Introduction to Stochastic Optimization Part 5: …homepage.univie.ac.at/georg.pflug/science/technical...Introduction to Stochastic Optimization Part 5: Examples of multistage models

Introduction to Stochastic OptimizationPart 5: Examples of multistage models

Georg Ch. Pflug

April 23, 2009

Georg Ch. Pflug Introduction to Stochastic Optimization Part 5: Examples of multistage models

An Example for a value-of-perfect-information process

A simple multi-stage asset-liability management problem.

I There are only two assets, interpreted as bonds and stocks.

I The buy-prices and the sell-prices coincide, i.e. there are no

transaction costs. The price processes are (ξ(1)t ) resp. (ξ

(2)t ).

I The initial capital is ζ0. The net income process is ζt . Ifζt < 0, it is a payment for liabilities.

I The decision variables are the amounts invested in bonds andstocks.

I The wealth Wt is the sum of all asset values at time t afterthe net payments ζt .

I The objective is to maximize the expected terminal wealth.

I The constraint is that the ruin probability, i.e. the probabilitythat Wt is negative, must be less or equal than α for all t.


The ruin constraint is formulated as

PWt < 0 ≤ α, t = 1, . . . ,T ,

or - slightly stronger -

V@Rα(Wt) ≥ 0.

In order to convexify the constraint set we replace them by theeven stronger constraint

AV@Rα(Wt) ≥ 0.

Notice that AV@Rα(Wt) ≥ 0 is equivalent to: there is an at suchthat

E(max(Wt − at , 0)) ≤ αat .


The original problem

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ζ0 = 100

ξ(1)0 = 1

ξ(2)0 = 1

ζ1 = −60

ξ(1)1 = 1.04

ξ(2)1 = 1.08

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ3 = −40

ξ(1)3 = 1.10

ξ(2)3 = 1.16

ζ4 = −10

ξ(1)4 = 1.09

ξ(2)4 = 1.00

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.90

ζ7 = −5

ξ(1)7 = 1.15

ξ(2)7 = 1.24

ζ8 = −20

ξ(1)8 = 1.16

ξ(2)8 = 1.10

ζ9 = 10

ξ(1)9 = 1.15

ξ(2)9 = 1.08

ζ10 = −5

ξ(1)10 = 1.14

ξ(2)10 = 0.94

ζ11 = 10

ξ(1)11 = 1.15

ξ(2)11 = 1.08

ζ12 = 5

ξ(1)12 = 1.14

ξ(2)12 = 0.98

ζ13 = −5

ξ(1)13 = 1.16

ξ(2)13 = 0.97

ζ14 = −10

ξ(1)14 = 1.15

ξ(2)14 = 0.90

t=0 t=1 t=2 t=3Georg Ch. Pflug Introduction to Stochastic Optimization Part 5: Examples of multistage models

The formulation of the decision problem of wealth maximizationunder a ruin constraint

Maximize ( in x(m)t , Zt) : E(WT )

subject to

ζ0 =M∑

m=1

x(m)0 ξ

(m)0 initial budget (1a)

Wt =M∑

m=1

x(m)t ξ

(m)t wealth equation (1b)

M∑

m=1

x(m)t−1ξ

(m)t + ζt =

M∑

m=1

x(m)t ξ

(m)t (1c)

rebalancing

E[Zt ] ≤ αat (1d)Zt ≥ Wt − at (1e)Zt , xt ≥ 0 (1f)


The same problem formulated in node-oriented form is a linearprogram

Maximize (in x(m)n , Zn) :

∑

n∈NT

πnWn

subject to ζ0 =M∑

m=1

x(m)0 ξ

(m)0 initial budget (2a)

Wn =M∑

m=1

x(m)n ξ

(m)n ; n 6= 0 wealth equation (2b)

M∑

m=1

x(m)n− ξ

(m)n + ζn =

M∑

m=1

x(m)n ξ

(m)n ; n 6= 0 (2c)

rebalancing (n− is the parent node of n)∑

n∈Nt

πnZn ≤ αat(n), t = 1, . . .T (2d)

Zn ≥ Wn − at(n); n 6= 0 (2e)Zn, xn ≥ 0 (2f)


The clairvoyant’sexpansion

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I0 = 10

ξ(1)0 = 1

ξ(2)0 = 1

ζ1 = −60

ξ(1)1 = 1.04

ξ(2)1 = 1.08

ζ1 = −60

ξ(1)1 = 1.04

ξ(2)1 = 1.08

ζ1 = −60

ξ(1)1 = 1.04

ξ(2)1 = 1.08

ζ1 = −60

ξ(1)1 = 1.04

ξ(2)1 = 1.08

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ3 = −40

ξ(1)3 = 1.10

ξ(2)3 = 1.16

ζ3 = −40

ξ(1)3 = 1.10

ξ(2)3 = 1.16

ζ4 = −10

ξ(1)4 = 1.09

ξ(2)4 = 1.00

ζ4 = −10

ξ(1)4 = 1.09

ξ(2)4 = 1.00

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.90

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.90

ζ7 = −5

ξ(1)7 = 1.15

ξ(2)7 = 1.24

ζ8 = −20

ξ(1)8 = 1.16

ξ(2)8 = 1.10

ζ9 = 10

ξ(1)9 = 1.15

ξ(2)9 = 1.08

ζ10 = −5

ξ(1)10 = 1.14

ξ(2)10 = 0.94

ζ11 = 10

ξ(1)11 = 1.15

ξ(2)11 = 1.08

ζ12 = 5

ξ(1)12 = 1.14

ξ(2)12 = 0.98

ζ13 = −5

ξ(1)13 = 1.16

ξ(2)13 = 0.97

ζ14 = −10

ξ(1)14 = 1.15

ξ(2)14 = 0.90

t=0 t=1 t=2 t=3


A conditional problem given node 2

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ζ0 = 100

ξ(1)0 = 1

ξ(2)0 = 1

ζ2 = −40

ξ(1)1 = 1.05

ξ(2)1 = 1.05

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.9

ζ11 = 10

ξ(1)11 = 1.15

ξ(2)11 = 1.08

ζ12 = 5

ξ(1)0 = 1

ξ(2)0 = 1

ζ13 = −5

ξ(1)13 = 1.16

ξ(2)13 = 0.97

ζ14 = −10

ξ(1)14 = 1.15

ξ(2)14 = 0.90

t=0 t=1 t=2 t=3


The clairvoyant’s expansion of the conditional tree given node 2

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ζ0 = 100

ξ(1)0 = 1

ξ(2)0 = 1

ζ2 = −40

ξ(1)2 = 1.05

ξ(2)2 = 1.05

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ5 = 20

ξ(1)5 = 1.11

ξ(2)5 = 1.00

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.90

ζ6 = −60

ξ(1)6 = 1.10

ξ(2)6 = 0.90

ζ11 = 10

ξ(1)11 = 1.15

ξ(2)11 = 1.08

ζ12 = 5

ξ(1)12 = 1.14

ξ(2)12 = 0.98

ζ13 = −5

ξ(1)13 = 1.16

ξ(2)13 = 0.97

ζ14 = −10

ξ(1)14 = 1.15

ξ(2)14 = 0.90

t=0 t=1 t=2 t=3


The value-of-perfect-information process

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A0 = 34.72C0 = 38.89D0 = 4.17

A1 = 18.68C1 = 22.58D1 = 3.90

A2 = 49.96C2 = 50.64D2 = 0.68

A3 = −0.36C3 = −0.36

D3 = 0

A4 = 43.21C4 = 43.21

D4 = 0

A5 = 98.88C5 = 98.88

D5 = 0

A6 = 1.0C6 = 1.0D6 = 0

A7 = 7.35C7 = 7.35D7 = 0

infeasible

A9 = 53.53C9 = 53.53

D9 = 0

A10 = 32.89C10 = 32.89

D10 = 0

A11 = 105.81C11 = 105.81

D11 = 0

A12 = 91.94C12 = 91.94

D12 = 0

A13 = 3.73C13 = 3.72D13 = 0

A14 = 0.0C14 = 0.0D14 = 0

t=0 t=1 t=2 t=3Georg Ch. Pflug Introduction to Stochastic Optimization Part 5: Examples of multistage models

The fundamental Theorem of Asset Pricing

Let St , t = 1, . . . , T be a positive m-dimensional stock-priceprocess on some probability space (Ω,A, P). St is adapted to thefiltration (Ft). S0 is the today’s price vector and is deterministic.There are not transaction costs.An arbitragist starts with no or negative initial capital and getsnonnegative, not identically zero capital at time T . Suppose thatyt is the (row) vector of holdings he has at time t. An arbitragestrategy yt must be Ft measurable and must a.s. satisfy

(i) y0S0 ≤ 0 start with no capital

(ii) yt−1St ≥ ytSt self financing condition, t=1, . . . , T-1

(iii) yTST ≥ 0 nonnegative wealth at maturity

We maximize EP(yTST ) under the constraints (i) - (iii). Introducethe set Λt of positive bounded Ft- measurable functions as thedual multipliers in a saddlepoint formulation of the optimizationproblem.


We see that

infλt∈Λt

EP(λt(yt−1 − yt)St) =

0 if yt−1St ≥ ytSt a.s.−∞ otherwise

(3)

Proposition. For the nonexistence of arbitrage possibilities, it isnecessary that there exists a positive process λt such thatcomponentwise

EP(St+1λt+1

λt|Ft) = St for all t (4)


Sketch of the proof. Write the optimization problem using thedual multipliers (3) and use the saddlepoint theorem forconcave-convex functions

supy

infλ,µ

E(yTST ) +

T∑

t=1

EP(λt(yt−1 − yt)St)− λ0y0S0 + E(µT yTST )

=

infλ,µ

supy

E(yTST +

T−1∑

t=0

EP(yt(λt+1St+1 − λtSt))− λ0y0S0 + E(µT yTST )

Thus for the finiteness of the problem it is necessary that for alladmissible y

E[yt(λt+1St+1 − λtSt)

]= 0.

The admissible y are all bounded Ft measurable functions. Thuswe may conclude that it is necessary that there is a process λt > 0such that (4) holds.


For the following, we choose one of the assets as ”numeraire”.Typically the risk free bond is used as numeraire, but one maychoose any other asset. Let the m-th asset be the numeraire. Set

Zt = St/S(m)t and γt = λtS

(m)t . We have γt > 0 and by (4)

EP(γt+1|Ft) = γt (5a)EP(Zt+1γt+1|Ft) = γtZt (5b)

One sees that both the process γt and the process γtZt aremartingales. Recall that a martingale is a stochastic process Xt

adapted to a filtration (Ft) which satisfies the equation

E(Xt+1|Ft) = Xt

for all t. Martingales are also referred to as fair games.We introduce now a new probability measure Q on (Ω,A) withdensity dQ

dP = γTγ0

. Notice that by (5a), Q is a probability measure,

since E(γTγ0

) = E(λtS(m)t

λ0S(m)0

) =S

(m)0

S(m)0

= 1.


We rewrite (5b) in terms of Q. To do so, we need the followingLemma:Lemma. Suppose that P and Q are equivalent. ThenEQ(X |F) = EP(X dQ

dP |F) dP|FdQ|F .

Proof. Let Y be bounded and F-measurable. Then EQ(X · Y ) =

EP(X · Y dQdP ) = EP [EP(X dQ

dP |F)Y ) = EQ [EP(X dQdP |F)Y dP|F

dQ|F ],

which implies the assertion (Reynolds property).Now, using the Lemma, we get

EQ(Zt+1|Ft) = EP(Zt+1dQ

dP|Ft)

dP|Ft

dQ|Ft

= EP(Zt+1γT

γ0|Ft)

γ0

γt= EP(Zt+1

γt+1

γ0|Ft)

γ0

γt= EP(Zt+1

γt+1

γt|Ft) = Zt

and see that Zt must be a martingale w.r.t. Q.


Let us summarize the result:

In order that a stochastic model for asset prices be reasonable, i.e.excludes absurd arbitrage possibilities, it is necessary and sufficientthat there exists at least one probability measure Q, which makes

the renormalized process Zt = St/S(m)t a martingale. Here S

(m)t

may be any positive component of the process St , which serves asnumeraire.

Let us check quickly that the martingale condition is sufficient.Suppose that there is a trading strategy, which leads to yTST ≥ 0,

EP(yTST ) > 0. Then also EQ(yTZT ) = EP(YTST/S(m)T

dQdP ) > 0.

Because of the martingale property however,

EQ(yTZT ) = y0Z0 = y0S0/S(m)0 > 0. Thus only a positive starting

capital may lead to a nonnegative, not identically zero final wealth.


The same Theorem for finite trees

Let a scenario tree be given. Let pn price (row) vector for theassets at node n. Let c = (cn) denote the vector of cash flows withone component for each node n. yn are the portfolio compositions.

−p0y0 = c0 for n = 0

pnyn− = cn for all terminal nodespnyn− −pnyn = cn otherwise.

These equations can be written in matrix form Ay = c , using thetrees structure. For a binary tree of height 3 e.g.

A =

−p0

p1 −p1

p2 −p2

p3

p4

p5

p6


An arbitrage opportunity exists if c = Ay ≥ 0 and c = Ay 6= 0, forsome vector y . Hence, if an arbitrage does not exist, then byStiemke’s Lemma (see next slide), there is a vector π = (πn) > 0such that πA = 0. Without loss of generality, assume π0 = 1, forthe root node. Hence we obtain

An arbitrage does not exist if and only if there exists a node priceπn > 0, for all n, independent of the original node probabilities, with

π0 = 1, such that πnpn =∑

m∈n+ πmpm ∀n 6∈ T .

Here n+ is the set of all child nodes of n.This proof is due to Kallio and Ziemba.


Fundamentals: Stiemke’s Lemma

Stiemke’s Lemma (1915). For any matrix A, either

Az ≥ 0, Az 6= 0, z ≥ 0

has a solution, orπA = 0, π > 0

has a solution, but never both.Stiemke’s Lemma is an extension of the earlier Farkas’ Lemma:Farkas’ Lemma (1902). For any matrix A, either

Ax = b, x ≥ 0

has a solution, ory>A ≥ 0, y>b < 0

has a solution, but never both.


A multi-stage inventory control problem

This Example generalizes the will known Newsboy problem to amulti-period setting. While the newsboy needs not to store theunsold copies (because they are worthless the next day), themulti-period inventory problem allows to store unsold merchandize.Suppose that the demand (say for grapefruits) at timest = 1, . . . ,T is given by a random process ξ1, . . . , ξT . The groceryshop has to place regular orders one period ahead. The costs forordering one piece is one. If the demand exceeds the inventory plusthe newly arriving order, the demands has to be fulfilled by rapidorders from the wholesaler (which are immediately delivered), for aprice of ut > 1 per piece. Unsold grapefruits may be stored in theinventory, but a fraction 1− `t is storage loss. The selling price isst > 1 and the final inventory KT has a value of `TKT . Noticethat all prices may change from period to period.


The problem formulation. Let Kt be the inventory volume rightafter all sales have been effectuated at time t. Let xt be the ordersize at time t. We have that K0 = 0 and

Kt = [`t−1Kt−1 + xt−1 − ξt ]+; t = 1, . . . ,T .

The shortage at time t is

Mt = [`t−1Kt−1 + xt−1 − ξt ]−; t = 1, . . . ,T .

These two equations can be merged into

`t−1Kt−1 + xt−1 − ξt = Kt −Mt ; Kt ≥ 0, Mt ≥ 0. (6)


The profit of the whole operation is

H(x0, ξ1, . . . , xT−1, ξT ) =T∑

t=1

stξt −T−1∑

t=0

xt −T∑

t=1

utMt + `TKT .

The problem is to maximize the expected profit

Maximize E[ T∑

t=1

(stξt − xt−1 − utMt) + `TKT

]

subject to xt ¢ Ft for t = 1, . . . ,T ;

and subject to (6)

.

Notice that E∑T

t=1 stξt does not depend on the decisions and canbe removed from the optimization problem. This problem leads inthe node-wise formulation on a tree to an LP.


Let

m∗ = maxE[ T∑

t=1

(stξt − xt−1 − utMt) + `TKT

]: x ¢ FF

This problem has a dual formulation which is

m∗ = min T∑

t=1

E[(−ξt) Zt) : Zt is Ft measurable for t = 1, . . . ,T ; (7)

E(Zt |Ft−1) = 1, t = 1, . . . ,T a.s.; `t ≤ Zt ≤ ut , t = 1, . . . ,T a.s.

To see this, let Zt ∈ L∞ and form the Lagrangian

L(x , K , M, Z ) = E[−

T∑

t=1

xt−1 −T∑

t=1

utMt + `TKT

]

−E[ T∑

t=1

Zt(Kt − `t−1Kt−1 + ξt − xt−1 −Mt)]

=T∑

t=1

E[xt−1(Zt − 1)] +T∑

t=1

E[(Zt − ut)Mt ]

+E[(`T − ZT )KT ] +T−1∑

t=1

E[(`tZt+1 − Zt)Kt ]−T∑

t=1

E(ξt Zt)

=T∑

t=1

E[xt−1(E(Zt |Ft−1)− 1)] +T∑

t=1

E[(Zt − ut)Mt ]

+E[(`T − ZT )KT ] +T−1∑

t=1

E[(`tE(Zt+1|Ft)− Zt)Kt ]−T∑

t=1

E(ξt Zt).


The dual problem

infZt

supL(x , K , M, Z ) : xt ¢Ft ,Kt ¢Ft , Mt ¢Ft , Kt ≥ 0,Mt ≥ 0

is only finite, if the following conditions are met

E(Zt |Ft−1) = 1 for t = 1, . . . , T

`t ≤ Zt ≤ ut for t = 1, . . . , T

which implies (7).We may compare (7) to the dual representation of AV@R

AV@Rα(ξ) = minE(ξ Z ) : E(Z ) = 1, 0 ≤ Z ≤ 1/α.

By setting Z ′t = Zt−`t1−`t

, the conditions on Z ′t are

E(Z ′t |Ft−1) = 1 for t = 1, . . . ,T

0 ≤ Z ′t ≤ut − `t

1− `tfor t = 1, . . . ,T .


This allows us to rewrite (7) as

m∗ =T∑

t=1

`tE(−ξt) +T∑

t=1

(1− `t)E[AV@Rαt (−ξt |Ft−1)] (8)

withαt = (1− `t)/(ut − `t), t = 1, . . . , T . (9)

Using the identities

AV@Rα(−Y ) =1− α

αAV@R1−α(Y )− 1

αEY ,

and

AV@Rβ(Y ) = Varβ(Y )− 1

βE([V@Rβ(Y )− Y ]+)

one may rewrite (8) as

m∗ =T∑

t=1

utE(−ξt) +T∑

t=1

(ut − 1)E[AV@Rβt (ξt |Ft−1)] (10)

withβt = 1− αt = (ut − 1)/(ut − `t). (11)


The optimal solution of the multi-period inventory problem is

xt = V@Rβt+1(ξt+1|Ft)− `tKt (12)

where βt = (ut − 1)/(ut − `t).Introduce Vt = V@Rβt+1(ξt+1|Ft), leading to xt = Vt − `tKt .Inserting this into (10) one gets

m∗ =T∑

t=1

utE(−ξt)+T∑

t=1

(ut−1)EVt−1−T∑

t=1

(ut−`t)E([Vt−1−ξt ]+).

(13)On the other hand, inserting the solution (12) into the constraints,one finds that

Kt = [Vt−1 − ξt ]+

Mt = [Vt−1 − ξt ]−

and thereforeMt = Kt + ξt − Vt−1.


The value of the objective for this choice becomes

E

[−

T−1∑

t=0

xt −T∑

t=1

utMt + `TKT

]

= E

[−

T∑

t=1

Vt−1 +t∑

t=1

`tKt −T∑

t=1

utKt −T∑

t=1

utξt +T∑

t=1

utVt−1

]

= E

[T∑

t=1

(ut − 1)Vt−1 −T∑

t=1

(ut − `t)[Vt−1 − ξt ]+ −

T∑

t=1

utξt

]

Since this value coincides with the maximal value (13), theassertion is shown.


For illustration, assume that the demand process (ξt) is lognormaland follows a multiplicative recursion. To this end, let

η0 ∼ N(µ0, σ20)

ηt = bηt−1 + εt ; t ≥ 1

with εt ∼ N(µ, σ2) (independent), where µ = µ0(1− b) andσ2 = σ2

0(1− b2). Then (ηt) is a stationary Markovian Gaussianprocess. Let ξt = exp(ηt). The conditional AV@R is

AV@Rβ(ξt |ξt−1) = [ξt−1]bAV@Rβ(exp(εt))

= [ξt−1]b 1

βexp(µ + σ2/2)Φ(Φ−1(β)− σ).


The expectation of the conditional AV@R is

E[AV@Rβt (ξt |ξt−1)]

= exp(bµ0 + b2σ20/2)

1

βexp(µ + σ2/2)Φ(Φ−1(βt)− σ)

= exp(µ0 + σ20/2)

1

βtΦ(Φ−1(βt)− σ).

Since E(ξt) = exp(µ0 + σ20/2) one may - with the help of duality -

calculate the optimal value in a totally analytic manner. Theoptimal value is

m∗ = − exp(µ0 + σ20/2)

T∑

t=1

ut

+T∑

t=1

(ut − 1) exp(µ0 + σ20/2)

1

βtΦ(Φ−1(βt)− σ).


Numerical Example

µ0 4σ2

0 0.36σ0 0.6b 0.8

µ = µ0(1− b) 0.8σ2 = σ2

0(1− b2) 0.1296σ 0.36ut 1.1`t 0.9st 1.2

Then E(ξt) = exp(µ0 + σ20/2) = exp(4.18) = 65.37 and βt = 0.5.

m∗ = −67.20 · T .The expected profit is 1.2 · exp(4.18) · T − 67.20 · T = 11.24 · T .For T = 4; 50 scenarios (170 nodes): optimal value=247.4; 500scenarios (1700 nodes): optimal value=269.9; 2000 scenarios(6800 nodes): optimal value=276.1; true value is 268.8.


Solution methods

I Iterative methodsI Stochastic gradient and quasigradient methodsI Stochastic decompostion methods

stepwise sampling of thescenario process

-

¾

recursive adaptation ofthe solution

I Approximation methods

Approximation of thestochastic scenarioprocess by a finite

stochastic tree

-numerical solution ofthe tree-structured

problem


Documents

Introduction to Stochastic Optimization Part 5: …homepage.univie.ac.at/georg.pflug/science/technical...Introduction to Stochastic Optimization Part 5: Examples of multistage models