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:01001111110010001110A00010，M00110，MM00000000100111001111區(qū)域劃分SiR(Si)A(Si)C(Si)E(Si)11,2,3，4,51122,3，41,2,5233，41,2,3,53441,2,3,4,54452,3，4，51,55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(S)P1,2,3,4

2、,5。12345111111201110M(P)300110400010501111級位劃分要素集合SiR(Si)A(Si)C(Si)E(Si)(P2)11,2,3,4,51122,3,41,2,52PL。33,41,2,3,53L14441,2,3,4,5445234,51,55PLoLi1235i,2,3,52,332,3,5ii,2,5i,2,3,5i,512353L23ii,2,5iiPLoLiL222i,2,522L3252,5i,55PLoLiL2ii,5iiL45L355i,555PLoL3LiL4L2iiiiiL5i(P)Li丄2丄3丄4丄54,3,2,5,14325iLi4i

3、0000L23ii000M(L),')L32iii00L45iiii0L5iiiiii提取骨架矩陣4325iLi4i0000L23ii000M(L)L320ii00L4500ii0L5i000ii4325iLi400000L23i0000AM(L)IL320i000L4500i00L5i000i0繪制多級遞階有向圖級一第級級級級二三四五第第第第實(shí)用方法:12345111111201110縮減矩陣MM30011040001050111143251Li4E40000L2311000M(L)L3211100L4511110L51111113-16b:規(guī)范方法:第一級第二級第三級第四級第五級0

4、01010111111000100010101000000001000A，M010001010101110000111111000000000001區(qū)域劃分SiR(sjA(Si)c(sjE(SJ1123,4,5,61,51,522,4,61,2,4,52,4331,3,53342,4,61,2,4,52,45123,4,5,61,51,5661,2,4,5,666A(S3)A(S6)1,3,51,2,4,5,61,5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(S)P1,234,5,6。12345611111112010101M(P)300100040101015111111600000

5、1級位劃分要素集合SiR(Si)A(Si)C(Si)E(Si)(P2)11,2,3,4,5,61,51,522,4,61,2,4,52,4331,3,533PL。42,4,61,2,4,52,4L13,651,2,3,4,5,61,51,5661,2,4,5,66611,2,4,51,51,5PL0L122,41,2,4,52,42L22,442,41,2,4,52,445124,51,51,5PL。L1L2151,51,51,51,51,51,515L31,5(P)Li丄2丄33,6,2,4,1,53624153100000%010000M(L)L2011100240111001L31111

6、115111111提取骨架矩陣36213621L131000Li31000M(L)60100，M(L)60100L220110L220110L311111L3110113 621L30000AM(L)I60000L220100L311010繪制多級遞階有向圖第一級第二級第三級實(shí)用方法: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：123611111縮減矩陣M201013001060'001362131000M(L)60100，2011011010第一級第二級第三級VVAAAVVAVP2VVAP3VV(A)AP4V(V)VP5VVAV(V)P7VP8P91234

7、5678910000001101010001110100001100010001110011101100000101111110111100000001100000000198146235798l|l00000000000000111100000041101000006110010000211010100031110001005111110010711111110Li1236789M(L)4M5繪制多級遞階有向圖: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。1234567110001012010000030010110(1)40101

8、000500001006001011070000101解：（1）規(guī)范方法:區(qū)域劃分12345678111010000201000000311110000(2)401010000501011000601011111701011011800000001SiR(Si)A(Si)C(Si)E(Si)11,5，711222,42233，5，63，63，642，444551，3，5,6，75563，5，63，63，675，71,77A(S2)A(S5)2,41,3,567所以系統(tǒng)可劃分為兩個(gè)相互獨(dú)立的區(qū)域，即（s）R,P22,4,1,3,5,6,7。2413567R21000000411000001001

9、0101M(P)30001110P2500001006000111070000101級位劃分要素集合SiR(Si)A(Si)C(Si)E(Si)(P)RL02422，42，44242L12PLoL144444L24(R)Li,L22,4要素集合SiR(S)A(Si)C(Si)E(Si)(P2)11,5,71133,5,63,63,6P2L。551,3,5,6,55L15763,5,63,63,675,71,7711,71133,63,63,63pL0L163,63,63,66L23,6,7771,777pL0L1L211111L31(P2)匚丄2丄35,3,6,7,12453671L12100

10、0000L241100000L150010000M(L)30011100L26001110070010010L310010011提取骨架矩陣M(L)67100000000000,M(L)-0-001001124537121000004 1100005 00100030011007001010100001124537120000004100000AM(L)I5000000300100070010001000010繪制多級遞階有向圖（2）規(guī)范方法:區(qū)域劃分SiR(Si)A(Si)C(Si)E(Si)11,2，41,31221,2,3,4,5,6,72231,2，3，43342，41,3,4,5,6

11、,7452,4，55,6,7562，4,5，6，7,86672,4,5,7,86,77886,7,888A(S2)A(S8)1,2,3,4,5676,7,86,7第一級第三級第二級所以系統(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(S)P123,4,5,6,7,8。12345678111010000201000000311110000M(P)401010000501011000601011111701011011800000001級位劃分要素集合SiR(S)A(Si)C(Si)E(Si)(P2)PL。12345L12,867811,41,3131,3,433PL0L144134,5,6,744

12、L2454,55,6,7564,5,6,76674,5,76,77111,31131,333pL0L1L2555,6,755L31,565,6,76675,76,77pL0L1L2L33636,736363L43,7776,777P1L0L1L2L3L466666L56(P)Li丄2丄3丄4丄52,8,4,1,5,3,7,628415376L1210000000L1801000000L2410100000M(L)L3110110000L3510101000L4310110100L4711101010L5611101011提取骨架矩陣28415376L1210000000L1801000000L

13、2410100000M(L)L3100110000L3500101000L4300010100L4701001010L56000000112L120L180L241AM(L)IL310L350L430L470L560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:123457110001120100003001010401010050000107000011縮減矩陣M3471M(L)53 04 17010000053671第二級第一級第三級第五級第一級第二級第三級第四級(2)實(shí)用方法:12345678284153761110100002100000002010000008010

14、00000311110000410100000M401010000M(L)11011000050101100051010100060101111131011010070101101171110101080000000161110101110.第四章系統(tǒng)動(dòng)力學(xué)9.已知如下的部分DYNAM方程:請畫出對應(yīng)的SD流（程）圖。L=+*DT（水準(zhǔn)方程）NS=10000（賦初值方程）R=*TSR（速率方程）CTSR=1（常量方程）L=+*DT（水準(zhǔn)方程）NT=1500（賦初值方程）R=*STR（速率方程）CSTR=（常量方程）11.TIME012345S10,00011,50013,50016,07519

15、,32523,378T1,5002,0002,5753,2504,0535,020<Time>vFINALTIME>SX12.(1)(2)15.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，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，滯銷的概率為。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，滯銷預(yù)測的準(zhǔn)確率為。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!，市場滯銷為2；設(shè)產(chǎn)品預(yù)測暢銷為捲，產(chǎn)品預(yù)測滯銷為X2，則由已知條件：P(J0.8

16、，P(2)0.2，P(X!)0.95，P(x22)0.90有：P(x2J0.05，P(x,2)0.10P(x,J0.76，P(x2J0.04，P(x,2)0.02，P(x22)0.18P(xJ0.78，P(x2)0.22P(1x1)0.9744，P(2x1)0.0256，P(1x2)0.1818，P(2x2)0.81821.50.51.50.51.50.5000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)

17、理的估計(jì)是:銷路差ei銷路一般e2銷路好e3概率P銷路與收益的關(guān)系如下表:eie2e3改變包裝-400600包裝不變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（Xi、X2、X3分別為試銷為差、一般和好的事件）：P(Xj|i)XiX2X3e10e2e30 畫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(Xji)XiX2X3Pi)ei0e2e30P(Xj)進(jìn)一步有,P(iXj)XiX2X3eie2e3決策樹為:402試銷298.41銷路好0.57銷路差0.22298.4銷路一般0.21改變包裝03不改變包裝4不改變包裝改變包裝135.240.727329.097銷路一般0.2727銷路差銷路好0.00000.1905銷路一般0.5714135.248銷路好0.2381銷路差不試銷47