When there is capital rationing in more than one year, and some or all of the projects require additional finance in each year where there will be capital rationing, a different method is needed to identify the combination of projects that will maximise total NPV, where the projects are divisible.
The mathematical technique used to identify the NPV-maximising combination of projects is called linear programming, and the technique of linear programming you are most likely to encounter is the simplex method.
For your examination, you are not required to solve a linear programming problem.
However, you should understand how to formulate a linear programming problem, and how the simplex method works.
The method will be explained using a simple example. (The example is simple because there is capital rationing in two years only, and there are only three projects. The same technique can be applied to problems when there is capital rationing in more than two years, and when there are many different divisible projects available for selection.)
Example
A company has $9 million to invest in Year 0 and $6 million to invest in Year 1.
There are three projects available for investment, all similar in terms of risk. The amount of investment required and the NPV of each project are as follows:
Project Capital required in Year 0
Capital required in Year 1
NPV
$m $m $m
X 6 3 + 4.0
Y 2 1 + 2.0
Z 5 2 + 2.0
Required
Which projects should be selected if they are all divisible?
Answer
There is capital rationing in both Year 0 and Year 1 because in each year the capital available for investment is less than the total needed to invest in all three projects.
Formulating the linear programming problem
A linear programming problem is stated as an objective function that is subject to certain constraints or limitations. The objective function is to maximise or minimise something. With capital rationing, the objective function should be to maximise the total NPV, and the objective function can be stated as follows:
Objective function: Maximise 4x + 2y + 2z
where x, y and z are the proportion of Projects X, Y and Z respectively chosen for investment.
There are five constraints on investment. These are the amounts of capital available in Year 0 and Year 1, and the fact that it is impossible to invest in more than 100% of a project.
The five constraints can be stated as follows:
Year 0 capital 6x + 2y + 5z ≤ 10.2
Year 1 capital 3x + y + 2z ≤ 6.0
Maximum investment in X x ≤ 1.0
Maximum investment in Y y ≤ 1.0
Maximum investment in Z z ≤ 1.0
(Note: The symbol ≤ means ‘is less than or equal to’.)
There are also non-negativity constraints, which mean that the value of x, y and z cannot be a negative amount.
If there is a more complex problem, with capital rationing in more years, there will be a constraint for each year in which the capital rationing will occur.
Formulating and interpreting the initial simplex tableau
The simplex method can be used to find the optimal solution to the linear programming problem. The method is to test feasible solutions to the linear programming, one feasible solution at a time, until the optimal solution is found that maximises (or minimises) the value of the objective function.
Each feasible solution is tested in a table or ‘tableau’. A tableau should have:
one column for each variable in the problem, one column for each constraint, and a total column
one row for each variable in the solution and a row for the objective function.
There will always be exactly as many variables in the solution as there are constraints in the linear programming problem. In our example, this is 5.
In addition, we need to introduce some new variables into the problem. There should be one variable for each constraint.
Let Y0 = the number of unused capital in Year 0
Let Y1 = the number of unused capital in Year 1.
Let Ux = the proportion of Project X not invested in.
Let Uy = the proportion of Project X not invested in.
Let Uz = the proportion of Project X not invested in.
The initial tableau for this problem is constructed as follows:
Variable in the
solution x y z Y0 Y1 Ux Uy Uz Total (million)
Y0 6 2 5 1 0 0 0 0 9
Y1 3 1 2 0 1 0 0 0 6
Ux 1 0 0 0 0 1 0 0 1
Uy 0 1 0 0 0 0 1 0 1
Uz 0 0 1 0 0 0 0 1 1
Objective function - 4 - 2 - 2 0 0 0 0 0 0 Notice that the figures in each row of the x, y and z columns are taken directly from the linear programme, and the objective function on the bottom row shows the NPV of each project with a minus sign.
The initial tableau tests the feasible solution that there is 9 million of unused capital in Year 0 and 6 million of unused capital in Year 1. The proportion of each project not invested in is 1.0 (100%) Y0, Y1, Ux, Uy, andUz are therefore the five variables in this feasible solution, and the values of x, y and z are 0. The total NPV (bottom row, total column) is $0 million.
Testing other feasible solutions
This is clearly not the optimal solution, and the simplex method now produces another feasible solution where the value of the objective function is higher. The second tableau might be as follows:
Variable in the
solution x y z Y0 Y1 Ux Uy Uz Total (million)
Y0 0 2 5 1 0 - 6 0 0 3.0
Y1 0 1 2 0 1 - 3 0 0 3.0
Project X (x) 1 0 0 0 0 1 0 0 1.0
Uy 0 1 0 0 0 0 1 0 1.0
Uz 0 0 1 0 0 0 0 1 1.0
Objective function 0 - 2 - 2 0 0 4.0 0 0 4.0 This tableau has introduced x into the solution in place of Ux.
In this solution, there will be an investment in 100% of project X (since x = 1.0 in the tableau). This will leave unused capital of $3.0 million in Year 0 and $3.0 million in Year 2. The total NPV will be $4 million.
This is not the optimal solution, because there are still some minus signs in the objective function row, and for every unit of y or z that is introduced into the solution, the total NPV can be increased by $2 million.
The final solution and its interpretation: dual prices or shadow prices
Variable in the
solution x y z Y0 Y1 Ux Uy Uz Total (million)
Project Z (z) 0 2 5 1 0 - 6 0 0 0.2
Y1 0 1 2 0 1 - 3 0 0 1.6
Project X (x) 1 0 0 0 0 1 0 0 1.0
Project Y (y) 0 1 0 0 0 0 1 0 1.0
Uz 0 0 1 0 0 0 0 1 0.8
Objective function 0 0 0 0.6 0 1.6 0.8 0 6.4 The final tableau is shown here, with the solution that maximises total NPV. The problem in this example is fairly simple, so the solution is quite straightforward.
To maximise total NPV, the company should invest in 100% of Project X, 100% of Project Y and 20% of Project Z (since x = 1.0, y = 1.0 and z = 0.2). This will leave unused capital of $1.6 million in Year 1 (since Y1 = 1.6). The proportion of Project Z not invested in is 0.8.
Total NPV will be $6.4 million.
The solution also shows the dual prices or shadow prices of the variables that are not in the solution. These are Y0, Ux and Uy.
All the available capital in Year 0 is used up by the solution. The dual price indicates that if $1 of extra capital could be made available in Year 0, the total NPV could be increased by $0.4.
The maximum investment is made in project X, so Ux = 0 in the final solution.
The dual price for Ux indicates that if the maximum investment in project X could exceed 100%, the total NPV could be increased by $1.6 million for every additional project X that is available.
The maximum investment is made in project Y, so Uy = 0 in the final solution.
The dual price for Uy indicates that if the maximum investment in project Y could exceed 100%, the total NPV could be increased by $0.8 million for every additional project Y that is available.
Dual prices for projects are not particularly significant. However, the dual price or shadow price for capital is significant. It shows by how much total NPV could be increased if more capital could be made available in that year (given no change in any other constraint in the problem).
For example, this solution indicates that if the available capital in Year 0 could be increased by, say, $1 million, from $9 million to $10 million, there would be a different optimal solution and the total NPV for this solution would be higher by
$400,000 ($1 million × 0.4).