## Sequential Compound Option

A sequential compound option exists when a project has multiple phases and latter phases depend on the success of previous phases. Figures 7.16 to 7.19 show the calculation of a sequential compound option. Suppose a project has two phases, where the first phase has a one-year expiration that costs \$500 million. The second phase's expiration is three years and costs \$700 million. Using Monte Carlo simulation, the implied volatility of the logarithmic returns on the projected expected future cash flows is calculated to be 20 percent. The risk-free rate on a riskless asset for the next three years is found to be yielding 7.7 percent. The static valuation of future profitability using a discounted cash flow model (that is, the present value of the future cash flows discounted at an appropriate market risk-adjusted discount rate) is found to be \$1,000 million. The underlying asset lattice is seen in Figure 7.16.

The calculation of this initial underlying asset lattice is similar to previous option types by first calculating the up and down factors and evolving the present value of the future cash flow for the next three years.

Figure 7.17 shows the second step in calculating the equity lattice of the second option. The analysis requires the calculation of the longer-term option first and then the shorter-term option because the value of a compound option is based on another option. At node V, the value is \$1,122.1

Binomial Approach - Step I: Binomial Approach - Step II: Equity Lattice

Maximum between Executing orO Execute = 1822.1 -Investment Cost2 = \$1122.1 This is the intermediate Equity Valuation Lattice required to solve the Compound Option

This is the intermediate Equity Valuation Lattice required to solve the Compound Option

Maximum Executing or Keeping the Option Open Executing = 670.3 - investment Cost 2 = -\$29.7 Keeping Option Open = [P(118.7)+(1-P)(0.0)] exp(-rf*dt) = \$71.33

FIGURE 7.17 Sequential Compound Option (Equity Lattice)

million because it is the maximum between zero and executing the option through \$1,822.1 — \$700 = \$1,122.1 million. The intermediate node W is \$71.3 million, its being the maximum between executing the option \$670.3 — \$700 = —\$29.7 million and keeping the option open with [(P)(\$118.7) + (1 — P)(\$0.0)]exp[(—rf )(Si)] = \$71.3 million, which is the maximum value. This calculation assumes a 7.7 percent risk-free rate rf, a time-step 8t of 1, and a risk-neutral probability P of 0.6488. Using this backward induction technique, this first equity lattice is back-calculated to the starting point to obtain the value of \$449.5 million.

Figure 7.18 shows the equity valuation of the first, shorter-term option. The analysis on this lattice depends on the lattice of the second, longer-term option as shown in Figure 7.17. For instance, node X has a value of \$121.3 million, which is the maximum between zero and executing the option \$621.27 — \$500 = \$121.27 million. Notice that \$621.27 is the value of the second, longer-term equity lattice as shown in Figure 7.17 and \$500 is the implementation cost on the first option.

Node Y on the other hand uses a backward induction calculation, where the value \$72.86 million is obtained through the maximization between executing the option \$449.5 — \$500 = —\$50.5 million and keeping the option open with [(P)(\$121.3) + (1 — P)(\$0.0)]exp[(—rf )(Si)] = \$72.86 million, which is the maximum value. The maximum value comes from keeping the option open. This calculation assumes a 7.7 percent risk-free rate rf, a time-step 8t of 1, and a risk-neutral probability P of 0.6488. Again notice that \$500 million is the implementation cost of the first option.

Binomial Approach - Step III: Option Valuation Lattice

Maximum between Executing or 0 Execute = 621.27 - Investment Cost 1 = \$121.3

Maximum Executing or Keeping the Option Open Executing = 449.5 - Investment Cost 1 = -\$50.5 Keeping Option Open = [P(121.3)+(1-P)(0)] exp(-rf*dt) = \$72.86.

This value of this Compound Option is \$72.86

Maximum Executing or Keeping the Option Open Executing = 449.5 - Investment Cost 1 = -\$50.5 Keeping Option Open = [P(121.3)+(1-P)(0)] exp(-rf*dt) = \$72.86.

FIGURE 7.18 Sequential Compound Option (Valuation Lattice)

Figure 7.19 shows the combined option analysis from Figures 7.17 and 7.18, complete with decision points on when to invest in the first and second rounds versus keeping the option to invest open for the future.

Binomial Approach - Step IV: Combined Option Valuation Lattice

SECOND OPTION

FIRST OPTION

FIRST OPTION 843.7 OPEN

SECOND OPTION

843.7 OPEN

351.9 OPEN

71.3 OPEN

INVEST 2nd ROUND

>521.4 INVEST 2nd ROUND

>118.7 INVEST 2nd ROUND

INVEST 2nd ROUND

351.9 OPEN

>521.4 INVEST 2nd ROUND

71.3 OPEN

>118.7 INVEST 2nd ROUND

DON'T INVEST