COMP331/557 Chapter 6: Optimisation in Finance: Cash-Flow - - PowerPoint PPT Presentation

COMP331/557 Chapter 6: Optimisation in Finance: Cash-Flow

(Cornuejols & Tütüncü, Chapter 3)

159

Cash-Flow Management Problem

A company has the following net cash flow requirements (in 1000’s of £): Month Jan Feb Mar Apr May Jun Net cash flow −150 −100 200 −200 50 300 E.g.: In January we have to pay £150k and in March we get £200k. Initially we have no cash but the following possibilities to borrow/invest money:

i a line of credit of up to £100k at an interest rate of 1% per month; ii in any one of the first three months, it can issue 90-day commercial paper bearing

a total interest of 2% for the three-month period;

iii excess funds can be invested at an interest rate of 0.3% per month.

Task: We want to maximise the companies wealth in June, while fulfilling all payments.

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Cash-Flow Management Problem – Modelling as LP

Decision Variables

◮ v .. wealth in June ◮ xi .. amount drawn from credit line in month i ◮ yi .. amount of commercial paper issued in month i ◮ zi .. excess funds in month i

LP formulation: max v s.t. x1 + y1 − z1 = 150 x2 + y2 − 1.01x1 + 1.003z1 − z2 = 100 x3 + y3 − 1.01x2 + 1.003z2 − z3 = −200 x4 − 1.02y1 − 1.01x3 + 1.003z3 − z4 = 200 x5 − 1.02y2 − 1.01x4 + 1.003z4 − z5 = −50 − 1.02y3 − 1.01x5 + 1.003z5 − v = −300 xi ≤ 100 ∀i xi, yi, zi ≥ ∀i

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Cash-Flow Management Problem – Modelling as LP

cashflow.lp

Maximize wealth: v Subject To Jan: x1 + y1 - z1 = 150 Feb: x2 + y2 - 1.01 x1 + 1.003 z1 - z2 = 100 Mar: x3 + y3 - 1.01 x2 + 1.003 z2 - z3 = -200 Apr: x4 - 1.02 y1 - 1.01 x3 + 1.003 z3 - z4 = 200 May: x5 - 1.02 y2 - 1.01 x4 + 1.003 z4 - z5 = -50 Jun:

• 1.02 y3 - 1.01 x5 + 1.003 z5 - v = -300

Bounds 0 <= x1 <= 100 0 <= x2 <= 100 0 <= x3 <= 100 0 <= x4 <= 100 0 <= x5 <= 100

• Inf <= v <= Inf

End

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Cash-Flow Management Problem – Modelling as LP

Gurobi Output

Solved in 5 iterations and 0.00 seconds Optimal objective 9.249694915e+01 v 92.4969491525 x1 0.0 y1 150.0 z1 0.0 x2 0.0 y2 100.0 z2 0.0 x3 0.0 y3 151.944167498 z3 351.944167498 x4 0.0 z4 0.0 x5 52.0 z5 0.0 Obj: 92.4969491525

Optimal Investment Strategy: Jan: Issue commercial paper for £150k. Feb: Issue commercial paper for £100k. Mar: Issue paper for ≈ £152k and invest ≈ £352k. Apr: Take excess to pay outgoing cashflow. May: Take a credit of £52k Jun: wealth ≈ £92k

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The Fundamental Theorem of Asset Pricing

(Cornuejols & Tütüncü, Chapter 4)

164

Derivative Securities

Derivative Securities

◮ also known as contingent claims ◮ price depend on the value of another underlying security ◮ e.g. European call options

European call option

Gives the holder the right to buy (or sell, “put option”)

◮ prescribed underlying security ◮ at expiration date (also known as maturity date) ◮ for prescribed amount (called strike price).

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Example

◮ Share of XYZ stock currently priced at $40. ◮ A month from today, we expect the share price to either double or halve with equal

probability.

◮ Consider a European call option on XYZ with strike price $50, which will expire in

a month.

◮ Assume that interest rates for cash (borrowing or lending) are zero. ◮ What would be a fair price for the option?

−10 10 30 50 70 90 110 120 −10 10 20 30 40 50 60 70 Share price of XYZ Payoff from option struck at 50 European call option payoff function 22:2362C9CC 23:586 86C6B 9CC5 5: 8 ,0 .256579CC 23:586 86 1:6B:C7/:6/:320C2CBD36CCC9623:5866C6B7DB6

S0 = $40 ✟✟✟

✯ ❍❍❍ ❥

80 = S1(u) 20 = S1(d) and C0 = ? ✟✟✟

✯ ❍❍❍ ❥

(80 − 50)+ = 30 (20 − 50)+ = 0 In Section 1.3.2 we obtained a fair price of $10 for the option using a replica-

22:2362C9CC 23:586 86C6B 9CC5 5: 8 ,0 .256579CC 23:586 86 1:6B:C7/:6/:320C2CBD36CCC9623:5866C6B7DB6

◮ identical future payoff ⇒ same value today ◮

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Arbitrage

Definition

Arbitrage is a trading strategy that:

i has positive initial cash flow and no risk of loss later (type A) ii requires no initial cash input, has no risk of loss and has positive probability of

making profit in the future (type B) In the example: any price other than $10 for the option would lead to a type A arbitrage. Notes:

◮ Prices adjust quickly so that arbitrage opportunities cannot persist in the marked. ◮ ⇒ Pricing arguments usually assume no arbitrage.

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Replication (in slightly more general setting)

◮ Consider portfolio of ∆ shares of underlying security and B cash. ◮ Two possible outcomes:

◮ Up-state: ∆ · S0 · u + B · R, where R = 1 + (risk-less interest rate) ◮ Down-state: ∆ · S0 · d + B · R

◮ For what values of ∆ and B does the portfolio have the same payoffs C u 1 and C d 1

as the call option? ∆ · S0 · u + B · R = C u

1

∆ · S0 · d + B · R = C d

1 ◮ Solving for ∆ and B gives

∆ = C u

1 − C d 1

S0(u − d) and B = uC d

1 − dC u 1

R(u − d)

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Replication (in slightly more general setting)

◮ Since portfolio is worth S0∆ + B today, this should also be the price for the

derivative security, i.e: C0 = C u

1 − C d 1

u − d + uC d

1 − dC u 1

R(u − d) = 1 R R − d u − d C u

1 + u − R

u − d C d

1

• ◮ no arbitrage ⇒ d < R < u

Definition: Risk-neutral probabilities

pu = R − d u − d and pd = u − R u − d Remark: If price = C0 then there is arbitrage opportunity.

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Generalisation of binomial (2-stage) setting

◮ Let ω1, . . . , ωm be a finite set of possible states. ◮ For securities Si, i = 0, . . . n:

◮ Si

.. current price (at time 0)

◮ Si

1(ωj) .. future price (at time 1) if in state ωj ◮ S0 .. risk-less security, i.e. S0 0 = 1 and S0 1(ωj) = R ≥ 1 for all j

Risk-neutral probabilities

A risk-neutral probability measure (RNPM) on the set Ω = {ω1, . . . , ωm} is a vector of positive numbers p1, . . . , pm such that m

j=1 pj = 1 and for every security Si,

i = 0, . . . n, Si

0 = 1

R  

m

• j=1

pjSi

1(ωj)

  .

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First Fundamental Theorem of Asset Pricing

Theorem 6.1 (First Fundamental Theorem of Asset Pricing).

A risk-neutral probability measure exists if and only if there is no arbitrage.

◮ Proof is a simple exercise in LP duality. ◮ We will make use of the following result of Goldman and Tucker on the existence of

strictly complementary optimal solutions:

Theorem 6.2 (Goldman-Tucker Theorem).

Consider the following primal-dual pair: min cTx max bTp s.t. Ax ≥ b s.t. ATp = c p ≥ 0 If both, the primal and the dual LP, admit a feasible solution, then there are primal and dual optimal solutions x∗, p∗ such that p∗ + (Ax∗ − b) > 0.

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Arbitrage Detection Using Linear Programming

Scenario:

◮ Portfolio (x1, . . . xn) of European call options S1, . . . , Sn of same underlying

security S.

◮ Payoff of portfolio: Ψ x(S1) := n i=1 Ψi(S1)xi, where

◮ Ψi(S1) = (S1 − Ki)+, and ◮ Ki is strike price of call option Si.

◮ cost of forming portfolio at time 0 is: n i=1 Si 0xi.

−10 10 30 50 70 90 110 120 −10 10 20 30 40 50 60 70 Share price of XYZ Payoff from option struck at 50 European call option payoff function 22:2362C9CC 23:586 86C6B 9CC5 5: 8 ,0 .256579CC 23:586 86 1:6B:C7/:6/:320C2CBD36CCC9623:5866C6B7DB6

Determine arbitrage possibility

◮ Negative cost of portfolio with non-negative payoff (type A). ◮ Zero cost and strictly positive payoff (type B).

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Detecting arbitrage

Checking for non-negative payoff:

◮ each Ψi is piecewise-linear in S1 with single breakpoint Ki ◮ thus Ψ x is piecewise-linear in S1 with breakpoints K1, . . . , Kn

(assume K1 ≤ . . . ≤ Kn)

◮ Ψ x is non-negative in [0, ∞), if and only if Ψ x is

◮ non-negative at 0, ◮ non-negative at all breakpoints, and ◮ right-derivative after last breakpoint Kn is non-negative.

Formally: Ψ x(0) ≥ 0 Ψ x(Kj) ≥ 0, ∀j, and Ψ x(Kn + 1) − Ψ x(Kn) ≥ 0.

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Detecting arbitrage – Linear Program

min

n

• i=1

Si

0xi

(6.1) s.t.

n

• i=1

Ψi(0)xi ≥ 0

n

• i=1

Ψi(Kj)xi ≥ 0, j = 1, . . . , n,

n

• i=1

(Ψi(Kn + 1) − Ψi(Kn)) xi ≥ 0.

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Detecting arbitrage

Theorem 6.3.

There is no type-A arbitrage if and only if the optimal objective value of (6.1) is zero.

Theorem 6.4.

Suppose that there is no type-A arbitrage. Then, there is no type-B arbitrage if and

• nly if the dual of (6.1) has a strictly positive feasible solution.

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European call options - Constraint matrix

◮ Ψi(Kj) = (Kj − Ki)+ ◮ Constraint matrix A of (6.1) has the form

A =   

K2−K1 ··· 0 K3−K1 K3−K2 ··· 0

. . . . . . . . . . . .

Kn−K1 Kn−K2 Kn−K3 ··· 0 1 1 1 ··· 1

  

Theorem 6.5.

Let K1 < K2 < . . . < Kn denote the strike prices of European call options written on the same underlying security with the same maturity. There are no arbitrage opportunities if and only if the prices Si

0 satisfy: i Si 0 > 0 for i = 1, . . . , n. ii Si 0 > Si+1

for i = 1, . . . , n − 1.

iii C(Ki) := Si 0 defined on {K1, . . . , Kn} is a strictly convex function.

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