Topic 5: Real Options: application to adaptation to climate change
Shauna Phillips
School of Economics
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Housekeeping
› Mid semester exams › Essay assignment
› Coming attractions:
› Today, Week 9 – Real options (continued) › Week 10: Tools for mitigating risk
› Week 11: Diversification and hedging
› Week 12: drought policy
› Week 13: Revision?
Irreversible decisions
› Some decisions may be irreversible and investment costs are likely sunk – for example:
– Undertaking an R&D programme
– Establishing irrigation canals
– Strip mining coal from the Liverpool plains
› Others may be reversible, even if the reversal incurs some cost – for example:
– Switching from extensive grazing enterprise to irrigated cotton enterprise
Real Options (RO)
› RO extends typical agricultural investment analyses based on concepts of cost- benefit analysis and net present values.
› RO better represent incomplete knowledge and uncertainty.
› Useful to take option values into account when some adaptation decisions can be
costly-to-reverse/irreversible consequences for farmers.
› RO can be used to analyse why farmers may fail to adopt new varieties or technologies : reluctance to adopt due to uncertainties about the impact on production.
› Possible that the risks > the benefits predicted by the science.
Decision making as exercising an option
› Investment decisions are often characterised by:
– Investments being partially or completely irreversible (sunk cost) i.e. there is an
opportunity cost of investing now rather than waiting
– Uncertainty over the future rewards from the investment
– Some leeway about the timing of the investment i.e. can delay or postpone action in order to get more information about the future
› Investment opportunity costs are option values
Option values
› Option values directly reflect the flexibility of the firm/farmer to delay investment commitments
– Option values can be very large, and increase with the degree of uncertainty
– Option value signals the decision-maker’s willingness to wait for some uncertainty to resolve
Option values
› The value of a firm is the sum of its capital asset value plus the value of its options
– Consider a farm in which the farmer must plant wheat (and only wheat) every April, in the second week of that month, every year
– The farmer has no flexibility
– This farm will be less valuable than a comparable farm which is able to consider the available and emerging information to guide planting and crop choice decisions
– Flexibility is valuable
– Option values quantify this flexibility
Why use real options?
› A common alternative is the Net Present Value (NPV) formula:
– Reconciliation of benefits and costs through time, subject to some discount rate – When uncertainty is present, the NPV is often just wrong (well, it’s incomplete)
– A real options approach will provide better (more complete) answers
– Real options valuations provide a way to think about conditions suitable to commitment to an action, which maximise returns to the decision-maker.
Simple NPV criterion for investment
NPV=−I−I1−I2 +3 +
239 (1+) (1+) (1+)
Simple NPV criterion for investment
› The schedule of benefits and costs off into the future assumes a fixed scenario
– Somehow, we know the value of π9 with certainty
– The reality is that prices and other factors will likely change
– The high degree of variability of output and input prices (especially in agriculture) means the NPV framework can provide an unreliable picture of the decision problem
› The NPV framework doesn’t include values related to the option to delay investment, stop before completion, abandon after completion, or temporarily suspend activities
Simple NPV criterion for investment
› Sometimes, we might adjust the NPV criterion to include risk – For example, we often use expected values
E ( N P V ) = − I − I 1 − I 2 + E ( 3 ) + … + E ( 9 )
239 (1+) (1+) (1+)
– This approach assumes we know something about the future states of nature and the associated probabilities
Example: NPV with simple option
› Project description:
– Have the option to build a production facility at a cost of $800, which is built
almost (i.e. effectively) instantaneously
– Price of output is currently $100, but, next year, it will either go up or down by 50%, and will remain at the new level indefinitely
– The facility can produce only one unit of output per year, and the discount rate is 10%
Example: NPV with simple option
P1 = $150 P2 = $150
0.5 P0 = $100
0.5 P1 = $50 P2 = $50
Example: NPV with simple option
› So, we can express the NPV for the project as:
NPV =−800+ 100 =−800+1100=$300$0
› NPV rule says, “invest now.” But, suppose we delay our decision by 1
year, and only invest if the price goes up:
−800 150
NPV =0.5(1+0.1) +(1+0.1)+0.50=0.5772.7$386
Example: NPV with simple option
› So, waiting is good
› Delaying the commitment to invest gives us a higher NPV than committing
upfront: $386 vs. $300
– The value of the flexibility is $86
– Suggests that holding the option to wait is worth $86
Example: using option pricing
› Let’s solve again – this time, using option pricing
› Next year, if the price goes to $150, we exercise our option by paying
$800, and we obtain an asset worth:
V= 150=$1650 1 (1+0.1)
› This is higher than the $800 investment cost, so we exercise the option
– If price fell to $50, the asset would be worth only $550, so we wouldn’t exercise the option
Example: using option pricing
› Let F0 = value of investment opportunity today
› Let F1 = value of investment opportunity next year
› If price rises to $150:
F =(1+0.1)−800=850
150 t=0
› If price falls to $50, the option to invest will not be exercised, which means F1
› So, we know all the possible values for F1
Example: using option pricing
› What is the value of F0? It is the value of the investment option today – Can solve for F0 by creating a portfolio that has two components:
– The investment opportunity (a)
– A certain number of units of output which will be short-sold (b)
– We will need to identify the number of units of output to be short-sold, so the
portfolio is risk-free
– Want the payoffs from (a) and (b) to move in opposite directions so the portfolio will be risk-free
Example: using option pricing
› The value of this portfolio today is:
=F −nP =F −100n
› The value of this portfolio next year depends on P1: = F − nP
› IfP1 =$150,soF1 =$850: › IfP1 =$50,soF1 =$0:
=850−150n 1
=0−50n 1
Example: using option pricing
› Now, we find a value for n such that the portfolio is risk-free
– In other words, such that φ1 is independent of what happens to price
› This is done by equating the two possible states of φ1:
850 −150n = −50n
› This gives us a value of n which is 8.5
– With this value of n, φ1 = -$425, whether the price goes up or down
Example: using option pricing
› Next, we calculate the return from holding this portfolio
– In other words, we are trying to identify the capital gain [φ1 – φ0] less any
payments that must be made to hold the short position on the output [0.1nP ] 0
› The expected rate of capital gain on output is zero, because P0 = $100, and E(P1) = $100, but we have a discount rate of 10%
– Short selling of output will require a payment of 10% of P0, which is $10 per unit per year
Example: using option pricing
› Because we have short sold 8.5 units of output, we must pay $85 › The return from holding this portfolio over the year is then:
− −0.1nP = −(F −nP)−85 100100
− −85=−425−(F −850)−85 100
− −85=340−F 100
Example: using option pricing
› Because our portfolio is risk-free, we should earn the risk- free rate (i.e. discount rate of 10%) multiplied by the initial portfolio value:
340−F =0.1(F −850) 00
Return on 10% of portfolio original value
› So, F0 = $386
› This is the total value of possessing the opportunity to build the factory now or next year
Changes in initial price
› Extending the example above, we can consider what happens if the initial price is some value P0
› We will say, whatever P0 is, P1 will be either: – Withequalprobabilityof0.5
P =1.5P P =0.5P 1010
P2 = 1.5P0
P1 = 1.5P0
P1 = 0.5P0
P2 = 0.5P0
Changes in initial price
› We go through the same process as described earlier for setting up the risk-free portfolio:
V= P=11P
› We only invest if V1 exceeds $800, which means the value of the option next year is:
F =max0,11P−800 11
Changes in initial price
› Then, the value of the portfolio next year, if the price goes up is:
=11P−800−nP 111
=11(1.5P)−800−n(1.5P) 100
=16.5P −800−1.5nP 100
› The value if the price goes down is:
=−0.5nP 10
Changes in initial price
› Equating these two possible states of φ1 gives:
16.5P −800−1.5nP =−0.5nP
› So, n is given by:
n = 16.5 − 800 P
› This means that φ1, insulated from price risk, will be:
=−8.25P +400 10
Changes in initial price
› Calculate the return on the portfolio, recalling that we need to make 10% payment for holding the short position
– Solving for F0 gives us the value of the option:
F =7.5P −363.5
› We can use this value to calculate the expanded set of investment conditions
– How high would the initial price need to be to invest in the first period (i.e. immediately)?
– How low would the initial price need to be in order to never invest?
Maths for above slide: Find F0 again (this time a function of P0)
› Procedure as above: 1.find the capital gain on the portfolio, 2.which should earn the risk-free rate (i.e. discount rate of 10%) multiplied by the initial portfolio value:
– ɸ1 – ɸ0 – 0.1n P0 = -8.25 P0 +400 – (F0 – n P0 ) – 0.1(n P0 )
– = -8.25 P0 +400 – (F0 – (16.5-800/ P0) P0 ) – 0.1(16.5-800/ P0) P0 ) – = -8.25 P0 +400 – (F0 – 16.5 P0 +800) – 1.65 P0 + 80
– = -8.25 P0 +400 – F0 + 16.5 P0 – 800 – 1.65 P0 + 80
– Solving for capital gain on portfolio: 6.6 P0 – 320 – F0
Maths for above slide: Find F0 again (this time a function of P0)
› This capital gain should equal the risk-free rate (i.e. discount rate of 10%) multiplied by the initial portfolio value:
› 6.6P0 –320-F0 =0.1(F0 -nP0 )
› =0.1(F0 -16.5P0+800)
› 6.6P0 –320-F0 =0.1F0 -16.5P0+80
› 8.25P0 -400=1.1F0
› Finally, F0 = 7.5 P0 – 363.5
How high would the initial price need to be to invest in the first period?
› We need to compare the option value, given by… F =7.5P −363.5
› To the obligation to invest, given by:
F = V − 8 0 0 = P
= 1 1 P − 8 0 0 0
(1+0.1) t=0 t
F =11P −800 00
Obligation to invest
-200 0 -400 -600 -800
F =11P −800 00
140 160 180
Initial price (Po)
Option Value
Option to invest
1000 800 600 400 200 0
F =7.5P −363.5 00
0 20 40 60 80 100 120 140 160 180 -200
Initial price (Po)
Option value
How high would the initial price need to be to invest in the first period?
› An important point occurs where the obligation to invest is equal to the option to invest
F =7.5P −363.5 00
11P −800=7.5P −363.5 00
3.5P = 436.5 0
P =124.71 0
F =11P −800 00
1200 1000 800 600 400 200 0
Obligation- don’t wait
Option- wait
0 20 40 60 80 100 120 140 160 180 Initial price (Po)
Option Value
How high would the initial price need to be to invest in the first period?
› So, at prices of $124.71 and above, you would be willing to invest in the first period
– In effect, the decision-maker is indifferent between waiting and committing at a price of exactly $124.71
– The total value of the option will be found at that point
– We can substitute into either equation to get a value for F0
F =7.5P −363.5=7.5(124.71)−363.5=$571.82 00
How low would the initial price need to be in order to never invest?
› In essence, this question is asking for the value of P0 which yields an option value of zero (i.e. F0 = 0)
F =7.5P −363.5=0 00
7.5P =363.5 0
P = 48.47 0
› So, if the initial price is $48.47 or less, we will never invest in this project
Initial prices
› So, given initial prices between P0 = $48.47 and P0 = $124.71, we will wait until the next period to make a decision about commitment
– For all values in between, the option value is given by:
F =7.5P −363.5 00
– Note: the higher the initial price, the higher the value of holding this option – Why?
– Because higher initial prices mean higher possible future prices
Extension to more than two periods
› If we continue to extend the number of periods, our F0 function becomes ‘funky’
– We calculate the option value for each waiting time horizon (we just did this for a waiting time horizon of 1 year)
– If we do this for two years, because the prices continue to evolve, we will get
another F0 function:
F1 =7.5P −363.5 00
Extension to more than two periods
P2 = 2.25P0
P3 = 2.25P0
0.5 P1 = 0.5P0
P1 = 1.5P0
P2 = 0.75P0
P2 = 0.25P0
P3 = 0.75P0
P3 = 0.25P0
Extension to more than two periods
› We can calculate option values in the same way as above
– Note: our option value will be the highest F0 at a given initial price P0
– Computationally, we go through the calculations and select the highest option values and plot them out as a continuous option value function
Extension to more than two periods
1200 1000 800 600 400 200 0
0 20 40 60 80 100 120 140 160 180 Initial price (Po)
Option Value
Extension to more than two periods
1200 1000 800 600 400 200 0
0 20 40 60 80 100 120 140 160 180 Initial price (Po)
Option Value
Option pricing
› Fortunately, computational approaches for option pricing problems are well-established
– Real Options for Adaptive Decisions (ROADs) has been developed to employ a number of stochastic processes
› There is at least one known analytical solution, which is the Black-Scholes option pricing model for Geometric Brownian Motion
Option pricing
› We can model the above using an appropriate stochastic process
– The previously described process is like Geometric Brownian Motion without drift
(α = 0) and with +/-50% movements in price (σ = 0.5)
dP =Pdt+PdW tttt
dP =0.5PdW ttt
– These processes are represented in continuous time, so we would expect answers to differ slightly from those above
60.00 40.00 20.00 0.00
-55 -30 -5 20 45 70 95 120 145 170 195
Recall: Extension to more than two periods
1200 1000 800 600 400 200 0
0 20 40 60 80 100 120 140 160 180 Initial price (Po)
Option Value
60.00 40.00 20.00 0.00
-55 -30 -5 20 45 70 95 120 145 170 195
80.00 60.00 40.00 20.00 0.00
-50 -10 30
Implementation of Real Options model
› In practical terms, we can implement an options analysis for decisions to enter an investment
– For example, taking money out of the bank and putting it into wheat farming
› We can also consider the switch between two risky alternatives – So far, we have been comparing a risk-free and risky assets
– For example, switch from wheat to sheep
Implementation of Real Options model
• Consideration of a sequence of production regimes: need to calculate the value of staying in a current regime whilst retaining the option to switch into a different regime.
• Uncertainty creates a trade-off between responding
now (with less information) versus retaining the option to respond later
(with fresh information that might reduce the uncertainty).
Wheat Livestock dominant dominant
Background
• Switching production regimes can be viewed as crossing a threshold from one regime to another & to do this, the farming system is transformed eg. from wheat cropping to extensive grazing (crossing back is very costly).
• Timing of regime switches depends on the risks and uncertainties associated with alternative regimes
• Farmer might choose to switch now, or never, depending on how the climate is changing and the variability associated with that change.
Goyder’s Line
Mapped in 1860s viewed as a demarcation for the suitability of land for cropping or extensive grazing
North or the line rainfall is too low to support cropping
Source: Wikipedia
Normalised Vegetation Index of a transect of rainfall from Clare to Orroroo with an indication of points that are 10%, 20%, 30% and 50% wetter than
Orroroo (diagram provided by SARDI/CSIRO)
Growing season rainfall (GSR)
Clare :GSR of 485.91mm, (std dev =126.87mm) Orroroo: GSR is 224.86mm (std dev =72.28mm)
Assuming climate change lowers GSR, then the future at Clare is modelled along a transect from Clare to Orroroo : compare optimal farming decisions at these two sites.
If Clare experiences 50% decline in GSR, then Orroroo describes the future of wheat dominant agriculture at Clare.
Farmers will monitor the climate and switch agricultural production accordingly.
Transect straddles Goyder’s line: Clare below the line, Orroroo on the line and Hawker above the line.
1. Clare: farmers are almost certain to adopt wheat, less likely to adopt merino grazing, with little/no chance of abandoning agriculture.
2. Orroroo: farmers are less likely to be in wheat and more likely to adopt merino grazing, & unlikely to abandon agriculture.
3. Hawker: small chance of adopting wheat, greater chance of adopting merino grazing & chance of abandoning agriculture.
Looking forward
Climate change:
South Australia will be hotter and drier Goyder’s line moves south
Clare will become more like Orroroo and Orroroo will become more like Hawker.
wheat becomes less dominant, merinos increasingly adopted and some farms leave agriculture.
APSIM simulations for South Australi
system dynamics
Phase diagram
› The first step is to represent the time-series of gross margins for each system as a phase diagram –plot the year-to-year variations in gross margins, dx, on the y-axis against the current gross margin, x, on the x- axis.
› A phase diagram indicates for a given point what the gross margin is for that year as well as the change in the gross margin that results in the observation for the following year.
› Important characteristics of any system are the rate at which the system can change year to year and whether or not there is a tendency toward equilibrium
Estimated Ornstein-Uhlenbeck Parameters: gross margins in ($/ha)
ROADs and TRIPS
software to calculate OV, threshold
– GMs and probabilities
Options pricing equations
Similar functions from last RO Lecture 7: value matching and smooth pasting functions –recall mathematics of this beyond scope AREC3005
Entering wheat production at Clare
-0.2 -0.1 0.0 0.1
0.2 0.3 0.4
GM ($/ha ‘000)
0.5 0.6 0.7 0.8
W ($/ha ‘000)
Entering wheat production at Clare
Initial Transition Invariant Threshold
f(s,x,t,y) 2.0 1.5 1.0 0.5 0.0
0.0 0.2 0.4
0.6 0.8 1.0
Entering wheat product
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