MGMT 1030 Chapter 12: MGMT 1030 Chapter 12 Notes
6 May 2018
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MGMT 1030 Chapter 12 Notes – Summary
Introduction
Floating Point in the Computer
• We must subtract it out to get the correct excess- result: − .
• The multiplication of two five-digit normalized mantissas yields a ten-digit result.
• Only five digits of this result are significant, however.
• To maintain full, five-digit precision, we must first normalize and then round the
normalized result back to five digits.
• EXAMPLE
• Multiply the two numbers 05220000 ×04712500.
• Adding the exponents and subtracting the offset results in a new, excess-50 exponent of
+ − =
• Multiplying the two mantissas, 0.20000 × 0.12500 = 0.025000000
• Normalizing the result by shifting the point one space to the right decreases the
exponent by one, giving a final result 04825000
• Checking our work, 05220000 is equivalent to 0.20000 × 102, 04712500 is equivalent to
. × − whih ultiplies out to . × −
• Noralizig ad roudig, . × − = . × − whih orrespods
to our previous result.
• The techniques discussed can be applied directly to the storage of floating point
numbers in the computer simply by replacing the digits with bits.
• Typically, 4, 8, or 16 bytes are used to represent a floating point number.
• In fact, the few differees that do eist result fro triks that a e plaed whe
ad are the ol optios.
• A typical floating point format might look like the diagram
• In this example, 32 bits (4 bytes) are used to provide a range of approximatel − to
10+38.
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