Valuing the Option to Switch Technologies with a Binomial Lattice
Summary
The document discusses how to value the flexibility to switch between two technologies at several future dates. It contrasts a formula that adds option values across dates with standard American option valuation, where each lattice node compares the exercise value with the discounted value of continuing. The response argues that adding separate date-by-date values can overstate flexibility because exercising early removes the later opportunity to switch.
For an exchange option, the suggested method is to build a binomial tree for the underlying values and work backward one step at a time, taking the greater of continuation and exercise at each node. If the ratio of the two underlying processes has a convenient structure, modeling that ratio may simplify the calculation. The explanation is conceptual: it gives no numerical example or proof, and the appropriate lattice still depends on assumptions about the underlying processes and exercise dates.
Key ideas
- An American switching option should account for the fact that exercising can eliminate future switching opportunities.
- At each lattice node, compare the exercise payoff with the discounted continuation value.
- Backward induction produces the option value at the root of the tree.
- Modeling the ratio of the two underlying values may simplify an exchange-option calculation.
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Full text
# Real Options: Calculating the "option to switch use" using binomial lattices # Real Options: Calculating the "option to switch use" using binomial lattices 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 ## Answer by Mark Joshi (score 1) https://quant.stackexchange.com/a/33688 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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