系統(tǒng)工程第四版習(xí)題解答

1、系統(tǒng)工程第四版習(xí)題解答第三章系統(tǒng)模型與模型化21.給定描述系統(tǒng)基本結(jié)構(gòu)的有向圖，如圖3-16a、b所示。要求:(1)寫出系統(tǒng)要素集合S及S上的二元關(guān)系集合Rb。(2)建立鄰接矩陣A、可達(dá)矩陣M及縮減矩陣M解：(2)3-16a：規(guī)范方法：01000010A=0001000001111000一10M=000111111100110,M'=M00101111SiR(Si)A(Si)C(Si)E(Si)112,3,4,51122,3,412,5233,41,2,3,53441,2,3,4,54452,3,4,51,550區(qū)域劃分所以系統(tǒng)無法劃分為兩個(gè)或兩個(gè)以上相互獨(dú)立的區(qū)域，即n(S)=P=1

2、,2,3,4,5。12M(P)=345123451111101110001100001001111要素集合SiR(Si)A(Si)C(Si)E(Si)口(Pz)11,2,3,4,511P一L022,3,41,2,52Li=433,41,2,3,53級位劃分4542,3,4,51,2,3,4,51,5454PL°-L112351,2,3,52,332,3,511,2,51,2,3,51,512353L2=3P-L0-L1-L21251,2,522,511,2,51,51252L3=2P_L0_L1_L2L3151,5511,5155L4=5PL0-L1L2-L3-L411111L5=1

3、Li4L23MO)12L45L51提取骨架矩陣432511000011000111001111011111L14L2M(L)：L3L45L51432511000011000011000011000011L14L23A=M(L)-I=12L32L45L514320001000100010005000011?！?000二(P)=Li,L2,L3,L4,L5=4,3,2,5,1繪制多級遞階有向圖第一級第二級第三級第四級第五級實(shí)用方法:123451一11111201110縮減矩陣M'=M=3001104000105-0111143251Li4'D0000一L231Q000M(L)=,L

4、eeL3211100L4511110L51:1111EL第一級第二級第三級第四級第五級3-16b:規(guī)范方法:q0i0i0"i"111111'000100010101000000001000A=,M=010001010101110000111111000000000001區(qū)域劃分SiR(SJA(Si)Cg)E(SJ1234561,2,3,4,5,62,4,632,4,61,2,3,4,5,661,51,2,4,51,3,51,2,4,51,51,2,4,5,61,52,432,41,5636A(S3)cA(S6)=1,3,5C1,2,4,5,6=1,5所以系統(tǒng)無法劃分

5、為兩個(gè)或兩個(gè)以上相互獨(dú)立的區(qū)域，即n(S)=P=1,2,3,4,5,6。121-11201M(P)=300401511600級位劃分3411011001110056110100011101要素集合SiR(Si)A(Si)C(Si)E(Si)n(P2)P-Lo1234561,2,3,4,5,62,4,632,4,61,2,3,4,5,661,51,2,4,51,3,51,2,4,51,51,2,4,5,61,52,432,41,5636L1=3,6P-Lo-L112451,2,4,52,42,41,2,4,51,51,2,4,51,2,4,51,51,52,42,41,524L2=2,4PLo-

6、L1-L2151,51,51,51,51,51,515L3=1,5,1_3=3,6,2,4,1,5L1M(L)=L23一10L35010提取骨架矩陣L1M(L)=一101L1一1A'：=M(L)-001一1繪制多級遞階有向圖第一級第二級第三級實(shí)用方法:1縮減矩陣M，：-2312361111010100106|00013621M繪制多級遞階有向圖:第一級第二級第三級22.請依據(jù)圖3-17建立可達(dá)矩陣，并用簡化方法建立其遞階結(jié)構(gòu)模型。解:VVAPPi(V)P7P8P912345678911000000201010004M=5678000100000111000100111001101000

7、0110110111110110000000019814M(L)=62357981462111000110100110010110101111000111110111111繪制多級遞階有向圖:357000000000000000000第一級第二級第三級第四級第五級23.已知下面的系統(tǒng)可適矩陣，分別用規(guī)范方法與實(shí)用方法建立其遞階結(jié)構(gòu)模型。12345671100010120100000111010000201000000300104|01011050000100600101107I0000101401015010160101701013111000010001111101180000000所以系統(tǒng)

8、可劃分為兩個(gè)相互獨(dú)立的區(qū)域，即n(S)=P1，P2=2,4,1,3,5,6,7。M(P)=P2I7一°413000100010001000001000567000000101110100110101解：（1）規(guī)范方法:區(qū)域劃分SiR(SJA(Si)Cg)E(SJ11,5,711222,42233,5,63,63,642,444551,3,5,6,75563,5,63,63,675,71，77A(S2)-A(S5)=2,4-1,3,5,6,7級位劃分要素集合SiR(Si)A(Si)C(Si)E(Si)n(P)P1-Lc2422,42,44242L1=2P1一L0L144444L2=4二

9、(Pi)uLi,I_2=2,4要素集合SiR(Si)A(Si)Cg)E(Si)rm)11,5,71133,5,63,63,6P2-L055135,6,755L1=563,5,63,63,675,71,7711,711P1_L0_L133,63,63,63L2=3,6,763,63,63,66771,777P1_L0_L1_L211111L31L2LiM(L)=415Io30L2L3607101?453600001000010001110111010001007100000000001011二(P2)=Li，L2,3=5,3,6,7,1提取骨架矩陣L1L2L1M(L)=L2L3245362110

10、00(41100050010030011-71000000006_0_0_1_1_1_0_07001001010010011M"(L)=50304010000500111030001007000011100000124532000041000A'=M"(L)I=50030000001000繪制多級遞階有向圖第一級第二級第三級(2)規(guī)范方法:區(qū)域劃分SiR(Si)A(Si)C(Si)E(Si)112,41,31221,2,3,4,5,6,72231,2,3,43342,41,3,4,5,6,7452,4,55,6,7562,4,5,6,7,86672,4,5,7,86

11、,77886,7,888A(S2)-A(S8)u1,2,3,4,5,6,7-6,7,8=6,7='所以系統(tǒng)不能劃分為兩個(gè)或兩個(gè)以上相互獨(dú)立的區(qū)域，即IKS)=P=1,2,3,4,5,6,7,8。123411101201003|11M(P)=4J015016101110010011011000000000111701011011800000001級位劃分要素集合SiR(Si)A(Si)C(Si)E(Si)n(P2)P-Lc1234Li=2,85678P1L0L11345671,41,3,444,54,5,6,74,5,71,33134,5,6,75,6,766,71345674L2=4P

12、1_L0_L1_L21356711,355,6,75,71,335,6,766,71356715L3=1,5P1-L0-L1-L2L336736,77366,736737L4=3,7P1-L0-L1-L2-L3-L466666L5=6二(P)=Li,L2,L3,L4,L5=2,8,4,1,5,3,7,62841L121000L180100L2411M(L)=L31I1L351L43|1L4710101010111L56J110101005000010113760001000000000000100010011提取骨架矩陣L1210L1801L24|10M"(L)=L3100L3500

13、00111L43000L47010L56000100010100537600000000000000001000010010100011A=M(L)28415376L12-00000000L1800000000L2410000000=L3100100000L3500100000L4300010000L4701001000L56-00000010繪制多級遞階有向圖第一級第二級第三級第四級第五級(1)實(shí)用方法:縮減矩陣M二30001M(L)7F第一級第二級第三級M(L)400110011100010101目第一級第二級第三級第四級第五級第四章系統(tǒng)動(dòng)力學(xué)9.已知如下的部分DYNAMO方程:請畫出對應(yīng)

14、的SD流（程）圖。10.LS.K=S.J+SR.JK*DTNS=10000RSR.KL=T.K*TSRCTSR=1LT.K=T.J+TR.JK*DTNT=1500RTR.KL=S.K*STRCSTR=0.0511.TIME012345S10,00011,50013,50016,07519,32523,378T1,5002,000e>2,575<FINA§,T50E>4,0535,02012.(1)家具銷售系數(shù)XSXL已買家具戶數(shù)YMS(2)15.第五章系統(tǒng)評價(jià)方法12.解：第六章決策分析補(bǔ)充題1某商店擬經(jīng)營一種高科技產(chǎn)品，若市場暢銷，可以獲利1萬5千元；若市場滯銷，

15、將虧損5千元；若不經(jīng)營，則不虧不賺。根據(jù)收集的市場銷售資料，該產(chǎn)品暢銷的概率為0.8,滯銷的概率為0.2。為了降低風(fēng)險(xiǎn)，可以聘請某咨詢公司進(jìn)行市場調(diào)查和分析，該咨詢公司對該產(chǎn)品暢銷預(yù)測的準(zhǔn)確率為0.95,滯銷預(yù)測的準(zhǔn)確率為0.90。畫出該決策問題的決策樹，并進(jìn)行決策分析。解：設(shè)市場暢銷為日,市場滯銷為&；設(shè)產(chǎn)品預(yù)測暢銷為Xi,產(chǎn)品預(yù)測滯銷為X2,則由已知條件：P(&)=0.8,P(e2)=0.2,P(x18)=0.95,P(x2|e2)=0.90有：P(x2|01)=0.05,P(x102)=0.10P(XiN)=0.76,P(x2)=0.04,P(xi±)=0.02

16、,P(x22)=0.18P(x1)=0.78,P(x2)=0.22P(2x1)=0.9744,P(仇x1)=0.0256,P(01x2)=0.1818,P(2x2)=0.81821.51.131暢銷0.9744-0.501.5-0.501.5-0.50貝葉斯行動(dòng)：如果市場預(yù)測結(jié)果為暢銷，應(yīng)該選擇經(jīng)營該高科技產(chǎn)品；若市場預(yù)測結(jié)果為滯銷，則不經(jīng)營。由決策樹可知，咨詢公司提供信息的價(jià)值為1.13-1.1=0.03萬元，因此要價(jià)超過300元不應(yīng)聘請。補(bǔ)充題2某公司擬改變產(chǎn)品的包裝，改變包裝后產(chǎn)品的銷路不能確定，公司經(jīng)理的估計(jì)是:銷路差01銷路一M也銷路好加概率P0.20.30.5銷路與收益的關(guān)系如下表

17、:01G263改變包裝-400600包裝不及000為了對銷路的估計(jì)更有把握，公司先在某個(gè)地區(qū)試銷改變了包裝的產(chǎn)品。根據(jù)以往的經(jīng)驗(yàn)，試銷的結(jié)果與產(chǎn)品在將來的實(shí)際銷路中有如下關(guān)系（xX2、X3分別為試銷為差、一般和好的事件）：P(xj|al)X1X2X3由0.80.20G20.20.40.40300.10.9回出該決策問題的決策樹；確定與各種試銷結(jié)果相應(yīng)的貝葉斯行動(dòng)；分析試銷費(fèi)用與是否試銷的關(guān)系。解：由已知條件有下表：p(Xjei)X1X2X3Pi)Qi0.160.0400.20.060.120.120.3300.050.450.5P(Xj)0.220.210.571.00進(jìn)一步有,P(QXj)X1X2X3910.72730.19050.00000.27270.57140.2105030.00000.23810.7895決策樹為:試銷298.4298.42不試銷銷路好0.57銷路差0.22銷路一般0.2129.09改變包裝030.0000不改變包裝銷路一般,0.2727銷路好/4不改變包裝改變包裝135.240.1905135.248銷路一般0.5