Chapter 1. Introduction

1. Chapter 1. Introduction • Ex : Diet Problem Daily requirements : energy(2000kcal), protein(55g), calcium(800mg)

2. Formulation: Subject to

3. Linear Programming Problem (선형계획법 문제)  objective function (목적함수) right hand side (우변상수)  Constraints (제약식) Subject to nonnegativity constraints (비음제약식) (may not exist for some variables, then they are called unrestricted or freevariables)

4. Unusual formulations • Cutting stock problem : (Refer Chapter 13) Rolls of papers (called raws) with width to be cut into small pieces(finals). pieces of width , need to be produced. How to cut the rolls to meet demands while minimizing wastes? minimize s. t. , and integer for all • Consider possible cutting patterns for a raw. : number of rolls to be cut using -th cutting pattern. denotes the total number of possible patterns which can be a very large number. if the number of finals produced in the j-th pattern is .

5. ex) raws W=100 in., need 97 finals of width 45 in. 610 finals of width 36 in. 395 finals of width 31 in. 211 finals of width 14 in. Min x1 + x2 + x3 + … + x37 • Note: • number of patterns grows fast as problem becomes large (We don’t solve the problem with all columns in the model. We start with a few columns and solve the LP, and then identify and add a new column to the model and solve the LP again, … (column generation method). • round down fractional optimal solution to LP to obtain integer solution, then use a few more raws to meet demands. • extensions to 2-dimensional cutting stock (nesting problem), 3-D packing

6. Piecewise linear convex function: Consider , (maximum of affine functions, called a piecewise linear convex function.)

7. Thm: Let be convex functions. Then is also convex. pf)   = 

8. , .  min s. t. , s.t., , • Minimization of piecewise linear convex function

9. ex)parallel processor scheduling problem There are processors and jobs to be processed on any one of the processors. : processing time of job on processor . Assign jobs to processors so that overall finish time (makespan) is minimized. Formulation as an integer programming problem. Let = 1 if job is assigned to processor , 0 otherwise. minimize {} s. t. , and integer for all  minimize s. t. , , and integer for all

10. Linear programming relaxation of an integer programming problem is obtained by dropping the integrality requirements on the variables and considering only the linear constraints. • The optimal value of the linear programming relaxation provides a lower bound (for the minimization problem) on the optimal value of the integer programming problem. Hence it can be used importantly in the algorithm for integer programming problem. ( Integer Program: min , , and integer. Let , be the optimal value of the integer program (IP) and its LP relaxation, respectively. Let be the optimal solution of IP, then is a feasible solution to LP relaxation. Hence .) • For the maximization problem, LP relaxation gives an upper bound on the optimal value of IP. • Note that maximize cannot be formulated as a linear programming problem. (maximizing a convex function)

11. Special case of piecewise linear objective function : separable piecewise linear objective function. • function is called separable if . If objective function is nonlinear, but separable  we may approximate it by piecewise linear function. (need some caution) • The cost structure may be progressive, e.g.: electricity cost. slope: 0

12. Ex: Suppose = 1, = 2, = 3, = 4 = 5, = 10, = 15 If = 7, we want to have = + + + = 5 + 2 + 0 + 0 and objective function value is = 15 + 22 + 30 + 4 0 = 9 • Replace in the constraints with , where 0   , 0   , 0   , 0  In the objective function, use : min Since we solve minimization problem, it is guaranteed that if we get in an optimal solution , have values at their upper bounds. • Ex(continued): = 7 may be expressed 5 + 0 + 2 + 0 with = 5, = 0, = 2, = 0. But then, objective function value is = 15 + 20 + 32 + 4 0 = 11, which is worse than the previous solution. (We have a minimization problem.)

13. Terminology • Feasible solution (가능해) : vector that satisfies all constraints • Optimal solution (최적해) : feasible solution that gives the largest objective function value. • Optimal value (최적값) : objective function value of an optimal solution. • Infeasible LP : LP that does not have a feasible solution • Unbounded : LP is called unbounded if there exists a feasible solution that gives optimal value > for any finite > 0. ex: {max , s.t.} • Every LP has one of the following three statuses: • There exists an optimal solution with finite optimal value (There may exist multiple optimal solutions) • Infeasible • Unbounded

14. Brief History of LP • Solving systems of linear inequalities : Fourier, 1826, not efficient (Chapter 16) • Simplex method : G. B. Dantzig, 1947 • Ellipsoid method : L. G. Khachian, 1979 First polynomial time algorithm (theoretically efficient algorithm) for LP, practically not good. • Interior point method : L. Karmarkar, 1984 polynomial time algorithm, many variations, practically good performance. Recently, used for some nonlinear programming problems successfully (convex program – min convex function, convex set) • Theory of LP provides important foundation for many other disciplines like integer programming, networks and graphs, nonlinear programming, etc .

15. Standard form • (In vector notation) maximize maximize s. t. , s.t. , • Any LP problem can be expressed as the standard form • Minimization problem : solve max , then take the negative of the optimal objective value. Optimal solution is the same • Equality constraint :  , • Unrestricted variable  substitute by (in the objective function and in the constraints where . (Simplex method finds an optimal solution with at most one of is positive) • Some people use min(max) as standard form.

16. ex) max max s. t.  s. t.

17. Formulation when absolute values appear in the objective function (Assume cj 0 for variables involving absolute values) min min s. t. , s.t., unrestricted for all for all • Express free variable as () in the constraints () Also in the objective function as ()(). We want = , if = , if We need in an optimal solution, i.e. at most one of is positive. If for all , this is guaranteed to hold at an optimal solution.

18. Ex: Suppose = 3, = 5 = 0 – 5 in an optimal solution, hence objective value is 3  (0 + 5) = 15. If we have = 2, = 7, then = 2  7 = – 5. But objective value is 3  (2 + 7) = 27, which is inferior compared to the previous solution. Hence at most one of is positive at an optimal solution. • Alternative formulation: minimize s. t. , , ,