Mathematicalanalysisofthehomogeneous ... · Modelling a dike ring Ht = dike height at time t, Pt = exceedance probability at time t, α = parameter exponential distribution for extreme

Mathematical analysis of the homogeneousdike height optim ization problem

Kees Roose-mail: [email protected]

URL: http://www.st.ewi.tudelft.nl/∼roos

Co-authors: Dick den Hertog, Guoyong Gu

Workshop Dike Height Optimization

Tilburg University

April 16, A.D. 2009

Optimization Group 1/29

Outline

• Dike rings in the Netherlands

• Modelling a dike ring

• Cost functions for investments and expected damage

• Stationarity with respect to τ = (t0; t1; t2 . . .)

• Stationarity with respect to u = (u0; u1; u2; . . .)

• Value of the objective function at the stationary point

• Finding t0, t1 and u0

• Solutions for eleven dike rings

• Graphs of Pt for these dike rings

• Graphs of St for these dike rings

• Dynamic Programming approach

• Concluding remarks and questions

• Some references

white


Dike rings in the Netherlands


Background

• About 70% of the Netherlands is submersible

• Protection against large-scale floods is necessary

• In total 3500 km primary dikes (lashers, embankments)

Main questions:

• What are economically efficient safety standards?

• What are the costs related to these safety standards?


Modelling a dike ring

Ht = dike height at time t,

Pt = exceedance probability at time t,

α = parameter exponential distribution for extreme water levels (1/cm),η = structural increase of the water level (cm/year),

Vt = loss by flooding at time t (million euros),γ = rate of growth of wealth in dike ring (per year),ζ = increase of loss per cm dike heightening (1/cm),

δ1 = δ2 + ρ (per year)δ2 = risk-free discount rate (per year),ρ = macro-economic risk premium.

The flood probability Pt and, if a flood occurs, the loss Vt at time t are given by

Pt = P0eαηte−α(Ht−H0), Vt = V0eγteζ(Ht−H0).

The expected loss at time t is therefore given by

St = PtVt = P0eαηte−α(Ht−H0) · V0eγteζ(Ht−H0) = S0eβte−θ(Ht−H0),

where S0 = P0V0, β = αη + γ, θ = α − ζ > 0.

The parameter β represents the growth rate of the damage costs; as the discount rates δ1and δ2, it is given per year.


Investments cost function

The dike height is a step function that increases at moments t0 ≥ 0, t1, t2, . . . in order to

guarantee a safety-standard. We denote the increment of the height at time tk as uk. The

related investments costs depend on uk and the previous dike updates, and are denoted as

Ik(uk), where uk denotes the vector (u0;u1; . . . ;uk). The function Ik(u

k) has the form

Ik(uk) = (δ0,uk

C + buk) eλhk, k = 0, 1, . . . , hk =k∑

i=0

ui,

where C, b and λ are positive constants. Moreover, δ0,uk= 1 − δ0,uk

, where δ0,ukde-

notes the Kronecker delta; so δ0,ukequals 0 if uk = 0 and 1 otherwise. Then the total

(compounded) investments costs are given by

I(u, τ) :=∞∑

k=0

(δ0,ukC + buk) eλhk−δ2tk, (1)

where

τ = (t0; t1; t2 . . .) , u = (u0;u1;u2; . . .) .

In the sequel we omit the ‘inverse’ Kronecker delta δ0,ukif the context requires that uk > 0.


Expected damage costs

The (discounted) expected damage costs are given by

E(u, τ) =

∫ ∞

0Ste

−δ1tdt = S0

∫ ∞

0eβte−θ(Ht−H0)e−δ1tdt = S0

∫ ∞

0eβ1te−θhtdt,

where

β1 = β − δ1 = αη + γ − δ1, ht = Ht − H0.

Due to the definition of u we have

ht =k∑

i=0

ui =: hk, t ∈ [tk, tk+1), k = 0, 1, . . . . (2)

Since dike updates occur at the moments tk, E(u, τ) can be reduced as follows:

E(u, τ) = S0

∫ t0

0eβ1t−θhtdt + S0

∞∑

k=0

∫ tk+1

tkeβ1t−θhtdt

= S0

∫ t0

0eβ1tdt + S0

∞∑

k=0

e−θhk

∫ tk+1

tkeβ1tdt

=S0

β1

[

eβ1t0 − 1]

+S0

β1

∞∑

k=0

[

eβ1tk+1 − eβ1tk]

e−θhk.


Total costs

The total costs are now given by f(u, τ) := I(u, τ) + E(u, τ):

f(u, τ) =S0

β1

[

eβ1t0 − 1]

+∞∑

k=0

[

(C + buk)eλhk−δ2tk +

S0

β1

[

eβ1tk+1 − eβ1tk]

e−θhk

]

Our aim is to find τ = (t0; t1; t2 . . .) and u = (u0;u1;u2; . . .) that minimize f(u, τ)

subject to the conditions

t0 ≥ 0, πk := tk+1 − tk ≥ 0, uk ≥ 0, k ≥ 0.

We could prove that if β1 + δ2 > 0 and θδ2 > λβ1 then

• there exists a stationary point (u, τ) with

u0 ≥ ν, uk = ν > 0, k = 1, 2, . . . ,

0 ≤ t0 ≤ p, πk = p > 0, k = 1, 2, . . . ,

where ν and p are positive numbers such that p = λ+θβ1+δ2

ν.

• f(u, τ) is strictly convex at this point.

It is not yet clear, however, if this stationary point is a global minimizer.

Note that β1 + δ2 > 0 ⇔ β − δ1 + δ2 > 0 ⇔ β − ρ > 0.


Graphical illustration of the solution

t

ht

0 t0 t1 t2 t3

u0

u1

u2

u3

p

p

p

ν

ν

ν

t

ht

t0 t1 t2 t3 t4

u0

u1

u2

u3

≤ p

p

p

p

≥ ν

ν

ν

ν

Case I: Healthy dike: solution is periodic Case II: Immediate update required


Stationarity with respect to t, I

f(u, τ) =S0

β1

[eβ1t0 − 1

]+

∞∑

k=0

[


S0

β1

[eβ1tk+1 − eβ1tk

]e−θhk

]

Some straightforward computations yield that

∂f(u, τ)

∂tj= −δ2(C + buj)e

λhj−δ2tj + S0eβ1tj−θhj(

eθuj − 1)

, j ≥ 0.

Hence f(u, τ) is stationary with respect to τ if and only if

S0eβ1tj−θhj(

eθuj − 1)

= δ2(C + buj)eλhj−δ2tj , j ≥ 0.

This can be written as

e(β1+δ2)tj−(λ+θ)hjS0

(

eθuj − 1)

= δ2(C + buj).

By taking logarithms at both sides we find that

∂f(u, τ)

∂tj= 0 ⇔ (β1 + δ2)tj − (λ + θ)hj = (uj),

where

(x) = lnδ2(C + b x)

S0

(

eθx − 1), x > 0.


Stationarity with respect to t, II

(∗) (β1 + δ2)tj − (λ + θ)hj = (uj), j = 0, 1, . . . .

Subtracting the relation for j ≥ 0 from the relation for j + 1 we obtain

(β1 + δ2)(tj+1 − tj) − (λ + θ)(hj+1 − hj︸︷︷︸

uj+1

) = (uj+1) − (uj), j = 0, 1, . . . .

Obviously, (∗) holds if there exist ν > 0 such that uj = ν for all j ≥ 0. Then hj+1−hj =

ν and we obtain a solution if tj+1 − tj = p, provided that

(β1 + δ2) p = (λ + θ) ν, (β1 + δ2)t0 − (λ + θ)h0 = (ν).

It will become clear that for suitable values of ν, t0 and h0 = u0 = ν this yields a stationarypoint with respect to all variables tj and uj. However, if (ν) < 0 then it may happen thatt0 is negative. Since t = 0 represents the present, this would correspond to a dike update inthe past. Therefore we assume that uj = ν only holds for j ≥ 1. As a consequence we have

hj = h0 + jν (j ≥ 0), tj = t1 + (j − 1)p (j ≥ 1).

So, after t = t1 this solution is periodic, and the period between two successive updates isp. From a physical point of view it is necessary that p is positive. This holds only if

β1 + δ2 > 0,

because λ + θ > 0. In the sequel we assume that this condition is satisfied.


Stationarity with respect to u, I

f(u, τ) =S0

β1

[eβ1t0 − 1

]+

∞∑

k=0

[


S0

β1

(eβ1tk+1 − eβ1tk

)e−θhk

]

After some straightforward computations one gets, for j ≥ 0,

∂f(u, τ)

∂uj= beλhj−δ2tj+

∞∑

k=j

[

λ(C + buk)eλhk−δ2tk −

θS0

β1

(

eβ1tk+1 − eβ1tk)

e−θhk

]

.

Due to stationarity with respect to τ we have

S0eβ1tj−θhj(

eθuj − 1)

= δ2(C + buj)eλhj−δ2tj , j ≥ 0.

Defining pk = tk+1 − tk, for k ≥ 0, we obtain

∂f(u,τ)∂uj

= beλhj−δ2tj + S0

∞∑

k=j

[

λ

δ2

(

eθuk − 1)

−θ

β1

(

eβ1pk − 1)]

eβ1tk−θhk, j ≥ 0.

If k ≥ 1 then we have uk = ν and pk = p, whence it follows that

∂f(u,τ)∂uj


[

λ

δ2

(

eθν − 1)

−θ

β1

(

eβ1p − 1)]

∞∑

k=j



Stationarity with respect to u, II

∂f(u,τ)∂uj


[λ

δ2

(eθν − 1

)−

θ

β1

(eβ1p − 1

)] ∞∑

k=j


Since hk = u0 + kν (k ≥ 0) and tk = t1 + (k − 1)p (k ≥ 1) we have for k ≥ 1:

β1tk − θhk = β1(t1 +(k − 1)p)− θ(u0 + kν) = β1(t1 − p)− θu0 + k(β1p − θν).

For each j ≥ 1 we have uj = ν. Hence

∞∑

k=j

eβ1tk−θhk = eβ1(t1−p)−θu0

∞∑

k=j

ek(β1p−θν).

The last sum is finite if and only if β1p − θν < 0, which holds if and only if θδ2 > λβ1,

which we assumed to hold. Since (β1 + δ2) p = (λ + θ) ν, it follows that

∞∑

k=j

ek(β1p−θν) =ej(β1p−θν)

1 − eβ1p−θν=

ej(λν−δ2p)

1 − eλν−δ2p,

Substitution gives, for j ≥ 1,

∂f(u,τ)∂uj

= beλhj−δ2tj + S0eβ1(t1−p)−θu0

[λ

δ2

(eθν − 1

)−

θ

β1

(eβ1p − 1

)]

ej(λν−δ2p)

1 − eλν−δ2p.


Stationarity with respect to u, III

∂f(u,τ)∂uj

= beλhj−δ2tj + S0eβ1(t1−p)−θu0

[λ

δ2

(eθν − 1

)−

θ

β1

(eβ1p − 1

)]

ej(λν−δ2p)

1 − eλν−δ2p.

We also have, for j ≥ 0,

λhj − δ2tj = λ(u0 + jν) − δ2(t1 + (j − 1)p) = λu0 − δ2(t1 − p) + j(λν − δ2p).

Thus we obtain that, for j ≥ 1,

∂f(u, τ)

∂uj

=

{

beλu0−δ2(t1−p) + S0eβ1(t1−p)−θu0

[λ

δ2

(eθν − 1

)−

θ

β1

(eβ1p − 1

)]

1

1 − eλν−δ2p

}

ej(λν−δ2p).

We conclude that we have stationarity with respect tot uj (j ≥ 1) if and only if

beλu0−δ2(t1−p)+S0eβ1(t1−p)−θu0

[

λ

δ2

(

eθν − 1)

−θ

β1

(

eβ1p − 1)]

1

1 − eλν−δ2p= 0.

Using (β1 + δ2) p = (λ + θ) ν again, we may write

λu0 − δ2(t1 − p) − [β1(t1 − p) − θu0] = (λ + θ)h0 − (β1 + δ2)(t1 − p)

= (λ + θ)(h0 + ν) − (β1 + δ2)t1

= (λ + θ)h1 − (β1 + δ2)t1 = −(ν);

the last equality is due to stationarity with respect to t1 and u1 = ν.


Stationarity with respect to u, IV

beλu0−δ2(t1−p) + S0eβ1(t1−p)−θu0

[λ

δ2

(eθν − 1

)−

θ

β1

(eβ1p − 1

)]

1

1 − eλν−δ2p= 0.

Dividing the above equation by eβ1t1−θu0−β1p we obtain

be−(ν) + S0

[

λ

δ2

(

eθν − 1)

−θ

β1

(

eβ1p − 1)]

1

1 − eλν−δ2p= 0.

Note that p depends on ν according to p = λ+θβ1+δ2

ν. This means that the last equationessentially is an equation in the single variable ν. It can be shown that this equation has apositive solution. (It is conjectured that this solution is unique.)

Given ν, we have shown that the point (u, τ) withτ = (t0; t1; t1 + p; t1 + 2p; . . .) , (u0; ν; ν; ν; . . .)

is stationary with respect to each uk and each tk for k ≥ 1. It also easily follows that whentaking u0 = ν then we have stationarity with respect to t0 and u0 if and only if

(β1 + δ2)t0 − (λ + θ)ν = (ν).

Only if (λ + θ)ν + (ν) ≥ 0 this gives t0 ≥ 0, yielding a feasible stationary point.Otherwise, the derivative of f(u, τ) with respect to t0 at t = 0 is positive. This means thatan immediate update of the dike is required. Then t0 = 0, u0 ≥ ν is optimal. This is thehard case in the analysis. It needs further investigation.


Value of the objective function at the stationary point if (λ + θ)ν + (ν) < 0, I

f(u, τ) =S0

β1

[eβ1t0 − 1

]+

∞∑

k=0

[


S0

β1

(eβ1tk+1 − eβ1tk

)e−θhk

]

We now have t0 = 0. We express f(u, τ) at the stationary point in u0, t1 and ν. One has

f(u, τ) =∞∑

k=0

[


S0

β1

(


e−θhk

]

= (C + bu0)eλu0 +

S0

β1

(

eβ1t1 − 1)

e−θu0

+∞∑

k=1

[


S0

β1

(


e−θhk

]

Denoting the last sum as Σ, using stationarity with respect to tj, for j ≥ 1, we get

Σ = S0

∞∑

k=1

[

1

δ2eβ1tk−θhk

(

eθuk − 1)

+1

β1

(


e−θhk

]

= S0

∞∑

k=1

[

eθν − 1

δ2eβ1tk−θhk +

eβ1p − 1

β1eβ1tk−θhk

]

= S0

[

eθν − 1

δ2+

eβ1p − 1

β1

]∞∑

k=1

eβ1tk−θhk.


Value of the objective function at the stationary point if (λ + θ)ν + (ν) < 0, II

The last sum has been computed before, namely:

∞∑

k=j

eβ1tk−θhk = eβ1(t1−p)−θu0

∞∑

k=j

ek(β1p−θν), j ≥ 1.

Hence we get

∞∑

k=1

eβ1tk−θhk = eβ1(t1−p)−θu0eβ1p−θν

1 − eβ1p−θν=

eβ1t1−θ(u0+ν)

1 − eλν−δ2p.

Thus we obtain

f(u, τ) = (C + bu0)eλu0 +

S0

β1

(

eβ1t1 − 1)

e−θu0

+S0

[

eθν − 1

δ2+

eβ1p − 1

β1

]

eβ1t1−θ(u0+ν)

1 − eβ1p−θν

= (C + bu0)eλu0 −

S0

β1e−θu0

+S0

{

1

β1+

[

eθν − 1

δ2+

eβ1p − 1

β1

]

e−θν

1 − eβ1p−θν

}

eβ1t1−θu0

The expression between accolades depends only on ν. We denote it as g(ν).


Value of the objective function at the stationary point if (λ + θ)ν + (ν) < 0, III

g(ν) = 1β1

+

[

eθν−1δ2

+ eβ1p−1β1

]

e−θν

1−eβ1p−θν

= 11−eβ1p−θν

(

1−eβ1p−θν

β1+

[

1−e−θν

δ2+ eβ1p−θν−e−θν

β1

])

= 11−eβ1p−θν

(

1β1

+

[

1−e−θν

δ2− e−θν

β1

])

= 1−e−θν

1−eβ1p−θν

(1β1

+ 1δ2

)

= β1+δ2β1δ2

1−e−θν

1−eλν−δ2p,

where we used again that (β1 + δ2) p = (λ + θ) ν. Thus we obtain the following expres-

sion for f(u, τ):

f(u, τ) = (C + bu0)eλu0 −

S0

β1e−θu0 + S0g(ν) eβ1t1−θu0.

Due to stationarity with respect to t1 we have

(β1 + δ2) t1 − (λ + θ)(u0 + ν) = (ν).

Hence, we can eliminate t1 from this expression, making clear that f(u, τ) can be considered

as a function of the single variable u0. We denote this function as F(u0).


Finding t0, t1 and u0, I

We did numerical experiments with eleven dike rings. The values of ν and (ν)+(λ+ θ)ν

for the eleven dike rings are as follows.

Dike ring ν (ν) + (λ + θ)ν

10 56.96 +1.40

11 62.42 +1.28

15 53.29 −0.17

16 52.59 +0.22

22 53.70 +0.81

45 41.49 −0.92

47 55.60 −0.31

48 50.84 −0.25

49 45.65 +0.97

51 40.49 +1.09

52 45.74 +0.11

In four cases we have (ν) + (λ + θ)ν < 0. The graphs of F(u0) for these dike rings areshown on the next sheet.


Finding t0, t1 and u0, II

The figure below shows the graphs of F(u0) for each of the four dike rings for which (ν)+

(λ + θ)ν < 0.

0 20 40 60 80 100 120 140600

700

800

900

1000

1100

1200

0 20 40 60 80 100 120 140−30

−20

−10

0

10

20

30

40

50

60

70

0 20 40 60 80 100 120 14045

50

55

60

65

70

75

80

85

90

0 20 40 60 80 100 120 140350

400

450

500

550

600

650

dike ring 15 dike ring 45 dike ring 47 dike ring 48

The red part of the each graph indicates the region where t1 < 0. The minimal value occurs

always in the region where F(u0) is convex. In fact, it can be shown that F(u0) is concave

for small values of u0, and for the larger values F(u0) is convex. This makes it easy to find

the minimizing value. The minimizing value of u0 is larger than ν, in each case. So, if an

immediate update is necessary, then u0 ≥ ν.


Solutions for the eleven dike rings, with δ1 = δ2 = 0.04

No. θδ2 − λβ1 β1 + δ2 ν p u0 t0 t1 Inv. Dam. Total

10 0.0012 0.031 56.96 57.12 56.96 45.80 102.92 10.20 29.84 40.04

11 0.0011 0.030 62.42 58.89 62.42 42.44 101.33 30.18 80.05 110.23

15 0.0017 0.058 53.29 51.54 55.96 0.00 51.20 415.50 129.68 545.18

16 0.0020 0.064 52.59 54.04 52.59 3.50 57.54 797.65 292.03 1089.68

22 0.0025 0.063 53.70 62.43 53.70 12.72 75.16 198.53 110.72 309.25

45 0.0013 0.031 41.49 50.94 61.80 0.00 45.69 21.91 11.81 33.72

47 0.0011 0.030 55.60 51.87 65.26 0.00 50.74 43.53 20.57 64.10

48 0.0009 0.031 50.84 42.47 58.24 0.00 40.57 227.88 175.12 403.00

49 0.0013 0.031 45.65 52.97 45.65 31.66 84.62 23.08 50.96 74.04

51 0.0013 0.031 40.49 51.49 40.49 35.48 86.97 15.27 38.91 54.18

52 0.0014 0.031 45.74 57.79 45.74 3.52 61.31 150.73 94.65 245.38


Graphs of Pt for the eleven solutions

0 50 100 150 200 250 3000

0.1372

0.2743

0.4115

0.5487

0.6859

0.823

0.9602

1.0974

1.2346

1.3717

1.5089

1.6461

1.7833x 10

−3

→ t

Pt

↑

0 50 100 150 200 250 3000

0.1634

0.3268

0.4902

0.6536

0.817

0.9804

1.1438

1.3072

1.4706

1.634

1.7974

1.9608

2.1242x 10

−4

→ t

Pt

↑

0 50 100 150 200 250 3000

0.1709

0.3419

0.5128

0.6838

0.8547

1.0256

1.1966

1.3675

1.5385

1.7094

1.8803

2.0513

2.2222x 10

−3

→ t

Pt

↑

0 38 76 114 152 190 228 2660

0.1709

0.3419

0.5128

0.6838

0.8547

1.0256

1.1966

1.3675

1.5385

1.7094

1.8803

2.0513

2.2222x 10

−3

→ t

Pt

↑


0 60 120 180 240 3000

0.7148

1.4296

2.1444

2.8592

3.574

4.2888

5.0036

5.7184

6.4332

7.148

7.8628

8.5777

9.2925x 10

−4

→ t

Pt

↑

0 60 120 180 240 3000

0.1806

0.3612

0.5419

0.7225

0.9031

1.0837

1.2643

1.4449

1.6256

1.8062

1.9868

2.1674

2.348x 10

−3

→ t

Pt

↑

0 50 100 150 200 250 3000

0.1286

0.2572

0.3858

0.5144

0.643

0.7717

0.9003

1.0289

1.1575

1.2861

1.4147

1.5433

1.6719x 10

−3

→ t

Pt

↑

0 60 120 180 240 3000

0.0964

0.1928

0.2892

0.3856

0.482

0.5784

0.6748

0.7712

0.8676

0.964

1.0604

1.1568

1.2532x 10

−3

→ t

Pt

↑


0 50 100 150 200 250 3000

0.2383

0.4766

0.715

0.9533

1.1916

1.4299

1.6682

1.9065

2.1449

2.3832

2.6215

2.8598

3.0981x 10

−3

→ t

Pt

↑

0 50 100 150 200 250 3000

0.2493

0.4986

0.7479

0.9972

1.2465

1.4958

1.7451

1.9944

2.2437

2.493

2.7423

2.9916

3.2409x 10

−3

→ t

Pt

↑

0 50 100 150 200 250 3000

0.1777

0.3554

0.5331

0.7107

0.8884

1.0661

1.2438

1.4215

1.5992

1.7769

1.9546

2.1322

2.3099x 10

−3

→ t

Pt

↑

dike ring 49 dike ring 51 dike ring 52


Graphs of St for the eleven solutions

0 50 100 150 200 250 3000

22.0606

44.1212

66.1818

88.2424

110.303

132.3637

154.4243

176.4849

198.5455

220.6061

242.6667

264.7273

286.7879

→ t

St

↑

0 50 100 150 200 250 3000

0.799

1.5979

2.3969

3.1959

3.9948

4.7938

5.5927

6.3917

7.1907

7.9896

8.7886

9.5876

10.3865

→ t

St

↑

0 50 100 150 200 250 3000

0.4821

0.9642

1.4463

1.9284

2.4106

2.8927

3.3748

3.8569

4.339

4.8211

5.3032

5.7853

6.2675

→ t

St

↑

0 38 76 114 152 190 228 2660

11.3399

22.6797

34.0196

45.3595

56.6993

68.0392

79.379

90.7189

102.0588

113.3986

124.7385

136.0784

147.4182

→ t

St

↑


0 60 120 180 240 3000

0.3846

0.7692

1.1538

1.5384

1.923

2.3076

2.6922

3.0768

3.4614

3.846

4.2306

4.6152

4.9998

→ t

St

↑

0 60 120 180 240 3000

0.7175

1.4351

2.1526

2.8701

3.5877

4.3052

5.0227

5.7403

6.4578

7.1753

7.8929

8.6104

9.3279

→ t

St

↑

0 50 100 150 200 250 3000

43.3492

86.6985

130.0477

173.397

216.7462

260.0955

303.4447

346.7939

390.1432

433.4924

476.8417

520.1909

563.5402

→ t

St

↑

0 60 120 180 240 3000

4.9478

9.8956

14.8434

19.7913

24.7391

29.6869

34.6347

39.5825

44.5303

49.4781

54.4259

59.3738

64.3216

→ t

St

↑


0 50 100 150 200 250 3000

1.056

2.112

3.1681

4.2241

5.2801

6.3361

7.3922

8.4482

9.5042

10.5602

11.6163

12.6723

13.7283

→ t

St

↑

0 50 100 150 200 250 3000

1.2172

2.4345

3.6517

4.8689

6.0862

7.3034

8.5206

9.7379

10.9551

12.1723

13.3896

14.6068

15.824

→ t

St

↑

0 50 100 150 200 250 3000

2.2486

4.4972

6.7458

8.9944

11.243

13.4916

15.7402

17.9888

20.2374

22.486

24.7346

26.9832

29.2318

→ t

St

↑

dike ring 49 dike ring 51 dike ring 52


Convexity at the stationary point

The Hessian matrix H of f(u, τ) with respect to (u, τ) has the form

H =

Hu,u Hu,τ

HTu,τ Hτ,τ

.

Some quite tedious computations reveal that at the stationary point we have

Hτ,τ = cτdiag(ξ), Hu,u − Hu,τH−1τ,τ HT

u,τ = cSchurdiag(ξ),

where cτ and cSchur are positive constants, and

ξ =

1

ς

ς2

ς3

...

ςℓ

...

, ς := eχ, χ := λν − δ2p = β1p − θν < 0.

Since diag(ξ) is a positive definite (diagonal) matrix, it follows from the Schur complementlemma that H is positive definite, which in turns implies that the stationary point is a localminimizer of f(u, τ).


Dynamic programming approach

1 2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

Each node represents a state (t, h) of the dike. The costs for moving from state (t, h1) to(t + 1, h2), with h2 ≥ h1, is given by

ct,h1,h2=

(

δh1h2C + b(h2 − h1)

)

eλh2−δ2(t+1) + E(t, t + 1, h1).

The (minimal) costs for reaching a state from the initial state are computed by using therecursive formula

Kt+1,h2= min

h2≥h1∈Ht

(

Kt,h1+ ct,h1,h2

)

, h2 ∈ Ht+1.

By adding a sink node to the network and an arc from each final state to this node, withcosts equal to the estimated costs after the planning horizon for that final state, the optimalsolution is obtained by finding a shortest path from the initial state to the virtual node.


Dynamic programming solutions for some dike rings, I

No. 10 11 15 16 22

Updates (tk : uk) 46 : 57.60 43 : 63.36 0 : 54.72 4 : 54.72 12 : 52.08

104 : 57.60 103 : 63.36 50 : 54.72 60 : 54.72 73 : 52.08

162 : 57.60 162 : 61.44 103 : 54.72 116 : 54.72 133 : 52.08

219 : 55.68 220 : 61.44 156 : 54.72 171 : 50.16 194 : 52.08

274 : 51.84 277 : 53.76 209 : 54.72 223 : 50.16 254 : 52.08

262 : 54.72 274 : 45.60

hT 280.32 303.36 328.32 310.08 260.4

Investment costs 10.16 29.70 413.39 796.31 202.09

Damage costs 29.87 80.54 131.95 294.13 107.33

Total costs 40.04 110.23 545.34 1090.44 309.41

OptimaliseRing hT 290 319 332 318 270

Investment costs 10.22 30.81 416.88 787.75 196.46

Damage costs 29.82 79.42 128.42 302.06 112.81

Total costs 40.04 110.23 545.30 1089.81 309.27


Dynamic programming solutions for some dike rings, II

No. 45 47 48 49 51 52

Updates (tk : uk) 0 : 61.44 0 : 64.44 0 : 56.54 32 : 45.60 36 : 40.57 4 : 45.60

46 : 42.24 50 : 55.85 39 : 50.59 85 : 45.60 87 : 40.57 61 : 45.60

98 : 42.24 102 : 55.85 82 : 50.59 138 : 45.60 139 : 40.57 119 : 45.60

150 : 42.24 155 : 55.85 124 : 50.59 191 : 45.60 191 : 40.57 177 : 45.60

201 : 42.24 207 : 55.85 166 : 50.59 243 : 43.78 242 : 38.81 234 : 43.78

255 : 46.08 259 : 58.00 208 : 50.59 292 : 32.83 289 : 29.99 287 : 34.66

250 : 47.62

288 : 38.69

hT 276.48 345.83 395.81 259.01 231.08 260.83

Investment costs 21.77 43.29 225.10 22.73 15.02 148.03

Damage costs 11.96 20.81 177.99 51.30 39.16 97.37

Total costs 33.72 64.10 403.08 74.04 54.18 245.40

OptimaliseRing hT 303 390 372 234 246 276

Investment costs 22.16 44.03 229.70 23.75 15.69 154.44

Damage costs 11.57 20.08 173.40 50.29 38.49 90.97

Total costs 33.73 64.11 403.10 74.05 54.18 245.41


Concluding remarks and questions

• The model is not convex. Is the solution that we found global optimal?

• Truncated versions of the model (with a time horizon of 300 years) have been solved

with dynamic programming. The optimal values are almost exactly the same. The DP

solution are not always periodic. This is due to two reasons: (1) inaccuracy is introduced

by discretizing both time and height, and (2) the DP solutions as quite sensitive to the

so-called rest term, which estimates the costs after the planning horizon.

• Truncated versions of the problem can also be modeled as a MINLP (Mixed Integer Non-

Linear Problem). This and the generalization to dike rings with more segments (e.g. sea

dikes, river dikes, street dikes, sluices and other artificial constructions) is the topic of

the next talk.


Some references

• S. Boyd and L. Vandenberghe. Convex optimization. Cambridge University Press, Cambridge, 2004.

• C.J.J. Eijgenraam. Optimal safety standards for dike-ring areas. CPB Discussion Paper 62, NetherlandsBureau for Economic Policy Analysis, The Hague, March 2006.

• G. Feichtinger and R.F. Hartl. Optimale Kontrolle okonomischer Prozesse. Anwendungen des Maxu-mumprinzips in den Wirtschaftswissenschaften. De Gruyter, Berlin, New York, 1986.

• D. den Hertog and C. Roos. Computing safe dike heights at minimal costs. Technical Report. Universityof Tilburg. CentER Applied Research. April, 2008.

• J.E. Prussing. The prinicpal minor test for semidefinite matrices. Engineering Notes, 9(1):121 – 122, 1986.

• D. van Dantzig. Economic decision problems for flood prevention. Econometrica, 24:376–287, 1956.

• H. G. Voortman. Risk-based design of large-scale flood defence systems. PhD thesis, TU Delft, The Nether-lands, 2003. Also published in the series Communications on Hydraulic and Geotechnical Engineering, DelftUniversity of Technology, Report no. 02-3.

• J.K. Vrijling, W. van Hengel, and R.J. Houben. Acceptable risk as a basis for design. Reliability Engineeringand System Safety, 59:141–150, 1998.


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Mathematicalanalysisofthehomogeneous ... · Modelling a dike ring Ht = dike height at time t, Pt = exceedance probability at time t, α = parameter exponential distribution for extreme