Linear Programming

1. Fundamentals

objective function and constraints:

min/max3x1+24x2+13x3+9x4...s.tlinearconstraintsmin/max\quad 3x_1+24x_2+13x_3+9x_4...\\s.t\quad\;\; linear\; constraints min/max3x1+24x2+13x3+9x4...s.tlinearconstraints
optimal solution: a feasible solution that has the min(or max) objective value
convex property: the optimal solution must fall in the corner of the feasible region(decided by constraints)
geometric interpretation:
- each constraint defines a half-space, the solutions are the intersection of half-spaces and are called simplex;
- objective functions form a hyperplane.

2. Standard and Slack forms

LP to standard form:
- convert a minimization into a maximization;
- add nonnegative constraints: all xxx variables are greater than zero;
  - convert some variables that do not have nonnegative constraints;
  - replace xjx_jxj with xj′−xj′′x_j'-x_j''xj′−xj′′
- convert equality constraints into inequality constraints; by replacing === with ≤\le≤ and ≥\ge≥
- convert ≥\ge≥ into ≤\le≤;
Standard form to slack form
- For example: 2x1+3x2+x3≤5−>2x1+3x2+x3+x4=52x_1+3x_2+x_3\le 5\quad->\quad 2x_1+3x_2+x_3+x_4=52x1+3x2+x3≤5−>2x1+3x2+x3+x4=5, where x4x_4x4 is the slack.
- Put all the slacks to right hand side and get the slack form, for example:
Basic Solutions: in slack form, set all RHS variables to 0, and the LHS variables are called basic, RHS variables are nonbasic.

3. Simplex algorithm
Convert into slack form;
find the variable in “z row” with biggest coefficient;
Do min test and replace the basic value with its corresponding non-basic value;
Repeat step 2 and 3 until all variables in “z row” become negative or sometimes LP is obviously unbounded.
Why would it terminate? If the number of basic solution(mmm) is finite then iterations would be: (m+nm)\binom{m+n}{m}(mm+n)
LP is Unbounded: when one variable increase, then all of other variables also increase. Non constraints are binding; LP is unbounded.

4. Degeneracy

The objective function’s value doesn’t change between iterations.

1.we need two out of three lines to find a corner which means that one line(constraint) is extra.
2.the reason why one of variable is equal to zero.

5. Cycling

If Simplex fails to terminate then it cycles;
two or more basic variables compete for leaving, then choose the one with smallest subscript to leave.

6. Infeasible First Basic Solution

Example:

7. Auxiliary Linear Programming

If original LP is feasible, then the slack form of the auxiliary LP
will yield a feasible basic solution to the original LP (and the
corresponding slack form).

8. Duality

LP in standard form and its dual
Weak Duality: x∗x^*x∗ feasible solution to the primal LP; y∗y*y∗ feasible solution to the dual LP.
If

∑j=1ncjxj∗=∑i=1mbiyi∗\sum_{j=1}^nc_jx_j^*=\sum_{i=1}^mb_iy_i^*j=1∑ncjxj∗=i=1∑mbiyi∗

then x∗x^*x∗ and y∗y^*y∗ are optimal solutions to the primal and to the
dual LPs, respectively.

9. Upper Bounds on Maximization LP

10. Additional Proof

10.1 If S IMPLEX fails to terminate in at most (n+mm)\binom{n+m}{m}(mn+m)iterations, then it cycles.

Proof Given a set BBB of basic variables(Lemma 29.4), the associated slack form is uniquely determined. There are n+mn+mn+m variables and ∣B∣=m|B|=m∣B∣=m, and therefore, there are at most (n+mm)\binom{n+m}{m}(mn+m) ways to determine BBB. Thus, there are at most (n+mm)\binom{n+m}{m}(mn+m) unique slack forms.

To be continue…

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Linear Programming线性规划(Introduction to Algorithms, 算法导论，CLRS)学习笔记

Linear Programming

1. Fundamentals

2. Standard and Slack forms

3. Simplex algorithm

4. Degeneracy

5. Cycling

6. Infeasible First Basic Solution

7. Auxiliary Linear Programming

8. Duality

9. Upper Bounds on Maximization LP

10. Additional Proof

10.1 If S IMPLEX fails to terminate in at most (n+mm)\binom{n+m}{m}(mn+m)iterations, then it cycles.

To be continue…

Linear Programming线性规划(Introduction to Algorithms, 算法导论，CLRS)学习笔记相关推荐

最新文章

热门文章

Linear Programming线性规划(Introduction to Algorithms, 算法导论，CLRS)学习笔记

Linear Programming

1. Fundamentals

2. Standard and Slack forms

3. Simplex algorithm

4. Degeneracy

5. Cycling

6. Infeasible First Basic Solution

7. Auxiliary Linear Programming

8. Duality

9. Upper Bounds on Maximization LP

10. Additional Proof

10.1 If S IMPLEX fails to terminate in at most (n+mm)\binom{n+m}{m}(mn+m​)iterations, then it cycles.

To be continue…

Linear Programming线性规划(Introduction to Algorithms, 算法导论，CLRS)学习笔记相关推荐

最新文章

热门文章

10.1 If S IMPLEX fails to terminate in at most (n+mm)\binom{n+m}{m}(mn+m)iterations, then it cycles.