2(x1,x2) = x1 +x2 -1 0
Th h KKT di i
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0,0:1 21 ll
as it is a minimization problem
0)1(
0)9(:3
212
2
2
2
11
+
+
xx
xx
l
l
1
9:4
21
2
2
2
1
+
+
xx
xx
1 1 1 2
1 2 2
2: 2 2 0
1 -2 0
x x
x
l l
l l
Thus the KKT conditions are:
0l 1l
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Now (from 2) gives01 l 12 l Not possible.
Hence 01 l and so 92
2
2
1+ xx
Assume . So (1st equation of ) (2) gives02 l
0)1(2 11 lx Since we getx1= 001 l
(5)
From (5), we get 32 x
2nd equation of (2) says (with )x2 = -3,01
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Problem-1
A monopolist can purchase up to 17.25 oz of a chemical for
$10/oz. At a cost of $3/oz, the chemical can be processedinto an ounce of product-1; or, at a cost of $5/oz, thechemical can be processed into an ounce of product-2. If x1oz of product-1 are produced, it sells for a price of $(30 x1)
per ounce. If x2 oz of product-2 are produced, it sells for aprice of $(50 2x2) per ounce. Determine how themonopolist can maximize profits.
P bl 2
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Problem-2
A power company faces demands during both peak and off-
peak times. If a price of p1 dollars per KWH is chargedduring the peak time customers will demand (60 0.5p1)KWH of power. If a price of p2 dollars per KWH ischarged during the off-peak time customers will demand (40
p2)KWH of power. The power company must havesufficient capacity to meet demand during both the peak andoff-peak time.
It costs $10 per day to maintain each KWH of capacity.Determine how the power company can maximize dailyrevenues less operating costs.