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

2. What is management Science ? Use of quantitative analysis in supporting large companies to choose the best decisions among many different possibilities The key to every management science application is a mathematical model: quantitative representation, or approximation, of a real problem. Why go through mathematical models: - growing complexity of systems - possibility of extract the essence of problem in concise form - growing availability of data - cheap and abundant computer power

3. Decision Support Systems Computer software designed for specific applications which may include the possibility of solve a model A decision support system includes: - data bases easily accessible - user friendly interface - mathematical model - automatic procedure for the solution of the model Easy interaction with managers or decision makers who need not to fully understand the mathematical aspects, but can concentrate on results for their specific application.

4. The concept of a model Many applications of science make use of models The term ‘model’ is usually used for a structure which has been built purposely to exhibit features and characteristics of some real object. Models Concrete: model aircraft for wind tunnel experiments Abstract: using mathematical symbolism Pure descriptive

5. The management scientist valuate the (possible more) alternatives obtained at the previous step Modelling process Define the problem Collecting data Formulate a mathematical model Verify the model, use for prediction Select a suitable alternativeValuate the resultImplement Feedback loops The management scientist defines the organization system including objctives Estimate the parameters that affect the problems The management scientist verifies whether the model is an accurate representation of reality dummy Relationships in the real world (technological relationships, physical laws, marketing constraints) translated into a set of equations, inequalities, logical dependencies Optimization of one (or more) objective function: maximizing a profit, minimizing a cost Use of software packages

6. Mathematical models Mathematical model: representation of a real problem in terms of mathematical expressions. benefit It reveals not apparent relationships, makes mathematical analysis possible Simulation of scenarios (what happens if …) Use of mathematical solution procedure drawback Lack of precision of data Impossibility of quantifying some data (social value)

7. Optimization models Optimization models plays a fundamental role in mathematical programming mathematical programming Different from computer programming “planning” Optimization model Involve maximize something or minimize something, choosing among different alternatives

8. Optimization model: formal definition Set of possible alternatives Feasible solutions x stays in a set F f: F   objective function One optimization criterion finite infinite or

9. An easy example Which is the shortest path to go from place A to place B ? Feasible solutionsF={x=all possible paths from A to B} Definition of the problem Objective functionMinimize the length f of the path

10. Production planning An engineering factory produces two types of tools: pliers and spanners. Pliers cost to the production 1.5 € each Spanners cost to the production 1 € each Pliers are sold at 6 € each Spanners are sold at 5 € each How much should the factory produce to maximize its profit ? Production costs Selling price Objective function

11. A possible scenario Assume that we are constructing P TOT = number of pliers * unit profit for pliers + number of spanners * unit profit for spanners 15000 pliers 15000 spanners Let us define the data of the problem 4,5 = (6 – 1,5) unit profit for pliers4 = (5 – 1) unit profit for spanners unit profit = selling price – unit cost Objective function = total profit P TOT

12. A possible scenario (2) CH = number of spanners to be produced PI = number of pliers to be produced P TOT = 4  CH + 4,5  PI = 4 * 15000 + 4,5 * 15000 = Equation of the overall profit Formally: variables output of our model Unit profit of spanners Unit profit of pliers 15000 127500 Is this the best value ?

13. First use of Excel for data storing We use an Excel table to summarize data and objective P TOT = 4  CH + 4,5  PI Profit equation C9 = C6 * C8 + D6 * D8 Excel formulaCH = D8 = number of pliers PI = C8 = number of spanners 4,5= unit profit pliers = D4 – D5 4 = unit profit spanners = C4-C5

14. Constraints Every point in the non negative quadrant is a possible feasible solution (a possible production planning) 246810121416 2 4 6 8 10 14 16 12 Spanners (thousands) Pliers (thousands) In practice: constraints exist that limit the production In principle: the more I produce the more I gain ! (25000) overall profit (68000) (127500)

15. A standard constraint: the budget one Budget constraint: the overall cost must not greater than € 18000 C TOT = overall cost CH = D8 = number of pliers PI = C8 = number of spanners C TOT = 1  CH +1,5  PI Unit cost spanner Unit cost pliers Cost equation C11 = C5 * C8 + D5 * D8 C TOT = 1  CH + 1,5  PI  budget =18000 Budget constraint

16. Geometric view of the constraint Let draw the set F of the feasible solutions In the plane (PI, CH), first draw the equation of the Feasible region All non negative points “below” the line budget constraint 1  CH + 1,5  PI = 18 C TOT = 20000 C TOT = 7000  18000: ammissibile > 18000: non ammissibile C TOT = 37500 1*CH + 1,5* PI = 18 12 CH PI 246810121416 2 4 6 8 10 14 16 18

17. Geometric view of the problem We need to find the “right” solution among the feasible ones “right” It satisfies the budget constraint It maximizes the profit Points in the red area satisfies the budget constraint, which is the better one ? CH 12 PI 246810121416 2 4 6 8 10 14 16 18 1*CH + 1,5* PI = 18

18. Scenarios with Excel Let us see with different scenarios feasible Not feasible feasible Best among the three feasible. Is the optimum ? By chance ?

19. Limit of the “What If” approach The scenarios are too many: infinite solutions ! It is not possible to look over all of them !! We may wonder if may be better when the number of solutions is finite This is not always true, let see with an example

20. A first example: capital budgeting A management firm is considering 3 possible capital projects to invest in during the coming year Definition of the problem Data Budget15 millions Investment (I) required for each project Earnings (E) for each project

21. Capital budgeting: a possible scenario Each Project Selected (YES = 1) Not selected (NO = 0) Investment required = 8 + 0 + 5 = 13 Earning obtained = 12 + 0 + 7 = 19 Is this the best ?

22. Capital budgeting: constructing the model Feasible solutions Definition of the possible alternatives Each Project Selected (YES = 1) Not selected (NO = 0)

23. Capital budgeting Budget15 millions All choices Not acceptable Are all compatible with the budget ?

24. A bad model: capital budgeting All possible choices F= Best value Which is the best with respect to the earnings ?

25. A bad model: capital budgeting Feasible solutionsExhaustive representation 2 n = huge number for large n Not independent from data If data changes, all the model must be change It may be even impossible to write it down

26. A good model: capital budgeting Implicit representation of feasible solutions Independent from data xi=xi= 1 if project i is selected 0 if project i is not selected Decision variables Budget constraint8 x 1 +6 x 2 +5 x 3 Investment for project 1 Investment for project 2 budget Investment for project 3

27. A good model: capital budgeting xi=xi= 1 if project i is selected 0 if project i is not selected earnings12 x 1 +8 x 2 +7 x 3 earnings for project 1 Earnings for project 2 Earnings for project 3 If data changes, only coefficients must be changed, but the model is the same

28. Construction of a good model: general rules xi=xi= 1 if project i is selected 0 if project i is not selected Definition of the objectives 12 x 1 +8 x 2 +7 x 3 Definition of the decision variables max earnings Definition of the constraints 8 x 1 +6 x 2 +5 x 3 budget i=1,2,3

29. Geometric view of the production problem The scenarios are too many 12 CH PI 246810121416 2 4 6 8 10 14 16 18 1*CH + 1,5* PI = 18 P TOT = 69 Actual best value Draw the line of the profit P TOT 4  CH + 4,5  PI = 69 We must find another method Depict the feasible region F (red area)

30. Geometric view of the problem 12 CH PI 246810121416 2 4 6 8 10 14 16 18 1*CH + 1,5* PI = 18 Growing direction P TOT = 72 Draw the line of the profit P TOT 4  CH + 4,5  PI For growing values of P TOT = P TOT = 18 18 P TOT = 0 0 P TOT = 56 56 Optimum !! Old best value

31. Another constraint: technology To produce 1000 pliers or 1000 spanners is required 1 hour of usage of a plant CH = D8= number of spanners PI = C8= number of pliers The plant can work for a maximum of 14 hours per day H TOT = total hours nedeed H TOT = 0.001  CH + 0.001  PI Equation of the total hours C11 = C7 * C8 + D7 * D8 Technological constraint H TOT = 0.001  CH + 0.001  PI  max_hours=14

32. 14 PI 12 2 4 6 8 10 14 CH 246810121618 1*CH + 1* PI = 14 hours 16 > 14: not feasible 16 Geometric view of the technological constraint 0.001  CH + 0.001  PI  14 1  CH + 1  PI  14000 In the plane (PI, CH), first draw the equation of the technological constraint Feasible region of the technological constraint  14: feasible hours 6

33. Mixing the two constraints We must put together budget and technological constraints. Both must be satisfied Spanner Plier Both violated budget violated technological satisfied Budget satisfied technological violated Both satisfied

34. Geometric view of both constraints CH PI 14 12 2 4 6 8 10 16 14246810121618 1*CH + 1* PI = 14 1*CH + 1,5* PI = 18 Hours: feasible Budget: not feasible Hours: not feasible Budget: feasible Feasible set F Hours: not feasible Budget: not feasible Hours: feasible Budget: feasible Not feasible points

35. Final feasible region Feasible solutions (CH,PI) satisfy: CH  0 PI  0 Non negativity 1  CH + 1,5  PI  18 1  CH + 1  PI  14 budget hours Which is the best value for (CH,PI)* among the feasible ones ? The best one (CH,PI)* maximizes the profit 4  CH + 4,5  PI 1  CH + 1,5  PI = 18 CH PI 14 12 2 4 6 8 10 16 14246810121618 1  CH + 1  PI = 14 F

36. Linear Programming 1  CH + 1,5  PI  18 1  CH + 1  PI  14 budget hours CH  0PI  0 Non negativity LINEAR PROGRAMMING (LP) Max or Min of one linear objective function Constraints are linear equalities (=) or inequalities (  =  ) max 4  CH + 4,5  PI CH, PI  R Objective function profit constraints CH,PI Real decision variables CH, PI  R

37. Construction of a good model: general rules Definition of the objectives Definition of the decision variables 4 x 1 +4,5 x 2 max profit Definition of the constraints x 1 + 1,5 x 2 budget CH, PI  R x 1, x 2  R x 1 + x 2 hours x 1, x 2 The name is not important

38. Graphical solution for LP PI CH 1  CH + 1,5  PI = 18 14 12 2 4 6 8 10 16 14246810121618 1  CH + 1  PI = 14 Let choice a feasible point The profit P TOT is 4  CH + 4,5  PI = 34 (34000 €) Better solution must have a value of P TOT  34 The idea is looking for solutions that satisfy also the “new fictitious” constrait 4  CH + 4,5  PI  34. (4,4) (4000 spanners e 4000 pliers) P TOT = 34

39. Graphical solution for LP PI 1  CH + 1,5  PI = 18 14 12 2 4 6 8 10 16 14246810121618 1  CH + 1  PI = 14 CH 4  CH + 4,5  PI = 34 Draw the new feasible region F’ Let choice (2,10). with Excel A better solution (if any) must be in the new feasible region F’ F’ The profit P TOT is now 4  CH + 4,5  PI = 53 (53000 €) P TOT = 53

40. Graphical solution for LP CH The feasible region F” is more and more narrow (violet region) PI 1  CH + 1,5  PI = 18 14 12 2 4 6 8 10 16 14246810121618 1  CH + 1  PI = 14 4  CH + 4,5  PI = 53 Increasing values of P TOT The region with the constraint P TOT > 60 is empty !!! Behind this point (6,8) with P TOT = 60, there are non more feasible better points P TOT = 60 The point (6,8) is the best feasible one

41. Graphical solution of LP: summary 1.Draw the constraints and the feasible region 2.Choice a feasible point 3.Draw the line of the objective function passing for this point 4.Parallel move the line of the objective function in the direction of better values 5.The last feasible point “touched” by the line is the optimal solution When the number of variables is two:

42. Solution of LP We use Excel Solver (www.frontsys.com) http://www.frontsys.com/ Graphical solution can be applied only when the number of variables is two Real problems has usually more than two variables Many standard software exist to solve LP problems of different level of complexity Computer must be used as a tool to tackle large quantities of data and arithmetic