最优控制课后习题

1、最优限制第四次作业4-2 一位公务员乘出租车要从机场赶到会场参加重要会议,交通网络如图所 示.图上数字为每条支路的驾驶时间,且全部支路都是单行线.试用动态规划找 出最短时间路线.会场+5解：机场用A表示,会场用0表示,从左到右,从上到下分别用字母 A-O表示, 采用逆序计算法,末端开始,终端为止.如上图.1N=14N 级JiN1 = 4J,N2 =5由于从N1到0以及从N2到0都只有一种可能,所以本级无决策问题.2N=13 M 级本级决策有三种选择,计算如下J2M1=dM1,N1 J1N16 4=10J2(M2) =mind(M2,N1) J1(N19 4=13d(M2,N2) 7(N2) =

2、 4 5 = 9决策变量S2M2 N2J2M3 =dM3,N2 JMN2 =2 5 = 73N=12 L 级本级决策有四种选择,计算如下J3(L1) =d(L1,M1) J2(M1)=5 10=15.d(L2,M1) + J2(M1) = 4 + 10 = 141J3(L2)=min d(L2,M2) J2(M2)=7 9=16 =14决策变量$(L2) = M2J3(L3) = mind(L3,M2) J2(M2)=5 34 £d(L3,M3) J2(M3) = 3 7 = 10决策变量$(L3) = M3J3(L4) =d(L4,M3) J2(M 3) = 4 7 =11(4)

3、N=11 (K 级)本级决策有五种选择,计算如下J4(KJ 二 d(K1, L1) J3(L1) = 3 15 = 18Jgrmin 呢丄° J3(»7 缺22=16d(K2,L2) J3(L2)=2 14=16 决策变量SdK2) = L2d(K3,L2) J3(L2) = 4 14=18J4(K3) = min16d(K3,L3) J3(L3)=6 10 = 16决策变量S4(K3)= L3.d(K4,L3)+J3(L3)=3 + 10 =13J4(K4)-min d(K4, L4) J3(L4) = 6 11 = 17 一"决策变量S4(K4) =L3J/K

4、5) =d(K5,L4) J3(L4) =8 11 =19(5) N=10 (J 级)本级决策有六种选择,计算如下J5(J1) = d(J1,K1) J4(K1) =5 18 = 23J5(J2) = mind(J2K1)心4 缺22 .21d(J2,K2) J4(K2) = 5 16 = 21决策变量Ss(J2) = K2.d(J3,K 2)+ J4(K2) =3 + 16=19J5(J3)=min d(j3,K3) J4(K3)=6 16 = 22 =19J5 (J4) = min决策变量S5(J3) = K 2d(J4,K3)J4(K3)m2.=2.d(J4,K4) J4(K4) =8

5、13 = 21决策变量S5(J4) = K3J5(J5)=min 昵5" J"4"烬15 七d(J5,K5) J4(K5)=3 19 = 22决策变量S5(J5) = K4J5(J6) =d(J6,K5) J4(K5)=4 19=23(6)N=9 (I 级)本级决策有七种选择,计算如下J6(l1)=d(l1,J1) J5(J1)=7 23 = 30J6(I 2) =mind(l2,J1) J5(J1) = 2 23 = 25d(I2,J2) J5(J2) =5 21 = 2625J6(l 3) = minJ6(l 4) =minJ6(l 5) = min决策变量S

6、s(l2) = J1d(l3,J2) + J5(J2) =4 + 21 =25 _d(l3,J3) J5(J3) = 2 19 = 21 _ 21决策变量&(I3) = J3d(l 4, J3) J5(J3) =7 19 = 26d(l4,J4) J5(J4) =5 20 = 25决策变量S6(I4) = J4d(I5J4) J5(J4"3 込23 仪d(I5,J5) J5(J5) =9 15 = 24决策变量S6 (I 5) = J 4.d(l6,J5) + J5(J5)=1+15 = 16J6(l6)=min d(l6,J6) J5(J6) =3 23 = 26 =16决

7、策变量(16) = J5J6(I7) =d(l7,J6) J5J6) =6 23 = 29(7)N=8 (H 级)本级决策有八种选择,计算如下J7(H1) =d(H1,11) J6(I1)=3 30 = 33J7(H2)=min d(H2,I1)6(小3 込33=31d(H 2,12) J6(I2) =6 25 = 31 决策变量S7(H2) = 12d(H3,l2) + J6(l 2) = 5 + 25=30、J7(H3) = min24d(H3,l3) J6(I3) = 3 21=24 决策变量Sy(H3) =13J7(H4)=min d(H4,I3) J6(l3)=4 2"26

8、 =26d(H4,l4) J6(I4)=4 25 = 29 决策变量S7(H4) = 13JWmin 吋,14) J6(5 込3° =28d(H5,l5) J6(I5)=5 23 = 28 决策变量Sy(H5) =15J7(H6)=min d(H6,I5) J6(I5) V30 =25d(H6,l6) J6(I6)=9 16 = 25J7(H7“min d(H7,I6) mH88 d(H7,l7) J6(I7)=4 29 =33决策变量S7(H6) = 16决策变量S7(H7) = 16J7(H8) =d(H8,l7) J6(I7) =3 29 =32决策变量S8(G7H7(8)N=

9、7 (G 级)本级决策有七种选择,计算如下J8(G1)=min 回心 J7(H"5 33=38 '5 d(G1,H2) J7(H2)=4 3仁35决策变量£(G1) = H2J8(G2)+n d(G2,H2) J7(H2M "37 .27d(G2,H3) J7(H3) = 3 24 = 27决策变量S8(G2 H3J8(G3) = mind(G3,H3) J7(H3) =2 24 = 26d(G3,H4) J7(H4) =9 25 =34-26J8(G4) = minJ8(G5) = minJ8(G6) =minJ8(G7) = min决策变量S8(G3)

10、 = H3d(G4,H4) J7(H4)=5 込3° &d(G4,H5) J7(H5) = 4 28 =32决策变量S8(G4H4d(G5,H5) J7(H5) = 3 28 f 3.d(G5,H6) J7(H6) =5 25 =30决策变量Q(G5) = H6d(G6,H6) J7(H6) = 9 25 = 34 =20 d(G6,H7 J7(H7) =2 18 =20决策变量S8(G6 H7d(G7,H7) J7(H7)25 £d(G7,H8) J7(H8) =7 32 =39(9)N=6 (F 级)本级决策有六种选择,计算如下J9(F1)=min *回 J8(

11、G1) = 2 込37.35d(F1,G2) J8(G2)=8 27 = 35决策变量Q(F1)=G2J9(F2)=min 卅2®)J8(G2)=5 “32 .32 d(F2,G3) J8(G3) = 7 26 =33决策变量S9(F2) =G2 d(F3,G3) + J8(G3)=8+26 = 341J9(F3)=min=34d(F3,G4) J8(G4)=6 30 =36决策变量S9(F3) =G3 d( F 4,G4) + J8 (G4) = 6 +30 = 36、J9 (F4) = min34d(F4,G5) J8(G5)=4 30 =34 决策变量S9(F4) =G5J9(

12、F5)=min 聊5®)J8(G5)=4 迹34 =23d(F5,G6) J8(G6)=3 20 = 23 决策变量S9(F5) =G6.fd(F6,G6) + J8 (G6) = 5 + 20 = 251J9 (F 6) = min25d(F6,G7) J8(G7)=3 25 = 28 决策变量S9(F6) =G6(10)N=5 ( E 级)本级决策有五种选择,计算如下d(E1,F1) J9(F1) = 6 35 = 41Jio(E1)-min. 36d(E1,F2) J9(F2)二4 32 二 36 决策变量Sw(E1) = F2Jmin 昨纤2)J9(F2) = 2 32 “4

13、 =34d(E2,F3) J9(F3) = 9 34 = 43J10(E3)=min 昨3® J9(F3)" 34° =40 d(E3,F4) J9(F4) = 7 34 = 41决策变量S0(E2) = F2J10(E4)=min 昨仃4)J9(F4)" *38 =31''d(E4,F5) J9(F5) =8 23 = 31决策变量Sw(E3) = F3决策变量S°(E4) = F5 aE5,F5) + J9(F5) = 3 + 23 = 26J10(E5) =min=26d(E5,F6) J9(F6)=7 25 =32 决策

14、变量Sw(E5) = F5(11)N=4 ( D 级)本级决策有四种选择,计算如下d(D1,E1) J10(E1) = 7 36 = 43J11 (D1) = min38d(D1,E2) J10(E2) = 4 34 = 38决策变量S11(D1E2J(D2)=mln心®J"®"30=如d(D2,E3) J10(E3) =2 40 = 42 决策变量Sm(D2)=：E2Jn(D3)=mln 呻日)J10(E3)" 40 = 48 =35 d(D3,E4) J10(E4) =4 31 =35 决策变量(D3) = E4J11(D4"慌:

15、即縊:需蔦卜2决策变量Sii(D4) = E5(12)N=3 (C 级)本级决策有三种选择,计算如下d(C1,D1) J11(D1) = 5 38 = 43J12(C1) = min43d(C1,D2) J11(D27 447Ji2(C2)=min d(C2,D2) Jii(D2)" 4° = 46 =38 d(C2,D3) J/D3) =3 35 = 38决策变量S2(C1) = D1Ji2(C3)=min d(C3,D3)723)=3 3"38 .37d(C3,D4) Jn(D4) =5 32 =37决策变量S,2(C2) = D3决策变量S,2(C3) =

16、D4(13)N=2 ( B 级)本级决策有两种选择,计算如下Ji3(B1)=min d(B1,C1) J""4 43皿.44d(B1,C2) J12(C2)=6 38 = 44决策变量S3(B1) = C2J13(B2)d(B2,C2) J12(C2) =3 38 = 41 minQ(B2,C3) + J12(C3)=10 + 37 = 47：41决策变量Sn(B2) =C2(14)N=1 (A 级)本级决策是唯一的,计算如下Jg(A)口 d(A,B1) J13(b52*6d (A, B2) +J13(B2) =5+41 = 46,决策变量S14(A) = B2最后可知最短

17、时间路线是 A-B2-C2-D3-E4-F5-G6-H7-I6-J5-K4-L3-M3-N2-O最短时间是464.4设二阶离散系统试求使性能指标X1(k 1)=2(k) u(k),X1(0) =1 x2(k 1) = %(k) x2(k),x2(0) =01J 八2x；(k 1) 2u2(k)k =0为极小的最优限制u*(k)和最优轨线x*(k). 解：此题为二级最优决策问题,采用逆行递推(1)令 N=2, k=1 时J；x(1) = mjn f2x|(2) + 2u2(1) + j0x(2)其中J°x(2) =0那么X2(2)=为(1) X2(1)min：2Xi(ir X2(1)2 2u2(1)由于u(k)无约束,故令rJ；x(1)：:u(1)=4u(1) = 0求得u*(1) = 0将结果代入J；x2(2)可得川X(1) = mu(j)n2(%(1) X2(1)p(2)令 N=1, k=0 时J2x(0)H min '2x；(1) 2u2(0) J；x(1)P其中X2(1)p(0) X2(0)那么为(1)=2%(0) u(0)X1(1)X2(1) = 3n(0) X2 (0) u(0)将其代入J；X2(