35th international energy workshop (iew) 2016 - adapting ... · adapting long-lived infrastructure...

Outline Motivation Model Results Concluding Remarks References

35th International Energy Workshop (IEW) 2016

Adapting long-lived infrastructure to uncertain and transientchange

Marius Paschen, Klaus Eisenack

Department for Economics and LawCarl von Ossietzky University Oldenburg

Cork, Ireland

1.-3. June 2016

Marius Paschen, Klaus Eisenack 35th International Energy Workshop (IEW) 2016

Motivation

How should long-lived infrastructure be adapted to ongoingand uncertain change? ⇒

Problems:

1 High sunk investment costs

2 Current decisions shape effects of changes in uncertainexogenous conditions up to multiple future decades

Examples:

1 Uncertainty of climate change affects negatively infrastructure(airports, bridges, water pipes) due to extreme weather events(cf. IPCC (2014)).

2 Rising uncertain capacity of renewables affects electricity gridcomponents

Motivation

How should long-lived infrastructure be adapted to ongoingand uncertain change? ⇒Problems:

Examples:

Motivation

Examples:

Motivation

Examples:

Motivation

Examples:

Motivation

Examples:1 Uncertainty of climate change affects negatively infrastructure

(airports, bridges, water pipes) due to extreme weather events(cf. IPCC (2014)).

Motivation

Examples:1 Uncertainty of climate change affects negatively infrastructure

(airports, bridges, water pipes) due to extreme weather events(cf. IPCC (2014)).

Motivation

Irreversibility/uncertainty leads to option value ⇒ Abandoninginvestment might be delayed (cf. Dixit, R. K. and R. S. Pindyck (1994))

Shorter life-time might enable more efficient rolling adjustments totransient change (cf. Hallegatte, S. (2009)) ⇒What is optimal infrastructure life-time and technical design?

One central research question:

How reacts optimal life-time, if uncertainty rises? ⇒

1 Expecting rise, if option premium rises

2 Expecting fall, if risk higher than option premium increase

Illustration using grid example

Motivation

Shorter life-time might enable more efficient rolling adjustments totransient change (cf. Hallegatte, S. (2009)) ⇒

What is optimal infrastructure life-time and technical design?

Motivation

How reacts optimal life-time, if uncertainty rises? ⇒1 Expecting rise, if option premium rises

Motivation

How reacts optimal life-time, if uncertainty rises? ⇒1 Expecting rise, if option premium rises

Application

Long-lived infrastructure: Components of electricity grid:Power lines

Influenced by rising uncertain future capacity of wind energy(technological/political uncertainty) ⇒Assumption: Positive relation capacity/peak generation

High capacity lines face rising wind energy, but expensive

What is optimal design of power lines (line material,quality, capacity, thickness)?

What is optimal life-time of lines, if wind capacity couldbe higher than selected line capacity?

Application

Application/Model

Optimal stochastic control model maximizes expected netpresent value of benefit of electricity grid components(discount rate r), current value at time t: a− 1

Capacity change dxt of wind capacity x over time:Stochastic process with geometric brownian motion ⇒Certain change rate µ, standard deviation σ included

Two design components: Size a, robustness b,costs of size quadratic function with cost factor c, robustnesscosts linear with cost factor c2

Current value rises if design elements rise, falls over time ifwind capacity rises

Larger b: Value for broader interval of renewable feed-in,larger a: Larger value, broader interval with positive value

Application/Model

Capacity change dxt of wind capacity x over time:Stochastic process with geometric brownian motion ⇒

Certain change rate µ, standard deviation σ included

Application/Model

Decision structure, Model solving

Choose optimal life-time T ? of components ⇒Optimal stopping

Choose optimal technical design a? and b? at beginning,constant over time

1 Decide on the long-term design components2 Decide on the life-time (design irreversible)3 Decide to end life-time (capacity change uncertain)

1 Decide on the long-term design components

2 Decide on the life-time (design irreversible)3 Decide to end life-time (capacity change uncertain)

1 Decide on the long-term design components2 Decide on the life-time (design irreversible)

3 Decide to end life-time (capacity change uncertain)

Stylized model as a stochastic dynamic control problem

Step 2,3:

h(x0, a, b, µ, σ2) = maxE

(∫ T

{(a− 1

bx(t)) e−rtdt

})w .r .t.T

such that:

dxt = µxt dt + σxt dzt (2)

µ > 0 , σ > 0 , a > 0 , b > 0, r > 0 , (zt) stand. WienerProcess

⇒ Yields expected stopping time T ?(x0, a, b, µ, σ2)

Stylized model as a stochastic dynamic control problem

Step 1:

max h(x0, a, b, µ, σ2)− ca2− c2b w .r .t. (a, b), c , c2 > 0 (3)

⇒ Yields optimal size a?(x0, µ, σ2, b) and planned life-time

T ??(x0, µ, σ2, b) = T ?(x0, a

?, µ, σ2, b) as well as optimalrobustness b?(x0, µ, σ

2, a) and planned life-timeT ??(x0, µ, σ

2, a) = T ?(x0, b?, µ, σ2, a)

Solution for step 2,3: Optimal expected stopping time,capacity level

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

h: Expected (optimal) benefit of grid components withoutoptimal a and b

Value matching, smooth pasting conditions:h(x?) = hx(x?) = 0 (cf. Seierstad, A. (2009))

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

E [x(T ?)] = x0eµT? = x? ⇒ T ? = ln x?−ln x0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

h(x , a, µ, σ2) = A x r1 +1

b(µ− r)x +

r1 =2µ−σ2+

√(2µ−σ2)2+8rσ2

2σ2 > 0

A = − 1r1

1b(µ−r) x

?1−r1

x? = ab(r−µ)r

r1r1−1 > ab > 0

Solution for step 1: Optimal technical design of robustness

Maximize expected optimal benefit of power line minus costs ofdesign with respect to design of robustness, when expected optimalstopping time, capacity level is known, at beginning of project t = 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

First, second order conditions: Hb(x0, b?, a, µ, σ2, r , c , c2) =

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

Solution exists, specific restrictions for (µ, σ2, a, r , x0, c , c2) ⇒Solution with x? > x0 and reasonable parameter values

Solution for optimal size also exists

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

x? ≥ x0 ⇒ b? ≥ b0 = x0r(r1−1)a(r−µ)r1

b? = v x0r(r1−1)a(r−µ)r1 , v ≥ 1⇒ x? ≥ x0

H(x0, a, b, µ, σ2, r) = h(x0, a, b, µ, σ

2, r)− ca2− c2b (5)

0,Hbb(x0, a, b, µ, σ2, r , c , c2) < 0.

Comparative statics

Focus on statics including optimal robustness

Focus on reaction of optimal technical robustness due to:

1 Rising uncertainty

Derive statics at beginning of project, b = b?,x = x0

v ≶ e1r1 ⇒ db?

dσ2 ≷ 0

Comparative statics

v ≶ e1r1 ⇒ db?

dσ2 ≷ 0

Comparative statics

v ≶ e1r1 ⇒ db?

dσ2 ≷ 0

Comparative statics

v ≶ e1r1 ⇒ db?

dσ2 ≷ 0

Comparative statics including optimal b?

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

Effect depends on optimal planned life-time

Uncertainty rises ⇒ range of possible realizations of x(t) rises ⇒unfavorable conditions if life-time small ⇒ small option premiumfrom waiting to end project ⇒ rise in optimal robustness to getpower line more robust ⇒ rise in net benefit if robustness costssmall enough

Uncertainty rises ⇒ range of possible realizations of x(t) rises ⇒favorable conditions if life-time large ⇒ large option premium fromwaiting to end project ⇒ rise in robustness might lead to lower netbenefit instead of “just waiting” if robustness costs too large ⇒optimal robustness decreases

Current simulation results show only negative effects, when discountrate exceeds certain rate

Current simulation results show both positive as well as negativeeffects, when certain rate exceeds discount rate

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ db?

dσ2 ≷ 0

Focus again on reaction of optimal (planned) life-time due to:

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

Higher uncertainty ⇒ More information might appear ⇒ Increase inoption premium for waiting to end

Robustness rises also if life-time small ⇒ higher net benefit iflife-time rises

Negative effect: Negative indirect robustness effect outweighs directeffect, might be due to more extreme wind capacity changes

Current simulation results show only positive effects, when discountrate exceeds certain rate

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

v < e1r1 ⇔ T ?(b?) < 1

µr1⇒ dT??

dσ2 = dT?(b?,σ2)dσ2 =

∂2σT?(σ2, b?) + ∂bT

?(σ2, b?)db?(σ2)dσ2 > 0

Concluding Remarks

Results:

1 All results without optimal design as expected i.e. risingexpected life-time due to rising uncertainty (increase in optionpremium)

2 Optimal robustness:Different reactions of optimal robustness/ planned life-timedue to uncertainty depending on optimal planned life-time ⇒simulated positive and negative effects of planned life-time dueto uncertainty possible, when certain rate exceeds discount rateOnly positive effects, when discount rate exceeds certain rate

Concluding Remarks

Results:

1 All results without optimal design as expected i.e. risingexpected life-time due to rising uncertainty (increase in optionpremium)

2 Optimal robustness:Different reactions of optimal robustness/ planned life-timedue to uncertainty depending on optimal planned life-time ⇒simulated positive and negative effects of planned life-time dueto uncertainty possible, when certain rate exceeds discount rateOnly positive effects, when discount rate exceeds certain rate

Concluding Remarks

Results may be important for long-lived infrastructure investmentplanning decisions:

1 Private2 Public

Basis for applied numerical computations

General message: Uncertainty does not necessarily require morerobust investment or longer life-time

Flexibility in terms of infrastructures, shorter life-times may pay off

Possible further research:

1 Alternative stochastic processes (arithmetic brownian motion)2 Subsequent investment cycles (adaptation between investment

periods)

Concluding Remarks

1 Private2 Public

periods)

Concluding Remarks

1 Private2 Public

periods)

Thanks for your attention!

References I

Dixit, R. K. and R. S. Pindyck (1994). Investment under uncertainty. PrincetonUniversity Press, Princeton.

Hallegatte, S. (2009). Strategies to adapt to an uncertain climate change. GlobalEnvironmental Change, 19(2):240–247.

Seierstad, A. (2009). Stochastic control in discrete and continuous time. Springer,New York.

IPCC (2014). Climate change 2014: impacts, adaptation, and vulnerability.Intergovernmental Panel on Climate Change, Cambridge University Press,Cambridge, UK.

Appendix1: Stochastic dynamic optimization problem (step2,3)

Solve Hamilton-Jacobi-Bellman Equation, second order ODE:

0 = −rh + a− 1bx + µxhx + 1

2σ2x2hxx ⇔

x2hxx + 2µσ2 xhx − 2r

σ2 h = 2σ2 ( 1

bx − a)

Solution: h(x) = A x r1 + B x r2 + 1b(µ−r) x + a

r , A,B ∈ <

If B = 0 ∧ h(x?) = hx(x?) = 0 ⇒

A = − 1r1

1b(µ−r) x

?1−r1 ∧ x? = ab(r−µ)r

r1r1−1 > ab > 0

hxx = A(r1 − 1)r1xr1−2 > 0 if x > 0 ⇒

h(x) > 0 if x ∈ (0, x?), (x?,∞) ⇒⇒ x?,T ? optimal

Appendix2: Optimal technical design of profitability (step1)

maxa H(a) = (− 1r1

1b(µ−r) x

?1−r1) x r10 + 1b(µ−r) x0 + a

r − ca2− c2b ⇒

maxa a1−r1 1r(r1−1)( x0r(r1−1)

b(r−µ)r1 )r1 + x0b(µ−r) + a

r − ca2 − c2b ⇒

Ha(a?) = −1r a?−r1( x0r(r1−1)

b(r−µ)r1 )r1 + 1r − 2ca? = 0

Haa(a) = a−(1+r1) r1r ( x0r(r1−1)

b(r−µ)r1 )r1 − 2c < 0⇒

If r = b2cww2x0

∧ a? = w x0r(r1−1)b(r−µ)r1 ∧ w > 1 ∧ w2 =

w r1 r(r1−1)(w r1−1)(r−µ)r1 ∧ (w > ( r1+1

1−r1)

1r1 ∧ c2 <

x0b2(µ−r)

∧ µ > r) ∨ (w >

(r1 + 1)1r1 ∧ w < ( (r1+1)(r−µ)

2(r1−1) )1r1 ∧ c2 <

x0r(r1−1)2b2(r−µ)r1 ∧ µ < r) ⇒

⇒ a? global maximum, if a? = w x0r(r1−1)b(r−µ)r1 ⇒ x? > x0, Scenario

with 0 < r ≤ 0.1 possible

Appendix3: Optimal technical design of robustness (step1)

maxb H(b) = b−r1 ar(r1−1)( x0r(r1−1)

a(r−µ)r1 )r1 + x0b(µ−r) + a

r − ca2 − c2b ⇒

Hb(b?) = − ar1r(r1−1)b

?−r1−1( x0r(r1−1)a(r−µ)r1 )r1 − x0

b?2(µ−r)− c2 = 0

Hbb(b) = b−(2+r1) ar1(r1+1)r(r1−1) ( x0r(r1−1)

a(r−µ)r1 )r1 + 2x0b3(µ−r)

< 0⇒

If c2 < − ar1r(r1−1)( x0r(r1−1)

a(r−µ)r1 )−1(( 2r1+1)

− 1+r11−r1 − ( 2

r1+1)− 2

1−r1 ) ⇒

⇒ b? global (inner) maximum with b? > b0 ⇒ x? > x0, Scenariowith 0 < r ≤ 0.1 possible,

H(b?) > 0 additional necessary and sufficient condition for b?

global maximum

Appendix4: Comparative statics T ?, x?

∂σ2 = − ab(r−µ)r(r1−1)2

∂r1∂σ2 > 0, (µ ≷ r ⇒ ∂r1

∂σ2 ≷ 0)⇒∂T?

∂σ2 = 1µ

∂σ2 > 0

∂µ = abr(r1−1)2

(−∂r1∂µ (r − µ)− r1(r1 − 1)) < 0, ∂r1∂µ < 0⇒

∂µ = 1µ( 1

x?∂x?

∂µ − T ?) < 0

∂a = b(r−µ)r1r(r1−1) > 0⇒

∂a = 1µ

∂a > 0

∂b = a(r−µ)r1r(r1−1) > 0⇒

∂b = 1µ

∂b > 0

Appendix5: Comparative statics a?

dp = −Hap(a?,p)Haa(a?,p)

, p ∈ (σ2, µ, b), −Haa(a?, p) > 0, if a? > a0 optimal

Haσ2 = −1r a?−r1( x0r(r1−1)

b(r−µ)r1 )r1 ∂r1∂σ2 (ln( x0r(r1−1)

b(r−µ)r1 )− ln(a?) + 1r1−1)

Haµ =

−1r a?−r1( x0r(r1−1)

b(r−µ)r1 )r1(∂r1∂µ (ln( x0r(r1−1)b(r−µ)r1 )−ln(a?))+ ∂r1

∂µ1

r(r1−1) + r1r(r−µ))

Hab = 1r a?−r1( x0r(r1−1)

(r−µ)r1 )r1r1b−r1−1

If a? > a0 ∧ µ > r

⇒ Haσ2 > 0 ∧ Haµ < 0 ∧ Hab > 0⇒ da?

dσ2 > 0 ∧ da?

dµ < 0 ∧ da?

db > 0

If a? > a0 ∧ µ < r

⇒ (w ≶ e1

r1−1 ⇔ T ?(a?) ≶ 1µ(r1−1) ⇒ Haσ2 ≷ 0) ∧ Haµ <

0∧Hab > 0⇒ (T ?(a?) ≶ 1µ(r1−1) ⇒

dσ2 ≷ 0)∧ da?

dµ < 0∧ da?

db > 0

Appendix6: Comparative statics T ?(a?)

If µ ≷ r ⇒ dT??

dµ = dT?(a?,µ)dµ = ∂µT

?(µ, a?) +∂aT?(µ, a?)da

?(µ)dµ <

0 ∧ dT??

db = dT?(a?,b)db = ∂bT

?(b, a?) + ∂aT?(b, a?)da

?(b)db > 0

If µ > r ⇒ dT??

dσ2 = dT?(a?,σ2)dσ2 =

∂σ2T ?(σ2, a?) + ∂aT?(σ2, a?)da

?(σ2)dσ2 > 0

If µ < r ⇒ dT??

dσ2 = dT?(a?,σ2)dσ2 =

∂σ2T ?(σ2, a?) + ∂aT?(σ2, a?)da

?(σ2)dσ2 =

− ab(r−µ)µx?r(r1−1)

∂r1∂σ2 ( 1

r1−1 +r1(ln(

1w)+ 1

r1−1)

(r1+1)(w r1−1)−r1w r1).

= −(r1ln( 1w ) + 1−w r1

1−r1)

⇒ (w ≶ w̄ > e1

r1−1 ⇔ T ?(a?) ≶ T̄ = 1µ ln(w̄)⇒ dT??

dσ2 ≷ 0)

Appendix7: Comparative statics b?

dp = −Hbp(b?,p)

Hbb(b?,p), p ∈ (σ2, µ, a), −Hbb(b?, p) > 0, if b = b?

Hbσ2 = − ar(r1−1)b

?−r1−1( x0r(r1−1)a(r−µ)r1 )r1 ∂r1

∂σ2 (r1(ln( x0r(r1−1)a(r−µ)r1 )− ln(b?)) +

= r1ln( 1v ) + 1

Hbµ = − ar(r1−1)b

?−r1−1( x0r(r1−1)a(r−µ)r1 )r1(r1

∂r1∂µ (ln( x0r(r1−1)

a(r−µ)r1 )− ln(b?)) +r21

r−µ + ∂r1∂µ ) + x0

b?2(µ−r)2=

ab?r(r1−1)(−v−r1 ∂r1

∂µ (r1ln( 1v ) + 1) + r1

r−µ(v−1 − r1v−r1))

Hba = r1b?r a

−r1( x0r(r1−1)b?(r−µ)r1 )r1

Appendix7: Comparative statics b?

If v ≶ e1r1 ⇔ T ?(b?) ≶ 1

µr1⇒ Hbσ2 ≷ 0⇒ db?

dσ2 ≷ 0

If e1r1 > r

1r1−1

1 ⇒ (v ≷ v̄ ⇔ T ?(b?) ≷ T̄2 = 1µ ln(v̄)⇒ Hbµ ≷ 0⇒

dµ ≷ 0) , if µ > r ⇒ e1r1 < v̄ < e

(r−µ) ∂r1∂µ ,

If e1r1 < r

1r1−1

1 ⇒ (v ≷ v̄ ⇔ T ?(b?) ≷ T̄2 = 1µ ln(v̄)⇒ Hbµ ≷ 0⇒

dµ ≷ 0) , if µ > r ⇒ 1 < v̄ < e1r1 , if µ < r ⇒ 1 < v̄ < r

1r1−1

Hba > 0⇒ db?

da > 0

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