# Decision tree with three decision nodes (Kaminski 2018)

### Shale gas

#### Andrew J. Sims

#### July 2020

# Introduction

Kaminski et al1 (Fig 7) provide an example of a decision tree with multiple decision nodes, including some that are descendants of another decision node. This vignette illustrates how `rdecision` can be used to model a complex decision tree, using the example from Figure 7 of Kamiński et al1.

# The problem

Kaminski et al1 state the problem as follows:

> Consider an investor owning a plot of land, possibly (a priori probability amounting to 70%) hiding shale gas layers. The plot can be sold immediately (800, all prices in $’000). The investor can build a gas extraction unit for a cost of 300. If gas is found, the profit will amount to 2,500 (if not there will be no profit, and no possibility of selling the land). Geological tests can be performed for a cost of 50, and will produce either a positive or negative signal. The sensitivity amounts to 90% and the specificity amounts to 70%. The installation can be built after the test or the land may be sold for 1000 (600) after a positive (negative) test result.

# Creating the model

## Constructing the tree

The model, comprising three decision nodes, four chance nodes, nine leaf nodes and 15 edges, is constructed as follows. Costs, benefits and probabilities are associated with each edge, which must be an Action or a Reaction object, see figure.

    # nodes
    d1 <- DecisionNode$new("d1")
    d2 <- DecisionNode$new("d2")
    d3 <- DecisionNode$new("d3")
    c1 <- ChanceNode$new("c1")
    c2 <- ChanceNode$new("c2")
    c3 <- ChanceNode$new("c3")
    c4 <- ChanceNode$new("c4")
    t1 <- LeafNode$new("t1")
    t2 <- LeafNode$new("t2")
    t3 <- LeafNode$new("t3")
    t4 <- LeafNode$new("t4")
    t5 <- LeafNode$new("t5")
    t6 <- LeafNode$new("t6")
    t7 <- LeafNode$new("t7")
    t8 <- LeafNode$new("t8")
    t9 <- LeafNode$new("t9")
    # probabilities
    p.sens <- 0.9
    p.spec <- 0.7
    p.gas <- 0.7
    p.nogas <- 1.0 - p.gas
    p.ptest <- p.sens * p.gas + (1.0 - p.spec) * p.nogas
    p.ntest <- (1.0 - p.sens) * p.gas + p.spec * p.nogas
    p.gas.ptest <- p.sens * p.gas / p.ptest
    p.gas.ntest <- (1.0 - p.sens) * p.gas / p.ntest
    # edges
    E <- list(
      Action$new(d1, t1, "sell", benefit = 800.0),
      Action$new(d1, c1, "dig", cost = 300.0),
      Reaction$new(c1, t2, p = p.gas, benefit = 2500.0, label = "gas"),
      Reaction$new(c1, t3, p = p.nogas, label = "no gas"),
      Action$new(d1, c2, "test", cost = 50.0),
      Reaction$new(c2, d2, p = p.ntest, label = "negative"),
      Action$new(d2, t4, "sell", benefit = 600.0),
      Action$new(d2, c3, "dig", cost = 300.0),
      Reaction$new(c3, t5, p = p.gas.ntest, benefit = 2500.0, label = "gas"),
      Reaction$new(c3, t6, p = (1.0 - p.gas.ntest), label = "no gas"),
      Reaction$new(c2, d3, p = p.ptest, label = "positive"),
      Action$new(d3, t7, "sell", benefit = 1000.0),
      Action$new(d3, c4, "dig", cost = 300.0),
      Reaction$new(c4, t8, p = p.gas.ptest, benefit = 2500.0, label = "gas"),
      Reaction$new(c4, t9, p = (1.0 - p.gas.ptest), label = "no gas")
    )
    # tree
    V <- list(d1, d2, d3,  c1, c2, c3, c4,  t1, t2, t3, t4, t5, t6, t7, t8, t9)
    DT <- DecisionTree$new(V, E)

## Tree diagram

Decision tree used in the shale gas problem

# Evaluating the strategies

There are a total of 12 possible strategies (3 choices from node `d1` \(\times\) 2 choices at node `d2` \(\times\) 2 choices at node `d3`). But some of these are not unique. For example if the choice at node `d1` is “sell”, the choices at nodes `d2` and `d3` (4 possible combinations) are unimportant; all four such strategies are identical.

Method `evaluate` calculates the expected cost, benefit and utility of each traversable path for each strategy, and aggregates by strategy. The results for the gas problem are computed as follows. Pay-off is defined as benefit minus cost.

    # find optimal strategies
    RES <- DT$evaluate()
    RES[, "Payoff"] <- RES[, "Benefit"] - RES[, "Cost"]

This gives the following pay-off for each strategy:

d1 | d2 | d3 | Cost | Benefit | Payoff
--- | --- | --- | --- | --- | ---
dig | dig | dig | 300 | 1750 | 1450
dig | dig | sell | 300 | 1750 | 1450
dig | sell | dig | 300 | 1750 | 1450
dig | sell | sell | 300 | 1750 | 1450
sell | dig | dig | 0 | 800 | 800
sell | dig | sell | 0 | 800 | 800
sell | sell | dig | 0 | 800 | 800
sell | sell | sell | 0 | 800 | 800
test | dig | dig | 350 | 1750 | 1400
test | dig | sell | 134 | 895 | 761
test | sell | dig | 266 | 1743 | 1477
test | sell | sell | 50 | 888 | 838

The optimal strategy is test;sell;dig, i.e., test, sell if negative and dig otherwise. The expected pay-off from this strategy is 1477.

# References

1.
Kamiński, B., Jakubczyk, M. & Szufel, P. A framework for sensitivity analysis of decision trees. Central European Journal of Operational Research 26, 135–159 (2018).