1. Engineering Computation
Part 2
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2. Solve for x, given that f(x) = 0 -or-
Roots of Equations
Objective:
Solve for x, given that f(x) = 0
-or-
Equivalently, solve for x such that
g(x) = h(x) ==> f(x) = g(x) – h(x) = 0
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3. Roots of Equations Chemical Engineering (C&C 8.1, p. 187):
van der Waals equation; v = V/n (= volume/# moles)
Find the molal volume v such that
p = pressure,
T = temperature,
R = universal gas constant,
a & b = empirical constants
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4. Roots of Equations Civil Engineering (C&C Prob. 8.17, p. 205):
Find horizontal component of tension, H, in a cable that passes through (0,y0) and (x,y)
w = weight per unit length of cable
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5. Roots of Equations Electrical Engineering (C&C 8.3, p. 194):
Find the resistance, R, of a circuit such that the charge reaches q at specified time t
L = inductance,
C = capacitance,
q0 = initial charge
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6. Roots of Equations Mechanical Engineering (C&C 8.4, p. 196):
Find the value of stiffness k of a vibrating mechanical system such that the displacement x(t) becomes zero at t= 0.5sec. The initial displacement is x0 and the initial velocity is zero. The mass m and damping c are known, and λ = c/(2m).
in which
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7. Determine real roots of :
Roots of Equations
Determine real roots of :
Algebraic equations (including polynomials)
Transcendental equations such as f(x) = sin(x) + e-x
Combinations thereof
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8. Roots of Equations Engineering Economics Example:
A municipal bond has an annual payout of $1,000 for 20 years. It costs $7,500 to purchase now. What is the implicit interest rate, i ?
Solution: Present-value, PV, is:
in which:
PV = present value or purchase price = $7,500
A = annual payment = $1,000/yr
n = number of years = 20 yrs
i = interest rate = ? (as a fraction, e.g., 0.05 = 5%)
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9. Roots of Equations Engineering Economics Example (cont.):
We need to solve the equation for i:
Equivalently, find the root of:
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10. Roots of Equations
Excel
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11. Roots of Equations Graphical methods:
Determine the friction coefficient c necessary for a parachutist of mass 68.1 kg to have a speed of 40 m/seg at 10 seconds.
Reorganizing.
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12. Roots of Equations Two Fundamental Approaches
1. Bracketing or Closed Methods
- Bisection Method
- False-position Method
2. Open Methods
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
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13. Bracketing Methods f(x)xl xu a) b) c) d) x
In Figure a) we have the case of f(xl) and f(xu) with the same sign, and there is no root in the interval (xl,xu).
In Figure b) we have the case of f(xl) and f(xu) With different sign, and there is a root in the interval (xl,xu).
In Figure c) we have the case of f(xl) and f(xu) with the same sign, and there are two roots.
In Figure d) we have the case of f(xl) and f(xu) with different sign, and there is an odd number of roots.
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14. Bracketing Methods
Though the cases above are generally valid, there are cases in which they do not hold.
In Figure a) we have the case of f(xl) and f(xu) with different sign, but there is a double root.
f(x) xl xu a) b) c) x
In Figure b) We have the case of f(xl) and f(xu) With different sign, but there are two discontinuities.
In Figure c) we have the case of f(xl) and f(xu) with the same sign, but there is a multiple root.
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15. Bracketing Methods (Bisection method)x
f(x)
f(xu)
(xu)
(x1)
f(x1)
f(xr)
f(x)
f(xu)
f(x1)
(x1)
(xu)
(xr)
f(xr)
f(x1) f(xr) > 0
xr => x1 x
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16. Bracketing Methods (Bisection method)
Given lower and upper bounds, xl and xu, which bracket the root: f(xl) f(xu) < 0
1) Estimate the Root by midpoint:
2) Revise the bracket: f(xl) f(xr) < 0, xr –> xu,
f(xl) f(xr) > 0, xr –> xl
3) Repeat steps 1-2 until:
a) |f(xr)| < k, b) a < s , with a =
c) | xl – xu | <  d) maximum # of iterations is reached. (Always do this in iteration algorithms.)
E. T. S. I. Caminos, Canales y Puertos

17. Bracketing Methods (Bisection method)
Engineering Economics Example:
We need to solve the equation for i:
Equivalently, find the root of:
Make conservative guesses at the upper and lower bounds:
100% interest rate, f(1.0) = 6,500
0% interest rate, f(0.0) = -12,500
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18. Bracketing Methods (Bisection method)
Excel
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19. Bracketing Methods (Bisection method)
Engineering Economics Example:
Score Sheet for Rootfinding Example:
Method Initial Est(s). s = 2 E-2 s = 2 E-7
Bisection (0.00, 1.00) 9 26
(0.05, 0.15) 6 22
Convergence guaranteed:
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20. Bracketing Methods (Bisection method)
One important advantage of this method is that one can
calculate the number of required iterations for a given error.
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21. Bracketing Methods (Bisection method)
Parachutist Example:
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22. Bracketing Methods (Bisection method)
Parachutist Example:
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23. Bracketing Methods (Bisection method)
Advantages:
1. Simple
2. Good estimate of maximum error:
3. Convergence guaranteed
Disadvantages:
1. Slow
2. Requires two good initial estimates which define an interval around root:
use graph of function,
incremental search, or
trial & error
E. T. S. I. Caminos, Canales y Puertos

24. Bracketing Methods (False-position Method)x
False-position Method
f(x)
f(x)
f(xu)
f(x1)
(x1)
(xu)
(xr)
f(xr)
f(xu)
(xu)
(x1)
f(x1)
f(x1) f(xr) > 0
x1 = xr
f(xr)
E. T. S. I. Caminos, Canales y Puertos

25. Bracketing Methods (False-position Method)
Similar to bisection. Uses linear interpolation to approximate the root xr: 1)
2) Revise the bracket: f(x1) f(xr) < 0, xr –> xu,
f(x1) f(xr) > 0, xr –> x1
3) Repeat steps 1-2 until:
a) |f(xr)| < k,
b) a < s , with a =
c) |xu – x1 | = d
d) maximum # of iterations is reached.
(Always do this in iteration algorithms.)
E. T. S. I. Caminos, Canales y Puertos

26. Bracketing Methods (False-position Method)
Excel
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27. Bracketing Methods (False-position Method)
E. T. S. I. Caminos, Canales y Puertos

28. Bracketing Methods (False-position Method)
Score Sheet for False-Position Example:
Method Initial Est(s). s = 2 E-2 s = 2 E-7
Bisection (0.00, 1.00) 9 26
(0.05, 0.15) 6 22
False-pos. (0.00, 1.00) 11 28
(0.05, 0.15) 3 14
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29. Bracketing Methods (False-position Method)
Parachutist Example:
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30. Bracketing Methods (False-position Method)
Parachutist Example:
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31. Bracketing Methods (False-position Method)
There are some cases in which the false position method is very slow, and the bisection method gives a faster solution.
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32. Bracketing Methods (False-position Method)
Summary of False-Position Method:
Advantages:
1. Simple
2. Brackets the Root
Disadvantages:
1. Can be VERY slow
2. Like Bisection, need an initial interval around the root.
E. T. S. I. Caminos, Canales y Puertos

33. Roots of Equations - Open Methods
Characteristics:
1. Initial estimates need not bracket the root
2. Generally converge faster
3. NOT guaranteed to converge
Open Methods Considered:
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
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34. Open Methods
An alternative method consists of separating the function into two parts.
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35. Open Methods (Fixed point method)
predict a value of xi+1 as a function of xi.
Convert f(x) = 0 to x = g(x)
iteration steps: xi+1 = g(xi )
x(new) = g(x(old) )
E. T. S. I. Caminos, Canales y Puertos

36. Open Methods (Fixed point method)
Example I:
Example II:
x = sin(x) –> xi+1 = sin(xi) OR
x = arcsin(x) –> xi+1 = arcsin(xi)
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37. Open Methods (Fixed point method)
Convergence:
Does x move closer to real root (?)
Depends on:
1. nature of the function
2. accuracy of the initial estimate
Interested in:
1. Will it converge or will it diverge?
2. How fast will it converge ?
(rate of convergence)
E. T. S. I. Caminos, Canales y Puertos

38. Open Methods (Fixed point method)
Convergence of the Fixed point Method:
Root satisfies: xr = g(xr)
The Taylor series for function g is:
xi+1 = g(xr) + g'(x)(xi - xr) xr < x < xi
Subtracting the second equation from the first yields
(xr – xi+1) = g'(x) (xr – xi) or
1. True error for next iteration is smaller than the true error in the previous iteration if |g'(x)| < 1.0 (it will converge).
2. Because g'(x) is almost constant, the new error is directly proportional to the old error (linear rate of convergence).
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39. Open Methods (Fixed point method)
Further Considerations:
Convergence depends on how f(x) = 0 is
converted into x = g(x)
So . . .
Convergence may be improved
by recasting the problem.
E. T. S. I. Caminos, Canales y Puertos

40. Open Methods (Fixed point method)
Convergence Problem:
For slowly converging functions
can be small, even though xnew is not close to root.
Remedy:
Do not completely rely on ea to ensure that the problem is solved.
Check to make sure |f(xnew) | < k .
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41. Open Methods (Fixed point method)
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42. Open Methods
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43. Roots of Equations Two Fundamental Approaches
1. Bracketing or Closed Methods
- Bisection Method
- False-position Method
2. Open Methods
- One Point Iteration
- Newton-Raphson Iteration
- Secant Method
E. T. S. I. Caminos, Canales y Puertos

44. Open Methods (Newton-Raphson Method)
Geometrical Derivation:
Slope of tangent at xi is
Solve for xi+1:
[Note that this is the same form as the generalized one-point iteration, xi+1 = g(xi)]
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45. Open Methods (Newton-Raphson Method)
Tangent w/slope=f '(xi )
f(x)
f(x)
f(xi)
f(xi)
f(xi+1)
f(xi+1)
xi+1 x xi x
(xi)
xi+1
xi = xi+1
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46. Open Methods (Newton-Raphson Method)
First order Taylor Series Derivation:
0 = f(xr)  f(xi) + f '(xi) (xr – xi)
solve for xr to yield next guess xi+1:
This has the form xi+1 = g(xi) with:
E. T. S. I. Caminos, Canales y Puertos

47. Open Methods (Newton-Raphson Method)
Newton-Raphson iteration:
This iteration process is repeated until:
1. f(xi+1)  0, i.e., | f(xi+1) | < k,
with  = small number 2.
3. Maximum number of iterations is reached.
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48. Open Methods a) Inflection point in the neighboor of a root.
b) Oscilation in the neighboor of a maximum or minimum.
c) Jumps in functions with several roots.
d) Existence of a null derivative.
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49. Open Methods (Newton-Raphson Method)
Bond Example:
To apply Newton-Raphson method to:
We need the derivative of the function:
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50. Open Methods (Newton-Raphson Method)
Score Sheet for Newton-Raphson Example:
Method Initial Est(s). s = 2 E-2 s = 2 E-7
Bisection (0.00, 1.00) 9 26
(0.05, 0.15) 6 22
False-pos. (0.00, 1.00) 11 28
(0.05, 0.15) 3 14
N-R 1.0 diverges diverges
0. 5 2, but wrong 48
EXCEL
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51. Open Methods (Newton-Raphson Method)
Error Analysis for N-R :
Recall that
Taylor Series gives:
where xr  x  xi and f(xr) = 0
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52. Open Methods (Newton-Raphson Method)
Dividing through by f '(xi) yields OR
Ei+1 is proportional to Ei2 ==> quadratic rate of convergence.
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53. Open Methods (Newton-Raphson Method)
Summary of Newton-Raphson Method:
Advantages:
Can be fast
Disadvantages:
May not converge
2. Requires a derivative
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54. Open Methods (Secant Method)
Approx. f '(x) with backward FDD:
Substitute this into the N-R equation:
to obtain the iterative expression:
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55. Open Methods (Secant Method)
f(x)
f(x)
f(xi-1)
f(xi)
f(xi-1)
f(xi)
xi+1
xi-1 x
xi+1 xi xi
xi-1 x
xi = xi+1
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56. Open Methods (Secant Method)
1) Requires two initial estimates: xi-1 and xi
These do NOT have to bracket root !
2) Maintains a strict sequence:
Repeated until:
a. | f(xi+1) | < k with k = small number b.
c. Max. number of iterations is reached.
3. If xi and xi+1 were to bracket the root, this would be the same as the False-Position Method. BUT WE DON'T!
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57. Open Methods (Secant Method)
In the secant method, the values are replaced in a strict sequence, xi+1 to xi, and this to xi-1. Thus, the new values can be on the de same sode of the root, and sometimes diverge.
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58. Open Methods (Secant Method)
Score Sheet for Secant Example:
Method Initial Est(s). s = 2 E-2 s = 2 E-7
Bisection (0.00, 1.00) 9 26
(0.05, 0.15) 6 22
False-pos. (0.00, 1.00) 11 28
(0.05, 0.15) 3 14
N-R 1.0 diverges diverges
0.5 2, but wrong 48
Secant (0, 1) diverges diverges
(0.00, 0.50) 4, but wrong (chaotic) 27
(0.05, 0.15) 3 6
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59. Open Methods Why do open methods fail?
Function may not look linear.
Remedy: recast into a linear form. For example,
Is a poorly constrained problem in that there is a large, nearly flat zone for which the derivative is near zero. Recast as:
i f(i) = 0 = 7,500 i [ 1 - (1+i)-20 ]
E. T. S. I. Caminos, Canales y Puertos

60. Open Methods
Recast as: i f(i) = 0 = 7,500 i [ 1 - (1+i)-20 ]
The recast function, "i f(i) will have the same roots as f(i) plus an additional root at i = 0.
It will not have a large, flat zone, thus:
h(i) = i f(i) = 7,500 i – 1000 [ 1 – (1+ i)–20]
To apply N-R we also need the first derivative:
h'(i) = 7, ,000 (1+ i)-21
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61. Open Methods Score Sheet for Open Methods:
Method Initial Est(s). s = 2 E-2 s = 2 E-7
  N-R 1.0 diverges diverges
0. 5 2, but wrong 48
Secant (0.00, 0.50) 4, but wrong 27
(0.05, 0.15) 3 6
N-R
[as i f(i)]
0.047 crazy results **
0.03 converges to i=0 **
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62. Open Methods Cases of Multiple Roots Multiple Roots:
f(x) = (x – 2)2 (x – 4)
x = 2 represents two of the
three roots.
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63. Open Methods Problems and Approaches:
Cases of Multiple Roots
1.Bracketing Methods fail locating x = 2.
Note that f(x) f(xr) > 0.
2. At x = 2, f(x) = f '(x) = 0.
Newton-Raphson and Secant methods may experience problems.
Rate of convergence drops to linear.
Luckily, f(x)  0 faster than f '(x)  0
3. Other remedies, recasting problem:
Find x such that u(x) = 0 where:
Note that u(x) and f(x) have same roots.
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64. Summary -- Rates of Convergence
m = 1: linear convergence
m = 2: quadratic convergence
Method m
Bisection 1
False Position 1
Secant, mult. root 1
NR, mult. root 1
Secant, single root "super linear"
NR, single root 2
Accel. NR, mult. root (f(x)/f'(x)=0) 2
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65. Multivariate (Multidimensional) Equations
Solve
fi(x1, ..., xn) = 0 for i = 1,...,n
Let X = (x1, ..., xn)T
· Given intial guess Xt, try to solve
where:
· Obtain X = (Xi+1 – Xi) from linear equations:
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66. Alternative Stopping Criteria
Always limit number of iterations using an outer DO loop. The problem may not converge and could try to go on forever.
Absolute error criteria for "small" differences: | xt - xt-1 | < d
3. Relative error criteria for "relatively small" changes
| xt – xt-1 | < e | xt |
4. Can combined error criteria 2 & 3 for large and small x-values:
| xt – xt-1 | < d + e | xt |
5. Converge on zero residual
| f(xt) | < k
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67. Three Performance Criteria
Stopping Criteria:
| xi – xi-1 | <  +  |xi|
or | f(xi) | < 
or Max. iterations
Convergence Criteria:
and | f(xi) | < 
N-R and Secant Confirmation of Convergent Behavior:
x in feasible region
and | f(xi) | ≤ 0.5 | f(xi-1) |
and | xi – xi-1 | ≤ 0.6 | xi-1 – xi-2 |
otherwise, do Bisection for a while.
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68. Three Phase Rootfinding Strategy
A real rootfinding problem can be viewed
as having three phases:
1) Opening moves: One needs to find the region of the parameter space in which desired root can be found.
Understanding of problem, physical insight, and common sense are valuable.
2) Middle Game: Use robust algorithm to reduce initial region of uncertainty.
3) End game: Generate a highly accurate solution in a few iterations.
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