# What is Newton's Method in Commerce

## Newton's approximation method

The **Newtonian approximation method** (*Newton method*) works like this:

One takes a point P the **Net present value curve** out (instead of two as with linear interpolation),

forms the **tangent** to the capital value curve at the point $ \ P = (i_1, C_0 (i_1)) $ and

uses the **Intersection** the tangent with the abscissa as $ \ i_2 $ and thus as *Approximation* for the** internal rate of return i *.**

The **Slope of the tangent** can be determined by tan $ \ \ alpha = {C_0 (i_1) \ over i_2-i_1} $, the angle here is the angle that the tangent forms with the positive direction of the abscissa.

Since the slopes of the tangents and the slope of the capital value curve at point P are identical, one can also write $ \ C_0´ (i_1) = {{C_0 (i_1) \ cdot P} \ over i_2-i_1} $.

The idea is made clear by the following picture:

### Formula for calculation

Solved for $ \ i_2 $ you get the *Approximation formula* Newton's approximation method. So note the**Derivation** in the denominator!

### Scheme for the Newton method

Here, too, a scheme for better manageability:

**Newton's approximation method:**

1. Find an interest rate $ \ i_1 $ for which the net present value $ \ C_0 (i_1) $ is already quite close to 0. Keep this net present value $ \ C_0 (i_1) $ for further calculations and name the point $ \ P = (i_1, C_0 (i_1)) $.

2. Form the capital value function as a function of the discount rate i.

3. Derive this according to the variable i.

4. Then insert the special discount rate $ \ i_1 $, i.e. form $ \ C_0´ (i_1) $.

5. Put the values from step 1 and step 4 into the following formula $$ \ i_2 = i_1 - {C_0 (i_1) \ over C_0´ (i_1)} $$ The result $ \ i_2 $ is an approximation for the internal rate of return i *.

### Example for the Newton method

We count this scheme on **example** by:

$ A_0 $ = € 5,000, $ E_1 $ = € 2,000, $ E_2 $ = € 3,000, $ E_3 $ = € 1,000.

$ \ I_1 $ = 10% is ideal. The associated net present value is

$$ \ C_0 (i_1) = + 48.8355 € $$ that is "quite close" to zero. The net present value function for general i is $$ \ C_0 (i) = - 5,000 + {2,000 \ over 1 + i} + {3,000 \ over (1 + i) ^ 2} + {1,000 \ over (1 + i ) ^ 3} $$ The derivation of this net present value function is: $$ \ C_0 (i) = {-2,000 \ over (1 + i) ^ 2} - {6,000 \ over (1 + i) ^ 3} - {3,000 \ over (1 + i) ^ 4} $$ At the point $ \ i_1 = 0.1 $ this is specifically $$ \ C_0´ (0.1) = - 8.209.82 $$ Inserting it into the formula results in $ $ \ i_2 = i_1 - {C_0 (i_1) \ over C_0´ (i_1)} = 0.1 - ({48.8355 \ over -8.209.82}) = 0.1059 = 10.59 \ \% $$

**Newtonian approximation method**is so

*more accurate*and so all the more

*better*as the

**linear interpolation**, the closer the point P, i.e. ($ \ i_1, C_0 (i_1) $), is to the abscissa. The Newtonian approximation method is to be preferred to linear interpolation, since it allows more precise solutions to be calculated more quickly.

**Example 19:**

The following investment is given.

year | 0 | 1 | 2 | 3 | 4 |

Payment series | -800 | 30 | 100 | -50 | 2.000 |

**Approximate** the **internal rate of return** the next investment with the **Newton's method**. With the Newton method one calculates as follows: With $ \ i_1 = 25 $% one has $ \ C_0 = 81.60 € $. The net present value is generally $$ \ C_0 = -800 + {30 \ over (1 + i)} + {100 \ over (1 + i) ^ 2} + -50 \ cdot {1 \ over (1 + i ) ^ 3} +2,000 \ cdot {1 \ over (1 + i) ^ 4} $$ The derivation of this capital value function is $$ \ C_0 = {-30 \ over (1 + i) ^ 2} - { 200 \ over (1 + i) ^ 3} + {150 \ over (1 + i) ^ 4} - {8,000 \ over (1 + i) ^ 5} $$ At the point $ \ i_1 = 0.25 $ one gets $ \ C_0 (0.25) = -322,304 $. So you use:

$$ \ i_2 = i_1 - {C_0 (i_1) \ over C_0´ (i_1)} = 0.25 - {81.60 \ over (-2,681.60)} = 0.2804 = 28.04 \ \% $ $

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