Linear Mathematical Programming (LP)

Transcription

1

2 Linear Mathematical Programming (LP) A MP is LP if : The objective function is linear where The set is defined by linear equality or inequality constraints c f T ) = ( ],..., [ n T c c c = = n b A where = b m b b = mn m n a a a a A

3 Linear Mathematical Programming (LP) An LP problem can be espressed as (LP) ma f ( ) = c T s. t. A b 0 c b A The quantities are two vectors and a matri of constant coefficients representing the parameters of the problem

4 Linear Mathematical Programming (LP) The set of feasible solutions of LP can be denoted as n = { R : A b, 0} Types of LP problems: n = { R : A b, 0} Continuous LP (LP) n = { Z : A b, 0} nteger LP (P) n n = { R, y Z : A + Dy b, 0, y Mied nteger LP (MP) 0}

5 Linear Mathematical Programming (LP) How can a decision problem to be modeled as a LP problem? What corresponds to the set? How can we find a solution to our decision problem? We see these aspects considering an eample...

6 LP an eample A company producing paints wants to plan its daily production Two types of paints are considered, a paint for interior () and a paint for eterior () Paint production uses two raw materials indicated with A and B The daily availability of raw material is: A = 6 tons, B = 8 tons The quantity of A and B consumed to produce one ton of paint and is known: Paints Raw material A B We assume that all the paint produced is sold The seling price per ton is K for and K for Further information obtained by the company through a market survey: the daily demand of paint never eceeds more than ton that of paint the maimum daily demand of paint is -ton Decision problem: determine the quantities to produce daily for the two paints in order to make the maimize the gain from sales

7 LP problem formulation MP formulation of a decision problem Definition of variables Mathematical formulation Definition of objective Definition of constraints Variables: define the solution of the problem Objective: quantifies the quality of a solution (as a function of the variables) Constraints: identify the set of allowed values for the variables defining the feasible solution set (e.g., technological constraints, budget, operational constraints, etc.)

8 LP an eample (cont.) Problem variables Two variables representing the quantity (tons) of paint of the two types produced (and sold) daily: Production of paint for eternal: Production of paint for internal : Continuous and non negative variables Problem objective function Maimize the daily gain (K ) from selling the produced paint types Z = + t is a linear epression

9 LP an eample (cont.) Problem constraints Technological constraints: the use of raw materials cannot eceed the material availability Consumption for unit of product Raw material ( A) ( B) Availability Constraints due to the market survey Variable positivity constraints (variable lower bounds) 0 0

10 LP an eample (cont.) The complete problem formulation ma Z = (5) 0 (6) () () () (4) Objective function Constraints defining the set of feasible solutions t is a LP problem

11 LP Graphic solution The (, ) plane (5) 8 7 () 6 () Non negativity constraints 0 (5) 0 (6) 5 4 ntoducing the constraints () () () (4) (6) + () (4)

12 LP Polyedra The feasibility set includes the inner points and the sides (in case of strictly inequality the problem is non linear) is a Polyedron obtained from the intersection of the semispaces defined by hyperplanes (lines in the eample) What s the difference between LP and P polyedra? A polyedron for an LP problem include feasible solutions n = { R : A b}

13 LP - Polyedra The polyhedron for an P problem has a finite number of feasible solutions n = { Z : A b} Red dots on the sides and in the polyedron correspond to the integer feasible solutions

14 LP - Graphic solution (5) The obj function is drawn in the plane (, ) as a parametric line that depends on the value fied for Z Z = + + f Z=0 = 0 = () () (4) (6) f Z=6 ()

15

16 LP - Graphic solution 4 (5) F () D () C (4) Optimal solution A=(0,0) B=(4,0) C=(0/,4/) D=(,) =(,) F=(0,) - - A B (6) () Z=0 Z=6 Z=8/

17 LP - Graphic solution 4 (5) F () D () C Property: The optimal solution lies on the frontier of polyedron (4) Z * Optimal solution = + = 8 C=(0/,4/) - - A B (6) ()

18 LP - Graphic solution Linearity implies that optimal (finite) solutions are in the polyedron vertices More than a single (finite) equivalent solutions (infinite number) if the objective function is parallel to a constraint ven in these cases the search for optimal solution is limited to polyedron vertices All the points on DC are optimal D F C A B 4

19 LP Graphic solution and post-optimality n the paint production problem it can be interesting to know how the raw materials (available resources) are used: s there any advantage from an increase of the resource availability? s it possible to decrease the resource availability without changing the optimal objective value? Another interesting point is to analyse how the optimal solution would change if the selling prices change

20 LP - Graphic solution and post-optimality Post-optimality Sensitivity analysis w.r.t changes in resources availability All the constraints are so they can be viewed as Quantity of resource used Resource availability Constraints () and () are satisfied as equalities by the optimal solution Then both raw materials A and B are used up to their availability Constraints () and () are saturated and raw materials A and B are said scarce resources Saturated constraint Scarse resource Non Saturated constraint Abundant resource

21 LP - Graphic solution and post-optimality First case: increase the availability of a scarse resource What is the maimum increment that is worth considering for resource A? z=8/ z= D K (4) ncreasing resource A availability moves constraints () and so F C () the optimal point A B K=(,) 4 ()

22 LP - Graphic solution and post-optimality ncrease beyond K=(,) is no profitable (why?) D (4) The new ( ) F C ( ) () + 7 (') A B Availability of A 4 increases from 6 to 7 ()

23 LP - Graphic solution and post-optimality Same behaviour for resource B F A z=8/ D C B (4) () z=8 ncreasing availability of B move () and optimal point L (6) () Beyond L=(6,0) no reason for further increase of B. New availability for B is + (')

24 LP - Graphic solution and post-optimality Opposite case: decrease of abundant resources Assume () and (4) associated with (abundant) resources How much can we reduce their avalability without affecting the current optimal solution? Constraint () + D () (4) + F A C B () 4 ()

25 LP - Graphic solution and post-optimality Opposite case: decrease of abundant resources Assume () and (4) associated with (abundant) resources How much can we reduce their avalability without affecting the current optimal solution? Constraint (4) F D C (4) () 4 A B 4 ()

26 LP - Graphic solution and post-optimality Which among resource A and B is worth increasing first? (the company may have a limited budget to invest) The Unit Value of a resource y i (also called Shadow Price): y i = ma Z variation ma resourcei variation y i = the increase of objective for a unitary increase of resource i availability

27 LP - Graphic solution and post-optimality amples Resource A: y A = = 9 8 = (K /ton) Resource B: 8 8 y B = 8 = = 4 (K /ton) Resource B is the most convenient to increase

28 LP - Graphic solution and post-optimality Consider a possible variation in the product selling prices How the optimal solution change? Changing coefficient c and c the slope of objective function changes () c increase c decrease F D C slope c c = A B 4

29 LP - Graphic solution and post-optimality F D C Changing c ec point C remains the optimal solutionc slope until the = slope of the objective c function equals the one of either () or () A B 4 () c decrease c increase

30 LP - Graphic solution and post-optimality Compute the ranges for c and c First case: change c with fied c = c c = = = slope of () c = = slope of () c = 4 c 4 f c = C and D are optimal. f c becomes < only D is optimal f c = 4 C and B are optimal. f c becomes > 4 only B is optimal

31 LP - Graphic solution and post-optimality Compute the ranges for c and c Second case: change c with fied c = c c = = = slope of () = slope of () c c = = c 6 f c = / C and B are optimal. f c becomes < / only B is optimal f c = 6 C and D are optimal. f c becomes > 6 only D is optimal