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.
The decision tree for the problem is shown below.
Note here that in that decision tree we have included an alternative relating to not building a plant (since it may not be profitable to build a plant). Below we carry out step 1 of the decision tree solution procedure which involves calculating the total profit for each of the paths from the initial node to the terminal nodes. We have also numbered all the nodes in the decision tree from 0 to 13.
Total revenue = 10 x 5 x 25 (10 years at 5,000 tonnes a year at £25
per tonne)
Total cost = 30 + 200 + (10 x 100) (development program cost + capital
cost + 10 years running cost)
Total profit = 20 (all figures in £K)
Total revenue = (1 x 5 x 30) + (9 x 5 x 25)
Total cost = 30 + 200 + (10 x 100)
Total profit = 45
Total revenue = (2 x 5 x 30) + (8 x 5 x 25)
Total cost = 30 + 200 + (10 x 100)
Total profit = 70
Total revenue = (1 x 2 x 50) + (9 x 2 x 30)
Total cost = 30 + 100 + (10 x 50)
Total profit = 10
Total revenue = (2 x 2 x 50) + (8 x 2 x 30)
Total cost = 30 + 100 + (10 x 50)
Total profit = 50
Total revenue = (3 x 2 x 50) + (7 x 2 x 30)
Total cost = 30 + 100 + (10 x 50)
Total profit = 90
Total revenue = 0
Total cost = 30
Total profit = -30
Total revenue = 0
Total cost = 30
Total profit = -30
Total revenue = 0
Total cost = 0
Total profit = 0
Hence we can form the table below indicating, for each branch, the total profit involved in that branch.
Terminal node Total profit (£K) 5 20 6 45 7 70 8 10 9 50 10 90 11 -30 12 -30 13 0
We can now carry out the second step of the decision tree solution procedure where we work from the right-hand side of the diagram back to the left-hand side.
Consider chance node 2 (with branches to terminal nodes 5, 6 and 7 emanating from it). The expected monetary value (EMV) for this chance node is given by
0.25 x (20) + 0.5 x (45) + 0.25 x (70) = 45 node 5 node 6 node 7
Consider chance node 4 then the EMV for this chance node is given by
0.25 x (10) + 0.5 x (50) + 0.25 x (90) = 50 node 8 node 9 node 10
Then for the decision node (node 4) relating to whether to build a large/small/no plant we have the three alternatives
(3) large plant EMV = 45 (calculated above)
(4) small plant EMV = 50 (calculated above)
(5) no plant EMV = -30 (terminal node 11)
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 continue the process. The EMV for chance node 1 is given by
0.7 x (50) + 0.3 x (-30) = 26 plant decision node node 12
Hence at the initial decision node relating to whether to invest 30K in the development program or not we have the two alternatives
(1) invest 30K EMV = 26 (calculated above)
(2) don't invest 30K EMV = 0 (terminal node 13)
Therefore our recommendation is that Nova should invest £30K in the development program with an expected monetary value of £26K. We would also anticipate (at this stage) building a small production plant (recall the alternative we chose at the decision node relating to the size of plant to be built). However it is plain that we will need to review the plant size decision once the development program has been completed.
Note that the actual monetary outcome (if we invest £30K in the development program) will be one of the terminal nodes 5 to 12 i.e. a value chosen from [20, 45, 70, 10, 50, 90, -30, -30] so that the downside risk is that we will lose £30K (depending upon chance events and future decisions).