THEORY EXAMINATION (SEM–VI) 2016-17 OPTIMIZATION TECHNIQUES IN ENGINEERING

OPTIMIZATION TECHNIQUES IN ENGINEERING (EME023)

Section-wise Solved Answers & Notes

SECTION – A (10 × 2 = 20 Marks)

Short, direct answers (2 marks each)

(a) Why optimization is required?

Optimization is required to obtain the best possible solution (minimum cost, maximum profit, or best performance) under given constraints.

(b) Heuristic approach

A heuristic approach uses rule-of-thumb or experience-based methods to find near-optimal solutions when exact methods are difficult.

(c) Multivariable function (with example)

A function having more than one independent variable.
Example:                                             f(x,y)=x2+y2f(x,y) = x^2 + y^2f(x,y)=x2+y2 

(d) Complex variable (with example)

A variable having real and imaginary parts.
Example:                                             z=a+ibz = a + ibz=a+ib 

(e) On-line real-time optimization

It is optimization performed continuously during system operation, adjusting parameters in real time based on current data.

(f) Genetic Algorithm (GA)

GA is an evolutionary optimization technique based on selection, crossover, and mutation.
Example: used in design optimization problems.

(g) Least square optimization

A method that minimizes the sum of squares of errors between observed and predicted values.

(h) Newton Algorithm

An iterative method used to find optimum or roots of equations using first and second derivatives.

(i) Cutting plane methods

Techniques used in integer programming to remove fractional solutions by adding linear constraints.

(j) Unconstrained optimization

Optimization problems with no equality or inequality constraints.

SECTION – B (Any 5 × 10 = 50 Marks)

(a) Local minimum and maximum

A local minimum is a point where the function value is smaller than nearby points.
A local maximum is a point where the function value is larger than nearby points.

Example:

f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x 

(b) Sequential Quadratic Programming (SQP)

SQP solves non-linear problems by converting them into a sequence of quadratic sub-problems.
It is efficient for constrained optimization.

(c) Gomory’s Cutting Plane Method – considerations

• Choose constraints that remove fractional solutions           • Maintain feasibility
• Avoid excessive constraints                                                   • Ensure integer solution convergence

(d) Integer Programming

Optimization problem where decision variables take integer values.

Example:
Maximize                       Z=5x+4yZ = 5x + 4yZ=5x+4y

subject to

x,y≥0 and integersx, y \ge 0 \text{ and integers}x,y≥0 and integers 

(e) Random variable

A variable that takes values based on outcomes of a random experiment.

(f) Runge-Kutta Method

A numerical technique for solving ordinary differential equations.
It improves accuracy over Euler’s method.

(g) Non-linear programming

Optimization where objective function or constraints are non-linear.

Example:

Minimize f(x)=x2+3x+2\text{Minimize } f(x) = x^2 + 3x + 2Minimize f(x)=x2+3x+2 

(h) Direct search method

Optimization without derivatives, using trial-and-error search.

Example: Pattern search method.

SECTION – C (Any 2 × 15 = 30 Marks)

Q3. Dynamic Programming

Dynamic programming solves complex problems by breaking them into smaller sub-problems.

Principle of optimality

An optimal solution consists of optimal sub-solutions.

Engineering example

Shortest path problem, resource allocation, inventory control.

Q4. Methods for Multivariable Optimization with Equality Constraints

• Lagrange multiplier method                                • Penalty method
• Gradient projection method

Example (Lagrange method):

Minimize f(x,y)\text{Minimize } f(x,y)Minimize f(x,y)

subject to                                                                g(x,y)=0g(x,y) = 0g(x,y)=0 

Q5. Genetic Algorithm Numerical

Given fitness values:                                            20, 20, 50, 10

Step 1: Total fitness

20+20+50+10=10020 + 20 + 50 + 10 = 10020+20+50+10=100

Step 2: Probability of 4th design point                P=10100=0.1P = \frac{10}{100} = 0.1P=10010​=0.1

Step 3: Expected number of copies

=4×0.1=0.4= 4 \times 0.1 = \boxed{0.4}=4×0.1=0.4​