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运筹学作业

• 运用逆序解法：

• 问题可分为三个阶段,状态变量sk为第k阶段xk 第三阶段可能的最大取值范围,决策变量uk为第k阶段xk 的取值,状态转移方程sk+1=sk-uk=sk-xk；最优指标函数fk(sk)表示第k阶段，初始状态为sk时，从第k到第三阶段所得的最大值， f1(s1)即为所求的最大值。

• 递推方程为：

fk(sk) = max {gk(xk)+fk+1(sk+1)}

0≤xk≤sk

f4(s4) =12

• K=3时，f3(s3) = max {2x32+f4(s4)}

• 0≤x3≤s3

• 当x3*=s3时，取得极大值2s32,即

• f3(s3) = max {2x32+f4(s4)}=2s32+12

• 0≤x3≤s3

• K=2时，

• f2(s2) = max {-x22+f3(s3)}

• 0≤2x2≤s2

• = max {-x22+2s32+12}

• 0≤2x2≤s2

• = max {-x22+2（s2-2x2）2 +12}

• 0≤2x2≤s2

• h2(s2,x2)=-x22+2(s2-2x2)2+12

• =2s22-8s2x2+7x22+12

• dh2/dx2=14x2-8x2=0(0≤x2≤s2)→x2=4/7s2

• dh2/dx2=14>0 当x2*=0,取得极大值f2(0)=2s22+12

• 当k=1时，

• f1(s1) = max {4x12+f2(s2)}

• 0≤3x1≤s1

• = max {4x12+2s22+12}

• 0≤3x1≤s1

• = max {4x12+2(s1-3x1)2+12}

• 0≤3x1≤s1

• 解得当x1*=0时 f1(9)=2•92+12=184

• 当x1=0,x2=0,x3=9 函数最大值为maxZ=184

• 运用逆序解法：

• 将问题分成三个阶段,状态变量Sk为第K阶段到第3阶段的最大可能的取值,决策变量Uk为第K阶段Xk的取值,状态转移方程为Sk+1＝Sk－ Xk最优指标函数fk（Sk）表示第K阶段,初始状态为Sk时，从第K到第三个阶段的最大值,f1（S1）即为所求最大值.

• 递推方程：

• fk(Sk)＝ MAX { gk(xk)+fk+1(sk+!)}

• 0≤Xk≤Sk

• f4（S4）＝0

• k＝3时,f3(S3)＝MAX {2x32 }=2S32/9

• 0≤Xk≤Sk

• k＝2时，f2(S2)＝MAX { 9x2+ f3(S3) }

• 0≤4X2≤S2

• h2(s2,x2)=9x2+2/9(s2-x2) 2

• dh2/dx2=9+4/9(s2-x2)(-1)=0

• x2=s2-81/4

• d2h2/dx22=4/9>0 x2=s2-81/4是极小点

• 极大值只可能在[0,s/4]端点取得

• f2 (0)＝2s22/9, f(s2/4)=9s2/4

• S2>81/8时,f(0)>f2 (s2/4), x*2 =0.

• S2<81/8时,f(0)<f2 (s2/4), x*2=s2/4

• K=1时,f1(s1)=MAX {4x1+f2 (s2) }

• 0≤2X1≤S1

• 当f2 (s2)=9s2/4时,

• f1 (10)=

• MAX {4x1+9(s1-2x1)/4 }=9s1/4=90/4

• 0≤2X1≤10

*

• f2 (s2)=2s22/9时，

• f1(10)=MAX {4x1=2(s1-2x1) 2}

• 0≤2X1≤10

• 极值在[0,5]两个端点.

• x1=0时,f1 (10)=2s1/9=200/9<90/4.

• x1=5时,f1 (10)=40.

• 得到x1*=0.

• s2 =s1-x1*=10-0=10, x2*=10/4=5/2, x3*=0

• 当x1=0,x2=5/2,x3=0时, MAX f ＝90/4.