深入理解動(dòng)態(tài)規(guī)劃的一系列問(wèn)題(13)
今天介紹的這個(gè)問(wèn)題還是一個(gè)任務(wù)調(diào)度的問(wèn)題(Flowshop問(wèn)題),每個(gè)過(guò)程i有兩個(gè)任務(wù)A和B,其中有個(gè)約束,那就是A一定要在B之前完成,任務(wù)A和B分別由兩個(gè)處理器完成,編號(hào)為1和2,每個(gè)過(guò)程i有對(duì)應(yīng)自己任務(wù)的完成時(shí)間,選擇如何調(diào)度這些過(guò)程,會(huì)產(chǎn)生不同的任務(wù)總體完成時(shí)間,問(wèn)題的目標(biāo)就是要找到最佳完成時(shí)間的策略。舉例來(lái)講,有4個(gè)過(guò)程為{0,1,2,3},每個(gè)過(guò)程對(duì)應(yīng)的A任務(wù)完成時(shí)間為{3,4,8,10},B任務(wù)完成時(shí)間為{6,2,9,15},如果處理器按照這個(gè)順序去處理任務(wù),依據(jù)約束條件,完成過(guò)程如下表表示:
可以看出這個(gè)方案的調(diào)度時(shí)長(zhǎng)需要40。
如果用DP去求解這個(gè)問(wèn)題,首先定義狀態(tài),我們定義過(guò)程d的A任務(wù)完成時(shí)間為pd,B任務(wù)完成時(shí)間為qd,那么引入一個(gè)虛擬階段號(hào)k,k表示上一次被確認(rèn)調(diào)度的A和B任務(wù)各自完成中間那段流失的時(shí)間(可能不是很好理解,這里解釋一下,為什么取這個(gè)值作為狀態(tài),首先對(duì)于pd和qd這都是固定計(jì)算成本,那么導(dǎo)致任務(wù)不能連續(xù)運(yùn)轉(zhuǎn)的原因就是有A在B前完成這個(gè)約束,對(duì)于上面的例子,我們看到B3任務(wù)并不能在B2任務(wù)結(jié)束后馬上開始,而一定要等到A3任務(wù)結(jié)束才可以,B2結(jié)束后到A3結(jié)束前這段時(shí)間即15-11這4個(gè)時(shí)間單位的時(shí)間就是這里的k了,定義了這個(gè)k后,我們就可以知道目標(biāo)狀態(tài)是要使k=0且未被調(diào)度的任務(wù)集合),定義S為剩余的過(guò)程集合,所以狀態(tài)為(k,S),這個(gè)狀態(tài)對(duì)于一個(gè)過(guò)程的A和B任務(wù)如果沒有延遲(k<pd),那么k=qd,反之有延遲則延遲時(shí)間為max(k-pd,0),所以一個(gè)調(diào)度過(guò)程d開始后的時(shí)間成本會(huì)記為A任務(wù)時(shí)間pd+延遲時(shí)間max(k-pd,0)+B任務(wù)時(shí)間qd。所以DPFE即f(k,S)=min d∈S{pd+f(max(k-pd)+qd,S-wmqeeuq)}。初始狀態(tài)對(duì)于S是空集,那么f(k,S)=k。于是對(duì)于上面的具體問(wèn)題,方程求解最優(yōu)時(shí)間是38,決策序列為d1=0,d2=2,d3=3,d4=1。
過(guò)程源碼如下:
1: /*
2: * Copyright (C) 2013 changedi
3: *
4: * Licensed under the Apache License, Version 2.0 (the "License");
5: * you may not use this file except in compliance with the License.
6: * You may obtain a copy of the License at
7: *
8: * http://www.apache.org/licenses/LICENSE-2.0
9: *
10: * Unless required by applicable law or agreed to in writing, software
11: * distributed under the License is distributed on an "AS IS" BASIS,
12: * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
13: * See the License for the specific language governing permissions and
14: * limitations under the License.
15: */
16: package com.jybat.dp;
17:
18: import java.util.HashSet;
19: import java.util.Set;
20:
21: //Flowshop Problem
22: public class FLOWSHOP {
23:
24: private static int[] first = { 3, 4, 8, 10 }; // sum=25
25: private static int[] second = { 6, 2, 9, 15 }; // sum=32
26: // upper bound on final completion time is 25+32:
27: private static int sum = 57;
28: private static int m = first.length;
29:
30: private static Set<Integer> setOfAllItems = new HashSet<Integer>();
31:
32: static{
33: setOfAllItems.add(0);
34: setOfAllItems.add(1);
35: setOfAllItems.add(2);
36: setOfAllItems.add(3);
37: }
38:
39: private static int fct(int t, int d) {
40: return Math.max(t - first[d], 0) + second[d];
41: }
42:
43: public static double f(Set<Integer> set, int t){
44: if(set.isEmpty())
45: return t;
46: double min = Double.MAX_VALUE;
47: for(int d : set){
48: Set<Integer> tmpSet = copySet(set);
49: tmpSet.remove(d);
50: double tmp = first[d]+f(tmpSet,fct(t,d));
51: if(tmp<min){
52: min=tmp;
53: }
54: }
55: return min;
56: }
57:
58: private static Set<Integer> copySet(Set<Integer> set) {
59: Set<Integer> s = new HashSet<Integer>();
60: for(int x : set){
61: s.add(x);
62: }
63: return s;
64: }
65:
66: /**
67: * @param args
68: */
69: public static void main(String[] args) {
70: System.out.println(f(setOfAllItems,0));
71: }
72:
73: }
posted on 2014-06-03 19:24 changedi 閱讀(1525) 評(píng)論(0) 編輯 收藏 所屬分類: 算法