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
anch and
ound
● 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,
anch 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
anch and
ound use linprog or Excel Solver to find the partial solutions.
● Problem 2 is a discrete nonlinear optimisation problem. For
anch 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
anch and bound? How many partial and full evaluations were needed for
anch and bound?
○ Include your MATLAB code for exhaustive enumeration
○ Include any code or an image of any spreadsheets used for
anch and
ound
○ Draw the trees for
anch 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:
x x x x x xf = XXXXXXXXXX3x XXXXXXXXXX + 7 7
Subject to:
x x x x x 0g1 = XXXXXXXXXX 4 + x5 + x6 + 3 7 ≥ 5
x x x x 0g2 = 7 1 + 2x XXXXXXXXXX ≤ 7
x x x x x 0g3 = XXXXXXXXXX 4 + x XXXXXXXXXXx7 ≥ 4
, x , x , x 1, 2, 3, 4}x XXXXXXXXXX ∈ {
, x , xx5 6 7 ≥ 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: x x x x xA = x XXXXXXXXXX − XXXXXXXXXXmc 2
Section modulus: (x x )S = x XXXXXXXXXX
x x1 2 mc 3
Bending moment: 00M = 4 Nmk
Axial force: 30P = 1 Nk
Bending stress: σB = S
1000M PaM
Axial stress: σP = A
10P PaM
Stress constraint: 50 σB + σP − 2 ≤ 0 PaM
Buckling constraint: 45 x2
x1 − 1 √4 (1+ )σB
σP 2
XXXXXXXXXXσB
σP 2
≤ 0
And subject to the following constraints on plate thickness and width:
37, 39, 41x1 :
1.1, 1.2, 1.3x2 :
30, 32, 34x3 :
0.8, 1.0, 1.2x4 :