RECENT ASSIGNMENT

49928: Design Optimisation for Manufacturing
Assignment 2: Discrete Optimisation
Due: 9:00 am Monday 15/10/2018
? Solve the following two problems with both exhaustive enumeration and branch and bound
? The assignment is worth 15 marks in total (15% of your final mark for the subject)
? Exhaustive enumeration is worth 2.5 marks for each problem, branch and bound is worth 5 marks for each problem.
? Problem 1 is a mixed integer linear optimisation problem (the problem has both discrete and continuous variables). Do not use intlinprog (from MATLAB) to solve this problem, for exhaustive enumeration solve it by enumerating through the discrete variables and then use linprog to find the continuous variables. For branch and bound use linprog or Excel Solver to find the partial solutions.
? Problem 2 is a discrete nonlinear optimisation problem. For branch and bound use fmincon or Excel Solver to find the partial solutions.
? Write a report:
? Describe the process of finding the solution: how many evaluations were needed for exhaustive enumeration? What path did the search take for branch and bound? How many partial and full evaluations were needed for branch and bound?
? Include your MATLAB code for exhaustive enumeration
? Include any code or an image of any spreadsheets used for branch and bound
? Draw the trees for branch and bound. For each node state:
¦ Which variables are constrained
¦ The partial or full solution
¦ Whether or not the solution is feasible
¦ Whether or not the node has been pruned
Problem 1 (8 marks)
Minimise:
f = 4x1 + 5x2 + 3x3 + 6x4 + 4x5 + 5x6 + 7x7
Subject to:
g1 = 4x1 + 3x2 + 6x3 + 5x4 + x5 + x6 + 3x7 = 50 g2 = 7x1 + 2x2 + 2x3 + 6x4 + 3x7 = 70 g3 = 6x1 + 5x2 + 3x3 + 3x4 + x5 + 8x6 + x7 = 40
x1, x2, x3, x4 ? {1, 2, 3, 4} x5, x6, x7 = 0

Problem 2 (7 marks)
An I-beam is shown in the figure to the right. Given the following equations and constraints, develop a mathematical model and find the dimensions of a beam with a minimal cross sectional area.
Cross sectional area: A = x1x2 + 2x3x4 - 2x2x4 cm2
Section modulus: S = x1(x3x4 + x16x2) cm3
Bending moment: M = 400 kNm
Axial force: P = 130 kN
Bending stress: sB = 100S0M MPa
Axial stress: sP = MPa
Stress constraint: sB + sP - 250 = 0 MPa
Buckling constraint:
And subject to the following constraints on plate thickness and width:
x1 : 37, 39, 41 x2 : 1.1, 1.2, 1.3 x3 : 30, 32, 34 x4 : 0.8, 1.0, 1.2