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Min Soo KIM
Department of Mechanical and Aerospace EngineeringSeoul National University
Optimal Design of Energy Systems (M2794.003400)
Chapter 8. Lagrange Multipliers
Chapter 8. Lagrange Multipliers
8.1 Calculus Methods of optimization
- Differentiable function
- Equality constraints
Using calculus Using the other methods
2/18
- Inequaility constraints
- Not continuous function
Chapter 8. Lagrange Multipliers
8.2 Lagrange Multiplier equations
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Function : π¦ = π¦ π₯1, π₯2, β― , π₯π
Constraints : π1 π₯1, π₯2, β― , π₯π = 0
π2 π₯1, π₯2, β― , π₯π = 0
ππ π₯1, π₯2, β― , π₯π = 0
βββββββββββββββββββββββ
How to optimize the function subject to the constraints?
Chapter 8. Lagrange Multipliers
8.2 Lagrange Multiplier equations
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For an optimized point function f(x,y) and g(x,y) are parallel, which meansgradient of f(x,y) is tangent to that of g(x,y)
Thus, for a certain constant Ξ», equation below is established
For multiple m constraints
π»π π₯, π¦ = ππ»π π₯, π¦
π»π =
π=1
π
πππ»ππ = 0
Chapter 8. Lagrange Multipliers
8.3 The gradient vector
Gradient of scalar
π»π¦ =ππ¦
ππ₯1 π1 +
ππ¦
ππ₯2 π2 +β―+
ππ¦
ππ₯π ππ
π» = ππππππππ‘ π£πππ‘ππ π = π’πππ‘ π£πππ‘ππ βΆ βππ£π ππππππ‘πππ, π’πππ‘ ππππππ‘π’ππ
( π1, π2 , π3 πππ π‘βπ π’πππ‘ π£πππ‘ππ ππ π‘βπ π₯1, π₯2, π₯3 ππππππ‘πππ)
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* *
1 1,m nx x
at optimum point
unknowns
m
n
m n
m nif fixed values of xβs(no optimization)
Chapter 8. Lagrange Multipliers
8.4 Further explanation of Lagrange multiplier equations
n scalar equations
π1:ππ¦
ππ₯1β π1
ππ1
ππ₯1β ππ
πππ
ππ₯1= 0
βββββββββββββββββββββββββββββββββββββββ
ππ:ππ¦
ππ₯πβ π1
ππ1
ππ₯πβ ππ
πππ
ππ₯π= 0
+ m constraint equations m+n simultaneous equations
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optimum point can be found
8.5 Unconstrained optimization
Chapter 8. Lagrange Multipliers
π¦ = π¦ π₯1, π₯2,β― , π₯π
β π»π¦ = 0 no Οβ²π ππππ¦
ππ₯1=
ππ¦
ππ₯2= β― =
ππ¦
ππ₯π= 0
The state point where the derivatives are zero = critical point
βmay be max, min, saddle point, ridge, valley
7/18
From (1),
Example 8.1 :
Chapter 8. Lagrange Multipliers
Find minimum value of z for a given constraint
(Solution)
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π§ = π₯2 + π¦2 1 = π₯2 + π₯π¦ + π¦2
π»π π₯, π¦ = ππ»π π₯, π¦
2π₯ β π 2π₯ + π¦ = 0
π π₯, π¦ β 1 = 0
Β·Β·Β·Β·Β·(1)
Β·Β·Β·Β·Β·(2)
2π¦ β π 2π¦ + π₯ = 0π¦ = π₯ ππ π¦ = βπ₯
Substituting (2) from the result
1 = 3π₯2 or 1 = π₯2
Thus, minimum value z is at π₯, π¦ =1
3,Β±
1
3
2
3
Chapter 8. Lagrange Multipliers
Example 8.2,3 : Constrained/Unconstrained optimization
A total length of 100 m of tubes must be installed in a shell-and-tube heat
exchanger.
πππ‘ππ πππ π‘ = 900 + 1100π·2.5πΏ + 320π·πΏ (π)
where L is the length of the heat exchanger and D is the diameter of the shell,
both in meters. 200 tubes will fit in a cross-sectional area of 1 m2 in the shell.
Determine the D and L of the heat exchanger for minimum first cost.
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Chapter 8. Lagrange Multipliers
(Solution) β Unconstrained Optimization
Have to put 100m tubes in the shell
: ππ·2
4m2 200 tubes/m2 β πΏ m = 100 m
β 50ππ·2πΏ = 100 π
Substitute (b) into (a)
β πππ‘ππ πππ π‘ = 900 +2200
ππ·0.5 +
640
ππ·
π(πππ‘ππ πππ π‘)
π(π·)=1100
ππ·0.5β640
ππ·2
π·β = 0.7 m, πΏβ = 1.3 m, πππ‘ππ πππ π‘β = $1777.45
10/18
Chapter 8. Lagrange Multipliers
(Solution) β Constrained Optimization
πππ‘ππ πππ π‘ = 900 + 1100π·2.5πΏ + 320π·πΏ (π)
50ππ·2πΏ = 100 π
π»y = 2.5 1100 π·1.5πΏ + 320πΏ π1 + 1100π·2.5 + 320π· π2
π»π = 100ππ·πΏ π1 + 50ππ·2 π2
π1: 2750π·1.5πΏ + 320πΏ β π100ππ·πΏ = 0 (π)
π2: 1100π·2.5 + 320π· β π50ππ·2 = 0 π
Using (b), (c), (d), find π·β, πΏβ, π
Result is the same as unconstrained optimization
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2x
1x
3y y
2y y
1y y
1dx
2dx
Chapter 8. Lagrange Multipliers
8.7 Gradient vector
- gradient vector is normal to contour line
π¦ = π¦ π₯1, π₯2
ππ¦ =ππ¦
ππ₯1ππ₯1 +
ππ¦
ππ₯2ππ₯2
Along the line of π¦ = ππππ π‘.
ππ¦ = 0 β ππ₯1 = βππ₯2ππ¦/ππ₯2ππ¦/ππ₯1
Substitution into arbitrary unit vector
ππ₯1 π + ππ₯2 π
(ππ₯1)2 + (ππ₯2)
2
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Tangent vector
Gradient vector
0T G
Chapter 8. Lagrange Multipliers
8.7 Gradient vector
π =ππ₯1 π + ππ₯2 π
(ππ₯1)2 + (ππ₯2)
2=
π2 βππ¦/ππ₯2ππ¦/ππ₯1
π1
ππ¦/ππ₯2ππ¦/ππ₯1
2
+ 1
=β
ππ¦ππ₯2
π1 +ππ¦ππ₯1
π2
ππ¦ππ₯2
2
+ππ¦ππ₯1
2
πΊ =π»π¦
π»π¦=
ππ¦/ππ₯1 π1 + ππ¦/ππ₯2 π2
ππ¦/ππ₯12 + ππ¦/ππ₯2
2
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saddle pointridge & valley point
ridge
valley
Chapter 8. Lagrange Multipliers
8.9 Test for maximum or minimum
decide whether the point is a maximum, minimum, saddle point, ridge, or valley
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minimum
maximum
a1x2x
a
1x2x
Chapter 8. Lagrange Multipliers
8.9 Test for maximum or minimum
π₯ = π βΆ π‘βπ ππππππ’π ππ ππ₯ππππ‘ππ π‘π ππππ’π
π¦ π₯ = π¦ π +ππ¦
ππ₯π₯ β π +
1
2
π2π¦
ππ₯2(π₯ β π)2 +β― Taylor series expansion near π₯ = π
If ππ¦
ππ₯> 0 ππ
ππ¦
ππ₯< 0 π‘βπππ ππ ππ ππππππ’π
βππ¦
ππ₯= 0
Ifπ2π¦
ππ₯2> 0
Ifπ2π¦
ππ₯2< 0
π¦ π₯1 > π¦ π π€βππ π₯1 > ππ¦ π₯2 > π¦ π π€βππ π₯2 < π
π¦ π₯1 < π¦ π π€βππ π₯1 > ππ¦ π₯2 < π¦ π π€βππ π₯2 < π
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min
11 12
21 22
2
11 22 12
y yD
y y
y y y
max
Chapter 8. Lagrange Multipliers
8.9 Test for maximum or minimum
π1, π2 : π‘βπ ππππππ’π ππ π¦ π₯1, π₯2 ππ ππ₯ππππ‘ππ π‘π ππππ’π
π¦ π₯1, π₯2 = π¦ π1, π2 +ππ¦
ππ₯1π₯1 β π1 +
ππ¦
ππ₯2π₯2 β π2 +
1
2π¦11β²β² (π₯1 β π1)
2 +
π¦12β²β² π₯1 β π1 π₯2 β π2 +
1
2π¦22β²β² (π₯2 β π2)
2 +β―
π· > 0 πππ π¦11β²β² > 0 (π¦22
β²β² > 0)
π· > 0 πππ π¦11β²β² < 0 (π¦22
β²β² < 0)
π· < 0 πππ π¦11β²β² < 0 (π¦22
β²β² < 0)
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Chapter 8. Lagrange Multipliers
Example 8.4 : Test for maximum or minimum
Optimal values of π₯1 and π₯2, test max, min
π¦ =π₯12
4+
2
π₯1π₯2+ 4π₯2
(Solution)
ππ¦
ππ₯1=π₯12β
2
π₯12π₯2
ππ¦
ππ₯2= β
2
π₯1π₯22 + 4
π₯1β = 2, π₯2
β =1
2
π2π¦
ππ₯12 =
3
2,π2π¦
ππ₯22 = 16,
π2π¦
ππ₯1ππ₯2= 2
At π₯1 and π₯2
β3
22
2 16> 0 πππ π¦11
β²β² > 0 minimum
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In example 8.3, if the total length of the tube increases from 100m torandom value βHβ, what would be the increase in minimum cost ?
Extra meter of tube for the heat exchanger would cost additional $8.78
Chapter 8. Lagrange Multipliers
8.10 Sensitivity coefficients
- represents the effect on the optimal value of slightlyrelaxing the constraints
- variation of optimized objective function π¦β
50ππ·2πΏ = π» β π·β = 0.7π, πΏβ = 0.013π»
πππ‘ππ πππ π‘β = 900 + 1100π·β2.5πΏβ + 320π·βπΏβ = 900 + 8.78π»
ππΆ =π(πππ‘ππ πππ π‘β)
ππ»= 8.78 = π
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