第三次作业

page 100

5.由题可得a1 → 1,a2→ 2,a3→ 3

所以F_x(0)=0,F_x(1)=0.2,F_x(2)=0.5,F_x(3)=1,l⁽⁰⁾=0,U⁽⁰⁾=1

 

所以： a1 :  l⁽¹⁾=l⁽⁰⁾+( u⁽⁰⁾ - l⁽⁰⁾)F_X(0)=0

                  U⁽¹⁾=l⁽⁰⁾+( u⁽⁰⁾- l⁽⁰⁾)F_X(1)=0.2

 

        a1a1:  l⁽²⁾=l⁽¹⁾+( u⁽¹⁾ - l⁽¹⁾)F_X(0)=0

                  U⁽²⁾=l⁽¹⁾+( u⁽¹⁾- l⁽¹⁾)F_X(1)=0.04

 

        a1a1a3:  l⁽³⁾=l⁽²⁾+( u⁽²⁾ - l⁽²⁾)F_X(2)=0.02

                     U⁽³⁾=l⁽²⁾+( u⁽²⁾- l⁽²⁾)F_X(3)=0.04     

 

        a1a1a3a2:  l⁽⁴⁾=l⁽³⁾+( u⁽³⁾ - l⁽³⁾)F_X(1)=0.024

                       U⁽⁴⁾=l⁽³⁾+( u⁽³⁾- l⁽³⁾)F_X(2)=0.03

 

       a1a1a3a2a3:  l⁽⁵⁾=l⁽⁴⁾+( u⁽⁴⁾ - l⁽⁴⁾)F_X(2)=0.027

                          U⁽⁵⁾=l⁽⁴⁾+( u⁽⁴⁾- l⁽⁴⁾)F_X(3)=0.03

 

       a1a1a3a2a3a1:  l⁽⁶⁾=l⁽⁵⁾+( u⁽⁵⁾ - l⁽⁵⁾)F_X(0)=0.027

                          U⁽⁶⁾=l⁽⁵⁾+( u⁽⁵⁾- l⁽⁵⁾)F_X(1)=0.0276

 

所以a1a1a3a2a3a1的标签为 T_x(113231)=(U⁽⁶⁾⁺l⁽⁶⁾)/2=0.0273

     

a1 → 1,a2→ 2,a3→ 3

所以 F_x(0)=0,F_x(1)=0.2,F_x(2)=0.5,F_x(3)=1,l⁽⁰⁾=0,U⁽⁰⁾=1

有公式t^*=(tag-l^(k-1))/(u^(k-1) - l^(k-1))

所以： t^*=(0.63215699-0)/(1-0)=0.63215699

         F_x(2)=0.5≤ t^*≤F_x(3)=1

         l⁽¹⁾=l⁽⁰⁾+( u⁽⁰⁾ - l⁽⁰⁾)F_X(2)=0.5

        U⁽¹⁾=l⁽⁰⁾+( u⁽⁰⁾- l⁽⁰⁾)F_X(3)=1       →a3

      

          t^*=(0.63215699-0.5)/(1-0.5)=0.26431398

         F_x(1)=0.2≤ t^*≤F_x(2)=0.5

         l⁽²⁾=l⁽¹⁾+( u⁽¹⁾ - l⁽¹⁾)F_X(1)=0.6

        U⁽²⁾=l⁽¹⁾+( u⁽¹⁾- l⁽¹⁾)F_X(2)=0.75       →a2

 

         t^*=(0.63215699-0.6)/(0.75-0.6)=0.21437933333

         F_x(1)=0.2≤ t^*≤F_x(2)=0.5

         l⁽³⁾=l⁽²⁾+( u⁽²⁾ - l⁽²⁾)F_X(1)=0.63

        U⁽³⁾=l⁽²⁾+( u⁽²⁾- l⁽²⁾)F_X(2)= 0.675      →a2

 

         t^*=(0.63215699-0.63)/(0.675-0.63)=0.047933111111

         F_x(0)=0≤ t^*≤F_x(1)=0.2

         l⁽⁴⁾=l⁽³⁾+( u⁽³⁾ - l⁽³⁾)F_X(0)=0.63

        U⁽⁴⁾=l⁽³⁾+( u⁽³⁾- l⁽³⁾)F_X(1)=0.639     →a1

 

        t^*=(0.63215699-0.63)/(0.639-0.63)=0.23966555556

         F_x(1)=0.2≤ t^*≤F_x(2)=0.5

         l⁽⁵⁾=l⁽⁴⁾+( u⁽⁴⁾ - l⁽⁴⁾)F_X(1)=0.6318

        U⁽⁵⁾=l⁽⁴⁾+( u⁽⁴⁾- l⁽⁴⁾)F_X(2)=0.6345      →a2

 

        t^*=(0.63215699-0.6318)/(0.6345-0.6318)=0.132218518519

         F_x(0)=0≤ t^*≤F_x(1)=0.2

         l⁽⁶⁾=l⁽⁵⁾+( u⁽⁵⁾ - l⁽⁵⁾)F_X(0)=0.6318

        U⁽⁶⁾=l⁽⁵⁾+( u⁽⁵⁾- l⁽⁵⁾)F_X(1)=0.63234      →a1

 

         t^*=(0.63215699-0.6318)/(0.63234-0.6318)=0.661092592593

         F_x(2)=0.5≤ t^*≤F_x(3)=1

         l⁽⁷⁾=l⁽⁶⁾+( u⁽⁶⁾ - l⁽⁶⁾)F_X(2)=0.63207

        U⁽⁷⁾=l⁽⁶⁾+( u⁽⁶⁾- l⁽⁶⁾)F_X(3)= 0.63234     →a3

 

         t^*=(0.63215699-0.63207)/(0.63234-0.63207)=0.3221851851852

         F_x(1)=0.2≤ t^*≤F_x(2)=0.5

         l⁽⁸⁾=l⁽⁷⁾+( u⁽⁷⁾ - l⁽⁷⁾)F_X(1)=0.632124

        U⁽⁸⁾=l⁽⁷⁾+( u⁽⁷⁾- l⁽⁷⁾)F_X(2)= 0.632205    →a2

 

        

         t^*=(0.63215699-0.632124)/(0.632205-0.632124)=0.407283950617284

         F_x(1)=0.2≤ t^*≤F_x(2)=0.5

         l⁽⁹⁾=l⁽⁸⁾+( u⁽⁸⁾ - l⁽⁸⁾)F_X(1)=0.6321402

        U⁽⁹⁾=l⁽⁸⁾+( u⁽⁸⁾- l⁽⁸⁾)F_X(2)=0.6321645     →a2

 

       

         t^*=(0.63215699-0.6321402)/(0.6321645-0.6321402)=0.690946502057

         F_x(2)=0.5≤ t^*≤F_x(3)=0.1    →a3

 所以结果是a3a2a2a1a2a1a3a2a2a3