1. Chapter 1. Introduction Ex : Diet Problem
Daily requirements : energy(2000kcal), protein(55g), calcium(800mg)
Food
Serving
size
Energy
(kcal)
Protein
(g)
Calcium
(mg)
Price per
serving (cents)
Max serving
allowed
Oatmeal
28g
110 4 2 3
Chicken
100g
205 32 12 24
Eggs
2 large
160 13 54
Wholemilk
237cc 8
285 9
Cherry pie
170g
420 22 20
Pork with beans
260g
260 14 80 19 OR

2. Formulation:
Subject to OR

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

4. Unusual formulations Cutting stock problem :
Rolls of papers with width W to be cut into small pieces(finals).
bi pieces of width wi , i =1, 2, …, m need to be produced.
How to cut the rolls to meet demands while minimizing wastes?
Subject to
Consider possible cutting patterns for a raw.
xj : number of rolls to be cut using j-th cutting pattern.
n denotes the total number of possible patterns which can be a very large number.
aij = k if the number of i-th piece produced in the j-th pattern is k.
(Refer Chapter 13 later) OR

5. need 97 finals of width 45 in. 610 finals of width 36 in.
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 x x x … x37
Note:
number of patterns grows fast as problem becomes large (don’t have all columns initially, but generate columns when needed)
round down fractional optimal solution to LP to obtain integer solution, then use a few more raws to meet demands.
extension to 2-dimensional cutting stock (nesting problem), 3-D packing OR

6.  Minimization of piecewise linear convex function c3’x+d3 c2’x+d2
Subject to
Subject to
c3’x+d3 f
c2’x+d2
c1’x+d1 x OR

7. ex) parallel processor scheduling problem
There are m processors and n jobs to be processed on any one of the processors.
aij : processing time of job j on processor i.
Assign jobs to processors so that overall finish time (makespan) is minimized. Formulation as an integer programming problem
Let xij = 1 if job j is assigned to processor i, 0 otherwise.
minimize
s. t.
 minimize z OR

8. cannot be formulated as a linear programming problem.
Linear programming relaxation of an integer programming problem is obtained by dropping the integrality requirement 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.
Note that
cannot be formulated as a linear programming problem.
(maximizing a convex function) OR

9. function f: Rn  R is separable if f(x) = f1(x1) + f2(x2) + … + fn(xn)
Special case of piecewise linear objective function : separable piecewise linear objective function.
function f: Rn  R is separable if f(x) = f1(x1) + f2(x2) + … + fn(xn)
If obj. fn. is nonlinear, but separable  we may approximate it by piecewise linear fn. (need some caution)
c1 < c2 < c3 < c4
fi(xi) c4 c3
slope: ci c2 c1 xi a1 a2 a3
x1i
x2i
x3i
x4i OR

10. If xi = 7, we want to have xi = x1i + x2i + x3i + x4i = 5 + 2 + 0 + 0
Ex: Suppose c1 = 1, c2 = 2, c3 = 3, c4 = 4
a1 = 5, a2 = 10, a3 = 15
If xi = 7, we want to have xi = x1i + x2i + x3i + x4i =
and objective function value is f(xi) = 15 + 22 + 30 + 4 0 = 9
Express xi in the constraints as xi  x1i + x2i + x3i + x4i , where
0  x1i  a1, 0  x2i  a2 - a1 , 0  x3i  a3 - a2, 0  x4i
In the objective function, use : min c1x1i + c2x2i + c3x3i + c4x4i
Since we solve min problem, it is guaranteed that if we get
xki > 0 in an optimal solution --> xji , j < k have values at their upper bounds.
Ex(continued): xi = 7 may be expressed with x1i = 5, x2i = 0, x3i = 2, x4i = 0.
But then, objective function value is f(xi) = 15 + 20 + 32 + 4 0 = 11, which is worse than the previous solution. (We have minimization prob.) OR

11. Terminology
Feasible solution (가능해) : vector x Rn 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 > M for any finite M > 0.
ex:
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 OR

12. 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 efficient
Interior point method : L. Karmarkar, 1984
polynomial time algorithm, many variations, practically good performance. Recently, used for some nonlinear programming problems successfully (convex optimization)
Theory of LP provides important foundation for many other disciplines like integer programming, networks and graphs, nonlinear programming, etc . OR

13. Standard form Any LP problem can be expressed as the standard form
Minimization problem : solve max (-c)’x, then take negative of the optimal objective value. Optimal solution is the same
Equality constraint : ai’x=bi  ai’x  bi,, (- ai)’x  (-bi)
Unrestricted variable xj  substitute xj by in the objective function and in the constraints where
(Simplex method find solution with at most one of is positive)
Some people use min(max) c’x, Ax = b, x 0 as standard form.
subject to OR

14. Formulation when absolute values are involved
ex)
Formulation when absolute values are involved
(Assume cj  0 for variables involving absolute values)
Subject to
Subject to OR

15. express free xj = xj+ - xj-, xj+  0, xj-  0
want xj = xj+ , if xj  0
= xj- , if xj < 0
need xj+  xj- = 0 in an optimal solution, i.e. at most one is positive.
If cj > 0, this is guaranteed to hold at an optimal solution.
Ex: Suppose cj = 3, xj = -5 = 0 – 5, hence objective value is 3  (0 + 5) = 15. If we have xj+ = 2, xj- = 7, then xj = = – 5. But objective value is 3  (2 + 7) = 27, which is inferior compared to the previous solution.
Alternative formulation:
subject to OR