assignment

1. Process Optimization
Term assignment
Submitted to : Prof . Amit Kumar
By : (Group VI)
19bch049 - Punit patel
19bch050 - Raj patel
19bch051 - Riya patel
19bch052 - Sahil patel
19bch053 - Yash patel

2. Bracketing Method
 In bracketing methods, the method starts with an interval that
contains the root and a procedure is used to obtain a smaller interval
containing the root.
 Such methods are always convergent.
 Examples of bracketing methods:
 Bisection method
 False position method

3. Bisection Method

4. ABOUT BISECTION METHOD
 Assumptions:
Given an interval [a, b]
f(x) is continuous on [a, b]
f(a) and f(b) have opposite signs.
 These assumptions ensure the existence of at least one zero in the interval [a, b] and the
bisection method can be used to obtain a smaller interval that contains the zero.
 For that we perform the following steps:
1. Compute the mid point c = (a + b) / 2
2. Evaluate f(c)
3. If f(a) f(c) < 0 then new interval [a, c]
If f(a) f(c) > 0 then new interval [c, b]
4. Repeat the procedure until we get convergence.
a
b
f(a)
f(b)
c
a C1 C2

5. CONTD..
 If f(c) > 0, let anew = a and bnew = c and repeat
process.
 If f(c) < 0, let anew = c and bnew = b and repeat
process.
 This reassignment ensures the root is always
bracketed!! initial point ‘a’
root ‘d’
initial point ‘b’
a
c b
d
 Bisection is an iterative process, where the initial interval is halved
until the size of the interval decreases below some predefined
tolerance :|a - b|   or f(x) falls below a tolerance :|f(c ) – f(c-1)| 
.

6. Example 1: Determine the real root of
𝒇 𝒙 = 𝟓𝒙𝟑
− 𝟓𝒙𝟐
+ 𝟔𝒙 − 𝟐 = 𝟎 using Bisection Method.
Solution: Lets find the interval first,
𝑥 = 0 ⟹ 𝑓 0 = −2
𝑥 = 1 ⟹ 𝑓 1 = 4
∴ 𝑥 ∈ [0, 1]
Lets start the Bisection iterations now,
𝑥1 =
0+1
2
= 0.5, 𝑓 𝑥1 = 𝑓 0.5 = 0.375 > 0∴ 𝑥 ∈ [0, 0.5]
𝑥2 =
0 + 0.5
2
= 0.25, 𝑓 𝑥2 = 𝑓 0.25 =-0.7344 < 0
∴ 𝑥 ∈ [0.25, 0.5]
𝑥3 =
0.25 + 0.5
2
= 0.375, 𝑓 𝑥3 = 𝑓 0.375 =-0.1895<0
∴ 𝑥 ∈ [0.375, 0.5]
< 0
> 0

7. 𝑥4 =
0.375 + 0.5
2
= 0.4375, 𝑓 𝑥4 = 𝑓 0.4375 = 0.0867 > 0∴ 𝑥 ∈ [0.375, 0.4375]
𝑥5 =
0.375 + 0.4375
2
= 0.4063, 𝑓 𝑥5 = 𝑓 0.4063 = -0.0522 < 0
∴ 𝑥 ∈ [0.4063, 0.4375]
𝑥6 =
0.4063 + 0.4375
2
= 0.4219,𝑓 𝑥6 = 𝑓 0.4219 = 0.0169 > 0
∴ 𝑥 ∈ [0.4063, 0.4219]
𝑥7 =
0.4063 + 0.4219
2
= 0.4141,𝑓 𝑥7 = 𝑓 0.4141 = -0.0177 < 0
∴ 𝑥 ∈ [0.4141, 0.4219]
𝑥8 =
0.4141 + 0.4219
2
= 0.418, 𝑓 𝑥8 = 𝑓 0.418 = -0.00044 < 0
∴ 𝑥 ∈ [0.418, 0.4219]
𝑥9 =
0.418 + 0.4219
2
= 0.42, 𝑓 𝑥9 = 𝑓 0.42 = 0.0084 > 0 ∴ 𝑥 ∈ [0.42, 0.4219]
|0.42 – 0.4219| = 0.0019 ≈ 0
∴ 𝒙 = 𝟎. 𝟒𝟐 is one of the approximate real root of the given equation
correct up to 2 decimal places.

8. Example 2: Determine the negative real root of 𝒇 𝒙 = 𝒙𝟑
+ 𝟐𝟏𝒙 + 35 using Bisection Method correct up to 2
decimal places.
Solution: Lets find the interval first, 𝑥 = 0 ⟹ 𝑓 0 = 35
𝑥 = −1 ⟹ 𝑓 −1 =13
∴ 𝑥 ∈ [−2, −1]
Lets start the Bisection iterations now,
𝑥1 =
−2−1
2
= −1.5, 𝑓 𝑥1 = 𝑓 −1.5 = 0.125 > 0 ∴ 𝑥 ∈ [−2, −1.5]
𝑥2 =
−2 − 1.5
2
= −1.75, 𝑓 𝑥2 = 𝑓 −1.75 = -7.12 < 0
∴ 𝑥 ∈ [−1.75, −1.5]
𝑥3 =
−1.75 − 1.5
2
= −1.625, 𝑓 𝑥3 = 𝑓 −1.625 = -3.42 < 0
∴ 𝑥 ∈ [−1.625, −1.5]
> 0
> 0
𝑥 = −2 ⟹ 𝑓 −2 = -15
< 0

9. 𝑥4 =
−1.625 − 1.5
2
= −1.563, 𝑓 𝑥4 = 𝑓 −1.563 = -1.641 < 0 ∴ 𝑥 ∈ [−1.563, −1.5]
𝑥5 =
−1.563 − 1.5
2
= −1.532, 𝑓 𝑥5 = 𝑓 −1.532 = -0.768 < 0 ∴ 𝑥 ∈ [−1.532, −1.5]
𝑥6 =
−1.532 − 1.5
2
= −1.516, 𝑓 𝑥6 = 𝑓 −1.516 = -0.32 < 0 ∴ 𝑥 ∈ [−1.516, −1.5]
𝑥7 =
−1.516 − 1.5
2
= −1.508, 𝑓 𝑥7 = 𝑓 −1.508 = -0.097 < 0∴ 𝑥 ∈ [−1.508, −1.5]
𝑥8 =
−1.508 − 1.5
2
= −1.504, 𝑓 𝑥8 = 𝑓 −1.504 = 0.014 > 0 ∴ 𝑥 ∈ [−1.508, −1.504]
|(-1.508) – (-1.504)| = 0.004 ≈ 0
∴ 𝒙 = −𝟏.504 is one of the approximate negative real root of the give
correct up to 2 decimal places.

10. False Position Method
(Regula -Falsi)

11. ABOUT REGULA-FALSI METHOD
This technique is similar to the bisection
method except that the next iteration is
taken as the line of interception between
the pair of x-values and the x-axis rather
than at the midpoint.
(a, f(a))
(b, f(b))
By two point line
formula,
𝒚 − 𝒚𝟏
𝒚𝟏 − 𝒚𝟐
=
𝒙 − 𝒙𝟏
𝒙𝟏 − 𝒙𝟐
⟹
𝒇(𝒙) − 𝒇(𝒂)
𝒇(𝒂) − 𝒇(𝒃)
=
𝒙 − 𝒂
𝒂 − 𝒃
⟹
−𝒇(𝒂)
𝒇(𝒂) − 𝒇(𝒃)
=
𝒙𝟏 − 𝒂
𝒂 − 𝒃
𝒇𝒐𝒓 𝒙 = 𝒙𝟏, 𝒇 𝒙 = 𝟎 ⟹ 𝒙𝟏 = 𝒂 −
(𝒂 − 𝒃)𝒇(𝒂)
𝒇(𝒂) − 𝒇(𝒃)
⟹ 𝒙𝟏 =
−𝒂𝒇(𝒃) + 𝒃𝒇(𝒂)
𝒇(𝒂) − 𝒇(𝒃)
⟹ 𝒙𝟏 =
𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)
𝒇(𝒃) − 𝒇(𝒂)
Iterative formula for
Method of False Position
X
f(x)
Root
a
b
𝒙𝟏 𝒙𝟐

12. CONTD..
Assumptions:
Given an interval [a, b]
f(x) is continuous on [a, b]
f(a) and f(b) have opposite signs.
 These assumptions ensure the existence of at least one zero in the interval [a, b] and
the false position method can be used to obtain a smaller interval that contains the
zero.
 For that we perform the following steps:
1. Compute the in between point on x-axis,
2. Evaluate f(𝒙𝟏)
3. If f(a) f(𝒙𝟏) < 0 then new interval [a, 𝒙𝟏]
If f(a) f(𝒙𝟏) > 0 then new interval [𝒙𝟏, b]
4. Repeat the procedure until 𝒇(𝒙𝒊) < tolerance value.
𝒙𝟏 =
𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)
𝒇(𝒃) − 𝒇(𝒂)

13. Solution: Lets find the interval first,
𝑥 = 0 ⟹ 𝑓 0 = −2 < 0
𝑥 = 1 ⟹ 𝑓 1 = 4 > 0
∴ 𝑥 ∈ [0, 1]
Lets start the iterations now,
Example 1: Determine the real root of
𝒇 𝒙 = 𝟓𝒙𝟑 − 𝟓𝒙𝟐 + 𝟔𝒙 − 𝟐 = 𝟎 using Method of False
Position.
𝒙𝟏 =
𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)
𝒇(𝒃) − 𝒇(𝒂)
=
𝟎𝒇 𝟏 − 𝟏𝒇(𝟎)
𝒇(𝟏) − 𝒇(𝟎)
=
𝟎 − 𝟏(−𝟐)
𝟒 − (−𝟐)
=
𝟐
𝟔
= 𝟎. 𝟑𝟑𝟑
𝒇 𝒙𝟏 = 𝒇 𝟎. 𝟑𝟑𝟑 = -0.372 < 0 ∴ 𝒙 ∈ [𝟎. 𝟑𝟑𝟑, 𝟏]

14. 𝒙𝟐 =
𝒙𝟏𝒇 𝒃 − 𝒃𝒇(𝒙𝟏)
𝒇(𝒃) − 𝒇(𝒙𝟏)
=
𝟎. 𝟑𝟑𝟑𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟑𝟑𝟑)
𝒇(𝟏) − 𝒇(𝟎. 𝟑𝟑𝟑)
=
𝟎. 𝟑𝟑𝟑(𝟒) − 𝟏(−𝟎. 𝟑𝟕𝟐)
𝟒 − (−𝟎. 𝟑𝟕𝟐)
= 𝟎. 𝟑𝟖𝟗𝟖
𝒇 𝒙𝟐 = 𝒇 𝟎. 𝟑𝟖𝟗𝟖 = -0.125 < 0 ∴ 𝒙 ∈ [𝟎. 𝟑𝟖𝟗𝟖, 𝟏]
𝒙𝟑 =
𝒙𝟐𝒇 𝒃 − 𝒃𝒇(𝒙𝟐)
𝒇(𝒃) − 𝒇(𝒙𝟐) =
𝟎. 𝟑𝟖𝟗𝟖𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟑𝟖𝟗𝟖)
𝒇(𝟏) − 𝒇(𝟎. 𝟑𝟖𝟗𝟖)
=
𝟎. 𝟑𝟖𝟗𝟖(𝟒) − 𝟏(−𝟎. 𝟏𝟐𝟓)
𝟒 − (−𝟎. 𝟏𝟐𝟓) = 𝟎.4083
𝒇 𝒙𝟑 = 𝒇 𝟎. 𝟒𝟎𝟖𝟑 = -0.043 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟎𝟖𝟑, 𝟏]
𝒙𝟒 =
𝒙𝟑𝒇 𝒃 − 𝒃𝒇(𝒙𝟑)
𝒇(𝒃) − 𝒇(𝒙𝟑)
=
𝟎. 𝟒𝟎𝟖𝟑𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟎𝟖𝟑)
𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟎𝟖𝟑)
=
𝟎. 𝟒𝟎𝟖𝟑(𝟒) − 𝟏(−𝟎. 𝟎𝟒𝟑)
𝟒 − (−𝟎. 𝟎𝟒𝟑)
= 𝟎.415
𝒇 𝒙𝟒 = 𝒇 𝟎. 𝟒𝟏𝟓 = -0.014 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟓, 𝟏]
𝒙𝟓 =
𝒙𝟒𝒇 𝒃 − 𝒃𝒇(𝒙𝟒)
𝒇(𝒃) − 𝒇(𝒙𝟒)
=
𝟎. 𝟒𝟏𝟓𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟓)
𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟓)
=
𝟎. 𝟒𝟏𝟓(𝟒) − 𝟏(−𝟎. 𝟎𝟏𝟒)
𝟒 − (−𝟎. 𝟎𝟏𝟒)
= 𝟎.417
𝒇 𝒙𝟓 = 𝒇 𝟎. 𝟒𝟏𝟕 = -0.005 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕, 𝟏]

15. 𝒙𝟔 =
𝒙𝟓𝒇 𝒃 − 𝒃𝒇(𝒙𝟓)
𝒇(𝒃) − 𝒇(𝒙𝟓)
=
𝟎. 𝟒𝟏𝟕𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟕)
𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟕)
=
𝟎. 𝟒𝟏𝟕(𝟒) − 𝟏(−𝟎. 𝟎𝟎𝟓)
𝟒 − (−𝟎. 𝟎𝟎𝟓)
= 𝟎.4177
𝒇 𝒙𝟔 = 𝒇 𝟎. 𝟒𝟏𝟕𝟕 = -0.0018 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕𝟕, 𝟏]
𝒙𝟕 =
𝒙𝟔𝒇 𝒃 − 𝒃𝒇(𝒙𝟔)
𝒇(𝒃) − 𝒇(𝒙𝟔)
=
𝟎. 𝟒𝟏𝟕𝟕𝒇 𝟏 − 𝟏𝒇(𝟎. 𝟒𝟏𝟕𝟕)
𝒇(𝟏) − 𝒇(𝟎. 𝟒𝟏𝟕𝟕)
=
𝟎. 𝟒𝟏𝟕𝟕(𝟒) − 𝟏(−𝟎. 𝟎𝟎𝟏𝟖)
𝟒 − (−𝟎. 𝟎𝟎𝟏𝟖)
= 𝟎.4179
𝒇 𝒙𝟕 = 𝒇 𝟎. 𝟒𝟏𝟕𝟗 = -0.0009 < 0 ∴ 𝒙 ∈ [𝟎. 𝟒𝟏𝟕𝟗, 𝟏]
∴ 𝒙 = 𝟎. 𝟒𝟏𝟕𝟗 is one of the approximate real root of the given equatio
correct up to 3 decimal places.
|f(0.4179)| = 0.0009 ≈ 0

16. Example 2: Determine the real root of 𝒍𝒏 𝒙𝟒 = 𝟎. 𝟕 using
Regula – Falsi Method correct up to three decimal places.
Solution: Lets find the interval for,
𝑥 = 1 ⟹ 𝑓 1 = −0.7
𝑥 = 2 ⟹ 𝑓 2 = 2.073
∴ 𝑥 ∈ [1, 2]
Lets start the iterations now,
𝒇 𝒙 = 𝒍𝒏 𝒙𝟒 − 𝟎. 𝟕
𝒙𝟏 =
𝒂𝒇 𝒃 − 𝒃𝒇(𝒂)
𝒇(𝒃) − 𝒇(𝒂)
=
𝟏𝒇 𝟐 − 𝟐𝒇(𝟏)
𝒇(𝟐) − 𝒇(𝟏)
=
𝟐. 𝟎𝟕𝟑 − 𝟐(−𝟎. 𝟕)
𝟐. 𝟎𝟕𝟑 − (−𝟎. 𝟕)
= 𝟏. 𝟐𝟓𝟐
𝒇 𝒙𝟏 = 𝒇 𝟏. 𝟐𝟓𝟐 = 0.199 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟐𝟓𝟐]
< 0
> 0

17. 𝒙𝟐 =
𝒂𝒇 𝒙𝟏 − 𝒙𝟏𝒇(𝒂)
𝒇(𝒙𝟏) − 𝒇(𝒂)
=
𝟏𝒇 𝟏. 𝟐𝟓𝟐 − 𝟏. 𝟐𝟓𝟐𝒇(𝟏)
𝒇(𝟏. 𝟐𝟓𝟐) − 𝒇(𝟏)
=
𝟎. 𝟏𝟗𝟗 − 𝟏. 𝟐𝟓𝟐(−𝟎. 𝟕)
𝟎. 𝟏𝟗𝟗 − (−𝟎. 𝟕)
= 𝟏.196
𝒇 𝒙𝟐 = 𝒇 𝟏. 𝟏𝟗𝟔 = 0.016 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟔]
𝒙𝟑 =
𝒂𝒇 𝒙𝟐 − 𝒙𝟐𝒇(𝒂)
𝒇(𝒙𝟐) − 𝒇(𝒂)
=
𝟏𝒇 𝟏. 𝟏𝟗𝟔 − 𝟏. 𝟏𝟗𝟔𝒇(𝟏)
𝒇(𝟏. 𝟏𝟗𝟔) − 𝒇(𝟏)
=
𝟎. 𝟎𝟏𝟔 − 𝟏. 𝟏𝟗𝟔(−𝟎. 𝟕)
𝟎. 𝟎𝟏𝟔 − (−𝟎. 𝟕)
= 𝟏.192
𝒇 𝒙𝟑 = 𝒇 𝟏. 𝟏𝟗𝟐 = 0.0025 > 0 ∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟐]
𝒙𝟒 =
𝒂𝒇 𝒙𝟑 − 𝒙𝟑𝒇(𝒂)
𝒇(𝒙𝟑) − 𝒇(𝒂)
=
𝟏𝒇 𝟏. 𝟏𝟗𝟐 − 𝟏. 𝟏𝟗𝟐𝒇(𝟏)
𝒇(𝟏. 𝟏𝟗𝟐) − 𝒇(𝟏)
=
𝟎. 𝟎𝟎𝟐𝟓 − 𝟏. 𝟏𝟗𝟐(−𝟎. 𝟕)
𝟎. 𝟎𝟎𝟐𝟓 − (−𝟎. 𝟕)
= 𝟏.1913
𝒇 𝒙𝟒 = 𝒇 𝟏. 𝟏𝟗𝟏𝟑 = 0.00002 > 0∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟏𝟑]
𝒙𝟓 =
𝒂𝒇 𝒙𝟒 − 𝒙𝟒𝒇(𝒂)
𝒇(𝒙𝟒) − 𝒇(𝒂)
=
𝟏𝒇 𝟏. 𝟏𝟗𝟏𝟑 − 𝟏. 𝟏𝟗𝟏𝟑𝒇(𝟏)
𝒇(𝟏. 𝟏𝟗𝟐) − 𝒇(𝟏)
=
𝟎. 𝟎𝟎𝟎𝟎𝟐 − 𝟏. 𝟏𝟗𝟏𝟑(−𝟎. 𝟕)
𝟎. 𝟎𝟎𝟎𝟎𝟐 − (−𝟎. 𝟕)
= 𝟏.19129
𝒇 𝒙𝟓 = 𝒇 𝟏. 𝟏𝟗𝟏𝟐𝟗 = 0.00001 > 0
∴ 𝒙 ∈ [𝟏, 𝟏. 𝟏𝟗𝟏𝟐𝟗]
∴ 𝒙 = 𝟏. 𝟏𝟗𝟏𝟐𝟗 is the approximate real root of the given equation correct up to 3