Primera Practica Calificada de Metodos Numericos
7
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=1 2 Conversión de 0.00001 a sistemas binario: 0.00001 x 2 = 0.000 02 0 0.00002 x 2 = 0.000 04 0 0.00004 x 2 = 0.000 08 0 0.00008 x 2 = 0.000 16 0 0.00016 x 2 = 0.000 32 0 0.00032 x 2 = 0.000 64 0 0.00064 x 2 = 0.001 28 0 0.00128 x 2 = 0.002 56 0 0.00256 x 2 = 0.005 12 0 0.00512 x 2 = 0.010 24 0 0.01024 x 2 = 0.020 48 0 0.02048 x 2 = 0.040 96 0 0.04096 x 2 = 0.081 92 0 0.08192 x 2 = 0.163 84 0 0.16384 x 2 = 0.327 68 0 0.32768 x 2 = 0.655 36 0 0.65536 x 2 = 1.310 72 1 0.31072 x 2 = 0.621 44 0 0.62144 x 2 = 1.242 88 1 0.24288 x 2 = 0.485 76 0 0.48576 x 2 = 0.971 52 0 0.97152 x 2 = 1.943 04 1 0.94304 x 2 = 1.886 08 1 0.88608 x 2 = 1.772 16 1 0.77216 x 2 = 1.544 32 1 0.54432 x 2 = 1.088 64 1 0.08864 x 2 = 0.177 28 0 0.17728 x 2 = 0.354 56 0 0.35456 x 2 = 0.709 12 0 0.70912 x 2 = 1.418 24 1
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Transcript of Primera Practica Calificada de Metodos Numericos
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
