204-工业C102-大学生缺少体育锻炼的原因(李晓-孙丽-徐嘉)

系统工程之解释结构分析

大学生缺少体育锻炼的原因小组成员：李晓107383孙丽107389徐嘉107391近年来，学生缺少体育锻炼的现象越来越严重，也越来越受到重视。免疫力低下、肥胖问题等等都严重的影响着学生们心理、身体的健康发展。因此我组针对此情况，以大学生为目标人群，通过学习结构化模型技术，利用解释结构模型法对大学生缺少体育锻炼的原因进行了分析，并提出相关解决建议。步骤1：提出问题，组建ISM实施小组。步骤2：设定问题步骤3：选择构成系统的要素。步骤3：组长李晓组员孙丽徐嘉选题大学生缺少体育锻炼的原因要素1S1:缺乏锻炼意识要素2S2：没有合适的器材要素3S3：自己的时间安排不合理要素4S4:学校没有做相关的宣传工作要素5S5：活动的场所不多要素6S6：学生课程多，课业繁忙要素7S7:自身懒惰，缺乏积极性要素8S8：运动的方式太少要素9S9:各种考试，占用时间和精力要素10S10:缺乏意志力，坚持不下来要素11S11:缺少相关课程和教练，给予学生体育锻炼方面的指导要素12S12:学生多宅在家里，沉迷于电视与网络要素13S13:不确定的天气因素，很难做出运动安排要素14S14:学校不组织或很少组织运动活动和比赛要素15S15：无统一时间进行集体活动项目。要素16S16:不投入和运作资金。步骤4：要素明细表→上三角矩阵→邻接矩阵→可达矩阵步骤4.1：要素明细表→上三角矩阵设要素为S1、S2、S3、S4、S5、S6、S7、S8、S9、S10、S11、S12、S13、S14、S15、S16。要素间四种关系符号：Si×Sj，即Si与Sj和Sj与Si互有关系，即形成回路；Si○Sj，即Si与Sj和Sj与Si均无关系；Si∧Sj，即Si与Sj有关，而Sj与Si无关；Si∨Sj，即Sj与Si有关．而Si与Sj无关。

S1S2S3S4S5S6S7S8S9S10S11S12S13S14S15S16S1

○○∧○○×○○×∨×○∨∨○S2

○○∨○○×○○○○○∧∧∨S3

○○∨∨○∨○○×○○○○S4

○○○○○○∧○○∧∧∨S5

○○∧○○∧○○∧∧∨S6

○○∧○∧○○∧∧○S7

○○×○×○○○○S8

○○∨○∨○○∨S9

○○○○○∧○S10

○∧○○○○S11

○○○○∨S12

○○○○S13

○○○S14

×∨S15

○S16

表2上三角矩阵根据上三角矩阵建立有向连接图。化简可以连通的路线得出如下结果（如图一所示）初步建立有向连接图图一有向连接图S1S2S3S4S5S6S7S8S9S10S11S12S13S14S15S16S10000001001010000S20000000100000110S30000000000010000S41000000000100110S50100000100100110S60010000010100110S71010000010100000S80100000000000000S90010000000000010S101000001000010000S111000000100000000S121010001000000000S130000000100000000S140000000000000010S150000000000000100S160101100100100100步骤4.2：上三角矩阵→邻接矩阵

根据要素明细表作构思模型（上三角矩阵），并建立邻接矩阵和可达矩阵。A=S1S2S3S4S5S6S7S8S9S10S11S12S13S14S15S16S11110001111110110S20100000100000110S31110001111110110S41111001111110110S51110101111110110S61110011111110110S71110001111110110S80100000100000110S91110001111110110S101110001111110110S111110001111110110S121110001111110110S130100000100001110S140000000000000110S150000000000000110S161111101111110111步骤4.3：邻接矩阵→可达矩阵

利用邻接矩阵加上单位矩阵，经过至多（n-1）次演算后得到可达矩阵。求可达矩阵M：（A+I）≠（A+I）2=≠（A+I）3≠（A+I）4≠（A+I）5≠（A+I）6

=（A+I）7，得：该系统最长通路为6M=可达集R（S1）={s1s2s3s7s8s9s10s11s12s14s15}R（S2）={s2s8s14s15}R（S3）={s1s2s3s7s8s9s10s11s12s14s15}R（S4）={s1s2s3s4s7s8s9s10s11s12s14s15}R（S5）={s1s2s3s5s7s8s9s10s11s12s14s15}R（S6）={s1s2s3s6s7s8s9s10s11s12s14s15}R（S7）={s1s2s3s7s8s9s10s11s12s14s15}R（S8）={s2s8s14s15}R（S9）={s1s2s3s7s8s9s10s11s12s14s15}R（S10）={s1s2s3s7s8s9s10s11s12s14s15}R（S11）={s1s2s3s7s8s9s10s11s12s14s15}R（S12）={s1s2s3s7s8s9s10s11s12s14s15}R（S13）={s2s8s13s14s15}R（S14）={s14s15}R（S15）={s14s15}R（S16）={s1s2s3s4s5s6s7s8s9s10s11s12s14s15s16}步骤5：将可达矩阵进行分解、缩约、简化处理，得到反映系统递阶结构的骨架矩阵。步骤5.1：要素之间的关系分为可达与不可达，并且判断哪些要素是连通的，即把系统分为几个部分或子部分。先行集A（S1）={s1s3s4s5s6s7s9s10s11s12s16}A（S2）={s1s2s3s4s5s6s7s8s9s10s11s12s13s16}A（S3）={s1s3s4s5s6s7s9s10s11s12s16}A（S4）={s4s16}A（S5）={s5s16}A（S6）={s6s16}A（S7）={s1s3s4s5s6s7s9s10s11s12s16}A（S8）={s1s2s3s4s5s6s7s8s9s10s11s12s13s16}A（S9）={s1s3s4s5s6s7s9s10s11s12s16}A（S10）={s1s3s4s5s6s7s9s10s11s12s16}A（S11）={s1s3s4s5s6s7s9s10s11s12s16}A（S12）={s1s3s4s5s6s7s9s10s11s12s16}A（S13）={s13}A（S14）={s1s2s3s4s5s6s7s8s9s10s11s12s13s14s15s16}A（S15）={s1s2s3s4s5s6s7s8s9s10s11s12s13s14s15s16}A（S16）={s16}共同集C（S1）={s1s3s7s9s10s11s12}C（S2）={s2s8}C（S3）={s1s3s7s9s10s11s12}C（S4）={s4}C（S5）={s5}C（S6）={s6}C（S7）={s1s3s7s9s10s11s12}C（S8）={s2s8}C（S9）={s1s3s7s9s10s11s12}C（S10）={s1s3s7s9s10s11s12}C（S11）={s1s3s7s9s10s11s12}C（S12）={s1s3s7s9s10s11s12}C（S13）={s13}C（S14）={s14s15}C（S15）={s14s15}C（S16）={s16}起始集B（S13）={s13}B（S16）={s16}B(S)={S13,S16}C（Si）=｛Si∈S｜R（Si）∩A（Si）｝B（S）中的要素在有向图中只有箭头流出，而无箭头流入，是系统的输入要素。其定义式为B（S）=｛Si｜Si∈S，C（Si）=A（Si）｝SiR（Si）A（Si）C（Si）B（Si）11,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,1222,8,14,151,2,3,4,5,6,7,8,9,10,11,12,13,162,831,2,3,7,8,9,10,11,12,14,151.3.4.5.6.7.9.10.11.12.16.1,3,7,9,10,11,1241,2,3,4,7,8,9,10,11,12,14,154.16451,2,3,5,7,8,9,10,11,12,14,155,16,561,2,3,6,7,8,9,10,11,12,14,156,16671,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,1282,8,14,151,2,3,4,5,6,7,8,9,10,11,12,13,162,8表3：可达集、先行集、共同集和起始集91,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12101,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12111,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12121,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12132,8,13,14,151313131414,151,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1614,151514,151,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1614,15161,2,3,4,5,6,7,8,9,10,11,12,14,15,16161616B(S)={S13,S16}R（S13）∩R（S16）={2,8,14,15}≠Φ（Φ为空集），则S13、S16及R（S13）、R（S16）中的要素属于同一区域。区域不可分。因此，该可达矩阵的块对角矩阵就是其可达矩阵，记为M(P)。矩阵只有一个有向连接图。S1S2S3S4S5S6S7S8S9S10S11S12S13S14S15S16S11110001111110110S20100000100000110S31110001111110110S41111001111110110S51110101111110110S61110011111110110S71110001111110110S80100000100000110S91110001111110110S101110001111110110S111110001111110110S121110001111110110S130100000100001110S140000000000000110S150000000000000110S161111101111110111M(P)=P终止集E（S1）={s2s8s14s15}E（S2）={s14s15}E（S3）={s2s8s14s15}E（S4）={s1s2s3s7s8s9s10s11s12s14s15}E（S5）={s1s2s3s7s8s9s10s11s12s14s15}E（S6）={s1s2s3s7s8s9s10s11s12s14s15}E（S7）={s2s8s14s15}E（S8）={s14s15}E（S9）={s2s8s14s15}E（S10）={s2s8s14s15}E（S11）={s2s8s14s15}E（S12）={s2s8s14s15}E（S13）={s2s8s14s15}E（S14）={}E（S15）={}E（S16）={s1s2s3s4s5s6s7s8s9s10s11s12s14s15}步骤5.2级位划分

系统中的所有要素，以可达矩阵为准则，划分成不同级（层）次。1）找出整个系统要素集合的最高级要素（终止集要素）后，即可将其从可达矩阵中划去相应的行和列；要素集合SiR（Si）A（Si）C（Si）=R(Si)交A(Si)C（Si）=R(Si)TT（P）P-L01414,151,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1614,15√L1={S14,S15}1514,151,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1614,15√P-L0-L122,8,14,151,2,3,4,5,6,7,8,9,10,11,12,13,162,8√L2={S2,S8}82,8,14,151,2,3,4,5,6,7,8,9,10,11,12,13,162,8√P-L0-L1-L211,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12√L3={S1,S3，S7,S9，S10,S11,S12,S13}31,2,3,7,8,9,10,11,12,14,151.3.4.5.6.7.9.10.11.12.14.16.1,3,7,9,10,11,12√71,2,3,7,8,9,10,11,12,14,151.3.4.5.6.7.9.10.11.12.16.1,3,7,9,10,11,12√91,2,3,4,7,8,9,10,11,12,14,154.161,3,7,9,10,11,12√SiR（Si）A（Si）C（Si）C（Si）=R(Si)TT（P）101,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12√111,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12√121,2,3,7,8,9,10,11,12,14,151,3,4,5,6,7,9,10,11,12,161,3,7,9,10,11,12√132,8,13,14,151313√P-L0-L1-L2-L341,2,3,4,7,8,9,10,11,12,14,154.164√

L4={S4,S5，S6}51,2,3,5,7,8,9,10,11,12,14,155,16,5√61,2,3,6,7,8,9,10,11,12,14,156,166√√P-L0-L1-L3-L3-L4161,2,3,4,5,6,7,8,9,10,11,12,14,15,161616√L5={S16}PΠ（P）=L1,L2,L3,L4L5区域块三角矩阵S14S15S2S8S1S3S7S9S10S11S12S13S4S5S6S16M（L）=PL1S141111111111111111S151111111111111111L2S20011111111111111S80011111111111111L3S10000111111101111S30000111111101111S70000111111101111S90000111111101111S100000111111101111S110000111111101111S120000111111101111S130000000000010000L4S40000000000001001S50000000000000101S60000000000000010L5S160000000000000001步骤5.3骨架矩阵提取

在同一区域内同级要素相互可达的要素就称为强连通

块，要素间构成回路，选择其中一个为代表要素。S14S15S2S8S1S3S7S9S10S11S12S13S4S5S6S16M（L）=PL1S141111111111111111S151111111111111111L2S20011111111111111S80011111111111111L3S10000111111101111S30000111111101111S70000111111101111S90000111111101111S100000111111101111S110000111111101111S120000111111101111S130000000000010000L4S40000000000001001S50000000000000101S60