拉格朗日松弛算法(LR)

dual
z LD max zLR ( )
0
1.寻找到下界 2.得到一个初始的解
illustration for dual problems
原问题：寻找使得 目标值最小的x
对偶问题：寻找使得目标 函数最大的
Steps of SG
Step1:选一个初始拉格朗日乘子如 ＝0,t=1; Step2：对于 从中任选一个次梯度s t；若
yes
end
Simulation results
eg1: 12 jobs and 2 identical machines ; Eg2:25 jobs and 4 identical machines ; Result: Eg1 bound=31.82 with best heuristic result 34; Eg2 bound=37.74 with best heuristic result 38;
Pai (what)
It tells us How important is the time K How important is the machine of time k Conclusion The cost of using the resource of time k
Mathematical sense (pai)
Flow chart of heuristic
List 按照（LG start） K=1,i=1
Reform the list Refresh start Set K= k+1
refresh start no
Mk=0?
no i++;调度下一个job i=job_num?
yes
有任务在k完成吗
yes Mk++;(k+1 N)
{ Bi } i k i
k M k Min iTi k ik { Bi } k i k
Ci Bi 1 ti
k 0
i 1, 2,…，N
(2)
decomposition
Min ω T + π δ i i k ik {B } i k i
Ci Bi 1 ti
k 0
i 1, 2,…，N
(2)
Dual problem
Max L with

L Min iTi k ( ik M k )
{ Bi } i k i
k M k Min iTi k ik { Bi } k i k
pai的 值
3 2 2 1 1 0 0
5
10
15
20
25 k个 时 刻
30
35
40
45
50
Illustration for pai
7 6 5
5
4
4
3
3
2
2
1 1 0
0
5
10
15
20
25
30
35
40
45
50
Pai !=0

Pai !=0;
原问题一定跟对偶问题很 不一样

How can the solution of a completely different problem gives you a good primal solution???
1
0.5
步长的曲线
0
0
500
1000
1500 counter
2000
2500
3000
3500
upper
40 35 蓝线的期望值高于紫线 30 期望值不同对迭代造成的影响 期望值接近最优解，造成迭代次数增加 25 好处？
value
20来自百度文库
15
10
5
0
0
500
1000
1500 counter
2000
2500
{ Bi }
J iTi
i

i
ik
Mk
k 1,2,…，K i 1,2,…，N
（1） (2)
Ci Bi 1 ti
Steps of LG method
step1 .relaxation of the hard constrain step2 .solve the L-G dual problem using S-G method step3. get feasible solution using heuristic alg0rithm
{ Bi } i k i
Good beginning promises success!!
Thanks for listening!!
Outline
Deep into SG algorithm。 Solve parallel machine scheduling using L-G method Simulation analysis
Deep into SG method
Widely used in non-differentiable optimization; Min f(x)
analysis
Deep into pai—the LG multiplier What is the meaning of pai in physical sense? How it acts and changes during the itaration? The LG solution is a good primal solution though non-feasible. Why the LG solution is a good primal solution?
给定 π 将上述问题分解成N个子问题
Min Li {1 B k-t i +1} i s.t Ci Bi 1 t i with Li ω T + π δ i i k ik k i 1, 2, …，N
Solve dual problem
原问题 给定一个 sub1 sub2
分解成N个子问题
将 Bi 带回原问题 得到关于 的函数 用次梯度法 得到一个 新的
求解N个子问 题得到一组 B
i
heuristic
Get feasible solution LG-solution is a good primal solution The sequence is good but violate the capacity constrain. The one who cost much goes first,so it goes on.
Optimize subproblem2 For given to obtain B2
Optimize subproblemN For given to obtain Bn
Dual problem
Max L with

L Min iTi k ( ik M k )
{ Bi } i k i
k M k Min iTi k ik { Bi } k i k
Ci Bi 1 ti
k 0
i 1, 2,…，N
(2)
My explanation
Max L

with
L Min iTi k ( ik M k )
Why using LG method

Good primal sequence Good bound.

First step relaxation
Min J
{Bi }
with
J iTi k (ik M k )
i k i

i
ik
Mk
k 1, 2,…，K
（ 1）
Basic parallel machine scheduling
job1 job2
M1
1. 目
标
2.能力约束 3.自身约束
job3
job4 Pro_time, weight, duedate, start_time
M2
Formulation

problem formulation
Min J with
w Lemma2： f *

and
k

for all k, then
limi mink i f ( xk ) f * ( f * w) /(2 )

Halve the dual gap
Illustration for gama
迭代次数－步长 1.5
与 lmd的 曲 线
3000
3500
lower
不同的下界开始的时间 40 36 35 从 下 界 36， 上 届 40开 始 搜 索
30 从 下 界 0， 上 届 40开 始 搜 索 25
value
20
15
10
5
0
0
100
200
300
400
500 counter
600
700
800
900
1000
description
Ci Bi 1 ti
k 0
i 1, 2,…，N
(2)
Max min problem (LSS)
Coordinator Max f( )
t 1 max{ t t st ,0}
Optimize subproblem1 For given to obtain B1
t
s t ＝0则达到了最优解而停止计算；否则，
t=t+1,重复STEP2；
t 1 max{ t t st , 0}
t
uper lower st
2
with
0 2
Illustration of SG
upper1
上升很慢
Gama

Lemma1：when w f *
Formulation

problem formulation
Min J with
{ Bi }
J iTi
i

i
ik
Mk
k 1,2,…，K i 1,2,…，N
（1） (2)
Ci Bi 1 ti
Dual problem
Max L with

L Min iTi k ( ik M k )
4 3.5
不同时刻的pai在整个迭代过程中的趋向 K在 前 端
3
2.5 中后端 2
1.5
1
0.5 后端 0
0
100
200
300
400
500
600
700
800
900
1000
Illustration for pai
7 5 6 pai向 量 在 迭 代 过 程 中 的 趋 向
5
4
3 4
1.2.3.4.5分别代表不同的迭代时间 经过迭代pai[46]最终的值
Max L

with
L k M k Min iTi k ik { Bi } k i k

* k
dJ * dM k
Pai stands for the cost of resource at time k
Illustration for pai