MS-E2140. Lecture 1. (course book chapters )

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1 Linear Programming MS-E2140 Motivations and background Lecture 1 (course book chapters ) Linear programming problems and examples Problem manipulations and standard form Graphical representation of linear programming problems Modeling absolute values Modeling piecewise linear convex functions v. 1.1

2 Motivations Linear Programming (LP) problems form an important class of optimization problems with many practical applications. Linear Programming has applications for example in Production planning, resource allocation, investment decisions, military operations, scheduling, transportation and logistics, inventory management, scheduling, game theory... Efficient LP solution methods have been developed, and nowadays routinely used within optimization packages to solve even very large problems The Simplex algorithm for solving Linear Programs (Dantzig, 1947) is considered one of the top 10 algorithms developed in the 20-th century ( LP theory and duality form the basis to the development of more sophisticated methods for solving hard combinatorial optimization problems

3 Hystorical background The rapid and systematic development of Linear Programming as a practical tool for modeling and solving optimization problems started with the invention of the Simplex algorithm in 1947 The Simplex algorithm was invented by Dantzig to solve military planning problems, and is one of the most practically effective methods for solving LPs (earlier methods are due to Fourier, 1824, and de la Vallee Poussin, 1910) Linear Programming models were also studied in economics in the late 30s by Leonid Kantorovich. Tjalling Koopmans (nobel prize in 1975 with Kantorovich) and Wassily Leontief (nobel prize in 1973) also played important roles The work of von Neumann in game theory (1928) and duality also proved to have strong connections with the fundamental theory of linear programming Paper: Dantzig, G. (2002). Linear Programming, reprinted in Operations Research, Vol. 50, No.1, pp ,

4 Linear programming problems Minimize or maximize a linear objective function subject to a set of m linear constraints Cost coefficient Minimize z = c 1 x 1 + c 2 x c n x n subject to Objective function Can be, or = a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b Constraints a m1 x 1 + a m2 x a mn x n = b m x 1, x 2, x n 0 Decision variables Non-negativity constraints In general, not all variables may be required to be 0, or some may be required to be 0

5 A Linear Programming problem (LP) can be expressed in matrix form Minimize z = c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n b a m1 x 1 + a m2 x a mn x n b m x 1, x 2, x n 0 Minimize z = c x subject to Ax b x 0 Notation: c = c 1, c 2,, c n R n x = (x 1, x 2,, x n ) R n b = (b 1, b 2,, b m ) R m A = a 11 a 1n R m n a m1 a mn Constraint matrix We will write a i to denote the i-th row of A so that the i-th constraint can be written as a i x b i a 1 Note: The symbol denotes transposition

6 Terminology. For an LP with a minimization objective function: A vector x that satisfies all the constraints is called feasible solution The set of all the feasible solutions is called feasible region A feasible solution x that minimizes the objective function (i.e., such that c x c x for any feasible solution x) is called optimal solution The value c x is called optimal cost x 2 (2) Minimize x 1 3x 2 s.t. x 1 x 2 6 constraint (1) (3) Feasible region (1) x 1 2x 2 8 x 1 + x 2 2 constraint (2) constraint (3) x 1 x 1, x 2 0

7 Examples of LPs Product blending. A manufacturer of plastics is planning to blend a new product by mixing four chemical compounds. Each compound contains three chemicals A, B, and C in different percentages. The composition and unit cost of the each compounds is: Each column gives the composition and cost of one compound Comp. 1 Comp. 2 Comp. 3 Comp. 4 % of A % of B % of C Cost/Kg The new product must contain 25% of element A, at least 35% of element B, and at least 20% of element C Moreover, to avoid side effects compounds 1 and 2 cannot exceed 25% and 30% of the total, respectively

8 What is the cheapest way of blending the new product? Decision variables x i : fraction of compound i (i = 1,, 4) used to produce one unit of the new product Mathematical formulation minimize z = 20x x x x 4 s.t. x 1 + x 2 + x 3 + x 4 = 1 30x x x x 4 = 25 20x x x x x x x x 4 20 x , x x 1, x 2, x 3, x 4 0 Cost to produce one unit of the new product % of element A % of element B % of element C Maximum % of comp. 1, 2 Non-negativity

9 The problem in matrix form is: x 1 Minimize z = c x s.t. A 1 x = b x 2 x 3 = 1 25 A 2 x b 2 A 3 x b 3 A 1 x 4 x x 1 b 1 x 0 where A 1 = b 1 = A 2 x 2 x 3 x 4 x b 2 A 2 = b 2 = x 1 x 2 x A 3 = b 3 = A 3 x 4 x b 3 c =

10 Production planning. A food firm is planning the production for the next 4 months, and can use a warehouse to stock food in each month. The maximum storage capacity of the warehouse is 100 tons, and each ton of stock at the end of any month has a cost of 5 EUR On month i, there is a production cost c i for each ton and a maximum production of p i tons is possible An extra production of q i tons is possible at extra cost: Each ton produced in excess of p i in month i has an additional cost of e i The firm has contracted to provide d i tons on each month i The warehouse is empty at the beginning of the first month, and must be empty at the end of the last month Finally, the regular production in each month must be at last 10% of the total production of the first three months (balanced production)

11 Row i gives the monthly production costs, demand, and capacities of month i Month (i) Production cost (c i ) Demand (d i ) Max. production (p i ) Max. extra production (q i ) Extra production cost (e i ) Objective: Minimize the costs over the four months Decision variables. x i : Regular production on month i = 1,, 4 s i : Extra production on month i = 1,, 4 y i : Warehouse stock at the end of month i = 1,, 3 Note: y 4 must always be 0, we don t need it explicitly in the model (the warehouse must be empty at the end of the last month)

12 Mathematical formulation 4 minimize z = (10x i + 16s i + 5y i ) Total production cost i=1 s.t. x 1 + s 1 = y 1 x 2 + s 2 + y 1 = y 2 x 3 + s 3 + y 2 = y 3 The total production in month i must equal the demand of that month plus the stock at the end of the month x 4 + s 4 + y 3 = 200 x i (s 1 + x 1 + s 2 + x 2 + s 3 + x 3 ) 0.1, i = 1,, 4 Balanced production x 1 140, x 2 150, x 3 140, x s 1 50, s 2 75, s 3 70, s 4 80 x i 0, s i 0, i = 1,, 4, y i 0, i = 1,, 3 Maximum regular and extra production Non-negativity

13 Problem manipulations LPs can be equivalently expressed in different forms: Minimize c x is equivalent to Maximize c x An equality constraint a 11 x a 1n x n = b 1 can be equivalently replaced by the two inequalities: a 11 x a 1n x n b 1 and a 11 x a 1n x n b 1 An inequality constraint a 11 x a 1n x n b 1 can be replaced by the equivalent inequality (a 11 x a 1n x n ) b 1 An inequality constraint a 11 x a 1n x n b 1 is equivalent to a 11 x a 1n x n + s 1 = b 1, where s 1 0 is a new variable called slack variable Similarly, a 11 x a 1n x n b 1 becomes a 11 x a 1n x n s 1 = b 1, where s 1 0

14 Any free variable x i (i.e., not restricted to be 0 or 0) can be replaced by the expression (x + i x i ) where x + i 0 and x i 0 Examples minimize z = 2x 1 + 5x 2 s. t 3x 1 + 2x 2 6 2x 1 + 9x 2 8 x 1 0 free Equivalent minimize z = 2x 1 + 5(x + 2 x 2 ) s. t 3x 1 + 2(x + 2 x 2 ) 6 2x 1 + 9(x + 2 x 2 ) 8 x 1, x + 2, x 2 0 minimize z = 2x 1 x 2 + 4x 3 minimize z = 2x 1 x 2 + 4x 3 s. t x 1 + x 2 + x 4 2 3x 2 x 3 = 5 x 3 + x 4 3 x 1 0, x 3 0 Equivalent s. t x 1 x 2 x 4 2 3x 2 x 3 5 3x 2 + x 3 5 x 3 +x 4 3 x 1 0, x 3 0

15 Standard form problems By using the previous transformations we can always express any LP in the following form, called standard form All variables must be non-negative Minimize z = c x s.t. Ax = b, x 0 All constraints are equality constraints Starting from any LP, we can put it in standard form by: 1. Replacing each free variable x i with (x i + x i ) where x i + 0, x i 0 2. Transforming any inequality constraint into an equality constraint by adding slack variables Any LP and its standard form are equivalent: Given a feasible solution to the original LP we can construct a feasible solution to its standard form with same cost, and vice versa

16 Example minimize z = 2x 1 + 4x 2 s. t x 1 + x 2 3 3x 1 + 2x 2 = 14 x 1 0 Standard form minimize z = 2x 1 + 4x + 2 4x 2 s. t x 1 + x + 2 x 2 s 1 = 3 3x 1 + 2x + 2 2x 2 = 14 x 1, x + 2, x 2, s 1 0 The Simplex algorithm is designed to solve LPs in standard form This is because it is based on the following operations: 1. multiply the coefficients and right hand side of a constraint by a nonnull real number 2. apply operation 1 to a constraint and sum the result to another one

17 The operations 1 and 2 when applied to a system of linear equations leave unchanged the set of feasible solutions However, when applied to a system of inequalities they do change the set of solutions Example (a) (b) x 1 2x 2 0 x 2 0 Multiply (b) by 2 and add to (a) x 1 0 x 2 0 x 2 x 2 x 1 x 1

18 Graphical representation of an LP By transforming any equality constraint into an inequality, we can also always rewrite an LP in the following general form where eventual non-negativity constraints are also included in the definition of A minimize z = 3x 1 2x 2 + x 3 s.t. 2x 1 x 2 1 x 2 + x 3 5 x 1 0 x 3 0 Minimize z = c x s.t. Ax b A = x = (x 1 x 2 x 3 ) b = ( ) Each constraint of an LP in this form with n variables defines a region of R n, called halfspace, containing all x R n satisfying the constraint The feasible region is then the intersection of all these halfspaces

19 For LPs in general form with two or three variables we can visualize the feasible region, and even solve them graphically: minimize z = x 1 3x 2 x 1 x 2 6 (1) x 1 2x 2 8 (2) x 1 + x 2 2 (3) x x 2 0 (5) Halfspace defined by x 1 + x 2 2 x 2 Halfspace defined by x 1 + x 2 2 x 1 + x 2 = 2 x 1 The feasible region is the intersection of the halfspaces defined by (1) (5) x 2 For any value z, the set of all solutions x with (3) (5) (2) feasible region (1) (4) cost z forms a line c x = x 1 3x 2 = z (sometimes called isoprofit line) This line is perpendicular to the vector c = ( 1, 3) c x 1

20 x 2 Optimal solution x = ( ), z = 38/3 c x = z x 1 3x 2 = 38/3 (z = 38/3) x 1 3x 2 = 12 (z = 12) c = ( 1 3) x 1 3x 2 = 6 (z = 6) x 1 3x 2 = 0 (z = 0) Isoprofit line for z = 0 The value z decreases along the direction c so minimizing z corresponds to moving the line z = x 1 3x 2 in the direction of c The minimum value of z is obtained when the line cannot be moved further without leaving the feasible region When this happens the line intersects a corner point of the feasible region which is an optimal solution

21 It is not always the case that an LP has a unique optimal solution Consider the following LP: minimize z = c 1 x 1 + c 2 x 2 x 1 + x 2 1 x 1 0 x 2 0 Depending on the choice of c, the following cases can occur: x 1 x 1 + x 2 = 1 Feasible region x 2 Unique optimal solution Bounded set of alternative optimal solutions x 1 x is the only optimal solution x 1 Any solution x = (0 x 2 ) with 0 x 2 1 is optimal (0 1) c = (1 1) x = (0 x 2 ) with 0 x 2 1 x = (0 0) c x = x 1 + x 2 = 0 (z = 0) x 2 (0 0) c = (1 0) x 2

22 Unbounded set of alternative optimal solutions Any solution x = (x 1 0) with x 1 0 is optimal Unbounded optimal cost (unbounded problem) There is not an optimal solution: z for x 1, x 2 the optimal cost is c = (0 1) c x = x 1 x 2 = z x x = (x 1 0) with x 1 0 c = ( 1 1) z Finally, the feasible region can be empty, in which case there are not feasible solutions and the LP is infeasible: minimize z = c 1 x 1 + c 2 x 2 x 1 + x 2 1 x 1 x 2 2 x 1 0 x 2 0 x 1 x 2 2 x 1 + x 2 1

23 Modeling absolute values LPs can be used to model situations where variables represent absolute values under some assumptions on the sign of their coefficients. Consider the problem: minimize n i=1 c i x i s.t. Ax b where x = (x 1,, x n ) and c i 0, i = 1,, n We can note that x i is the smallest number y i satisfying y i x i and y i x i, so we can rewrite the problem as the LP minimize n i=1 c i y i s.t. Ax b y i x i, i = 1,, n y i x i, i = 1,, n This is correct because c i 0 and we are minimizing: An optimal solution must have y i = max x i, x i = x i ; Otherwise we could reduce y i and obtain a feasible solution with lower cost

24 Another possibility to rewrite the same problem as an LP is to: replace x i with (x i + x i ) where x i + 0, x i 0 replace x i with (x i + + x i ) In this way we obtain the LP: Example: for x i = 4 we can write x + i = 0 and x i = 4 so that x i = x + i x i and x i = x + i + x i minimize n i=1 c i x i + + x i s.t. Ax + Ax b x +, x 0 where x + = (x + 1,, x + n ) and x = (x 1,, x n ) Since we are minimizing and c i 0, an optimal solution must have x + i = 0 or x i = 0, i, because otherwise we could reduce both x + i and x i by the same amount and obtain a better feasible solution

25 Example. Consider the problem: Minimize z = 2 x 1 + x 2 s.t. x 1 + x 2 4 The two alternative reformulations to transform it into an LP are: Minimize z = 2y 1 + x 2 s.t. x 1 + x 2 4 y 1 x 1 y 1 x 1 Minimize z = 2x x 1 + x 2 s.t. x + 1 x 1 + x 2 4 x x 1 0

26 Modeling piecewise linear functions A function f: R n R is called convex if for every x, y R n, and for every λ [0,1], we have f λx + 1 λ y λf x + (1 λ)f y f is concave if it satisfies the above with " " replaced by " Note: All points λx + 1 λ y lie on the line segment joining x and y; Informally, f is convex if its graph lies below the line connecting x and y f x λf x + (1 λ)f y, for λ [0,1] f x f y f y x y x y A convex function f λx + (1 λ)y, for λ [0,1] A concave function

27 A function of the form f x = max (c i x + d i ) is called piecewise i=1,,m linear convex (or piecewise linear concave if max is replaced by min). Example with m = 3, and x R 1 f x = max 1 x, 2 x + 2, x + 5 is piecewise linear convex Piecewise linear convex (or concave) functions can be used to approximate convex (or concave) functions f x f x x x x + 2 A piecewise linear convex function x A convex function approximated by a piecewise linear convex function x

28 LPs can be used to model: the minimization of a piecewise linear convex function (1) Minimize z = max (c i x + d i ) i=1,,m s.t. Ax b the maximization of a piecewise linear concave function (2) Maximize z = min (c i x + d i ) i=1,,m s.t. Ax b The idea is to introduce z explicitly as a variable and impose z c i x + d i or z c i x + d i, i. Problems (1), (2) become the LP: (1) Minimize z (2) Maximize z s.t. Ax b z c i x + d i, i = 1,, m s.t. Ax b z c i x + d i, i = 1,, m

29 Example. Consider a firm producing and selling a product. A typical strategy is to introduce economies of scale to incentivate bulk buys Suppose the unit selling price is set to 10 EUR for the first 1000 units, but is reduced to 7 EUR for any amount in excess to that Suppose an LP is used in planning the production with variables x i representing the amount produced (and sold) in month i = 1,, n, subject to a set of constraints Ax b (with x = (x 1,, x n )) The profit p i that can be made from the sales in month i is: p i (x i ) = 10x i, if 0 x i 1, , (x i 1,000), if x i 1,000 which is equivalent to = x i p i x i = min 10x i, 10, x i 1,000 = min 10x i, 7x i

30 Example (cont.) The problem of maximizing the profit over the n months can then be modeled as the maximization of a piecewise linear concave function: which can be rewritten as the LP Maximize i=1,,n p i (x i ) s.t. Ax b Maximize i=1,,n p i s.t. Ax b p i 10x i, i = 1,, n p i 7x i , i = 1,, n

MS-E2140. Lecture 1. (course book chapters )

MS-E2140. Lecture 1. (course book chapters ) Linear Programming MS-E2140 Motivations and background Lecture 1 (course book chapters 1.1-1.4) Linear programming problems and examples Problem manipulations and standard form problems Graphical representation