Raphson Method

Objective: 1. Derive the Newton-Raphson method formula, 2. Develop the algorithm of the Newton-Raphson method, 3. Use the Newton-Raphson method to solve a nonlinear equation (in Power System) Definition: Newton's method (also known as the Newton–Raphson method), named after and Joseph Raphson, is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. History: Raphson's most notable work is Analysis Aequationum Universalis, which was published in 1690. It contains a method, now known as the Newton–Raphson method, for approximating the roots of an equation. Isaac Newton had developed a very similar formula in his Method of , written in 1671, but this work would not be published until 1736, nearly 50 years after Raphson's Analysis. However, Raphson's version of the method is simpler than Newton's, and is therefore generally considered superior. For this reason, it is Raphson's version of the method, rather than Newton's, that is to be found in textbooks today. Raphson also translated Newton's Arithmetica Universalis into English.

Newton’s Method: Faster, requires one initial guess Convergent not always guaranteed, Divergence at inflection points, Division by zero, Oscillations near local maximum and minimum,

Intuition behind Newton Raphson Method: The Newton-Raphson method is based on the principle that if the initial guess of the root of ( ) = 0 is at , then if one draws the tangent to the curve at ( ), the point where the tangent crosses the x -axis is an improved estimate of the root. 𝑓𝑓 𝑥𝑥 𝑥𝑥𝑖𝑖 𝑓𝑓 𝑥𝑥𝑖𝑖 𝑥𝑥𝑖𝑖+1

Derivation: Using the definition of the slope of a function at =

𝑥𝑥 𝑥𝑥𝑖𝑖

So starting with an initial guess, one can find the next guess, . One can repeat this process until one finds the root within a desirable tolerance. 𝑥𝑥𝑖𝑖 𝑥𝑥𝑖𝑖+1 Animation: https://www.intmath.com/applications-differentiation/newtons-method- interactive.php Numerical Example: Find the roots of the equation:

Using MATLAB: p = [1 -0.165 0 0.0003993]; r = roots(p) 0.14636 0.062378 -0.043737

Let us assume the initial guess of the root of f(x)=0 is 0.05. The estimate of the root for iteration 1 is:

The estimate of the root for iteration 2 is:

References: 1. Newton-Raphson Method of Solving a Nonlinear Equation (Link) 2. P Kundur, Power System Stability and Control 3. ECE 476: Power system analysis, Professor Tom Overbye Application of Newton Raphson Method in Power System Power Flow Analysis: obtain complete voltages angle and magnitude information for each bus in a power system for specified load and generator real power and voltage conditions. For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assume that bus one is the slack and Sbase = 100 MVA.

Line Z = 0.1j

One 1.000 pu Two 1.000 pu

0 MW 200 MW 0 MVR 100 MVR

θ2 − jj10 10 xY=  bus =  V2 jj10− 10

General power balance equations n Pi =∑ VVi k( G ik cosθθ ik +=− B ik sin ik ) P Gi P Di k=1 n Qi =∑ VVi k( G ik sinθθ ik −=− B ik cos ik ) Q Gi Q Di k=1 Bus 2 power balance equations

VV21(10sinθ 2 )+= 2.0 0 2 VV21(− 10cosθ 2 ) +V 2 (10) += 1.0 0

P2 (x )=V 22 (10sinθ ) += 2.0 0 2 QV2(x )= 2 ( − 10cosθ 22 ) + V (10) += 1.0 0

Now calculate the power flow Jacobian

∂∂P(22xx ) P( )  ∂∂θ2 V J ()x = 2 ∂∂Q()22xx Q() ∂∂θ 2 V 2

10V22 cosθθ 10sin 2 =  10VV22 sinθθ−+ 10cos 2 20 2

(0) 0 Set x =  1 Calculate

V22(10sinθ )+ 2.0 2.0 (0) = = f(x ) 2  VV2(− 10cosθ 22 ) ++ (10) 1.0 1.0

(0) 10V22 cosθθ 10sin 2 10 0 Jx()= =  10VV22 sinθθ−+ 10cos 2 20 2 0 10 −1 (1) 0  10 0   2.0 −0.2 Solve x = −   =   1  0 10   1.0 0.9 

−+ (1) 0.9(10sin( 0.2)) 2.0 0.212 f(x ) = =  0.9(− 10cos( − 0.2)) + 0.92 ×+ 10 1.0 0.279

(1) 8.82− 1.986 Jx()=  −1.788 8.199 −1 (2) −−0.2  8.82 1.986 0.212   − 0.233  x = −=    0.9 − 1.788 8.199 0.279   0.8586  − (2) 0.0145 (3) 0.236 f(xx) = =  0.0190 0.8554

(3) 0.0000906 f(x ) =Done! V2 = 0.8554 ∠− 13.52 ° 0.0001175