#| purl = FALSE, #| include = FALSE # read vignette source chunks from corresponding testthat script knitr::read_chunk( file.path("..", "tests", "testthat", "test-model-ShaleGas.R") ) # read vignette build utility functions knitr::read_chunk(file.path("vutils.R"))
#| include = FALSE, #| purl = FALSE knitr::opts_chunk$set( collapse = TRUE, echo = FALSE, comment = "#>" )
library(rdecision) # nolint
Kaminski et al [-@kaminski2018] (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 KamiĆski
et al [-@kaminski2018].
Kaminski et al [-@kaminski2018] 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 $'000). The investor can build a gas extraction unit for a cost of 300. If gas is found, the profit will amount to 2,500 (if not there will be no profit, and no possibility of selling the land). Geological tests can be performed for a cost of 50, and will produce either a positive or negative signal. The sensitivity amounts to 90% and the specificity amounts to 70%. The installation can be built after the test or the land may be sold for 1000 (600) after a positive (negative) test result.
The model, comprising three decision nodes, four chance nodes, nine leaf nodes and 15 edges, is constructed as follows. Costs, benefits and probabilities are associated with each edge, which must be an Action or a Reaction object, see figure.
#| create-model, #| echo = TRUE
#| purl = FALSE, #| results = "hide", #| fig.keep = "all", #| fig.align = "center", #| fig.cap = "Decision tree used in the shale gas problem" DT$draw(border = TRUE)
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.
#| evaluate, #| echo = TRUE
This gives the following pay-off for each strategy:
#| echo = FALSE, #| results = "asis" RES[, "Run"] <- NULL RES[, "Probability"] <- NULL RES[, "Utility"] <- NULL RES[, "QALY"] <- NULL knitr::kable(RES, row.names = FALSE)
imax <- which.max(RES[, "Payoff"]) popt <- paste( RES[[imax, "d1"]], RES[[imax, "d2"]], RES[[imax, "d3"]], sep = ";" ) popt <- 42
The optimal strategy is r popt
, i.e., test, sell if negative and
dig otherwise. The expected pay-off from this strategy is
r RES[[imax, "Payoff"]]
.
Any scripts or data that you put into this service are public.
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.