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Chapter 3: Q44E (page 241)

Question: Write down the bit pattern in the fraction assuming a floating-point format that uses Binary Coded Decimal (base 10) numbers in the fraction instead of base 2. Assume there are 24 bits, and you do not need to normalize. Is this representation exact?

Short Answer

Expert verified

The value is 0011 0011 0011 0011 0011 0011 and it is not exact.

Step by step solution

01

Determine the floating-point format and BCD.

The value of the fraction will be the number with a decimal point. The number will be converted to a binary number. The binary coded decimal numbers are with the base 10 in the fraction format. Each decimal value will have a corresponding binary code, which will be presented without any conversions.

02

Determine the bit pattern of the fraction value 1/3 as BCD.

The given number is 1/3.

The decimal value of the given number is 0.333333.

The above number can be written as

Now, converting the number to its BCD, we have.

The above value is not exact, because the value does not represent the exponent and it is not normalized.

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Most popular questions from this chapter

IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.5625×10-1assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit floating point format compares to the single precision IEEE 754 standard.

What is 4365-3412when these values represent unsigned 12-bit octal numbers? The result should be written in octal. Show your work.

Question: Based on your answers to 3.35 and 3.36, does (3.41796875 10-3 X 6.34765625 X 10-3) X 1.05625 X 102 = 3.41796875 X 10-3 X (6.34765625 X 10-3 X 1.05625 X 102) ?

Calculate 3.984375×10-1+3.4375×10-1+1.771×103 by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit

floating point format and in decimal.

IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.5625×10-1assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit floating point format compares to the single precision IEEE 754 standard.
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