Chapter 2 Optimization Models

Optimization: – Aim: To find or choose a procedure or make decisions so as to achieve

some goal or to make some key performance indicators (KPI) to be optimal. The process of finding the best solution is known as optimization

– Optimization is widely used in practice: • Optimal design, Optimal decision, Optimal fitting, • Optimal urban planning, Optimal control, …….

Main Tools:– Linear programming, nonlinear programming– Dynamic programming, queuing, ….

One-variable optimization

An example: A pig weighing 200 pounds gains 5 pounds per day and costs 45 cents a day to keep. The market price for pigs is 65 cents per pound, but is falling 1 cent per day. When should the pig be sold?Solution– Variables:

• t --- time (days); • w(t) --- weight of pig (lbs)• p(t) --- price for pig ($/lb); • C(t) --- cost of keeping pig t days ($)• R(t) --- revenue obtained by selling pig ($)• P(t) --- profit from sale of pig ($)

An example

– Assumption and relation:• w(t)=200+5t• p(t)=0.65-0.01t• C(t)=0.45t• R(t)=p(t) w(t)• P(t)=R(t)-C(t), t>=0.

– Objective: Maximize P > Find t* s.t. P(t*)>=P(t)– Optimization problem: max ( ) (0.65 0.01 )(200 5 ) 0.45

with { : 0}t S

P t t t t

S t t

One-variable optimization

– The result:

One-variable optimization model

Solution method

8 1( ) 0 & ( ) * 8 & (8) 133.20$

10 10

tP t P t t P

max ( ) subject to

( )

f x

g x b

Multivariable optimization

An example: A carpenter makes tables and bookcases. He is trying to determine how many of each type of furniture he should make each week. The carpenter wishes to determine a weekly production schedule for tables and bookcases that maximizes his profits. It costs $5 and $7 to produce tables and bookcases, respectively. The revenues are estimated by expressions:

– The problem: To decide how many tables and bookcases to make every week???

21 1 1

22 2 2

50 0.2 , where is the number of tables produced per week &

65 0.3 , where is the number of bookcases produced per week

x x x

x x x

Another example: Carpenter schedule

– The solution:• Objective function is the net weekly profit to be realized

from selling the tables and bookcases. The profit is revenue minus costs.

• The profit function:

• The optimization problem – integer nonconstrained optimization

• The results

2 21 2 1 1 2 2 1 2( , ) 50 0.2 65 0.3 5 7f x x x x x x x x

1 21 2

( , )max ( , ) with {( , ) : , 0,1,2, }x x S

f x x S x y x y

1 2 1 2113& 97 or 112 & 97x x x x

Another example: Carpenter schedule

A variation of the example: The carpenter realizes a net unit profit of $25 per table and $30 per bookcase. He is trying to determine how many of each piece of furniture he should make each week. He has up to 600 board-feet of lumber to devote weekly to project and up to 40 hr of labor. He can use lumber and labor productively elsewhere if they are not used in the production of tables and bookcases. He estimates that it requires 20 board-feet of lumber and 5 hr of labor to complete a table and 30 board-feet of lumber and 4 hr of labor for a bookcase. Moreover, he has signed contracts to deliver four tables and two bookcases every week. The carpenter wishes to determine a weekly production schedule for tables and bookcases that maximizes his profits.

Variables

1

2

: the number of tables produced per week &

: the number of bookcases produced per week

x

x

Another example: Carpenter schedule

– The optimization model – constrained

– The solution – by using linear programming

1 2

1 2

1 2

1 2

max 25 30

subject to

20 30 600 (lumber)

5 4 40 (labor)

4 & 2 (contract)

x x

x x

x x

x x

Another example: Plant headquarters

Locating a plant’s headquarters: Suppose we are attempting to locate a plant’s headquarters in such a way as to minimize the distance traveled to subsidiary plants. Assume that all plants are visited with the same frequency and neglect vertical distances, giving a two-dimensional model. These assumptions might be reasonable where travel is by private aircraft and each plant and the headquarters have an airstrip or when the travel is by automobile and the terrain is flat. – Assumption: Suppose there are m plants and they are located at

Let (a,b) denote the headquarters location.

( , ) 1,2, ,i ix y i m

Another example: Plant headquarters

– The model: – The result:

A variation: A plant’s headquarters is defined in such a way as to minimize the square of the distance traveled to subsidiary plants.– The model:

– The result:

( , ) 0 & ( , ) 0a bD a b D a b

2 2

1

min ( , ) ( ) ( )m

i ii

D a b a x b y

2 2

1

min ( , ) ( ) ( )m

i ii

E a b a x b y

1 1

1 1

1( , ) 2( ) 0

1( , ) 2( ) 0

m m

a i ii i

m m

b i ii i

E a b a x a xm

E a b b y b ym

Another example: Investment Problem

Multiobjective optimization problem: An investor has $40,000 to invest. She is considering investments in saving at 7%, municipal at 9%, and stocks that have been consistently averaging 14%. Because there are varying degrees of risk involved in the investments, the investor listed the following goals for her portfolio:

– A yearly return of at least $5,000– An investment of at least $10,000 in stocks – The investment in stocks should not exceed the combined total in bonds and

savings– A liquid savings account between $5,000 and $15,000– The total investment must not exceed $40,000.

Another example: Investment Problem

The mathematical model– Variables:

• x ----- the investment in savings• y ---- the investment in bonds• z ---- the investment in stocks

– Goals: Goal 1: 0.07 0.09 0.14 5000

Goal 2: 10000

Goal 3:

Goal 4: 5000 15000

Goal 5: x+y

x y z

z

z x y

x

+ 40000z

Another example: Investment Problem

– Discussion of the problem:• The investor has more than one objective• Unfortunately, often the case with real-world problems,

not all goals can be achieved simultaneously!! In some case, there maybe no feasible solution!!

• In general, the investor will have to compromise on one or more of her goals to find a feasible solution.

• An example: She want the best return and less money in stocks!!

Another example: Investment Problem

– The model

– The solution – by linear programming method—simplex method

max ( , , ) 0.07 0.09 0.14

Subject to

0.07 0.09 0.14 5000

10000

5000 15000

f x y z x y z z

x y z

z

z x y

x

x+y+ 40000

, , 0

z

x y z