Tablas De Ensayos

2). Determine el ciclo o periodo de vida de los siguientes generadores congruenciales.

a) Xi+1 =(2xi +15) mod (31) con X0=21

1 X1= (2)(21)+15 mod (31) = 26 r1= 26/30 = 0.8666

2 X2= (2)(26)+15 mod (31) = 5 r2= 5/30 = 0.1666

3 X3= (2)(5)+15 mod (31) = 25 r3= 25/30= 0.8333

4 X4=(2)(25)+15 mod (31) = 3 r4= 3/30 = 0.1

5 X5=(2)(3)+15 mod (31) = 21 r5= 21/30 =0.7

Ciclo de vida es Hasta 4

b) Xi+1 = (13xi + 9) mod (128) con X0 =7

1 X1 =[13(7)+9]mod128=100 r1 = 100/127=0.7874

2 X2 =[13(100)+9]mod128=29 r2 = 29/127=0.2283

3 X3 =[13(29)+9]mod128=2 r3 = 2/127=0.0157

4 X4 =[13(2)+9]mod128=35 r4 = 35/127=0.2756

5 X5 =[13(35)+9]mod128=80 r5 = 80/127=0.6299

6 X6 = [13(80)+9]mod128=25 r6 =25/127=0.1968

7 X7 = [13(25)+9]mod128=78 r7 = 78/127=0.6141

8 X8 =[13(78)+9]mod128=127 r8 = 127/127= .9999

9 X9 = [13(127)+9]mod128=124 r9 = 124/127=0.9763

10 X10 = [13(124)+9]mod128=85 r10 = 85/127=0.6692

11 X11 =[13(85)+9]mod128=90 r11 = 90/127=0.7086

12 X12 =[13(90)+9]mod128=27 r12 = 27/127=0.2125

13 X13 =[13(27)+9]mod128=104 r13 = 104/127=0.8188

14 X14 =[13(104)+9]mod128=81 r14 = 81/127=0.6377

15 X15 =[13(81)+9]mod128=38 r15= 38/127=0.2992

16 X16 =[13(38)+9]mod128=119 r16 = 119/127=0.9370

17 X17 =[13(119)+9]mod128=20 r17 = 20/127=0.1574

18 X18 =[13(20)+9]mod128=13 r18 = 13/127=0.1023

19 X19 =[13(13)+9]mod128=50 r19 = 50/127=0.3937

20 X20 =[13(50)+9]mod128=19 r20 = 19/127=0.1496

21 X21 =[13(19)+9]mod128=0 r21 = 0/127=0

2). Determine el ciclo o periodo de vida de los siguientes generadores congruenciales.

a) Xi+1 =(2xi +15) mod (31) con X0=21

1 X1= (2)(21)+15 mod (31) = 26 r1= 26/30 = 0.8666

2 X2=

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