PRIMERA PRACTICA CALIFICADA DE MÉTODOS NUMÉRICOS

Profesor del curso: Ayala Mina, Jorge .G

Alumno: Huanca Arapa, Ronald .M

Código: 20110312A

Enunciado: sumar 1+ 0.00001 en sistema binaro .Para lo cual convertimos cada sumando a binaro.

Conversión de 1 a sistemas binario: 1=12

Conversión de 0.00001 a sistemas binario:

0.00001x 2 =

0.00002 0

0.00002x 2 =

0.00004 0

0.00004x 2 =

0.00008 0

0.00008x 2 =

0.00016 0

0.00016x 2 =

0.00032 0

0.00032x 2 =

0.00064 0

0.00064x 2 =

0.00128 0

0.00128x 2 =

0.00256 0

0.00256x 2 =

0.00512 0

0.00512x 2 =

0.01024 0

0.01024x 2 =

0.02048 0

0.02048x 2 =

0.04096 0

0.04096x 2 =

0.08192 0

0.08192x 2 =

0.16384 0

0.16384x 2 =

0.32768 0

0.32768x 2 =

0.65536 0

0.65536x 2 =

1.31072 1

0.31072 x 2 0.621 0

= 44

0.62144x 2 =

1.24288 1

0.24288x 2 =

0.48576 0

0.48576x 2 =

0.97152 0

0.97152x 2 =

1.94304 1

0.94304x 2 =

1.88608 1

0.88608x 2 =

1.77216 1

0.77216x 2 =

1.54432 1

0.54432x 2 =

1.08864 1

0.08864x 2 =

0.17728 0

0.17728x 2 =

0.35456 0

0.35456x 2 =

0.70912 0

0.70912x 2 =

1.41824 1

0.41824x 2 =

0.83648 0

0.83648x 2 =

1.67296 1

0.67296x 2 =

1.34592 1

0.34592x 2 =

0.69184 0

0.69184x 2 =

1.38368 1

0.38368x 2 =

0.76736 0

0.76736x 2 =

1.53472 1

0.53472x 2 =

1.06944 1

0.06944x 2 =

0.13888 0

0.13888x 2 =

0.27776 0

0.27776x 2 =

0.55552 0

0.55552x 2 =

1.11104 1

0.11104x 2 =

0.22208 0

0.22208x 2 =

0.44416 0

0.44416x 2 =

0.88832 0

0.88832x 2 =

1.77664 1

0.77664x 2 =

1.55328 1

0.55328x 2 =

1.10656 1

0.10656x 2 =

0.21312 0

0.21312x 2 =

0.42624 0

0.42624x 2 =

0.85248 0

0.85248x 2 =

1.70496 1

0.70496x 2 =

1.40992 1

0.40992x 2 =

0.81984 0

0.81984x 2 =

1.63968 1

0.63968x 2 =

1.27936 1

0.27936x 2 =

0.55872 0

0.55872x 2 =

1.11744 1

0.11744x 2 =

0.23488 0

0.23488x 2 =

0.46976 0

0.46976x 2 =

0.93952 0

0.93952x 2 =

1.87904 1

0.87904x 2 =

1.75808 1

0.75808x 2 =

1.51616 1

0.51616x 2 =

1.03232 1

0.03232x 2 =

0.06464 0

0.06464x 2 =

0.12928 0

0.12928x 2 =

0.25856 0

0.25856x 2 =

0.51712 0

0.51712x 2 =

1.03424 1

0.03424x 2 =

0.06848 0

0.06848x 2 =

0.13696 0

0.13696x 2 =

0.27392 0

0.27392x 2 =

0.54784 0

0.54784x 2 =

1.09568 1

0.09568x 2 =

0.19136 0

0.19136x 2 =

0.38272 0

0.38272x 2 =

0.76544 0

0.76544x 2 =

1.53088 1

0.53088x 2 =

1.06176 1

0.06176x 2 =

0.12352 0

0.12352x 2 =

0.24704 0

0.24704x 2 =

0.49408 0

0.49408x 2 =

0.98816 0

0.98816x 2 =

1.97632 1

0.97632x 2 =

1.95264 1

0.95264x 2 =

1.90528 1

0.90528x 2 =

1.81056 1

0.81056x 2 =

1.62112 1

0.62112x 2 =

1.24224 1

0.24224x 2 =

0.48448 0

0.48448x 2 =

0.96896 0

0.96896x 2 =

1.93792 1

0.93792x 2 =

1.87584 1

0.87584x 2 =

1.75168 1

0.75168x 2 =

1.50336 1

0.50336x 2 =

1.00672 1

0.00672x 2 =

0.01344 0

0.01344x 2 =

0.02688 0

0.02688x 2 =

0.05376 0

0.05376x 2 =

0.10752 0

0.10752x 2 =

0.21504 0

0.21504x 2 =

0.43008 0

0.43008x 2 =

0.86016 0

0.86016x 2 =

1.72032 1

0.72032x 2 =

1.44064 1

0.44064x 2 =

0.88128 0

0.88128x 2 =

1.76256 1

0.76256x 2 =

1.52512 1

0.52512x 2 =

1.05024 1

0.05024x 2 =

0.10048 0

0.10048x 2 =

0.20096 0

0.20096x 2 =

0.40192 0

0.40192x 2 =

0.80384 0

0.80384x 2 =

1.60768 1

0.60768x 2 =

1.21536 1

0.21536x 2 =

0.43072 0

0.43072x 2 =

0.86144 0

0.86144x 2 =

1.72288 1

0.72288x 2 =

1.44576 1

0.44576x 2 =

0.89152 0

0.89152x 2 =

1.78304 1

0.78304x 2 =

1.56608 1

0.56608x 2 =

1.13216 1

0.13216x 2 =

0.26432 0

0.26432x 2 =

0.52864 0

0.52864x 2 =

1.05728 1

0.05728x 2 =

0.11456 0

0.11456x 2 =

0.22912 0

0.22912x 2 =

0.45824 0

0.45824x 2 =

0.91648 0

0.91648x 2 =

1.83296 1

0.83296x 2 =

1.66592 1

0.66592x 2 =

1.33184 1

0.33184x 2 =

0.66368 0

0.66368x 2 =

1.32736 1

0.32736x 2 =

0.65472 0

0.65472x 2 =

1.30944 1

0.30944x 2 =

0.61888 0

0.61888x 2 =

1.23776 1

0.23776x 2 =

0.47552 0

0.47552x 2 =

0.95104 0

0.95104x 2 =

1.90208 1

0.90208x 2 =

1.80416 1

0.80416x 2 =

1.60832 1

0.60832x 2 =

1.21664 1

0.21664x 2 =

0.43328 0

0.43328x 2 =

0.86656 0

0.86656x 2 =

1.73312 1

0.73312x 2 =

1.46624 1

0.46624x 2 =

0.93248 0

0.93248x 2 =

1.86496 1

0.86496x 2 =

1.72992 1

0.72992x 2 =

1.45984 1

0.45984x 2 =

0.91968 0

0.91968x 2 =

1.83936 1

0.83936x 2 =

1.67872 1

0.67872x 2 =

1.35744 1

0.35744x 2 =

0.71488 0

0.71488x 2 =

1.42976 1

0.0000110=¿0.0000000000000000101001111100010110101100010001110001101101000111100001000010001100001111110011111000000011011100001100110111001000011101010100111100110111011101……b2 ; b=1 ó b=0 .

NORMALZACON

Usando la representación del punto flotante.

0.1010011111000101101011000100011100011011010001111000010000100011…. X 2-16

16 a binario

16 2

0 8 2

0 4 2

0 2 2

0 1

16=10002

PARA 04 bytes (32 bits)

0.1010011111000101101011000100011100011011010001111000010000100011…. X 2-10000

Matiza = 0.101001111100010110101100 010001110001101…

1 Bit para el Signo del exponente

0 1 0 1 0 0 0 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1 0 1 1 0 1 0 1 1 0 0

7 Bits para el exponente 24 Bits para la mantisa

1 Bit signo del número

Después del bit truncado no se guarda la información, esto dará lugar a otro valor al momento de volver al número decimal.

101001111100010110101100 x 10-16= 1 x 2-16+ 0x 2-17+1 x 2-18+0 x 2-19+0 x 2-20+1 x 2-21+1 x 2-

22+1 x 2-23+1 x 2-24+1 x 2-25+0 x 2-26+0 x 2-2+0 x 2-2+1 x 2-30+0 x 2-31+1 x 2-32+1 x 2-33+01x 2-34+0x 2-35+ 1x 2-36+0 x 2-37+1 x 2-38+1 x 2-39+0 x 2-40+0 x 2-41

= 0.000001762931..

El programa MATLAB trabaja con 8 byte(64 bits)

64 bits: 1de signo; 11 de exponente y 52 bits de mantisa.

En nuestro problema:

1+0.00001 = 1.00001

1.0000110=¿1.0000000000000000101001111100010110101100010001110001101101000111100001000010001100001111110011111000000011011100001100110111001000011101010100111100110111011101……b2 ; b=1 ó b=0 .

1.000000000000000010100111110001011010110001000111000 110110100011…….. X 2+1

Bit truncado

Bit truncado

1 Bit para el Signo del exponente

0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 …

11 Bits para el exponente 52 Bits para la mantisa

1 Bit signo del número

Después del bit truncado no se guarda la información, esto dará lugar a otro valor al momento de volver al número decimal.

Resultado que no da el programa MATLAB después de haber sumado 1+ 0.00001 es igual a 1:

De este resultado se puede colegir que debido al a la limitación de la memora de 64 bits se produce errores en el momento de guardar la información para después usarlo en la siguiente operación. Este error se genera mediante la perdida de cifras lo cual altera el resultado final.

1 + 0.00001 = 1