# Linear interpolation

A straight-line estimation method for determining an intermediate value.

## Contents

Example 1: Interpolation

Consider a set of cashflows which has:

Net present value (NPV) of +\$4m at a yield of 5%.

NPV of -\$4m at a yield of 6%.

Using linear interpolation, the estimated yield at which the cashflows have an NPV of \$0 is given by:

5% + ( +4 / ( +4 - -4) ) x (6 - 5)%

5% + ( +4 / +8 ) x 1%

5% + 0.5%

= 5.5%.

5.5% is the estimated internal rate of return (IRR) of the cashflows.

## Interpolation and Iteration

Interpolation is often used in conjunction with Iteration.

Using iteration, the straight-line estimated IRR of 5.5% would then be used, in turn, to recalculate the NPV at the estimated IRR of 5.5%, producing a recalculated NPV even closer to \$0.

5.5% and the recalculated NPV would then be used with interpolation once again to further refine the estimate of the IRR.

This iteration process can be repeated as often as required until the result converges on a sufficiently stable final figure.

## Extrapolation

Another closely related linear estimation technique is extrapolation.

This involves the straight-line estimation of values outside the range of the sample data used to do the estimation with.

Example 2: Extrapolation

Using the following data to estimate net present value (NPV) at a yield of 7%, using extrapolation:

NPV of +\$4m at a yield of 5%.

NPV of -\$4m at a yield of 6%.

Solution

Based on the sample data, for every 1% increase in the yield, the NPV moved by:

-\$4m - \$4m = -\$8m

Extrapolating this trend to a yield of 7%, this is a further increase in the yield of 7 - 6 = 1%.

The NPV would be modelled to fall from -\$4m to:

= -\$4m - \$8m

= -\$12m.