Integrated Risk-Hedging Control and Production Planning

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Integrated Risk-Hedging Control and Production Planning David Yao Columbia University, CityU/IAS

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

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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 max E[HT (Q)] = F −1 Q

 p  . p+c

F (·) denotes the distribution function of DT . LW/DY (Columbia, CityU/IAS)

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NV with a Shortfall (SF) Objective Established studies on corporate finance show that meeting or beating earnings targets are primary concerns for executives; in case of missing targets, how much below the target (i.e. shortfall) matters (more than a probability of such event.). Minimizing shortfall: with m > 0 being a given target level, QNV (m) := arg min E[m − HT (Q)]+ . Q

Solution:

m ∧ QNV . QNV (m) = p

• 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. LW/DY (Columbia, CityU/IAS)

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Motivation: Demand Dependent on Financial Market • “Deere story”: the firm manufactures equipment for planting or

harvesting corn, which is a tradable commodity. As the corn price fluctuates on the futures market, demand for corn, and hence for the firm’s product, will change accordingly. (The New York Times, 23 Nov 2016,“Wall St. Closes Mostly Higher.”) • “Caterpillar story”: the firm produces construction and mining

machineries. Anticipating a booming commodity market, the firm increases their capacity and production; then commodities slide and the 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 experienced

increasing demand. “Car maker story”: coming out of recession, car makers predict a huge increase in demand. LW/DY (Columbia, CityU/IAS)

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Motivation: Demand Dependent on Financial Market

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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 out dynamically over the horizon. (ϑ gives a cumulative wealth at T : χT =

RT 0

θt dXt .)

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

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A New Demand Model • Sources of uncertainty (building block for information dynamics): two

˜t , t ∈ [0, T ]. independent standard BM’s, Bt and B • Xt : the price of a tradable asset, or a broad market index as proxy for

the 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: geometric Brownian Motion. • Dt : (cumulative) demand up to t, ˜t , dDt = µ ˜(Xt )dt + σ ˜ dB where σ ˜ > 0; and µ ˜(x) ≥ 0 is a non-negative function. Hence, R ˜T , where AT := T µ D T = AT + σ ˜B ˜(Xt )dt. Note: µ ˜(x) = constant reverts 0 back to the traditional demand model. • Use DT+ (instead of DT ) to enforce non-negativity. LW/DY (Columbia, CityU/IAS)

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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 corn price) over 2011 - 2015. • Rate function: µ ˜(x) = ax + b. • For i-th month (δt is one month; ξi are i.i.d. standard normals): Z

ti+1

Di =

Z µ ˜(Xs )ds + σ ˜ (Bti+1 − Bti ) = a

ti

ti+1

ti

• To get a and b, regress Di (monthly sales) onto daily prices via trapezoidal rule).

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√ Xs ds + b · δt + σ ˜ δtξi

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R ti+1 ti

Xs ds (evaluated using

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Learning the Rate Function: Deere versus Corn

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Learning the Rate Function: GM versus SP500

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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 out dynamically over the horizon. (ϑ gives a cumulative wealth at T : χT =

RT 0

θt dXt .)

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

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From Variance to Shortfall • Wang and Yao (Operations Research, 2016) studied mean-variance

hedging: inf

Q,ϑ

Var[HT (Q) + χT (ϑ)]

s.t. E[HT (Q) + χT (ϑ)] = m0 • Here, we use shortfall minimization as objective. With

W := HT + χT denoting the total terminal wealth and m0 := E(W ): min E[(m0 − W )2 ]



min E[(m − W )+ ]

where m is any given target. • Same demand model, but with partial information and a budget

constraint (for hedging).

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Formulation: Shortfall Minimization • Two decisions: Q, production quantity; and ϑ := {θt }t∈[0,T ] , with θt

being the hedging position: number of shares of Xt held at t. • Q induces HT (Q) (payoff from production), and ϑ induces

χT (ϑ) :=

RT 0

θt dXt (terminal wealth from hedging).

• Shortfall-minimization under partial information:

h i+ inf E m − HT (Q) − χT (ϑ) Q,ϑ Z t s.t. χt := θs dXs ≥ −C, θt ∈ FtX ,

t ∈ [0, T ].

0

• First fix Q, solve for the optimal hedging strategy to obtain the

minimal shortfall value, s(m, Q). Then, s(m, Q∗ (m)) = inf s(m, Q) Q

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Q∗ (m). Aug 24/25, 2017

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Duality: Solution via a Lower-Bound Problem The hedging problem (given Q) is: + inf E m − HT − χT (ϑ) ϑ

Z s.t.

t

θs dXs ≥ −C,

χt :=

θt ∈ FtX ,

t ∈ [0, T ].

0

Applying Jensen’s inequality, we first turn the above hedging problem into a static optimization problem: min E m − HT − VT VT

+

s.t.

VT ≥ −C,

EM (VT ) ≤ 0 ,

where, the additional constraint follows from χt being a PM -supermartingale, and it serves the purpose of closing the duality gap (for o.w., a strategy that attains RT the lower bound may not be primal feasible). Then, VT∗ = χ∗T = 0 θt∗ dt.

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Solving the Hedging Problem The dual (lower-bound) problem is solved by a standard Lagrangian multiplier approach; we focus on the optimal terminal payoff χ∗T , as the strategy θt∗ is deduced by standard technique using Itˆ o’s Lemma.

Proposition For given m and Q satisfying m − pQ + C ≥ 0, the optimal hedging is: ˆ + (λ∗ )]+ − C, χ∗T = VT∗ = (m − pQ + C)1{λ∗ ZT ≤ 1} + (p + c)[Q − D T M

where ZT := dP is the R-N derivative with representation: R TdP RT µ(t,Xt ) ZT = exp{− 0 ηt dBt − 21 0 ηt2 dt} with ηt := σ(t,X , (“Market Price of t) Risk”); √ √ ˆ T (λ∗ ) := AT + σ D ˜ T Φ−1 (λ∗ ZT ), a surrogate for DT = AT + σ ˜ T ξ; and λ∗ satisfies the constraint EM (VT∗ ) = 0. (The solution exists and is unique. Note: VT∗ decreases in λ∗ .) • Turns out we don’t need to consider the range m + C − pQ < 0, in which case the shortfall will increase in Q. LW/DY (Columbia, CityU/IAS)

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Optimal Hedging = Put Option + Digital Option ˆ + )+ + (m − pQ + C)1{λ∗ ZT ≤ 1} − C VT∗ = (p + c)(Q − D T • m − HT (Q) = (p + c)(Q − DT+ )+ + (m − pQ) is the remaining gap

(from the target) after payoff from production. ˆ + )+ , tries to close the first part of • The “put option,” (p + c)(Q − D T

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

• The “digital option,” (m − pQ + C)1{λ∗ ZT ≤ 1}, aims to close the

other part of the gap (after subtracting C). • Under optimal hedging, the corresponding shortfall is

ˆ + − D+ ]+ + (m − pQ + C)P(λ∗ ZT ≥ 1). s(m, Q) = (p + c)E[Q ∧ D T T

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Optimal Production Quantity Recall, given m, the optimal production quantity is denoted Q∗ (m) = arg minQ s(m, Q), and we know Q∗ (m) ∈ [0, m+C p ].

Proposition ˆ + − D+ ]+ + (m − pQ + C)P(λ∗ ZT ≥ 1) • s(m, Q) = (p + c)E[Q ∧ D T T ∗ is convex in Q ∈ [0, m+C p ]. (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 ≥ supm Q∗ (m) is readily identified via a line search. Hence, for large m, the importance of the “NV+” strategy: optimal hedging combined with the NV production quantity. LW/DY (Columbia, CityU/IAS)

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Shortfall Reduction Recall, given m, Q∗ (m) is the optimal solution integrated with the optimal NV is the optimal solution to the NV shortfall hedging, and QNV (m) := m p ∧Q objective.

Proposition For a given target, m, the following range of Q guarantees shortfall reduction over 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 = R-N derivative, which follows a log-normal distribution.

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dPM dP

is the

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Impact of Budget Proposition • The shortfall is decreasing and convex in the budget C (given Q

and m); specifically, slide.)

∂s ∂C

= −ψ. (ψ := E(λ − ZT )+ as defined in the previous

• When C → 0, Q∗ (C) →

m p

∧ QNV , the NV optimal solution .

• When C → ∞, the shortfall s(Q) → 0 for any Q. • When C → ∞, Q∗ (C) → 0. • Furthermore, when C → ∞, we have (give m).

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s(Q) s(0)



m+cQ m

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> 1 for any Q > 0

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Production and Shortfall Risk Data: p = 1, c = 0.5, C = 400, µ = 0.2, σ = 0.15, σ˜ = 500; 1 . X0 = 1550, T = 63, N = 250, dt = 252 Results: production, shortfall and upper bound on production. µ ˜ (x), m

model NV NV+ optimal NV NV+ optimal

µ ˜(x) = 10x m = 4103 6

√ µ ˜(x) = 1.2×10 1+x m = 7659

Q∗ (Qu ) 4103 4103 3878 (4303) 7659 7659 7600 (7868)

shortfall 295 250 (-15%) 213 (-28%) 291 172 (-41%) 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 ≥ supm Q∗ (m). ˜t . Recall dDt = µ ˜(Xt )dt + σ ˜ dB LW/DY (Columbia, CityU/IAS)

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Efficient Frontier k0 = 5

k 0 = 10

k0 = 3

300 NV NV+ optimal

250

250

250

NV NV+ optimal

200

200

150

150

100

100

50

50

NV NV+ optimal

200

150

100

50

shortfall

0 3700

3800

3900

4000

4100

1700

k 1 = 1.2 × 106 300

1900

2000

0

2100

900

NV NV+ optimal

250

1000

1100

1200

1300

1900

2000

k 1 = 0.3 × 106

k 1 = 0.6 × 106

NV NV+ optimal

250

1800

NV NV+ optimal

250

200

200

150

150

100

100

50

50

200 150 100 50 0

7250

7350

7450

7550

7650

0

3500

3600

3700

3800

3900

0

1600

1700

1800

target

Figure: First row: µ ˜(x) = k0 x; second row: µ ˜(x) =

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√k1 . 1+x

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