Computer System floating point exercise. Please help, thank you very much! Pleas

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Computer System floating point exercise.

Please help, thank you very much!

Please just don't put the textbook contents on it, it won't help.

Consider a 9-bit floating-point representation based on the IEEE floating-point format, with one sign bit, four exponent bits (k = 4), and four fraction bits (n : 4). The exponent bias is 2 -17 4-1 In this exercise you need to add 2 floating point numbers. We outline the algorithm (for floating point addition) at each step-you need to carefully carry out each step that requires filling out the given blanks You may find this link helpfuhttp://www.cs.umd.edu/class/sum2003/cmsc311/Notes/BinMath/addFloat.html Assuming the given 9-bit IEEE floating-point format, add X-13 and Y = 096875. 1) First covert X and Y to their bit representations. Please specify the bit pattern within double quotation marks and without a space. (E.g.: "011100010") Bit Pattern 2) Using the binary representations convert X and Y to scientific form-ie., express these values as 1.(mantissa_part)* 2 A (exponent) where A represents "raised to". You need to fill out the mantissa_part and the exponent. The mantissa_part should be given as binary (Please specify the bit pattern within double quotation marks and without a space). The exponent part should be entered as a decimal value. Note that 1.(mantissa_part) or 0.(mantissa_part) is referred to as the mantissa.] 2 A Y=1 3) Now rewrite Y such that the exponent of Y is equal to whatever X's exponent is. This could result in Y not being normalized. Think about what this means! Note: The first blank below is multiple choice with options 0 or 1- to indicate if we have 0___or1._for our answer for Y. For second blank (mantissa_part), please specify the bit pattern within double quotation marks and without a space. Finally, note that the third blank (exponent) should be entered as a decimal 4) Add the two mantissas (not just mantissa_part!) of X (from step 2) and adjusted Y (from step 3) together-give the result of just that addition below 2 A (adjusted_exponent of_Y_from_step_3) 5) Convert Z to 9-bit floating point IEEE format. Please specify the bit pattern within double quotation marks and without a space. (E.g.: "011100010"] [IMP NOTE: To store result Z in 9-bit IEEE binary floating point representation- we need to do use rounding. Please use round-to-even mode]

Explanation / Answer

X=13 = 0 1010 1010

Y=0.96875=0 0110 1111 (This is half of 1.9375 whose mantissa part(0.9375) is obtained after all the values given in table below so from here we get that 1111 should be mantissa and we have divided it 1.9375 by 2 hence our exponent-7 =-1 because then only 2^(exponent-7)*1.9375=0.5*1.9375=0.96875 hence exponent is 6(0110))

If you keep this handy then it is no big deal to calculate the fraction part as you can observe that the terms obtained are sums of these only and we have to keep the exponent in such a way that the required output is obtained after multiplication with it.
0.5
0.25
0.125
0.0625

Part 2

X=1.625*2^3

Y=1.9375*2^-1

Part 3

If exponent has to be kept same then we will have to divide Y/16 in order to make there exponents same.

Y=0.12109375*2^3

Part 4

Z=1.625+0.12109375=1.74609375*2^3

Part 5

Z=0 1010 1100

(Rounding off Z to 0.75)

Please do give a thumbs up to my answer.

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