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Dr. Daniel Xing Email: [email protected]

EBUS-504

Operations Modelling and Simulation

Lecture 6

Introduction to optimization

University of Liverpool

Management School,

UK

Key learning outcomes

1. Concept of optimisation

2. Use charts in Witness

3. Use advanced experimenter for obtaining optimal solutions

Improve bottleneck

Run the sample model from Week 5 to 1000 minutes

Where is bottleneck?

Improve bottleneck

Use pie chart to help find bottleneck Go to Element States tab

Find your

target

element

Improve bottleneck

Create pie charts for all machines and run the model again

How do we interpret this result?

Improve bottleneck

See the demo on Witness for bottleneck analysis

Question:

Where is the end of our improvements?

How do we make such decisions in real world?

Optimisation

The field of “optimization” is concerned with how this process

can be quantitatively modelled, and, within the bounds of these

quantitative models, how the best decisions can be made.

▪ At the centre of every policy or planning decision are choices intended

to achieve one or more outcomes

▪ It is “the science of better.” This field is often known as operations

research, and has close ties with industrial or systems engineering.

Optimisation

What is an optimisation problem comprised of?

▪ An objective function: a single quantity to be either maximised or

minimised. E.g. the minimised costs, maximised safety etc.

▪ Decision variables: aspects of the problem that decision makers have

control over. E.g. number of machines, procurement frequencies etc.

▪ Constraints: Any kind of limitation on the values that the decision

variables they take. E.g. limited resources such as total amount of

budget, certain standards such as maintenance times, or some trivial

ones such as outputs can’t be negative.

A few examples

Example 1 – You have 60 feet of fence available, and wish to

enclose the largest rectangular area possible. What dimensions

should you choose for the fenced-off area?

Solution: The objective is clear from the problem statement: you wish to maximize the

area enclosed by the fence. The decision variables are not directly given in the problem.

Rather, you are told that you must enclose a rectangular area. To determine a rectangle,

you need to make two decisions: its length and its width. These are both decision

variables you can control directly, and there are no indirect decision variables because the

length and width directly determine its area. There is one obvious constraint — the

perimeter of the fence cannot exceed 60 feet — and two less obvious ones: the length and

width must be nonnegative. Since the length and width are independent of each other (the

perimetric constraint notwithstanding), there is no need to add a “consistency constraint”

linking them

A few examples

Simple mathematical formulation

L represents length

W represents width

Objective: max 𝐿,𝑊

𝐿𝑊

s.t. (subject to)

2𝐿 + 2𝑊 ≤ 60 𝐿,𝑊 ∈ 𝑅+

A few examples

Example 2 – Your company is selling A and B two products. Machine 1,

2, 3 are needed for processing them. Particularly, one final product A

needs M1 for 1 hour and M2 for 2 hours and one final product B needs M1

for 1 hour, M2 for 1 hour and M3 for 1 hour. M1 cannot be used over 300

hours per period, M2 cannot be used over 400 hours per period and M3

cannot be used over 250 hours per period. The market price for A is £50

and for B is £100. How do you plan your production per period to get the

best revenue?

Can you write the mathematical formulation?

A few examples

Answer:

𝑥1: number of A

𝑥2: number of B

Objective: max 𝑥1𝑥2

50𝑥1 + 100𝑥2

s.t.

𝑥1 + 𝑥2 ≤ 300 2𝑥1 + 𝑥2 ≤ 400

𝑥2 ≤ 250 𝑥1, 𝑥2 ∈ 𝑍

+

Optimisation

The next question is: how do we solve those problems?

Use experimenter in Witness to solve problem

Each A5 can be sold for £2000. A new M1-4 costs you £13000. Increasing

every 25% efficiency for M1-4 costs you £3800 and increasing the

efficiency for C1-C3 costs you £4000 per 10%. If you are given £30000 to

spend, how will you make your investment decisions?

Experimenter function

Investment decision on

machine efficiency Investment decision on

conveyor efficiency

Effects of decision variables

Experimenter function

Auxiliary variables used for

objective function and

constraints

Additional auxiliary variables

used for constraints

Experimenter function

Use function element to define

objective function

Experimenter function

Use advanced experiment mode

to find the optimal solution

Experimenter function Add new parameters

Parameters: all associated decision variables + some

auxiliary variables

Constraints: conditions that limits your optimisation

Responses: your objective function

Minimum & Maximum: The range for your

decision variable

Step size: How do you change the value

when search for optimal in each scenario

Suggested: Initial search value for your

decision variable

Experimenter function The full variable list

New machine decisions

Conveyor efficiency decisions

Machine efficiency decisions

Auxiliary variables

Experimenter function Constraint: all spendings are no more than £30000

Coefficient for each variable

Constraint condition

Click “Add” after the below

information is populated

Experimenter function Response: Function001

Since Function 001 is already

defined, so we just need to select it

as our objective function

You can also manually write your function in this box as well

Experimenter function Run your solver

Make sure your objective function is

selected here

Click it to run

Experimenter function Retrieve your solution

Red line shows the best optimal value and blue line shows

the actual objective value in each scenario Click here for results

Solution set for each scenario

Final

Can you find the optimal solution for our Example 2

(Page 12) with Witness experimenter function?

Dr. Daniel Xing Email: [email protected]

EBUS-504

Operations Modelling and Simulation

Lecture 6

Introduction to optimization

University of Liverpool

Management School,

UK