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In this computer, numbers are stored in 12-bits. We will also assume that for a floating point (real) number, 6 bits of these bits are reserved for the mantissa (or significand) with 2^(k-1)-1 as the exponent bias (where k is the number of bits for the characteristic).

011100100110010111110011

What pair of floating point numbers could be represented by these 24-bits?

I have gone this far:

As described above that each number is of 12 bit so we get each number

011100100110

First one is 0 bit so it is positive and

Mantissa will be 100110

Exponent will be 11100b=28

my unbiased exponent will be 2^(28-15)=2^13

How to find the floating point number from here?

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Usually the mantissa is considered to have a binary point after the first bit, so your mantissa would be $1.1100_2=\frac 74=1.75_{10}$. Sometimes a leading $1$ is assumed, so your mantissa would be $(1).11100_2=\frac{15}8=1.875_{10}$ This gives one more bit of precision. To find the exponent, you subtract the offset from the stored value. You probably meant $2^{k-1}-1$ as the offset. Can you do that?

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  • $\begingroup$ Yes I did a mistake there.It will be 2^(k-1)-1. $\endgroup$ – Wolverine Apr 5 '15 at 17:20
  • $\begingroup$ Can you please show the calculations for the first one? I will do it for the second one. $\endgroup$ – Wolverine Apr 5 '15 at 17:22
  • $\begingroup$ What is $k$? Can you compute $2^{k-1}-1$ and subtract it from $38$? That is the power of $2$ to multiply your mantissa by. $\endgroup$ – Ross Millikan Apr 5 '15 at 17:26
  • $\begingroup$ k is 5 here.So I need to multiply it (38-15) ie 23. $\endgroup$ – Wolverine Apr 5 '15 at 17:29
  • $\begingroup$ It looks to me like $k=6$-you have six bits of exponent. The point of the offset is that six bits of unsigned binary ranges from 0 to 63. With an offset of $2^{6-1}-1=31$ you get a range from $2^{-31}$ to $2^{32}$, which is (reasonably) logarithmically centered around $1$. You need to subtract this number from $38$ (the opposite sense of your last comment) to get $5$, so you multiply the mantissa by $2^5$ $\endgroup$ – Ross Millikan Apr 5 '15 at 17:33

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