Abstract

Introduction

Topic

The topic of project 5-7 is concurrent elevator simulation. The program read requests from stdin, simulate the elevator system, and print runtime infomation to stdout.

Request

<PersonID>-FROM-<FromLevel>-TO-<ToLevel>

  • On behalf of a pessenger
  • <PersonID> must be unique

Elevator

  • Function

    • Move: Move up or down to next level in <MoveTime>, and print ARRIVE-<Level>-<ElevatorName>
    • Open Door: Open Door in <OpenTime>, and print OPEN-<Level>-<ElevatorName> when begin opening
    • Close Door: Close Door in <CloseTime>, and print CLOSE-<Level>-<ElevatorName> when close finished
    • Pessenger in: Print IN-<PersonID>-<Level>-<ElevatorName>
    • Pessenger out: Print OUT-<PersonID>-<Level>-<ElevatorName>
  • Argument
    • Name
    • Time
      • OpenTime
      • CloseTime
      • MoveTime
    • Level
      • InitialLevel
      • SupLevel
      • InfLevel
      • ApprovedLevel
    • Capacity

Analysis

The major focus is concurrency: The program has to plan and simulate while reading requests, so it has to be concurrent. Thus we have to design a concurrent architechture and guarantee its thread safety.

Also, for multi-elevator scene, we need to design a coordinating thread to coordinate elevators.

The program's function can be classified as follows, and the threads can be designed correspondingly:

Reading Requests

Reading requests in real time, and pass them to coordinating thread.

Coordinating

Get requests from reading thread, set path for the person and pass it to scheduling thread.

Scheduling and controling

Get requests from reading thread, schedule and pass the action of elevator to elevator, and control the movement of passenger.

Elevator simulation

Get command from scheduling thread, open and close door, and go up or down stairs.

Course

This topic has three assignments.

P5-P6

Reading (and coordinating)

  • Read from given input interface
  • Analyse the request, put the person to corresponse location in concurrent two-dimensional map
    • Dimension 1: From level
    • Dimension 2: To level
  • Pass person to scheduling and controling thread through cuncurrent map: ConcurrentHashMap<MapKey, Set<Integer>>

Scheduling and controling

  • Move up and down again and again

    • Take in person whose destination direction is consistent with the elevator
    • Free person who arrived
    • U-turn if there's no person ahead
    • Stop if there's no person anywhere
  • Give command through TransferQueue<Command> to elevator

Elevator simulation

  • Execute command

    • Print information
    • Change level
    • Sleep for given time

P7

Reading and coordinating

  • Read from given input interface
  • Analyse and coordinate
    • If there's a single elevator approved to access both from and to floor

      • The person only have one request: from <FromLevel> to <ToLevel>
    • else
      • Because it takes at most two steps to go from <FromLevel> to <ToLevel>
      • Find the elevator who can access <FromLevel>
      • Find the elevator who can access <ToLevel>
      • Find the level which can be accessed by both elevator
      • The person has two request
        • From <FromLevel> to <TransferLevel>
        • From <TransgerLevel> to <ToLevel>
    • The coordinate algorithm can be refacted to recursive one to support any number of transfers
  • Pass
    • Pass person to scheduling and controling thread through cuncurrent map: ConcurrentHashMap<MapKey, Set<Integer>>
    • The mapkey presents the first request: new MapKey(<FromLevel>, <ToLevel>)
    • The rest requests stored in the person object

Scheduling and controling

  • Move up and down again and again

    • If current level is accessable

      • Take in person

        • Destination direction is consistent with the elevator
        • Destination level is accessable
      • Free person who arrived
        • If the person has rest requests, put him to correspensive location in reqMap, and delete its next request
    • U-turn if there's no person at accessable level ahead
    • Stop if there's no person anywhere
  • Give command through TransferQueue<Command> to elevator

Elevator simulation

  • Execute command

    • Print information
    • Change level
    • Sleep for given time

Class Design

Measurement

Complexity Metric

Method Complexity

Method ev(G) iv(G) v(G)
com.nyan.EleSysTest.main(String[]) 1 1 1
com.nyan.elesys.Commander.Commander(int,String,Integer[],Integer,Integer,Integer,ConcurrentHashMap<MapKey, Set<Person>>,Long,Long,Long) 1 1 1
com.nyan.elesys.Commander.closeDoor() 1 2 2
com.nyan.elesys.Commander.downHave() 3 2 3
com.nyan.elesys.Commander.inDownPsgers() 1 3 3
com.nyan.elesys.Commander.inPsgers(Integer) 4 4 5
com.nyan.elesys.Commander.inUpPsgers() 1 3 3
com.nyan.elesys.Commander.levelHave(int) 5 6 8
com.nyan.elesys.Commander.loop() 7 5 9
com.nyan.elesys.Commander.openDoor() 1 2 2
com.nyan.elesys.Commander.outPsgers() 1 4 4
com.nyan.elesys.Commander.run() 1 4 5
com.nyan.elesys.Commander.upHave() 3 2 3
com.nyan.elesys.Coordinator.Coordinator(ConcurrentHashMap<MapKey, Set<Person>>,Integer[][]) 1 2 2
com.nyan.elesys.Coordinator.run() 12 11 14
com.nyan.elesys.Elevator.Elevator(Long,Long,Long,Integer,String,TransferQueue<Command>) 1 1 1
com.nyan.elesys.Elevator.getLevelNum() 1 1 1
com.nyan.elesys.Elevator.run() 3 7 8
com.nyan.elesys.Level.Level(Integer) 1 1 1
com.nyan.elesys.Level.add(Integer) 1 2 2
com.nyan.elesys.Level.addOne() 1 1 2
com.nyan.elesys.Level.equals(Object) 2 2 2
com.nyan.elesys.Level.getLevelNum() 1 1 1
com.nyan.elesys.Level.sub(Integer) 1 2 2
com.nyan.elesys.Level.subOne() 1 1 2
com.nyan.elesys.Level.toString() 1 1 1
com.nyan.elesys.MapKey.MapKey(Integer,Integer) 1 1 1
com.nyan.elesys.MapKey.equals(Object) 2 3 3
com.nyan.elesys.MapKey.hashCode() 1 1 1
com.nyan.elesys.MapKey.toString() 1 1 1
com.nyan.elesys.Person.Person(Integer) 1 1 1
com.nyan.elesys.Person.addReq(Request) 1 1 1
com.nyan.elesys.Person.compareTo(Person) 1 1 1
com.nyan.elesys.Person.nextReq() 2 1 2
com.nyan.elesys.Person.toString() 1 1 1
com.nyan.elesys.PsgsList.PsgsList(Integer,Integer) 1 2 2
com.nyan.elesys.PsgsList.add(int,Set<Person>) 1 1 1
com.nyan.elesys.PsgsList.get(int) 1 1 1
com.nyan.elesys.PsgsList.getTotal() 1 2 2
com.nyan.elesys.PsgsList.isEmpty() 3 3 4
com.nyan.elesys.Request.Request(Integer,Integer) 1 1 1
com.nyan.elesys.Request.getFromLevel() 1 1 1
com.nyan.elesys.Request.getToLevel() 1 1 1

Class Complexity

Class OCavg WMC
com.nyan.EleSysTest 1 1
com.nyan.elesys.Command n/a 0
com.nyan.elesys.Commander 3.33 40
com.nyan.elesys.Coordinator 7 14
com.nyan.elesys.Elevator 3 9
com.nyan.elesys.Level 1.62 13
com.nyan.elesys.MapKey 1.25 5
com.nyan.elesys.Person 1.2 6
com.nyan.elesys.PsgsList 1.8 9
com.nyan.elesys.Request 1 3

Package Complexity

Package v(G)avg v(G)tot
com.nyan 1 1
com.nyan.elesys 2.67 112

Module Complexity

Module v(G)avg v(G)tot
P7Project 2.63 113

Project Complexity

Project v(G)avg v(G)tot
project 2.63 113

Bug Review

Level movement step

Door movement time

Test Strategy

Concurrent Knowledge

Acknowledge

Reference

BUAAOO P5-P7 Elevator Simulation的更多相关文章

  1. hdu分类 Dynamic Programming(这是一场漫长的旅途)

    下面是difficulty 1的题 1003   Max Sum 最长递增子序列.非常经典,最棒的解法是在线算法O(n)的复杂度. 贴的呢,是用dp做的代码. 先是一个高亮的dp递推式,然后找到最大处 ...

  2. [android]亲自破解Flappy Bird(去广告+永生)

    以下内容为原创,欢迎转载,转载请注明 来自天天博客:http://www.cnblogs.com/tiantianbyconan/p/3544785.html  听说最近Flappy Bird很火,但 ...

  3. BZOJ3679 : 数字之积

    设f[i][p2][p3][p5][p7][j][k]表示前i位,2,3,5,7的次数,前i位是否等于x,是否有数字的方案数 然后数位DP即可,ans=cal(r)-cal(l) #include&l ...

  4. ADF_Starting系列1_JDeveloper IDE开发环境简介

    2013-05-01 Created By BaoXinjian

  5. SOA_环境安装系列3_Oracle Weblogic安装和环境搭建(案例)

    2014-01-03 Created By BaoXinjian

  6. WLS_Oracle Weblogic安装和环境搭建(案例)

    2014-01-03 Created By BaoXinjian

  7. mysql颠覆实战笔记(一)--设计一个项目需求,灌入一万数据先

    版权声明:笔记整理者亡命小卒热爱自由,崇尚分享.但是本笔记源自www.jtthink.com(程序员在囧途)沈逸老师的<web级mysql颠覆实战课程 >.如需转载请尊重老师劳动,保留沈逸 ...

  8. poj3307

    可以证明,每个符合的数都由2,3,5,7相乘得到. 依据猜想:下一个出现的数是由前面某个数乘上这几个数之一得到的新的数. 假设之前的数均满足序列,则因为下一个数必有2,3,5,7相乘得到,而这个数之前 ...

  9. springmvc(六)——视图和视图解析器

    aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAoIAAAGrCAIAAADb2WEhAAAgAElEQVR4nOzdaVhTd78vfF8/z772c9 ...

随机推荐

  1. xadmin邮箱验证码 标题 EmailVerifyRecord object

    [修改users-models模块] 1.如果这样不生效 def __unicode__(self): return '{0}({1})'.format(self.code, self.email) ...

  2. python学习(十一)

  3. recyclerview 主活动里监听点击事件

    记性真的不行啊...贴上来有时间多复习复习 主活动 package com.example.com.webtext; import android.content.Intent; import and ...

  4. C#工作总结(一):Fleck的WebSocket使用

    一.引子(Foreword) 最近公司里面要做窗体和网页交互的功能.网上找了一下资料,这里做一个简单的扩充和整理,部分内容可能是摘自其他博客,这里会注明出处和原文地址供大家和自己日后查阅. 二.基础知 ...

  5. MySQL运用

    进入mysql 命令行: mysql -uroot -p查看所有数据库: show databases;创建数据库: create database niu charset utf8;删除数据库: d ...

  6. xcode打包苹果应用遇到的问题及解决方法

    1.手机升级到iOS 10之后,运行真机出现了Development cannot be enabled while your device is locked. 原因分析: 这里是你对这台电脑设置了 ...

  7. ubuntu16.04安装pycharm

    Ubuntu16.04下,默认安装了python2.7和python3.5,在终端下,输入“Python”  或“python3”可查看具体版本. 1.安装PyCharm前,先配置PyCharm的JD ...

  8. P2010 回文日期 题解

    这题其实就是纯暴力,暴力,再暴力,毫无技巧可言(总之您怎么乱搞都不会超时QAQ) 首先,根据题意,我们明白每年自多产生一个回文日期,因为对于每年的三百多天,前四位是固定的. 所以,我们只需要进行一个从 ...

  9. 浅谈java 之 Map

    先来一张Map的类继承图 Map :Hashtable .HashMap .LinkedHashMap .TreeMap 的比较   1.Hashtable的方法实现了synchronized 是线程 ...

  10. DevExpress v18.2新版亮点——DevExtreme篇(三)

    行业领先的.NET界面控件2018年第二次重大更新——DevExpress v18.2日前正式发布,本站将以连载的形式为大家介绍新版本新功能.本文将介绍了DevExtreme Complete Sub ...