I'm currently looking into calculating the "option to switch use" to determine the benefit of the ability to switch between two technologies at any point in time (american option). This is also called an "exchange option" for financial options.

Trigeorgis calculates the value of flexibility in his book (p173) as: $F(A \rightarrow B) = S_0(A \rightarrow B) + S_1(A \rightarrow B) + S_2(A \rightarrow B)$

where $S_t$ is the value of the option at each point in time when switching from technology A to technology B. The value of the option is therefore calculated as the sum(!) of the options values from each point in time backwards to $t=0$.

Unfortunately there is no explanation there why this is treated different than any other calculation of a binomial lattice I found. In any other case I found there is just one backwards pass calculated to find the value at the root node. This final value is then treated as the value of the option.

Edit: Question: Why is Trigeorgis calculating the option value differently than everyone else?

Any insights would be greatly appreciated!

regards, Bernhard

  • $\begingroup$ Interesting. While I'm not exactly sure what you are asking, I'd be very curious to learn how you came up with spot price, correlation and volatility for 2 pieces of technology. $\endgroup$ – amdopt Apr 14 '17 at 18:27
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    $\begingroup$ It is stated that the volatility needs to be correlated between both assets, in most documents you can find they are just assumed to be equal. The value of the underlying is the PV of the project and the parameters for the binomial lattice are determined by using monte carlo analysis $\endgroup$ – Bernie Apr 14 '17 at 18:39
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    $\begingroup$ @amdopt you probably want to take a look at: "Mun - Real Options Analysis Tools & Techniques" $\endgroup$ – Bernie Apr 14 '17 at 18:42
  • $\begingroup$ Thanks @Bernie. I have used exchange options in the past but never with anything outside the realm of financial products. $\endgroup$ – amdopt Apr 14 '17 at 18:44
  • $\begingroup$ @amdopt I'm interested in everything related to that matter, do you know of any literature about valuation of exchange options by using binomial trees? $\endgroup$ – Bernie Apr 14 '17 at 18:46

hmm, Trigeorgis seems to be saying that the value of being able to switch at one of the times 1,2 and 3 is the same as the sum of being able to switch at each of the times.

This seems wrong to me since if you switch at time 1, the value of the ability to switch at time 2 becomes worthless.

Valuing an exchange option on a binomial tree is pretty easy -- all you do is a build up a tree of underlying values and discount back one step at a time, and the value at each node is the maximum of the discounted continuation value and the exercise value.

If the process for the ratio is nice, it is generally much easier to work with that than the two individual ones.

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  • $\begingroup$ well, with no switching costs associated, the option to switch back at a later time may not be worthless if there is more value to it. This could lead to switching all the time, which is fine in my example, but if you want to circumvent that you could introduce switching costs, which would create some kind of hysteresis. So do you think this has something to do if the option is worthless after execution or not? $\endgroup$ – Bernie Apr 15 '17 at 9:16
  • $\begingroup$ even if you could switch both ways for free, it wouldn't lead to the expression in Trigeorgis. I think he just had a brainstorm. $\endgroup$ – Mark Joshi Apr 15 '17 at 10:03

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