J E Beasley
OR-Notes are a series of introductory notes on topics that fall under the broad heading of the ﬁeld of
operations research (OR). They were originally used by me in an introductory OR course I give at
Imperial College. They are now available for use by any students and teachers interested in OR
subject to the following conditions.
A full list of the topics available in OR-Notes can be found here.
In many problems chance (or probability) plays an important role. Decision analysis is the general
name that is given to techniques for analysing problems containing risk/uncertainty/probabilities.
Decision trees are one speciﬁc decision analysis technique and we will illustrate the technique by
use of an example.
A company faces a decision with respect to a product (codenamed M997) developed by one of its
research laboratories. It has to decide whether to proceed to test market M997 or whether to drop it
completely. It is estimated that test marketing will cost £100K. Past experience indicates that only
30% of products are successful in test market.
If M997 is successful at the test market stage then the company faces a further decision relating to
the size of plant to set up to produce M997. A small plant will cost £150K to build and produce 2000
units a year whilst a large plant will cost £250K to build but produce 4000 units a year.
The marketing department have estimated that there is a 40% chance that the competition will
respond with a similar product and that the price per unit sold (in £) will be as follows (assuming all
Large plant Small plant
Competition respond 20 35
Competition do not respond 50 65
Assuming that the life of the market for M997 is estimated to be 7 years and that the yearly plant
running costs are £50K (both sizes of plant - to make the numbers easier!) should the company go
ahead and test market M997?
Although the above example is somewhat simpliﬁed it plainly represents the type of decision that
often has to be made about new products.
In particular note how we cannot separate the test market decision from any decisions about
the future proﬁtability (if any) of the product if test marketing is successful.
To enable us to see what is going on consider the ﬁgure below where we have drawn the decision
tree for the problem.
In that ﬁgure we have three types of node represented:
chance nodes; and
Decision nodes represent points at which the company has to make a choice of one alternative from
a number of possible alternatives e.g. at the ﬁrst decision node the company has to choose one of the
two alternatives "drop M997" or "test market M997".
Chance nodes represent points at which chance, or probability, plays a dominant role and reﬂect
alternatives over which the company has (effectively) no control.
Terminal nodes represent the ends of paths from left to right through the decision tree.
It is worth saying here that the difﬁcult part of the decision tree technique is drawing up a
diagram such as the ﬁgure above from the written description of the problem. Once that has
been done the solution procedure is quite straightforward. Note here that most, but not all,
decision trees start with a decision node. One tip that may help you to draw decision trees is to
ask yourself the question "What happens next?" at each point in the tree as you draw it.
Note here the inclusion of the "no plant" alternative at the plant size decision node. This is necessary
because it simply may not be proﬁtable to build any plant (large or small) even if the product is
successful in test market. It is common in decision tree problems to ﬁnd that at decision nodes we
need a "do nothing" alternative which is an implicit decision that can be taken.
Note that it is important for the decision tree to be drawn so that there is a unique path in the tree
from the initial node to each of the terminal nodes.
To ease the discussion of the decision tree we have numbered the nodes (decision/chance/terminal)
1,2,3,...,12. At each decision node we have also numbered the alternatives, at node 1 we have
alternatives 1 and 2 and at node 5 alternatives 3, 4 and 5.
Although the decision tree diagram does help us to see more clearly the nature of the problem it has
not, so far, helped us to decide whether to drop M997 or whether to test market it (the decision we
are trying to make!). To do this we have two steps as illustrated below.
In these steps we will need to use information (numbers) relating to future sales, prices, costs,
etc. Whilst we may not be able to give accurate ﬁgures for these we need to factor such ﬁgures
into our calculations if we are to proceed. Investigating how our decision to test market or not
might change as these ﬁgures change (i.e. sensitivity analysis) can be done once we have
carried out the basic calculations using our assumed ﬁgures.
In this step we, for each path through the decision tree from the initial node to a terminal node of a
branch, work out the proﬁt (in £) involved in that path. Essentially in this step we work from the
left-hand side of the diagram to the right-hand side.
path to terminal node 2 - we drop M997
Total revenue = 0
Total cost = 0
Total proﬁt = 0
Note that we ignore here (and below) any money already spent on developing M997 (that being
a sunk cost, i.e. a cost that cannot be altered no matter what our future decisions are, so
logically has no part to play in deciding future decisions).
path to terminal node 4 - we test market M997 (cost £100K) but then ﬁnd it is not successful
so we drop it
Total revenue = 0
Total cost = 100
Total proﬁt = -100 (all ﬁgures in £K)
path to terminal node 7 - we test market M997 (cost £100K), ﬁnd it is successful, build a
small plant (cost £150K) and ﬁnd we are without competition (revenue for 7 years at 2000
units a year at £65 per unit = £910K)
Total revenue = 910
Total cost = 250 + 7x50 (running cost)
Total proﬁt = 310
path to terminal node 8 - we test market M997 (cost £100K), ﬁnd it is successful, build a
small plant (cost £150K) and ﬁnd we have competition (revenue for 7 years at 2000 units a
year at £35 per unit = £490K)
Total revenue = 490
Total cost = 250 + 7x50
Total proﬁt = -110
path to terminal node 10 - we test market M997 (cost £100K), ﬁnd it is successful, build a
large plant (cost £250K) and ﬁnd we are without competition (revenue for 7 years at 4000
units a year at £50 per unit = £1400K)
Total revenue = 1400
Total cost = 350 + 7x50
Total proﬁt = 700
path to terminal node 11 - we test market M997 (cost £100K), ﬁnd it is successful, build a
large plant (cost £250K) and ﬁnd we have competition (revenue for 7 years at 4000 units a
year at £20 per unit = £560K)
Total revenue = 560
Total cost = 350 + 7x50
Total proﬁt = -140
path to terminal node 12 - we test market M997 (cost £100K), ﬁnd it is successful, but decide
not to build a plant
Total revenue = 0
Total cost = 100
Total proﬁt = -100
Note that, as mentioned previously, we include this option because, even if the product is successful
in test market, we may not be able to make sufﬁcient revenue from it to cover any plant construction
and running costs.
Hence we can form the table below indicating, for each branch, the total proﬁt involved in that
branch from the initial node to the terminal node.
Terminal node Total profit (£K)
So far we have not made use of the probabilities in the problem - this we do in the second step
where we work from the right-hand side of the diagram back to the left-hand side.
Consider chance node 6 with branches to terminal nodes 7 and 8 emanating from it. The branch to
terminal node 7 occurs with probability 0.6 and total proﬁt 310K whilst the branch to terminal node
8 occurs with probability 0.4 and total proﬁt -110K. Hence the expected monetary value (EMV) of
this chance node is given by
0.6 x (310) + 0.4 x (-110) = 142 (£K)
Essentially this ﬁgure represents the expected (or average) proﬁt from this chance node (60% of the
time we get £310K and 40% of the time we get -£110K so on average we get (0.6 x (310) + 0.4 x
(-110)) = 142 (£K)).
The EMV for any chance node is deﬁned by "sum over all branches, the probability of the branch
multiplied by the monetary (£) value of the branch". Hence the EMV for chance node 9 with
branches to terminal nodes 10 and 11 emanating from it is given by
0.6 x (700) + 0.4 x (-140) = 364 (£K)
node 10 node 11
We can now picture the decision node relating to the size of plant to build as below where the
chance nodes have been replaced by their corresponding EMV's.
Hence at the plant decision node we have the three alternatives:
Alternative 3: build small plant EMV = 142K
Alternative 4: build large plant EMV = 364K
Alternative 5: build no plant EMV = -100K
It is clear that, in £ terms, alternative number 4 is the most attractive alternative and so we can
discard the other two alternatives, giving the revised decision tree shown below.
We can now repeat the process we carried out above.
The EMV for chance node 3 representing whether M997 is a success in test market or not is given
0.3 x (364) + 0.7 x (-100) = 39.2 (£K)
plant decision node node 4
Hence at the decision node representing whether to test market M997 or not we have the two
Alternative 1: drop M997 EMV = 0
Alternative 2: test market M997 EMV = 39.2K
It is clear that, in £ terms, alternative number 2 is preferable and so we should decide to test market
Let us be clear then about what we have decided as a result of the above process:
we should test market M997 and this decision has an expected monetary value (EMV) of
if M997 is successful in test market then we anticipate, at this stage, building a large plant
(recall the alternative we chose at the decision node relating to the size of plant to build).
However it is plain that in real life we will review this once test marketing has been
Note here that the EMV of our decision (39.2 in this case) DOES NOT reﬂect what will
actually happen - it is merely an average or expected value if we were to have the tree many
times - but if fact we have the tree once only. If we follow the path suggested above of test
marketing M997 then the actual monetary outcome will be one of [-100, 310, -110, 700, -140,
-100] corresponding to terminal nodes 4,7,8,10,11 and 12 depending upon future decisions and
Upside and downside
Conceptually one can think of the set of terminal nodes that can be reached as a result of our test
market decision as the set of possible futures. The best possible future corresponding to the decision
to test market M997 is called the upside and the worst possible future corresponding to the decision
to test market M997 is called the downside.
We need to be slightly careful here since, as noted above, the actual monetary outcomes will depend
both upon future decisions and chance events.
If we are committed to a large plant (assuming we are successful in test market) then the set of
possible futures is given by [-100, 700, -140] corresponding to terminal nodes 4,10 and 11 and
hence the upside is 700 (£K) and the downside is -140 (£K).
If we are not committed to a large plant (assuming we are successful in test market) then the set of
possible futures is given by [-100, 310, -110, 700, -140, -100] corresponding to terminal nodes
4,7,8,10,11 and 12 and hence the upside is 700 (£K) and the downside is -140 (£K).
Whilst in this particular example the upside and downside are the same irrespective of whether we
are committed to a large plant or not note how the possible futures are different depending upon
whether we are committed to a large plant or not.
The above problem was solved using the package, the input being shown below. The description of
the input is as follows. All data is entered by nodes in the tree. Each node has a type (d for decision
nodes and c for chance nodes), or is left blank for terminal nodes. All following nodes in the
decision tree are listed.
If a node comes from a chance node then it has a non-zero probability value assigned. If the node is
a terminal node then a payoff/cost is also assigned.
In the output shown below we have for each decision node the expected value and the appropriate
decision and for each chance node the expected value.
It is clear that the package agrees with the calculation we carried out before.
Note here that this package is also capable of drawing the tree for use from the input, so we can
check we have entered it correctly), as below.
Note here that since the decision tree calculation is so straightforward it is relatively easy to conduct
sensitivity analysis to see how the recommended course of action changes (if at all) in response to
Consider the decision tree given above. It is plain that the decision to test market is inﬂuenced by the
proﬁt of 700 (£K) that we will get if test marketing is successful and we choose to build a large plant
and we ﬁnd that we have no competition. Colloquially we could say that we are "following the
money". Hence we may vary this ﬁgure of 700 (and/or vary the probability that this outcome occurs)
to see if it changes the test market decision.
For example, suppose the probability that we have no competition with a large plant is no longer 0.6
but is instead 0.45. This implies that the probability that we have competition is hence 1-0.45=0.55.
Redoing the decision tree calculation (here using the package) we get:
Hence we can see that the initial decision (to test market) is still the optimal decision.
We can also conduct sensitivity analysis using a more systematic algebraic basis (i.e. assign a
symbol p to a given probability and work out algebraic expressions for EMVs). To see this suppose
that the probability of competition with a large plant is no longer 0.6 but is instead p. This implies
that the probability that we have competition is hence 1-p. Assume that we can leave the
probabilities for a small plant of competition/no competition unaltered. It is clear that as p decreases
we will at some stage prefer a small plant over a large plant (e.g. if p=0 then a small plant with EMV
of 142 would be preferable to a large plant with EMV of -140). We can therefore ask the logical
question "How small does p have to be before we prefer a small plant?"
To answer this question we have that we will be indifferent between a small and a large plant when
their EMVs are equal, i.e. when
p(700) + (1-p)(-140) = 142
i.e. when 840p - 140 = 142, so p = 282/840 = 0.3357
Hence if p drops below 0.3357 we would prefer a small plant to a large plant. This type of
systematic sensitivity analysis can be preferable in some circumstances to simply trying different
numbers and redoing the calculation to see what the effect is.
The basic decision tree technique presented above can be applied to any problem for which a
decision tree can be drawn. There are a number of extensions to the technique and we brieﬂy
consider four such extensions below. These relate to:
We consider each in turn.
In the example given above we were concerned with money received over 7 years. It is plain that £1
received in seven years time is worth less than £1 received now and a technique called discounting,
or discounted cash ﬂow, (involving ﬁnding the net present value of any sum of money) can be used
to overcome this difﬁculty. Applying discounting merely alters the numbers which are fed into the
decision tree so that we are dealing with monetary values on an equivalent (present-day) basis. It
does not effect the processing of the tree (which remains exactly as indicated above).
In the example given above we calculated a value for each chance node. Whilst we have used EMV
as the value of a chance node this choice is, in many respects, somewhat arbitrary and other ways of
calculating a value for a chance node have been suggested. To put it another way whilst it is a law of
the universe (in this particular corner of the space-time continuum) that E=mc2 it is not a law of the
universe that the value of a chance node must be the equal to the EMV value!
Reﬂect that EMV is an average value - but at a chance node we never see the average - something
happens once only (e.g. at chance node 6 we either see competition or not). Hence perhaps the
average value is misleading and we need to look at a chance node a different way.
If we were adverse to losing money and wished to take a conservative attitude to decision making
(risk) we might calculate the value of a chance node as the worst possible outcome that might occur
at that node. Such a strategy is often called a pessimistic strategy (e.g. such a strategy would assign
chance node 6 a value of -110 compared with an EMV of 142).
An alternative strategy would be the optimistic strategy of calculating the value of a chance node as
the best possible outcome that might occur at that node (e.g. such a strategy would assign chance
node 6 a value of 310 compared with an EMV of 142).
Yet another strategy would be to take the value of a chance node as equal to the most likely (highest
probability) outcome that might occur at that node (e.g. such a strategy would assign chance node 4
a value of 310 compared with an EMV of 142).
Alternatively we might take the value of a chance node to be some weighted combination of the
EMV and the values given by the optimistic and pessimistic strategies. The literature contains a
number of variations on this theme of changing the value of a chance node.
At each decision node we choose one of the alternative decisions at that node based upon an implicit
rule ("choose the alternative with the highest EMV"). Other rules are equally plausible (e.g. "choose
the alternative with the highest ROI" (ROI = return on investment = proﬁt divided by total
For example consider the small and large plants above. We saw that a small plant led to an EMV
(actually an expected net proﬁt) of 142. That involved a total investment of 100 for test marketing
plus 150 to build, so a ROI of 142/(100+150) = 56.8%
A large plant led to an EMV (actually an expected net proﬁt) of 364. That involved a total
investment of 100 for test marketing plus 250 to build, so a ROI of 364/(100+250) = 104%
Whilst, in this case ROI would lead to the same decision relating to the size of plant to build it could
have led to a different decision. For example had the EMV at chance node 9 been 175 then on an
EMV basis we would still have chosen at decision node 5 to build a large plant. But on a ROI basis
[175/(100+250) = 50%] we would chose to build a small plant.
Using monetary values in the decision tree implies, for example, that a loss of 200K is only twice as
bad as a loss of 100K. If the company doesn't have 200K to lose, but does have 100K, then it is plain
that they will regard losing 200K as much worse than losing 100K. Moreover note that often
decisions are made by people within the company. The company makes the proﬁt/loss, not the
people concerned with the decision.
Hence the idea of utility is to replace the monetary values at each terminal node in the decision tree
by utilities (points) which reﬂect the view of the decision maker (or company) to that sum of money
(e.g. a loss of 100K might get a utility value of -5 but a loss of 200K a utility value of -500). In
simple terms you can think of utilities as replacing £'s by points.
Trying to capture the decision-maker's translation of £'s to points is an imprecise process, but it does
enable the decision-maker's valid view of the problem in terms of the importance of £'s to them to be
incorporated. Once utility values have been found then we proceed as before.
Some more decision tree examples can be found here.