Third time job

5.

My answer is as follows:

By title: FX (0) =0,FX (1), FX (2) =0.5,FX (3) =1 l0=0,u0=1

by formula

lk=lk+ (UK-LK) Fx (xk-1)

uk=lk+ (UK-LK) Fx (XK)

and a sequence of 113231

(1) K=1,input 1

l1=l0+ (u0-l0) Fx (0) =0

u1=l0+ (u0-l0) Fx (1) =0.2

(2) K=2,input 1

l2=l1+ (U1-L1) Fx (0) =0

u2=l1+ (U1-L1) Fx (1) =0+ (1-0) *0.2=0.04

(3) K=3,input 3

l3=l2+ (U2-L2) Fx (2) =0+ (0.04-0) *0.5=0.02

u3=l2+ (U2-L2) Fx (3) =0+ (0.04-0) *1=0.04

(4) K=4,input 2

l4=l3+ (U3-L3) Fx (1) =0.02+ (0.04-0.02) *0.2=0.024

u4=l3+ (U3-L3) Fx (2) =0.02+ (0.04-0.02) *0.5=0.03

(5) K=5,input 3

l5=l4+ (U4-L4) Fx (2) =0.024+ (0.03-0.024) *0.5=0.027

u5=l4+ (U4-L4) Fx (3) =0.024+ (0.03-0.024) *1=0.03

(6) K=6,input 1

l6=l5+ (U5-L5) Fx (0) =0.027+ (0.03-0.027) *0=0.027

u6=l5+ (U5-L5) Fx (1) =0.027+ (0.03-0.027) *1=0.03

T (113231) = (L6+u6)/2= (0.027+0.03)/2=0.0285

6.

Solution: by Title: FX (0) =0,FX (1), FX (2) =0.5,FX (3) =1 l0=0,u0=1

(1) k=1

Because t*= (label-lk-1)/(Uk-1-lk-1) = (0.63215699-0)/(1-0) =0.63215699

Then Fx (2) =0.5<T*<1=FX (3) can have input 3

l1=l0+ (u0-l0) Fx (2) =0+ (1-0) *0.5=0.5

u1=l0+ (u0-l0) Fx (3) =0+ (1-0) *1=1

(2) k=2

Because t*= (label-lk-1)/(Uk-1-lk-1) = (0.63215699-0.5)/(1-0.5) =0.26431398

then Fx (1) =0.2<t*<0.5=fx (2) can have input 32

l2=l1+ (U1-L1) Fx (1) =0.5+ (1-0.5) *0.2=0.6

u2=l1+ (U1-L1) Fx (2) =0.5+ (1-0.5) *0.5=0.75

(3) K=3

Because t*= (label-lk-1)/(Uk-1-lk-1) = (0.63215699-0.6)/(0.75-0.6) =0.21437993

then Fx (1) =0.2<t*<0.5=fx (2) can have input 2

l3=l2+ (U2-L2) Fx (1) =0.6+ (0.75-0.6) *0.2=0.63

u3=l2+ (U2-L2) Fx (2) =0.6+ (0.75-0.6) *0.5=0.675

.........

(9) K=9

Because t*= (label-lk-1)/(Uk-1-lk-1) = (0.63215699-0.632124)/(0.632205-0.632124) =0.000081

then Fx (0) =0<t*<0.2=fx (1) can have input 322121321

l9=l8+ (U8-L8) Fx (0) =0

u9=l8+ (U8-L8) Fx (1) =0.632124+ (0.632205-0.632124) *0.2=0.6321402

(Ten) k=10

Because t*= (label-lk-1)/(Uk-1-lk-1) = (063215699-0.632124)/(0.6321402-0.632124) =0.20364198

then Fx (1) =0.2<t*<0.5=fx (2) can have input 3221213212

The last available sequence code is: A3A2A2A1A2A1A3A2A1A2

Third time job