Integrated Risk-Hedging Control and Production Planning · Integrated Risk-Hedging Control and...

Integrated Risk-Hedging Controland Production Planning

David Yao

Columbia University, CityU/IAS

Joint work with Liao Wang, Columbia/IEOR and HKU/Business

Aug 24/25, 2017

LW/DY (Columbia, CityU/IAS) Production Planning with Risk Hedging Aug 24/25, 2017 1 / 21

Classical Production Planning: Newsvendor Model

Payoff function (net return):

HT (Q) := p(Q ∧DT )− c(Q−DT )+ = pQ− (p+ c)(Q−DT )+,

p: unit profit, c: unit net cost (minus salvage value), ∧ := min, (x)+ := max{x, 0}.

NV solution:

QNV := arg maxQ

E[HT (Q)] = F−1( p

p+ c

).

F (·) denotes the distribution function of DT .


NV with a Shortfall (SF) Objective

Established studies on corporate finance show that meeting or beating earningstargets are primary concerns for executives; in case of missing targets, how muchbelow the target (i.e. shortfall) matters (more than a probability of suchevent.).

Minimizing shortfall: with m > 0 being a given target level,

QNV(m) := arg minQ

E[m−HT (Q)]+.

Solution:QNV(m) =

m

p∧QNV.

• Optimal production quantity will never exceed QNV.

• sNV(m) := minQ E[m−HT (Q)]+ = E[m−HT (QNV(m))]+ is increasing(and convex) in m, which constitutes an efficient frontier.

• These are readily verified by considering two cases: m ≤ pQ and m ≥ pQ.


Motivation: Demand Dependent on Financial Market

• “Deere story”: the firm manufactures equipment for planting orharvesting corn, which is a tradable commodity. As the corn pricefluctuates on the futures market, demand for corn, and hence for thefirm’s product, will change accordingly. (The New York Times, 23Nov 2016,“Wall St. Closes Mostly Higher.”)

• “Caterpillar story”: the firm produces construction and miningmachineries. Anticipating a booming commodity market, the firmincreases their capacity and production; then commodities slide andthe firm suffers from decreasing demand. (The Wall Street Journal,16 Oct 2016, “How Caterpillar’s Big Bet Backfired.”)

• “Wal-Mart story”: during the last recession, Wal-Mart experiencedincreasing demand.

“Car maker story”: coming out of recession, car makers predict ahuge increase in demand.


Motivation: Demand Dependent on Financial Market


Production Planning with Risk Hedging

• In addition to the production quantity decision Q made at t = 0,there’s a hedging/trading strategy ϑ := {θt}t∈[0,T ] to be carried outdynamically over the horizon.

(ϑ gives a cumulative wealth at T : χT =∫ T0θtdXt.)

• Hence, the terminal wealth at t = T is HT (Q) + χT (ϑ).


A New Demand Model

• Sources of uncertainty (building block for information dynamics): twoindependent standard BM’s, Bt and Bt, t ∈ [0, T ].

• Xt: the price of a tradable asset, or a broad market index as proxy forthe economy, with µ(t, x) and σ(t, x) being functions in (t,Xt),

dXt = µ(t,Xt)dt+ σ(t,Xt)dBt,

Note Xt is a general Markov diffusion process; special case: geometricBrownian Motion.

• Dt: (cumulative) demand up to t,

dDt = µ(Xt)dt+ σdBt,

where σ > 0; and µ(x) ≥ 0 is a non-negative function. Hence,

DT = AT + σBT , where AT :=∫ T0µ(Xt)dt. Note: µ(x) = constant reverts

back to the traditional demand model.

• Use D+T (instead of DT ) to enforce non-negativity.


Rate Function: Deere versus Corn

“ Lower farm commodity prices directly affect farm incomes, which could

negatively affect sales of agricultural equipment.” — Disclosure of Risk Factors,

John Deere’s 2016 10k.

• Data: Monthly sales (in units) of Deere combines over 2011 - 2015; source:Deere’s data release. Daily prices (Xt) of CORN (an ETF tracking cornprice) over 2011 - 2015.

• Rate function: µ(x) = ax+ b.

• For i-th month (δt is one month; ξi are i.i.d. standard normals):

Di =

∫ ti+1

ti

µ(Xs)ds+ σ(Bti+1 −Bti) = a

∫ ti+1

ti

Xsds+ b · δt+ σ√δtξi

• To get a and b, regress Di (monthly sales) onto∫ ti+1

tiXsds (evaluated using

daily prices via trapezoidal rule).


Learning the Rate Function: Deere versus Corn


Learning the Rate Function: GM versus SP500


Production Planning with Risk Hedging

• In addition to the production quantity decision Q made at t = 0,there’s a hedging/trading strategy ϑ := {θt}t∈[0,T ] to be carried outdynamically over the horizon.

(ϑ gives a cumulative wealth at T : χT =∫ T0θtdXt.)

• Hence, the terminal wealth at t = T is HT (Q) + χT (ϑ).


From Variance to Shortfall

• Wang and Yao (Operations Research, 2016) studied mean-variancehedging:

infQ,ϑ

Var[HT (Q) + χT (ϑ)]

s.t. E[HT (Q) + χT (ϑ)] = m′

• Here, we use shortfall minimization as objective. WithW := HT + χT denoting the total terminal wealth and m′ := E(W ):

min E[(m′ −W )2] → min E[(m−W )+]

where m is any given target.

• Same demand model, but with partial information and a budgetconstraint (for hedging).


Formulation: Shortfall Minimization

• Two decisions: Q, production quantity; and ϑ := {θt}t∈[0,T ], with θtbeing the hedging position: number of shares of Xt held at t.

• Q induces HT (Q) (payoff from production), and ϑ induces

χT (ϑ) :=∫ T0 θtdXt (terminal wealth from hedging).

• Shortfall-minimization under partial information:

infQ,ϑ

E[m−HT (Q)− χT (ϑ)

]+s.t. χt :=

∫ t

0θsdXs ≥ −C, θt ∈ FX

t , t ∈ [0, T ].

• First fix Q, solve for the optimal hedging strategy to obtain theminimal shortfall value, s(m,Q). Then,

s(m,Q∗(m)) = infQs(m,Q) ⇒ Q∗(m).


Duality: Solution via a Lower-Bound Problem

The hedging problem (given Q) is:

infϑ

E(m−HT − χT (ϑ)

)+s.t. χt :=

∫ t

0

θsdXs ≥ −C, θt ∈ FXt , t ∈ [0, T ].

Applying Jensen’s inequality, we first turn the above hedging problem into a staticoptimization problem:

minVT

E(m−HT − VT

)+s.t. VT ≥ −C, EM (VT ) ≤ 0 ,

where, the additional constraint follows from χt being a PM -supermartingale, andit serves the purpose of closing the duality gap (for o.w., a strategy that attains

the lower bound may not be primal feasible). Then, V ∗T = χ∗T =∫ T0θ∗t dt.


Solving the Hedging Problem

The dual (lower-bound) problem is solved by a standard Lagrangian multiplierapproach; we focus on the optimal terminal payoff χ∗T , as the strategy θ∗t isdeduced by standard technique using Ito’s Lemma.

Proposition

For given m and Q satisfying m− pQ+ C ≥ 0, the optimal hedging is:

χ∗T = V ∗T = (m− pQ+ C)1{λ∗ZT ≤ 1}+ (p+ c)[Q− D+T (λ∗)]+ − C,

where ZT := dPM

dP is the R-N derivative with representation:

ZT = exp{−∫ T0ηtdBt − 1

2

∫ T0η2t dt} with ηt := µ(t,Xt)

σ(t,Xt), (“Market Price of

Risk”);DT (λ∗) := AT + σ

√TΦ−1(λ∗ZT ), a surrogate for DT = AT + σ

√Tξ; and λ∗

satisfies the constraint EM (V ∗T ) = 0.(The solution exists and is unique. Note: V ∗T decreases in λ∗.)

• Turns out we don’t need to consider the range m+ C − pQ < 0, in whichcase the shortfall will increase in Q.


Optimal Hedging = Put Option + Digital Option

V ∗T = (p+ c)(Q− D+T )+ + (m− pQ+ C)1{λ∗ZT ≤ 1} − C

• m−HT (Q) = (p+ c)(Q−D+T )+ + (m− pQ) is the remaining gap

(from the target) after payoff from production.

• The “put option,” (p+ c)(Q− D+T )+, tries to close the first part of

the gap, (p+ c)(Q−D+T )+, but needs to use DT as a surrogate for

DT due to partial information.

• The “digital option,” (m− pQ+ C)1{λ∗ZT ≤ 1}, aims to close theother part of the gap (after subtracting C).

• Under optimal hedging, the corresponding shortfall is

s(m,Q) = (p+ c)E[Q ∧ D+T −D

+T ]+ + (m− pQ+ C)P(λ∗ZT ≥ 1).


Optimal Production Quantity

Recall, given m, the optimal production quantity is denotedQ∗(m) = arg minQ s(m,Q), and we know Q∗(m) ∈ [0, m+C

p ].

Proposition

• s(m,Q) = (p+ c)E[Q ∧ D+T −D

+T ]+ + (m− pQ+ C)P(λ∗ZT ≥ 1)

is convex in Q ∈ [0, m+Cp ]. (Thus, finding Q∗(m) is a convex

minimization problem, solvable by a simple line search.)

• s(m,Q∗(m)) is increasing in m, hence constitutes an efficient frontier.

• Q∗(m)→ 0 as m→ 0.

• Q∗(m)→ QNV as m→∞.

Furthermore, a finite upper-bound Qu ≥ supmQ∗(m) is readily identified

via a line search.

Hence, for large m, the importance of the “NV+” strategy: optimal hedgingcombined with the NV production quantity.


Shortfall Reduction

Recall, given m, Q∗(m) is the optimal solution integrated with the optimal

hedging, and QNV(m) := mp ∧Q

NV is the optimal solution to the NV shortfallobjective.

Proposition

For a given target, m, the following range of Q guarantees shortfall reductionover the production-only decision (i.e., the minimum NV shortfall):

Q ∈ [Q∗(m) ∧QNV(m), Q∗(m) ∨QNV(m)];

and the reduction is at least (i.e., lower-bounded by):

β∗(m− pQNV)+ + Cψ∗,

where β∗ = E(ZT − λ∗)+ and ψ∗ = E(λ∗ − ZT )+. Recall, ZT = dPM

dP is theR-N derivative, which follows a log-normal distribution.


Impact of Budget

Proposition

• The shortfall is decreasing and convex in the budget C (given Qand m);specifically, ∂s

∂C = −ψ. (ψ := E(λ− ZT )+ as defined in the previousslide.)

• When C → 0, Q∗(C)→ mp ∧Q

NV, the NV optimal solution .

• When C →∞, the shortfall s(Q)→ 0 for any Q.

• When C →∞, Q∗(C)→ 0.

• Furthermore, when C →∞, we have s(Q)s(0) →

m+cQm > 1 for any Q > 0

(give m).


Production and Shortfall Risk

Data: p = 1, c = 0.5, C = 400, µ = 0.2, σ = 0.15, σ = 500;X0 = 1550, T = 63, N = 250, dt = 1

252 .

Results: production, shortfall and upper bound on production.

µ(x), m model Q∗ (Qu) shortfall

µ(x) = 10xm = 4103

NV 4103 295

NV+ 4103 250 (-15%)

optimal 3878 (4303) 213 (-28%)

µ(x) = 1.2×106√1+x

m = 7659

NV 7659 291

NV+ 7659 172 (-41%)

optimal 7600 (7868) 170 (-42%)

Table: The target is set at m = pQNV; “shortfall” reports the objective value,and the percentage in parentheses is the reduction from the NV case;Qu ≥ supmQ

∗(m).

Recall dDt = µ(Xt)dt+ σdBt.


Efficient Frontier

3700 3800 3900 4000 4100

shor

tfall

50

100

150

200

250

300

k0

= 10

NV

NV+

optimal

1700 1800 1900 2000 21000

50

100

150

200

250

k0

= 5

NV

NV+

optimal

900 1000 1100 1200 13000

50

100

150

200

250

k0

= 3

NV

NV+

optimal

7250 7350 7450 7550 76500

50

100

150

200

250

300

k1= 1.2 × 106

NV

NV+

optimal

target

3500 3600 3700 3800 39000

50

100

150

200

250

k1= 0.6 × 106

NV

NV+

optimal

1600 1700 1800 1900 20000

50

100

150

200

250

k1= 0.3 × 106

NV

NV+

optimal

Figure: First row: µ(x) = k0x; second row: µ(x) = k1√1+x

.


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