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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”

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Blending problems A food is manufactured by refining row oils and blending them together. VEG1, VEG2 cost (per tons) respectively 110 € and 120 € OIL1,OIL2,OIL3 cost (per tons) respectively 130 €, 110 € and 115 € Production costs The final food is sold at 150 € per ton.Selling price How much the food manufacturer make his product so to maximize his profit ? Objective function The raw oils came in two categories: vegetables (VEG1, VEG2) and Non-vegetables oils (OIL1,OIL2,OIL3)

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Mathematical model The decision variables represents the quantities of oils to be blended togheter We assume that there is no loss of weight in the refining process and blending VEG1 = x 1, VEG2 = x 2, OIL1 = x 3, OIL2 = x 4, OIL3 = x 5 Total product = x 1 + x 2 + x 3 + x 4 + x 5 x 1, x 2, x 3, x 4, x 5 >= 0 Objective function: maximize profit = selling price - costs

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Mathematical model Selling price = 150 * total product= 150 * (x 1 + x 2 + x 3 + x 4 + x 5 ) Max 40 x 1 + 30 x 2 + 20 x 3 + 40 x 4 + 35 x 5 x 1, x 2, x 3, x 4, x 5 >= 0 Unit cost of VEG1 Costs = 110 x 1 + 120 x 2 + 130 x 3 + 110 x 4 + 115 x 5 Overall problem

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Excel table =SOMMA(C7:G7) cost=C5*C7+D5*D7+E5*E7+F5*F7+G5*G7 Selling price = C10*E10 data

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Setting the Solver in Excel Only non negative constraints

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Final result with Excel There is no optimal solution (the algorithm does not converge) Unboundedness The model is not well defined !

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Refining the model of blending problem In any month it is not possible to refine more than 200 tons of VEGETABL OILS and 250 tons of NON-VEGETABL OILS Availability

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Mathematical model Availability constraints Max 40 x 1 + 30 x 2 + 20 x 3 + 40 x 4 + 35 x 5 x 1 + x 2 <= 200 x 3 + x 4 + x 5 <= 250 x 1, x 2, x 3, x 4, x 5 >= 0 VEG1 + VEG2 <= 200 OIL1 + OIL2 + OIL3 <= 250 The new problem

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Excel table new data x 1 +x 2 = C8+D8=C9 x 3 +x 4 +x 5 = E8+F8+G8=F9

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Setting the Solver constraints Objective function = profit

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Final result with Excel

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Adding hardness constraint Technological restriction on the hardness of the final product. In the percentage it must lie between 3 and 6. It is assumed that hardness behaves linearly In blending problem there is a typical constraint on the “quality” of the final product new data = hardness % of raw oils

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Modelling hardness constraint Hardness constraints: 8.8 VEG1 + 6.1 VEG2 + 2 OIL1 + 4.2 OIL2 +5 OIL3 Contribution in “hardness” of the raw oils Requirement in hardness of the total product 6 (VEG1 + VEG2 + OIL1 + OIL2 + OIL3)max 3 (VEG1 + VEG2 + OIL1 + OIL2 + OIL3)min

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Hardness constraints Hardness constraints: 8.8 x 1 + 6.1 x 2 + 2 x 3 + 4.2 x 4 + 5 x 5 <= 6(x 1 + x 2 + x 3 + x 4 + x 5 ) 8.8 x 1 + 6.1 x 2 + 2 x 3 + 4.2 x 4 + 5 x 5 >= 3(x 1 + x 2 + x 3 + x 4 + x 5 ) but it is not linear ! These constraints can be written also in the form Hardness in % of the final product

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Mathematical model Max 40 x 1 + 30 x 2 + 20 x 3 + 40 x 4 + 35 x 5 x 1 + x 2 <= 200 x 3 + x 4 + x 5 <= 250 8.8 x 1 + 6.1 x 2 + 2 x 3 + 4.2 x 4 + 5 x 5 <= 6(x 1 + x 2 + x 3 + x 4 + x 5 ) 8.8 x 1 + 6.1 x 2 + 2 x 3 + 4.2 x 4 + 5 x 5 >= 3(x 1 + x 2 + x 3 + x 4 + x 5 ) x 1, x 2, x 3, x 4, x 5 >= 0 The final blendig mathematical model

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Excel table Hardness constraints Data on the final product

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Setting the Solver constraints Objective function = profit

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Final result with Excel Final food features

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A multi plant problem A company consists of two factories A and B. Each factory makes two products: standard and deluxe Each factory use two processes, grinding and polishing for producing its product Each unit of product yields the following profit

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A multi plant problem The grinding and polishing times in hours for a unit of each type of product in each factory are Factory A has a grinding capacities of 80 hours per week and polishing capacity of 60 hours per week Factory B has a grinding capacities of 60 hours per week and polishing capacity of 75 hours per week

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A multi plant problem Availability of raw material Each product (standard or deluxe) requires 4 kg of a raw material The company has 120 kg of raw material per week 120 kg. Factory A is allocated 75 Kg Factory B is allocated 45 Kg A possible scenario

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Mathematical model for factory A The two type of products are the decision variables for FACTORY A Objective function is the profit to be maximize standard = x 1, deluxe = x 2 max 10 x 1 + 15 x 2 x 1, x 2 >= 0 Constraints:Availability of raw material 4 x 1 + 4 x 2 <= 75 Kg of raw material for unit of standard product Kg of raw material for unit of deluxe product Unit profit of standard product Unit profit of standard deluxe

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Mathematical model for factory A (2) Constraints: 4 x 1 + 2 x 2 <= 80 Technological constraints Grinding process 2 x 1 + 5 x 2 <= 60 Polishing process max 10 x 1 + 15 x 2 4 x 1 + 4 x 2 <= 75 4 x 1 + 2 x 2 <= 80 2 x 1 + 5 x 2 <= 60 x 1, x 2 >= 0 Overall model for factory A

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4 x 1 + 4 x 2 = 75 Geometric representation of F Let draw the set F of the feasible solutions for factory A In the plane ( x 1, x 2 ), draw the equations of the constraints 5 10 15 20 25 30 35 40 45 510152025303540 45 x1x1 x2x2 The constraint 4 x 1 + 2 x 2 = 80 does not play any role in defining the feasible region: removing it does not change F 2 x 1 + 5 x 2 = 60 4 x 1 + 2 x 2 = 80 Bad use of resources ! Feasible region All non negative pointsconstitutes the

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4 x 1 + 4 x 2 = 75 Geometric representation of the profit In the plane ( x 1, x 2 ), draw the equation of the profit P TOT for increasing values 510152025303540 45 x1x1 5 10 15 20 25 30 35 40 45 x2x2 2 x 1 + 5 x 2 = 60 P TOT = 10 x 1 + 15 x 2 =0 =150 =300 P TOT =0 P TOT = 150 P TOT = 300 They are parallel lines Find the value of P TOT such that the corresponding line “touch” the points P TOT =300 does not touch any point in F

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4 x 1 + 4 x 2 = 75 Geometric solution In the plane ( x 1, x 2 ), draw the parallel lines to the equation P TOT = 10 x 1 + 15 x 2 =0 until the last point is found that “touches” the feasible region 510152025303540 45 x1x1 5 10 15 20 25 30 35 40 45 x2x2 2 x 1 + 5 x 2 = 60 P TOT =0 Raw material hours 2 x 1 + 5 x 2 = 60 4 x 1 + 4 x 2 = 75 Optimal solution P TOT = 10 x 1 + 15 x 2 = 112.5 + 112.5 = 225

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Excel table for factory A data x 1 =C9, x 2 =D9 Decision variables = level of production Profit = C4*C9+D4*D9 Raw constraint = C5*C9+D5*D9 Grinding constraint = C6*C9+D6*D9 Polishing constraint = C7*C9+D7*D9

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Using the Solver constraints Objective function = profit Decision variables

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Mathematical model for factory B The two type of products are the decision variables for FACTORY B Objective function is the profit to be maximize standard = x 3, deluxe = x 4 max 10 x 3 + 15 x 4 x 3, x 4 >= 0 Constraints: Availability of raw material 4 x 3 + 4 x 4 <= 45

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Mathematical model for factory B (2) Constraints: Technological constraints 5 x 3 + 3 x 4 <= 60 Grinding process 5 x 3 + 6 x 4 <= 75 Polishing process max 10 x 3 + 15 x 3 4 x 3 + 4 x 3 <= 45 5 x 3 + 3 x 4 <= 60 5 x 3 + 6 x 4 <= 75 x3, x 3 >= 0 Overall model for factory B

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Geometric representation of F Let draw the set F of the feasible solutions for factory B In the plane ( x 3, x 4 ), draw the equations of the constraints 5 x 3 + 3 x 4 = 60 5101520304050 x3x3 5 10 15 20 30 40 50 x4x4 4 x 3 + 4 x 4 = 45 5 x 3 + 6 x 4 = 75 Feasible region All non negative pointsconstitutes the Two constraints 5 x 3 + 6 x 4 = 75 and 5 x 3 + 3 x 4 = 60 do not play any role in defining the feasible region: removing them does not change F Bad use of resources !

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Geometric solution 51020304050 x3x3 5 10 15 20 30 40 50 x4x4 4 x 3 + 4 x 4 = 45 In the plane ( x 3, x 4 ), draw the parallel equations of the profit P TOT for increasing values P TOT = 10 x 3 + 15 x 4 =0 =100 Find the value of P TOT such that the corresponding line “touch” the points P TOT =0 P TOT = 100 Raw material x 3 = 0 4 x 3 + 4 x 4 = 45 P TOT = 112.5 Optimal solution =

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Excel table for factory B data x 3 =C9, x 4 =D9 Decision variables = level of production Profit = C4*C9+D4*D9 Raw constraint = C5*C9+D5*D9 Grinding constraint = C6*C9+D6*D9 Polishing constraint = C7*C9+D7*D9 Note: the excel formulae are the same for factory A and B. The model is independent from data

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Look at the company in this scenario Overall production = sum of the production of factory A and factory B Profit of the company = sum of the profits of factory A and factory B This solution has been obtained with arbitrary allocation of resources

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Changing the scenario Factory A is allocated 90 Kg Factory B is allocated 30 Kg The solution has been obtained with arbitrary allocation of raw material, we can see what happens when allocation change 120 kg. Total raw material

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Changing the scenario: geometric view 51015203040 50 x3x3 5 15 20 30 40 50 x4x4 5101520304050 x1x1 4 x 1 + 4 x 2 = 90 5 10 15 20 30 40 45 50 x2x2 2 x 1 + 5 x 2 = 60 4 x 1 + 2 x 2 = 80 Factory A Factory B 5 x 3 + 3 x 4 = 60 4 x 3 + 4 x 4 = 30 5 x 3 + 6 x 4 = 75 5101520 x1x1 5 10 15 20 x2x2 5101520 x3x3 5 10 15 20 x4x4 new optimum for A new optimum for B P TOT = 250 P TOT = 112.5

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Changing the scenario: excel view Factory A Profit is higher than the preceding scenario Factory B Profit is lower than the preceding scenario

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Look at the company in the new scenario Overall production = sum of the production of factory A and factory B Profit of the company = sum of the profits of factory A and factory B This solution is worst than the preceding one

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Mathematical model for the company The two type of products produced in FACTORY A and B are the decision variables Objective function is the overall profit to be maximize max 10 x 1 + 15 x 2 + 10 x 3 + 15 x 4 standard in factory A= x 1, deluxe in factory A = x 2 standard in factory B= x 3, deluxe in factory B= x 4 x 1, x 2, x 3, x 4 >= 0

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Mathematical model for the company (2) Constraints: 4 x 1 + 2 x 2 <= 80 5 x 3 + 3 x 4 <= 60 Technological constraints Grinding process 2 x 1 + 5 x 2 <= 60 5 x 3 + 6 x 4 <= 75 Polishing process Constraints:Availability of raw material Factory A Factory B Factory A 4 x 1 + 4 x 2 + 4 x 3 + 4 x 4 <= 120 Common constraint

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Mathematical model for the company max 10 x 1 + 15 x 2 + 10 x 3 + 15 x 4 4 x 1 + 2 x 2 <= 80 5 x 3 + 3 x 4 <= 60 2 x 1 + 5 x 2 <= 60 5 x 3 + 6 x 4 <= 75 4 x 1 + 4 x 2 +4 x 3 + 4 x 4 <= 120 x 1, x 2, x 3, x 4 >= 0 More than two variables: we can solve it with the Solver

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Excel table for the company x 1 =C10, x 2 =D10, x 3 =E10, x 4 =F10 Decision variables = level of production Profit = C4*(C10+E10)+D4*(D10+F10) Raw constraint = C5*(C10+ E10 )+D5*(D10+F10)

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Setting the solver

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Optimal solution for the company Optimal production: deluxe = 20.8, standard = 9.17 Profit = 404.16 Better than 393.75 obtained with the arbitrary allocation

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References H.P. Williams, Model building in mathematical programming, John Wiley, 1999 W. L Winston and S. C. Albright, Practical Management Science, Duxbury Press, 1997 L. Palagi, Electronic version of the lectures (2004) http://www.dis.uniroma1.it/~palagi

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