Corso MAE Metodi Quantitativi per il Management Quantitative methods for Management

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Corso MAE Metodi Quantitativi per il Management Quantitative methods for Management. Prof. Gianni Di Pillo Prof. Laura Palagi Dipartimento di Informatica e Sistemistica Universita` di Roma “La Sapienza”. Roma, 18 settembre - 24 ottobre 2003. x 1 , x 2 R. CH, PI R.

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## Corso MAE Metodi Quantitativi per il Management Quantitative methods for Management

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Corso MAE

Metodi Quantitativi per il Management

Quantitative methods for Management

Prof. Gianni Di Pillo

Prof. Laura Palagi

Dipartimento di Informatica e Sistemistica

Universita` di Roma “La Sapienza”

Roma, 18 settembre - 24 ottobre 2003

x1 , x2 R

CH, PIR

Decision variables

Objective function

max

4 x1+4,5 x2

profit

x1+ 1,5 x2

budget

constraints

x1+ x2

hours

x1 , x2

Mathematical model of the production problem

LINEAR PROGRAMMING (LP)

Linear Programming (LP)

Both objective function and constraints are linearexpressions

Proportionality

When the level of ony activity is multiplied by a constant factor, then any contribution to the objective function or to any of the constraints is multiplied by the same factor

The value of the objective function and of the constraints is the sum of contribution from the various activity

Divisibility

Both integer and fractional levels of the activities allowed

PI

16

1CH + 1PI = 14

14

12

10

8

F

1CH + 1,5PI = 18

6

4

CH

2

2

4

6

8

10

12

14

16

18

Final feasible region

The best value for(CH,PI)*among the feasible ones is (6,8) which corresponds to the profit

PTOT= 4  CH + 4,5  PI = 60

Real decision variables c8,d8

data

Objective function

c6*c8+d6*d8

Cost: c5*c8+d5*d8 Hours: c7*d8+d7*d8

Equation of the Constraints

Production problem with Excel

Let us consider the Excel table

1 if project i is selected

Decision variables

xi=

i=1,2,3

0 if project i is not selected

Objective function

max

12 x1+8 x2+7 x3

earnings

constraints

8 x1+6 x2+5 x3

budget

x1, x2 , x3

Mathematical model for Capital budgeting

INTEGER LINEAR PROGRAMMING (ILP)

Integer Linear Programming (ILP)

Both objective function and constraints are linearexpressions

Proportionality

When the level of ony activity is multiplied by a constant factor, then any contribution to the objective function or to any of the constraints is multiplied by the same factor

The value of the objective function and of the constraints is the sum of contribution from the various activity

Indivisibility

Only integer levels of the activities allowed

data

Integer decision variables c7,d7,e7

Objective C5*C7+D5*D7+E5*E7

Constraint C4*C7+D4*D7+E4*E7

Capital Budget with Excel
Solution of mathematical models

Graphical solution (of LP) can be applied only when the number of variables is two

Real problems has usually more than two variables

Computer must be used as a tool to tackle large quantities of data and arithmetic

Many standard software exist to solve LP problems of different level of complexity

We useExcel Solver (www.frontsys.com)

http://www.frontsys.com/

Tools

Solver

Excel: an easy platform to optimization

Excel has an optimization toolbox: Solver

Solving PL with Excel

In the main menù select Tools (Strumenti) and then Solver (solutore)

Objective function

Decision variables

Constraints

Tipo di problema (max o min)

Solving PL with Excel

It will appear a dialog window like below

Let now fill in

Objective function PTOT = c9

The value can be set easily by clicking the corresponding cell (it puts the address \$c\$)

Setting the objective function
Setting the initial guess

Cells C8 and D8 contains the value of the variables. At the end of the optimization process they contain the optimal value

English

Italian

Address of the cell ora constant

Address of the cell

Constraint can be of the typeA   B

A Int (integer value)

A bin (binary value 0,1)

Setting the constraints

Clicking Options (Opzioni)the window of parameters appears

We must Assume Linear Model (use simplex method) and non-negative variables (in alternative we can define the additional constraints c8, d8 0).

Setting the options

Maximum time allowed to obtain a solution

Maximum iterations of the algorithm to obtain a solution

It uses an algorithm for linear problems (simplex)

More complex models (non linear)

Setting the options
Solve LP con Excel

We can start optimization

Click the button Solve (Risolvi)

Final result with Excel

The “algorithmic” solution is the same obtained with the graphical solution

Reducing time

Reducing iterations

Reducing or increasing tolerance

Same solution

Changing the options for LP

Change the model

Same solution

Changing the options for LP

In general this is not true

Reducing time

Reducing iterations

same solution but the Solver is not able to certify optimality

Changing the options for ILP

Increasing Tolerance

SOLUTION CHANGES

Optimality declared, but it is not true

Changing the options for ILP

For LP problems optimality can always be certified

For LP problems sub-optimal solutions do not exist

For ILP problems optimality is difficult to be certified

For ILP problems many sub-optimal solution may exist

Is LP behavior different from ILP ?

Each unit of product requires a certain time on each process

Another production problem

An engineering factory can produce five type of productsPROD1, PROD2, PROD3, PROD4, PROD5

Two production process must be used: grinding and drilling

The factory has 3 grinding machines and 2 drilling machines that works a 6-day week with 2 shifts of 8 hours each day

After deducing raw material costs, each unit of product yields the following profit

How much to make of each product so to maximize the total profit ?

Objective function

Another production problem

Final assembly of each unit of product uses 20 hours of a workman’s time

8 man are employed in assembly each working one shift a day

Hours assembly for unit

Mathematical model

The five type of products are the decision variables

PROD1 = x1, PROD2 = x2, PROD3 = x3, PROD4 = x4, PROD5 = x5

x1 , x2, x3, x4, x5 >= 0

Objective function is the profit to be maximize

max (2.5x1 + 6 x2 + 3.5 x3 + 4 x4 + 2 x5 )*100

Constraints:

Only 8 man * 1 shift * 6 days for assembly

20x1 + 20 x2 + 20 x3 + 20 x4 + 20 x5 <= 384

Mathematical model (2)

Constraints:

Technological constraints

Grinding process

Only 3 machines * 2 shift * 6 days

12x1 + 20 x2 + 25 x4 + 15 x5 <= 288

Drilling process

Only 2 machines * 2 shift * 6 days

10x1 + 8 x2 + 16 x3 <= 192

20 x1 + 20 x2 + 20 x3 + 20 x4 + 20 x5

12 x1 + 20 x2 + 25 x4 + 15 x5

10 x1 + 8 x2 + 16 x3

x1 , x2, x3 , x4, x5

Mathematical model (3)

max2.5 x1 + 6 x2 + 3.5 x3 + 4 x4 + 2 x5

x1 , x2, x3 , x4, x5 integer

384=8 men * 8 hours *6days

288= 3 machines * 16 hours * 6days

192= 2 machines * 16 hours * 6days

Production problem with Excel

data

Objective function

constraints

(Real) decision variables

Model LP con Excel

Fractional solution

Solve LP con Excel

We need to insert the integer constraint

constraints

Integer decision variables

Model ILP con Excel

Integer solution

Solve ILP con Excel

The solution obtained is better than the “approximating one”