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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.
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 $’000). The investor can build a gas extraction unit for a cost of 300. If gas is found, the profit will amount to 2,500 (if not there will be no profit, and no possibility of selling the land). Geological tests can be performed for a cost of 50, and will produce either a positive or negative signal. The sensitivity amounts to 90% and the specificity amounts to 70%. The installation can be built after the test or the land may be sold for 1000 (600) after a positive (negative) test result.
The model, comprising three decision nodes, four chance nodes, nine leaf nodes and 15 edges, is constructed as follows. Costs, benefits and probabilities are associated with each edge, which must be an Action or a Reaction object, see figure.
# nodes
<- DecisionNode$new("d1")
d1 <- DecisionNode$new("d2")
d2 <- DecisionNode$new("d3")
d3 <- ChanceNode$new("c1")
c1 <- ChanceNode$new("c2")
c2 <- ChanceNode$new("c3")
c3 <- ChanceNode$new("c4")
c4 <- LeafNode$new("t1")
t1 <- LeafNode$new("t2")
t2 <- LeafNode$new("t3")
t3 <- LeafNode$new("t4")
t4 <- LeafNode$new("t5")
t5 <- LeafNode$new("t6")
t6 <- LeafNode$new("t7")
t7 <- LeafNode$new("t8")
t8 <- LeafNode$new("t9")
t9 # probabilities
<- 0.9
p.sens <- 0.7
p.spec <- 0.7
p.gas <- 1.0 - p.gas
p.nogas <- p.sens * p.gas + (1.0 - p.spec) * p.nogas
p.ptest <- (1.0 - p.sens) * p.gas + p.spec * p.nogas
p.ntest <- p.sens * p.gas / p.ptest
p.gas.ptest <- (1.0 - p.sens) * p.gas / p.ntest
p.gas.ntest # edges
<- list(
E $new(d1, t1, "sell", benefit = 800.0),
Action$new(d1, c1, "dig", cost = 300.0),
Action$new(c1, t2, p = p.gas, benefit = 2500.0, label = "gas"),
Reaction$new(c1, t3, p = p.nogas, label = "no gas"),
Reaction$new(d1, c2, "test", cost = 50.0),
Action$new(c2, d2, p = p.ntest, label = "negative"),
Reaction$new(d2, t4, "sell", benefit = 600.0),
Action$new(d2, c3, "dig", cost = 300.0),
Action$new(c3, t5, p = p.gas.ntest, benefit = 2500.0, label = "gas"),
Reaction$new(c3, t6, p = (1.0 - p.gas.ntest), label = "no gas"),
Reaction$new(c2, d3, p = p.ptest, label = "positive"),
Reaction$new(d3, t7, "sell", benefit = 1000.0),
Action$new(d3, c4, "dig", cost = 300.0),
Action$new(c4, t8, p = p.gas.ptest, benefit = 2500.0, label = "gas"),
Reaction$new(c4, t9, p = (1.0 - p.gas.ptest), label = "no gas")
Reaction
)# tree
<- list(d1, d2, d3, c1, c2, c3, c4, t1, t2, t3, t4, t5, t6, t7, t8, t9)
V <- DecisionTree$new(V, E) DT
Decision tree used in the shale gas problem
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
<- DT$evaluate()
RES "Payoff"] <- RES[, "Benefit"] - RES[, "Cost"] RES[,
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 42, i.e., test, sell if negative and dig otherwise. The expected pay-off from this strategy is 1477.
These binaries (installable software) and packages are in development.
They may not be fully stable and should be used with caution. We make no claims about them.