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Data, Information, Knowledge, IntelligenceIntelligence Knowledge Information Data房间温度高 解决温度 高的办法温度高原因通风量不足增大通风量房间温度 32℃理想温度 23℃Data, Information, Knowledge, IntelligenceIntelligence KnowledgeInformation Data传统控制面临的挑战 实际系统由于存在复杂性、非线性、时变 性、不确定性和不完全性等，一般无法获得精 确的数学模型。
应用传统控制理论进行控制必须提出并遵循 一些比较苛刻的线性化假设，而这些假设在应 用中往往与实际情况不相吻合。
传统控制面临的挑战 传统控制方法在解决大范围变工况、异常 工况等问题方面往往不尽人意。
环境和被控对象的未知和不确定性，导致无 法建立模型。
9 传统控制往往不能满足某些系统的性能要 求。
控制科学发展过程进展方向最优控制 确定性反馈控制 开环控制 智能控制 自学习控制自组织控制 自适应控制 鲁棒控制 随机控制对象的复杂性智能控制的发展¾ 1985 年 8月，IEEE在纽约召开第一届智能控制学术 研讨会，主题：智能控制原理和智能控制系统。
会议 决定在 IEEE CSS 下设 IEEE 智能控制专业委员会。
这 标志着智能控制这一新兴学科研究领域的正式诞生。
¾ 1987 年 1 月 ， 美 国 费 城 ， 第 一 次 智 能 控 制 国 际 会 议，IEEE CSS与CS两学会主办； ¾ 1987 年以来，一些国际学术组织，如 IEEE 、 IFAC 等定期或不定期举办各类有关智能控制的国际学术会 议或研讨会，一定程度上反映了智能控制发展的好势 头。
智能控制的发展¾ 1991年7月，中国人工智能学会成立。
¾ 1993年7月，成都，中国人工智能学会智能机器人专 业委员会成立大会暨首届学术会议。
Assignment11.We will consider the routing problem shown in Figure1.We aim at finding a route from S to T,where S is the starting point and T is the goal point.The number on every edge refers to the cost(e.g.,time/distance).An example path is S→A→D→T and the cost of this path is70+80+300=450.Figure11.1Please write down the points to be visited in order using Breath-First Search (BFS),then calculate the cost of the solution.(12marks)[0]{[S,0]}[1]{[S→A,70];[S→B,40];[S→C,90]}[2]{[S→B,40];[S→C,90];[S→A→D,150]}[3]{[S→C,90];[S→A→D,150];[S→B→F,200]}[4]{[S→A→D,150];[S→B→F,200];[S→C→E,210]}[5]{[S→B→F,200];[S→C→E,210];[S→A→D→T,450]}[6]{[S→C→E,210];[S→A→D→T,450];[S→B→F→T,230]}[7]{[S→A→D→T,450];[S→B→F→T,230];[S→C→E→T,490]}The length of the shortest path is230.The path is S→B→F→T.1.2Please write down the points to be visited in order using Depth-First Search (DFS),then calculate the cost of the solution.(12marks)[0]{[S,0]}[1]{[S→A,70];[S→B,40];[S→C,90]}[2]{[S→A→D,150];[S→B,40];[S→C,90]}[3]{[S→A→D→T,450];[S→B,40];[S→C,90]}[4]{[S→A→D→T,450];[S→B→F,200];[S→C,90]}[5]{[S→A→D→T,450];[S→B→F→T,230];[S→C,90]}[6]{[S→A→D→T,450];[S→B→F→T,230];[S→C→E,210]}[7]{[S→A→D→T,450];[S→B→F→T,230];[S→C→E→T,490]}The length of the shortest path is230.The path is S→B→F→T.1.3Please write down the points to be visited in order using Uniform-Cost Search(UCS),then calculate the cost of the solution.(12marks)[0]{[S,0]}[1]{[S→A,70];[S→B,40];[S→C,90]}[2]{[S→A,70];[S→B→F,200];[S→C,90]}[3]{[S→A→D,150];[S→B→F,200];[S→C,90]}[4]{[S→A→D,150];[S→B→F,200];[S→C→E,210]}[5]{[S→A→D→T,450];[S→B→F,200];[S→C→E,210]}[6]{[S→A→D→T,450];[S→B→F→T,230];[S→C→E,210]}[7]{[S→A→D→T,450];[S→B→F→T,230];[S→C→E→T,490]}The length of the shortest path is230.The path is S→B→F→T.2.Consider the Travelling Salesman Problem(TSP).Given a set of cites and the distances between each pair of cities,we aim at finding the shortest route that visits each city once and only once and returns to the origin city.Figure2is an example solution to a TSP problem,the solution is1→2→3→4→1,which leads to the shortest distance890.Figure2Figure3Consider a TSP problem illustrated in Figure3.2.1How will you represent solutions to TSP in an evolutionary algorithm?(5 marks)The ordinal representation.parent1(125643):113321parent2(135246):1231112.2What would be an appropriate fitness function for measuring the quality of solutions?(5marks)Subtract each tour length to the maximum tour length found in the current population.2.3Design an appropriate crossover operator.Justify your design.(10 marks)The two parent chromosomes encode the tours125643and135246,respectively. After cutting at position2,a feasible offspring(125346)is created.parent1(125643):11|3321parent2(135246):12|3111offspring(125346):1131112.4Design an appropriate mutation operator.Justify your design.(10marks)The offspring is created by first replacing the substring31in parent2by the substring33.Hence,2replaces3,and1replaces city3(step1)(125364).Outside of the cut points,1replaces2,and2replaces1(step2)(125346).parent1(125643):11|33|21parent2(135246):12|31|11offspring(step1)(125364):113121(step2)(125346):1131123.For the tours13654287and14236578(path representation),the initial edge map is shown in the following figure.Here,we consider the Edge recombination crossover(ER)in the adjacency representation.What is the final tour if city1is selected as the starting city?Write down the derivation process and justify your conclusion.(34marks)(a)From city1,we can go to cities3,4,7or8.Cities3,4,7and8have two active edges.Hence,a random choice is made between cities3,4,7and8.(b)Assume that city3is selected.From3,we can go to cities2and6.(c)City2has two active edges and city6only one,so city6is selected.From city6, there is no choice but to go to city5.(d)City5is selected.From City5,cities4and7have two active edges.(e)Assume that city4is randomly selected.From city4,there is no choice but to go to city2.(f)City2is selected.From city2,there is no choice but to go to city8.(g)City8is selected.From city8,we must go to city7.(h)City7is selected.So the final tour is13654287and all edges are inherited from both parents.。
