PDF《第六章习题》

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第六章习题 1.

解:设~A表示 乘客满意 ,等车时间为论域 }

0 | { ≥ = t t U ,由分段函数法, 可得 高峰期: ? ? ? >

≤ ≤ ? =

3 0

3 0

3 /

1 ) ( ~ t t t t A μ ,非高峰期: ? ? ? >

≤ ≤ ? =

5 0

5 0

5 /

1 ) ( ~ t t t t A μ 设~B表示 公交公司满意 ,载客率为论域 }

0 | { >

= v v U ,由分段函数法,可得?????>

≤ = i x x x y i i i ,建立降水偏大模糊集

2 12 , A A )

12 , ,

2 ,

1 ,

0 ( min = >

= i x x x y i i i ,建立降水偏小模糊集

2 12 , B B )

12 , ,

2 ,

1 ,

0 ( min = ≠ ? ? ? = i x x x y i i i i i i α α α ,建立降水适中模糊集

2 12 , C C ,其中 i α 为第i 县降水量适度值. Matlab 程序如下: R=[276.2,324.5,158.6,412.5,292.8,258.4,334.1,303.2,292.9,243.2,159.7,331.2;

25 1.6,287.3,349.5,297.4,227.8,453.6,321.5,451.0,466.2,307.5,421.1,455.1;

192.7,433.2,2 89.9,366.3,466.2,239.1,357.4,219.7,245.7,411.1,357.0,353.2;

246.2,232.4,243.7,372.5, 460.4,158.9,298.7,314.5,256.6,327.0,296.5,423.0;

291.7,311.0,502.4,254.0,245.6,324. 8,401.0,266.5,251.3,289.9,255.4,362.1;

466.5,158.9,223.5,425.1,251.4,321.0,315.4,31 7.4,246.2,277.5,304.2,410.7;

258.6,327.4,432.1,403.9,256.6,282.9,389.7,413.2,466.5,1 99.3,282.1,387.6;

453.4,365.5,357.6,258.1,278.8,467.2,355.2,228.5,453.6,315.6,456.3, 407.2;

158.2,271.0,410.2,344.2,250.0,360.7,376.4,179.4,159.2,342.4,331.2,377.7;

324. 8,406.5,235.7,288.8,192.6,284.9,290.5,343.7,283.4,281.2,243.7,411.1];

A1=R./[ones(10,1)*max(R)];

B1=[ones(10,1)*min(R)]./R;

C1=[ones(10,1)*min(abs(R-ones(10,1)*mean(R)))]./[abs(R-ones(10,1)*mean(R) )];

L=[A1(:,2),A1(:,12),B1(:,2),B1(:,12),C1(:,2),C1(:,12)] 运行结果为:

22 表2.1 两地区降水量偏大、偏小、适中隶属度 降水量偏大 降水量偏小 降水量适中 年份 ~

12 ~

2 A A ~

12 ~

2 B B ~

12 ~

2 C C

1981 0.7491 0.7278 0.4897 1.0000 0.0605 0.0707

1882 0.6632 1.0000 0.5531 0.7278 0.0315 0.0679

1983 1.0000 0.7761 0.3668 0.9377 0.0063 0.1109

1984 0.5365 0.9295 0.6837 0.7830 0.0097 0.1379

1985 0.7179 0.7956 0.5109 0.9147 1.0000 0.1440

1986 0.3668 0.9024 1.0000 0.8064 0.0050 0.2281

1987 0.7558 0.8517 0.4853 0.8545 0.0493 1.0000

1988 0.8437 0.8947 0.4347 0.8134 0.0143 0.2802

1989 0.6256 0.8299 0.5863 0.8769 0.0189 0.3023

1990 0.9384 0.9033 0.3909 0.8056 0.0081 0.2233 发现问题如下:1981 年第二个县降水量为 324.5 小于第十二县的降水量 331.2, 但是其属于降水量偏大的隶属度为 0.7491 大于第十二县的隶属度 0.7278, 这显然不合理,原因在于两个县是分别考虑的,因此对该地区建立隶属度偏大、 偏小、适中的隶属度必须通盘考虑. (2)解: R=[276.2,324.5,158.6,412.5,292.8,258.4,334.1,303.2,292.9,243.2,159.7,331.2;

25 1.6,287.3,349.5,297.4,227.8,453.6,321.5,451.0,466.2,307.5,421.1,455.1;

192.7,433.2,2 89.9,366.3,466.2,239.1,357.4,219.7,245.7,411.1,357.0,353.2;

246.2,232.4,243.7,372.5, 460.4,158.9,298.7,314.5,256.6,327.0,296.5,423.0;

291.7,311.0,502.4,254.0,245.6,324. 8,401.0,266.5,251.3,289.9,255.4,362.1;

466.5,158.9,223.5,425.1,251.4,321.0,315.4,31 7.4,246.2,277.5,304.2,410.7;

258.6,327.4,432.1,403.9,256.6,282.9,389.7,413.2,466.5,1 99.3,282.1,387.6;

453.4,365.5,357.6,258.1,278.8,467.2,355.2,228.5,453.6,315.6,456.3, 407.2;

158.2,271.0,410.2,344.2,250.0,360.7,376.4,179.4,159.2,342.4,331.2,377.7;

324. 8,406.5,235.7,288.8,192.6,284.9,290.5,343.7,283.4,281.2,243.7,411.1];

A=R./ max(R(:)) %降水量偏大 B=min(R(:))./R %降水量偏小 C=min(min(abs(R-ones(10,12)*mean(R(abs(R-ones(10,12)*mean(R(:)))% 降水量适中