1班課程設(shè)計(jì)題目

1、1.畫(huà)出分段函數(shù)圖:答：在MATLAB中創(chuàng)建M文件，輸入以下代碼：function f=p(x,y) f=0.5457*exp(-0.75*y2 - 3.75*x2 - 1.5*x).*(x+y>1)+. 0.7575*exp(-y2 - 6*x2).*(x+y>-1)+. 0.5457*exp(-0.75*y2 - 3.75*x2 + 1.5*x).*(x+y<=-1); end在MATLAB工作空間中，輸入以下代碼：>> x,y=meshgrid(-5:0.1:5); %meshgrid用于生成網(wǎng)格采樣點(diǎn)n=size(x); %當(dāng)只有一個(gè)輸出參數(shù)時(shí)，返回一個(gè)行

2、向量，該行向量的第一個(gè)元素時(shí)矩陣的行數(shù)。f=zeros(n(1),n(2); %創(chuàng)建一個(gè)n(1)行n(2)列的0矩陣>> for i=1:n(1) for j=1:n(2) f(i,j)=p(x(i,j),y(i,j); %定義分段函數(shù) end end mesh(x,y,f); %把x,y,f之間用網(wǎng)格連接起來(lái)>>2.信號(hào)的產(chǎn)生與變換數(shù)字信號(hào)處理中y(n)=x(-n)的畫(huà)圖實(shí)現(xiàn);其中 答：在MATLAB工作空間中，輸入以下代碼：>> n=-5:5; x=zeros(1,length(n); for i=1:length(n) if n(i)<=3&am

3、p;n(i)>=-3 x(i)=2*n(i)+1; else x(i)=0; end end y=flipdim(x,2); %矩陣左右翻轉(zhuǎn)stem(n,y,'.'); %實(shí)現(xiàn)圖的繪制xlabel('n'),ylabel('y(n)'); >>3.財(cái)經(jīng)問(wèn)題一筆100000元的貸款要按每月等額償付d元的方式付清。利息是在未付余額的基礎(chǔ)上以每年10%的利率按月計(jì)入(如第一個(gè)月之后總債等于100000+(0.1/10)*100000)。求每月的償付額d，使得這筆貸款10年還清，并求在10年期間付款的總數(shù)目。在MATLAB工作空間中，輸

4、入以下代碼：total=100000; year=10; month=year*12; %總月數(shù)year_rate=0.1; month_rate=year_rate/12; %月利率pay_back=total*month_rate*(1+month_rate)month/(1+month_rate)month-1); %每月等額償付額pay_back_total=pay_back*month %付款總數(shù)目結(jié)果：pay_back = 1.3215e+003pay_back_total = 1.5858e+0054.計(jì)算平均學(xué)分積GPA大學(xué)某班，第一學(xué)期若干學(xué)生各科成績(jī)存在一個(gè)EXCEL文件中

5、，文件名為score.xls要求:matlab編程讀取這個(gè)EXCEL文件并求出各個(gè)學(xué)生的平均學(xué)分積，最后將平均學(xué)積分按名字寫(xiě)入每個(gè)同學(xué)后面。成績(jī)文件的位置如下: score.xls-Sheet1各科的學(xué)分分配如下:高數(shù):5英語(yǔ):4物理:3.5德育:2電路:4數(shù)字電路:3.5體育:2平均學(xué)分積計(jì)算公式:（成績(jī)×學(xué)分）/學(xué)分在MATLAB工作空間中，輸入以下代碼：a=xlsread('score.xls') %讀取文件 n=1:30;a(n,10)=(a(n,3)*5+a(n,4)*4+a(n,5)*3.5+a(n,6)*4+a(n,7)*3.5+a(n,8)*2+a(n

6、,9)*2)/(5+4+3.5+4+3.5+2+2)a(n,10)=round(a(n,10)xlswrite('score.xls', 'GPA','sheet1','J1') %寫(xiě)入excelxlswrite('score.xls', a(n,10),'sheet1','J2')結(jié)果：a = 1 NaN 98 93 97 98 81 72 77 2 NaN 81 80 71 90 79 72 72 3 NaN 92 77 75 87 80 64 64 4 NaN 96 94 80

7、 92 74 60 74 5 NaN 93 88 90 96 73 62 68 6 NaN 82 69 87 93 85 51 76 7 NaN 92 88 82 86 82 70 75 8 NaN 83 80 80 88 72 72 70 9 NaN 94 86 99 96 80 75 66 10 NaN 94 90 92 93 77 63 67 11 NaN 94 86 95 96 81 69 72 12 NaN 99 100 93 98 78 64 73 13 NaN 91 94 99 92 81 71 70 14 NaN 72 80 75 90 70 69 77 15 NaN 94 9

8、0 88 87 87 70 81 16 NaN 87 90 77 93 69 61 72 17 NaN 81 84 77 82 74 75 52 18 NaN 81 85 78 85 71 62 62 19 NaN 75 73 72 83 78 70 65 20 NaN 41 37 60 13 77 42 69 21 NaN 76 90 64 94 69 62 66 22 NaN 2 24 44 6 87 47 67 23 NaN 60 40 66 75 81 62 66 24 NaN 84 80 61 69 86 60 67 25 NaN 80 52 60 73 76 64 51 26 Na

9、N 71 32 80 68 81 60 76 27 NaN 47 38 44 48 60 60 72 28 NaN 67 50 31 74 78 51 71 29 NaN 88 79 72 84 76 62 48 30 NaN 86 70 67 86 74 60 60a = 1.0000 NaN 98.0000 93.0000 97.0000 98.0000 81.0000 72.0000 77.0000 90.6250 2.0000 NaN 81.0000 80.0000 71.0000 90.0000 79.0000 72.0000 72.0000 79.0833 3.0000 NaN 9

10、2.0000 77.0000 75.0000 87.0000 80.0000 64.0000 64.0000 79.7708 4.0000 NaN 96.0000 94.0000 80.0000 92.0000 74.0000 60.0000 74.0000 84.6250 5.0000 NaN 93.0000 88.0000 90.0000 96.0000 73.0000 62.0000 68.0000 84.6458 6.0000 NaN 82.0000 69.0000 87.0000 93.0000 85.0000 51.0000 76.0000 79.7500 7.0000 NaN 9

11、2.0000 88.0000 82.0000 86.0000 82.0000 70.0000 75.0000 84.1667 8.0000 NaN 83.0000 80.0000 80.0000 88.0000 72.0000 72.0000 70.0000 79.2917 9.0000 NaN 94.0000 86.0000 99.0000 96.0000 80.0000 75.0000 66.0000 87.7708 10.0000 NaN 94.0000 90.0000 92.0000 93.0000 77.0000 63.0000 67.0000 85.5625 11.0000 NaN

12、 94.0000 86.0000 95.0000 96.0000 81.0000 69.0000 72.0000 87.3333 12.0000 NaN 99.0000 100.0000 93.0000 98.0000 78.0000 64.0000 73.0000 89.9792 13.0000 NaN 91.0000 94.0000 99.0000 92.0000 81.0000 71.0000 70.0000 87.9583 14.0000 NaN 72.0000 80.0000 75.0000 90.0000 70.0000 69.0000 77.0000 76.6458 15.000

13、0 NaN 94.0000 90.0000 88.0000 87.0000 87.0000 70.0000 81.0000 87.1875 16.0000 NaN 87.0000 90.0000 77.0000 93.0000 69.0000 61.0000 72.0000 81.0000 17.0000 NaN 81.0000 84.0000 77.0000 82.0000 74.0000 75.0000 52.0000 77.1458 18.0000 NaN 81.0000 85.0000 78.0000 85.0000 71.0000 62.0000 62.0000 77.2708 19

14、.0000 NaN 75.0000 73.0000 72.0000 83.0000 78.0000 70.0000 65.0000 74.7500 20.0000 NaN 41.0000 37.0000 60.0000 13.0000 77.0000 42.0000 69.0000 46.1042 21.0000 NaN 76.0000 90.0000 64.0000 94.0000 69.0000 62.0000 66.0000 76.5625 22.0000 NaN 2.0000 24.0000 44.0000 6.0000 87.0000 47.0000 67.0000 34.0208

15、23.0000 NaN 60.0000 40.0000 66.0000 75.0000 81.0000 62.0000 66.0000 63.7708 24.0000 NaN 84.0000 80.0000 61.0000 69.0000 86.0000 60.0000 67.0000 74.3542 25.0000 NaN 80.0000 52.0000 60.0000 73.0000 76.0000 64.0000 51.0000 66.9167 26.0000 NaN 71.0000 32.0000 80.0000 68.0000 81.0000 60.0000 76.0000 66.2

16、708 27.0000 NaN 47.0000 38.0000 44.0000 48.0000 60.0000 60.0000 72.0000 50.2917 28.0000 NaN 67.0000 50.0000 31.0000 74.0000 78.0000 51.0000 71.0000 60.6875 29.0000 NaN 88.0000 79.0000 72.0000 84.0000 76.0000 62.0000 48.0000 76.2500 30.0000 NaN 86.0000 70.0000 67.0000 86.0000 74.0000 60.0000 60.0000

17、74.4792a = 1 NaN 98 93 97 98 81 72 77 91 2 NaN 81 80 71 90 79 72 72 79 3 NaN 92 77 75 87 80 64 64 80 4 NaN 96 94 80 92 74 60 74 85 5 NaN 93 88 90 96 73 62 68 85 6 NaN 82 69 87 93 85 51 76 80 7 NaN 92 88 82 86 82 70 75 84 8 NaN 83 80 80 88 72 72 70 79 9 NaN 94 86 99 96 80 75 66 88 10 NaN 94 90 92 93

18、77 63 67 86 11 NaN 94 86 95 96 81 69 72 87 12 NaN 99 100 93 98 78 64 73 90 13 NaN 91 94 99 92 81 71 70 88 14 NaN 72 80 75 90 70 69 77 77 15 NaN 94 90 88 87 87 70 81 87 16 NaN 87 90 77 93 69 61 72 81 17 NaN 81 84 77 82 74 75 52 77 18 NaN 81 85 78 85 71 62 62 77 19 NaN 75 73 72 83 78 70 65 75 20 NaN 4

19、1 37 60 13 77 42 69 46 21 NaN 76 90 64 94 69 62 66 77 22 NaN 2 24 44 6 87 47 67 34 23 NaN 60 40 66 75 81 62 66 64 24 NaN 84 80 61 69 86 60 67 74 25 NaN 80 52 60 73 76 64 51 67 26 NaN 71 32 80 68 81 60 76 66 27 NaN 47 38 44 48 60 60 72 50 28 NaN 67 50 31 74 78 51 71 61 29 NaN 88 79 72 84 76 62 48 76

20、30 NaN 86 70 67 86 74 60 60 745.供煤量分配問(wèn)題某兩個(gè)煤廠A1和A2，每月進(jìn)煤量分別為60噸和l00噸，聯(lián)合供應(yīng)3個(gè)居民區(qū)B1，B2和B3，每個(gè)居民區(qū)每月對(duì)煤的需求量依次分別為50噸，70噸，40噸。煤廠A1離3個(gè)居民區(qū)B1，B2和B3的距離分別為10km，5km和6km，煤廠A2離3個(gè)居民區(qū)B1，B2和B3的距離分別為4km, 8km和12km，問(wèn)如何分配供煤量使得運(yùn)輸量(即噸×km)達(dá)到最小?在MATLAB工作空間中，輸入以下代碼：>> f=10;5;6;4;8;12; %總運(yùn)輸量的目標(biāo)Aeq=1 1 1 0 0 0;0 0 0 1 1

21、1;1 0 0 1 0 0;0 1 0 0 1 0;0 0 1 0 0 1; beq=60;100;50;70;40; lb=0 0 0 0 0 0; ub=50 70 40 50 70 40; x,fval,exitflag=linprog(f,Aeq,beq,lb,ub) % fval是函數(shù)值，linprog功能：求解線性規(guī)劃問(wèn)題返回目標(biāo)函數(shù)最優(yōu)解x，和在x處的值：fval = f'*x，是否存在exitflag標(biāo)志Optimization terminated. %運(yùn)算正常終止結(jié)果為：x = 0.0000 20.0000 40.0000 50.0000 50.0000 0.000

22、0fval = 940.0000exitflag = 16.便利店問(wèn)題某便利店晚上2名職工值班，顧客不太多，是開(kāi)一個(gè)出口，一人收錢(qián)一人裝袋好？還是開(kāi)兩個(gè)出口，一人既收錢(qián)又裝袋好？假定，收款和裝袋都是一分鐘，顧客到達(dá)出口是隨機(jī)的，服從poisson分布，平均每分鐘40%沒(méi)有顧客，30%一個(gè)顧客，30%兩個(gè)顧客，試設(shè)計(jì)一個(gè)隨機(jī)模擬實(shí)驗(yàn)，分析這個(gè)問(wèn)題。假設(shè)：1.顧客的到達(dá)收款臺(tái)是隨機(jī)的，服從規(guī)律：40%的時(shí)間沒(méi)有人來(lái)，30%的時(shí)間有1個(gè)人來(lái)，30%的時(shí)間有2個(gè)人來(lái)。2. 收款裝袋的時(shí)間是相同的。  3. 此方案中，收款與裝袋同時(shí)進(jìn)行。參量，變量：n(t) 在時(shí)

23、刻t 到達(dá)收款臺(tái)人數(shù), L(t) 在時(shí)刻t 在收款臺(tái)等待人數(shù),  T1(t) 到時(shí)刻 t為止所有排隊(duì)顧客等待時(shí)間的總和。T2(t) 到時(shí)刻t為止，所有已交款顧客接受服務(wù)的總時(shí)間, t 收款或裝袋的時(shí)間。 平衡關(guān)系：當(dāng)L(t)=0 且n(t)=0 時(shí), L(t+rt)=L(t);  T1(t+r t)=T1(t);  T2(t+r t)=T2(t); 否則 L(t+r t)=L(t)+n(t)-1;

24、 T1(t+r t)=T1(t)+l(t);     T2(t+r t)=T2(t)+ t   取時(shí)間步長(zhǎng)rt=1 ，收款或裝袋的時(shí)間t=1 。在t時(shí)刻, 取隨機(jī)數(shù)r=rand, 當(dāng)0£r<0.4時(shí),  n(t)=0, 當(dāng)0.4£r<0.7時(shí), n(t)=1, 當(dāng)0.7£r £1時(shí),  n(t)=2.

25、60; 仿真30分鐘內(nèi)收款臺(tái)處的排隊(duì)情況，程序：>> clf L=zeros(1,31); %L 等待的顧客人數(shù)，T1=zeros(1,31); %T1等待時(shí)間的累加，T2=zeros(1,31); %T2服務(wù)時(shí)間的累加, L1=zeros(1,31);% L1到達(dá)顧客人數(shù)累加。t=1; tau=1; x=0:30; r=rand(1,30); for i=1:30; t=t+1; if 0<=r(i) & r(i)<0.4 n=0; elseif 0.4<=r(i) & r(i)<0.7 n=1; else n=2; end; if L(t

26、-1)=0 & n=0 L(t)=L(t-1);T1(t)=T1(t-1);T2(t)=T2(t-1);L1(t)=L1(t-1); else L(t)=L(t-1)+n-1;T1(t)=T1(t-1)+L(t);T2(t)=T2(t-1)+tau; L1(t)=L1(t-1)+n; end; end; r=0 r; a=x',r',L',L1',T1',T2' eL=T2(end)/tau %已被服務(wù)的人數(shù) L2=(find(L1>eL) L3=sum(L(L2)%未被服務(wù)的顧客等待時(shí)間總和g1=(T1(end)-L3)/eL %

27、平均等待時(shí)間g2=g1+tau %平均逗留時(shí)間g3=eL/30 %平均每分鐘服務(wù)的顧客人數(shù)結(jié)果：a = 0 0 0 0 0 0 1.0000 0.6822 0 1.0000 0 1.0000 2.0000 0.3028 0 1.0000 0 1.0000 3.0000 0.5417 0 2.0000 0 2.0000 4.0000 0.1509 0 2.0000 0 2.0000 5.0000 0.6979 0 3.0000 0 3.0000 6.0000 0.3784 0 3.0000 0 3.0000 7.0000 0.8600 1.0000 5.0000 1.0000 4.0000 8.

28、0000 0.8537 2.0000 7.0000 3.0000 5.0000 9.0000 0.5936 2.0000 8.0000 5.0000 6.0000 10.0000 0.4966 2.0000 9.0000 7.0000 7.0000 11.0000 0.8998 3.0000 11.0000 10.0000 8.0000 12.0000 0.8216 4.0000 13.0000 14.0000 9.0000 13.0000 0.6449 4.0000 14.0000 18.0000 10.0000 14.0000 0.8180 5.0000 16.0000 23.0000 1

29、1.0000 15.0000 0.6602 5.0000 17.0000 28.0000 12.0000 16.0000 0.3420 4.0000 17.0000 32.0000 13.0000 17.0000 0.2897 3.0000 17.0000 35.0000 14.0000 18.0000 0.3412 2.0000 17.0000 37.0000 15.0000 19.0000 0.5341 2.0000 18.0000 39.0000 16.0000 20.0000 0.7271 3.0000 20.0000 42.0000 17.0000 21.0000 0.3093 2.0000 20.0000 44.0000 18.0000 22.0000 0.8385 3.0000 22.00