NUMBER SYSTEMS. and DATA REPRESENTATION. for COMPUTERS (PROBLEM ANSWERS)

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1 NUMBER SYSTEMS and DATA REPRESENTATION for COMPUTERS (PROBLEM ANSWERS) 05 March 2008

2 Number Systems and Data Representation 2 Table of Contents Table of Contents... 2 Conversion Between Binary and Hexadecimal Answers... 3 Convert Single Digit Hexadecimal to Binary by Inspection...3 Convert Binary to Hexadecimal...3 Convert Hexadecimal to Binary...3 Data Representation Answers... 4 Character Code Problems...4 Jubilation Code and Gran Zeff Code Problems...6 Conversion to and from Decimal Answers... 8 Number Systems Conversion Answers...0 String Function Answers... Hexadecimal Addition Answers...3 Convert to IBM BCD Answers...4 Fixed Point Conversion Answers...5 Partitioned Numbers...6 Subnet Answers...6 German Roman Numeral Answers...8 Fibonacci Bases Answers...20 Phi Base Number System Answers...23 Continued Fraction Answer...24 Complement Arithmetic Answers...25 Fast Two s Complement Answers...25 Binary Multiplication Answer...26 Hexadecimal Multiplication Answer...26 Fixed Point Binary Division Answers...27 Prime Number Answers...27 Scientific Notation Conversion Answers...28 Scientific Notation Multiplication Answers...28 Floating Point Conversion Answers...29 There are three kinds of mathematicians: those who can count, and those who can t. Attributed to the famous mathematician E. Bombieri. Albert R. Meyer, Problems on Equational Proof Systems, M.I.T. (8 October 998). Visited 2 January 2008.

3 Number Systems and Data Representation 3 Conversion Between Binary and Hexadecimal Answers Convert Single Digit Hexadecimal to Binary by Inspection Problem Problem Problem Problem Problem Problem 6. 2 Convert Binary to Hexadecimal Problem 7. D 6 Problem Problem Problem Problem. 0 6 Problem 2. 6B85 6 Convert Hexadecimal to Binary Problem 3. 00:0 2 Problem 4. 0:00:0 2 Problem :00:00:000::00 2 Problem :0000:0:0:0 2

4 Number Systems and Data Representation 4 Data Representation Answers Character Code Problems Problem A. 2 Zoos Character 2 Z o o s EBCDIC F2 40 E A2 Binary : :0000 0:00 00:00 00:00 00:000 Character 2 Z o o s ASCII A 6F 6F 73 Binary 00: : :00 00: 00: 0:00 Problem A2. Bill Character B i l l EBCDIC F 40 C Binary : : : :00 00:00 00:00 Character B i l l ASCII C 6C Binary 00: : :000 00:00 00:00 00:00 Problem A3. Hi Ho 0 lo? Notes: Hi are letters. Ho are letters. 0 are numbers. lo are letters, followed by punctuation. Character H i H o EBCDIC C C Binary 00: :00 000: :000 00:00 000:0000 ASCII F 20 Binary 000:000 00:00 000: :000 00: 000:0000 Character 0 l O? EBCDIC F F D6 6F Binary :000 : : :00 0:00 00: ASCII C 4F 3F Binary 00:000 00: : :00 000: 00:

5 Number Systems and Data Representation 5 Problem A4. 2ZS50Oo li. 2ZS5 Number, Letter, Letter, Number 0Oo Number, Letter, Letter li. Number, Letter, Letter, Punctuation Character 2 Z S 5 0 O EBCDIC F2 E9 E2 F5 F0 D6 Binary :000 0:00 0:000 :00 :0000 0:00 ASCII 32 5A F Binary 00:000 00:00 00:00 00:00 00: : Character o l i. EBCDIC F Binary 00:00 000:0000 :000 00:00 000:00 0:00 ASCII 6F C 69 2E Binary 00: 000: :000 00:00 00:00 000:0 Problem A5. Code the following in hexadecimal using ASCII and then convert to binary. Character O v e n ASCII Hex 4F E ASCII Binary 000: 0:00 00:00 00:0 EBCDIC Hex D6 A EBCDIC Binary 0:00 00:00 000:00 00:00 Character H o t ASCII Hex F 74 ASCII Binary 000: :000 00: 0:000 EBCDIC Hex A3 EBCDIC Binary 000: :000 00:00 00:00

6 Number Systems and Data Representation 6 Problem A6. Code the following in hexadecimal using EBCDIC and then convert to binary. Character 0 i ASCII Hex ASCII Binary 00:000 00: : :00 EBCDIC Hex F F EBCDIC Binary :000 : : :00 Character s l o ASCII Hex C 6F ASCII Binary 0:00 000: :00 00: EBCDIC Hex A EBCDIC Binary 00: : :00 00:00 Problem A7. Convert the following to hexadecimal and then to characters using EBCDIC code. Binary : :0000 0:000 00:000 00:000 00:0 EBCDIC Hex F 40 D A Character J u m p Jubilation Code and Gran Zeff Code Problems Problem B. Code the word adelfos (brother) from SECRET SYMBOL to JUBILATION CODE hexadecimal, and then to JUBILATION CODE binary. a d e l f o s A A4 B C3 F2 D3 E2 00:000 00:000 0:000 00:00 :000 0:00 0:000 Problem B2. Code the word dwron (gift) from SECRET SYMBOL to GRAN ZEFF CODE hexadecimal, and then to GRAN ZEFF CODE binary. d w r o n :000 00:00 000:00 0:0 0:00

7 Number Systems and Data Representation 7 Problem B3. Code the word sofos (wise) from SECRET SYMBOL to UNICODE hexadecimal, and then to UNICODE binary. s o f o s 03C3 03BF 03C6 03BF 03C2 0000:00: 00: :00: 0: 0000:00: 00: :00: 0: 0000:00: 00:000 Problem B4. Code the word kardia (heart) from SECRET SYMBOL to UNICODE hexadecimal, and then to UNICODE binary. k a r d i a 03BA 03B 03C 03B4 03B9 03B 0000:00: 0: :00: 0: :00: 00: :00: 0: :00: 0:00 Problem B5. Code the JUBILATION CODE binary into JUBILATION CODE hexadecimal, and then to SECRET SYMBOL. (death) 0000:00: 0:000 0:000 00:000 0:000 00:000 0:000 0:00 0:000 B4 A D A E4 D3 E2 q a n a t o s Problem B6. Code the GRAN ZEFF CODE binary into GRAN ZEFF CODE hexadecimal, and then to SECRET SYMBOL. (wicked) 0:000 0:0 0:00 00:0 000:00 0:0 000: p o n h r o s Problem B7. Code the UNICODE binary into UNICODE hexadecimal, and then into SECRET SYMBOL. (nation) 0000:00: 0: :00: 0: :00: 0:0 0000:00: 0: 0000:00: 00:000 03B5 03B8 03BD 03BF 03C2 e q n o s

8 Number Systems and Data Representation 8 Conversion to and from Decimal Answers Conversion to Decimal Answers Problem A. 6 0 Problem A Problem A Problem A Problem A Problem A6.,359 0 Problem A Problem A Common Answer Table for Conversion from Decimal: Problems B through B6 Between Base 4, Binary, and Hexadecimal: Problem B7 Between Octal and Binary: Problem B8

9 Number Systems and Data Representation 9 Column A B C D Row Decimal Binary Hexadecimal Base 4 Octal 3 : F : :00 A :00 A :0000: :0:0 BD :0: DF :00:000 2C Column A B C D Row Decimal Binary Hexadecimal Base 4 Octal 3 B B3 B5 B B B3 B5 B B B3 B5 B B B3 B5 B B4 B2 B7 B B4 B2 B7 B B4 B2 B7 B B4 B2 B7 B8

10 Number Systems Conversion Answers Problem. Convert the following decimal numbers to binary. Problem Decimal Hexadecimal Binary a H b 240 F0H 0000 c 6 3DH 0 d 202 CAH Problem 2. Convert the following decimal numbers to hexadecimal. Problem Decimal Hexadecimal Binary 2a 37 25H b 222 DEH 0 0 2c 68 A8H d 959 3BFH 0 Problem 3. Convert the following numbers to decimal. Problem Binary Hexadecimal Decimal 3a 0 0 D7H 25 3b H 2 3c 0 0 EBH 235 3d 7FH 27

11 Number Systems and Data Representation String Function Answers Problem. Length. Let x = B l e s s e d a r e t h e p o o r l Find k = LEN(x) = 20 Problem 2. Concatenation. Let x = B e l y = P r e p a r e d l Concatenate x and y. z = x & y. z = B e P r e p a r e d l Problem 3. Left. Let x = B l e s s e d a r e t h e p o o r l Find y = Left(x,5). y = B l e s s l Problem 4. Right. Let x = B l e s s e d a r e t h e p o o r l Find y = Right(x,8). y = t h e p o o r l Problem 5. MID. Let x = B l e s s e d a r e t h e p o o r l Find y = MID(x,4,2) y = h e l Problem 6. Comparison. Let x = B l e s s e d a r e t h e p o o r l y = B l e s s e d a r e t h e m e e k l If x < y Answer is False. Problem 7. Search. x = B e h o l d, I b r i n g g l a d l y = h o l d l k = FIND(x,y) = 3

12 Number Systems and Data Representation 2 Problem 8. Search. x = t i d i n g s. Y o u s h a l l l y = y o u l Search only for whole words; permit upper or lower case. k = FIND(x,y) = 0 Problem 9. Compound Function x = B l e s s e d a r e t h e p o o r l y = P r e p a r e d l Find z = LEFT(x,5) & MID(x,3,3) & y z = B l e s s t h e P r e p a r e d l Problem 0. Compound Function x = B e b r a v e. B e s t r o n g. B e r e a d y. B e a l e r t. l y = B e l Find z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, FIND(x,y, FIND(x,y)+LEN(y))+LEN(y))) z = Solution: z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, FIND(x,y, FIND(x,y)+LEN(y))+LEN(y))) j = FIND(x, y) = k = LEN(y) = 2 z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, FIND(x,y, +2)+LEN(y))) z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, FIND(x,y, 3)+LEN(y))) FIND(x,y, 3) = z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, +LEN(y))) z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, +2)) z = RIGHT(x, LEN(x) LEN(y) FIND(x,y, 3)) m = LEN(x) = 40 z = RIGHT(x, 40 2 FIND(x,y, 3)) = RIGHT(x, 38 FIND(x,y, 3)) FIND(x,y, 3) = 22 z = RIGHT(x, 38 22) = RIGHT(x, 6) z = r e a d y. B e a l e r t. l

13 Number Systems and Data Representation 3 Hexadecimal Addition Answers Problem. 0 6 Throw away Problem Throw away Problem 3. F96H No throw away Problem 4. 86ECH No throw away Problem 5. E00D 6 No throw away Problem 6. 0C63 6 No throw away

14 Number Systems and Data Representation 4 Convert to IBM BCD Answers Problem. Convert ASCII -736 to IBM Unpacked BCD Z d Z d S d Problem 2. Convert ASCII -736 to IBM Packed BCD d d d S

15 Number Systems and Data Representation 5 Fixed Point Conversion Answers Problem. Problem 2. Problem = Problem = = 3B5.34H FED.CAB 6 = B Problem 5. AD.E 6 = Problem = AE.C2H

16 Number Systems and Data Representation 6 Partitioned Numbers Problem A. a. Left partition = 60 b. Right partition = 22 c. Sum = 82 Problem A2. a. Left partition = 96 b. Right partition = 2 c. Sum = 08 Subnet Answers Problem B. Convert the dotted decimal for IP address to binary and hexadecimal, using the table below A 7 F 0 5 Problem B2. How many subnet addresses are possible in the subnet mask shown below? x x x x x x x x x y y y y y y y y y y y y y y y x = FF = = 5. The number of subnet addresses is 52, where 0 is one of the possible addresses. (0 5) Problem B3. How many host machine addresses are possible for each subnet in the subnet mask shown below? x x x x x x x x x y y y y y y y y y y y y y y y y = 7FFFH = = The number of host machine addresses is Recall that 0 is one of the possible addresses. ( ). The number of host machines that can be on that subnet is two less than the number of possible addresses, or = Problem B4. How many bits in the below mask need to be set aside to make sure each subnet can have at least 283 host machines? x x x x x x x x x x x x x x x y y y y y y y y y 9 bits are needed. 8 bits gives only 256 addresses (0 255). 9 bits gives 52 addresses, which is at least is the smallest power of two greater than or equal to 283.

17 Number Systems and Data Representation 7 Problem B5. How many subnet addresses are possible in the below mask if enough bits are set aside to make sure each subnet can have at least 283 host machines? x x x x x x x x x x x x x x x y y y y y y y y y 5 bits are available for subnet addresses if enough bits (9) are set aside for host machine addresses. This yields different addresses ( ). Problem B6. What is the dotted decimal notation for the subnet mask needed for ensuring enough bits are set aside to ensure each subnet can have at least 283 host machines? x x x x x x x x x x x x x x x The bits in the subnet mask coded with and x are turned on. Bits coded 0 are turned off. The subnet mask in dotted decimal notation is

18 Number Systems and Data Representation 8 German Roman Numeral Answers Problem. Write 390 B.C. in Roman numerals. B.C. CCCXC Zero and negative numbers do not exist in Roman numerals. We attach Another representation is XCD B.C. 500 is D. 400 is CD, which is 00 less than is XCD, which is 0 less than 400. Ambiguity: This is ambiguous because Roman numerals are not associative. XC is 90 (XC)D is 40 Only if you scan from right to left is your answer unique. Problem 2. Write 407 A.D. in Roman numerals. CDVII A.D. We attach A.D. 500 is D. 400 is CD, which is 00 less than is CDVII, which is 7 more than 400. Problem 3. Write 255 in Roman numerals. CCLV. 200 is CC. 250 is CCL. 255 is CCLV. Problem 4. Add XI and XXXII. Group similar letters together, while maintaining the relative position of letters. Simplify by substituting aggregate symbols. XXXX III XLIII The purpose of this problem is to illustrate that addition was fairly simple for the Romans. Problem 5. XXV is 25. Time to get the clock fixed. Problem 6. Zero cannot be represented by Roman numerals. It lacks nothing. Problem 7. Divide XXX by VI. I do not know of an easy way of doing division using Roman numerals. Roman engineers used a calculating board or an abacus for doing arithmetic and translated the answers into Roman numerals. One approach would be to expand both to all ones, and group the ones. Likewise, they needed some way of doing multiplication. You can see that calculation with Roman numerals is not as easy as calculation using our positional number system. XXX = 30, VI = 6. 30/6 = 5 = V. So, the Roman numeral system was good for enumeration, counting, and addition.

19 Number Systems and Data Representation 9 Problem 8. MDCCLXXVI Americans: You should recognize the answer after you decode it. Problem 9. I(VL) = 54 = 44 Problem 0. (IV)L = 50 4 = 46

20 Number Systems and Data Representation 20 Fibonacci Bases Answers Problem. Convert 50 0 to Minimal Fibonacci Base (Zeckendorf) Representation Remainder R F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values digits decimal values Remainder R Problem 2. Convert 53 0 to Minimal Fibonacci Base (Zeckendorf) Representation Remainder R F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values digits decimal values Remainder R If you add the Zeckendorf Representation numbers for , you get: F 5 F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values X Y Answer decimal values

21 Number Systems and Data Representation 2 Converting this Fibonacci Base Representation to a Minimal Fibonacci Base Representation: F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values Answer Rule Rule 0 0 Rule Rule Rule Rule Rule Rule Rule Rule 2 The answer is Fib To check the answer, this is ( x 89) + ( x 3) + ( x ) = 03 0 Problem 3. Add the two Zeckendorf Representation numbers: Fib Fib F 5 F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values X Y Answer decimal values The sum is 67 0 Converting this Fibonacci Base Representation to a Minimal Fibonacci Base Representation:

22 Number Systems and Data Representation 22 F 5 F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values Answer Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule Rule 2 The answer is Fib Converted to decimal, it is: ( x 60) + ( x 5) + ( x 2) = 67 0 Problem 4. Restore Fibonacci Base Representation 002 Fib to Minimal Fibonacci Base Representation. F 4 F 3 F 2 F F 0 F 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 Fibonacci Number place values Answer Rule Rule Rule Rule Rule Rule Rule 2 The answer is Fib Converted to decimal, it is ( x 55) + ( x 5) + ( x 2) = 62 0

23 Number Systems and Data Representation 23 Phi Base Number System Answers Problem. Express A = Φ 4 + Φ 3 + Φ -3 as phigits: F 6 F 5 F 4 F 3 F 2 F F 0. F - F -2 F -3 F -4 F -5 F -6 A = Problem 2. Express B = Φ 4 + Φ + Φ -4 + Φ -6 as phigits: F 6 F 5 F 4 F 3 F 2 F F 0. F - F -2 F -3 F -4 F -5 F -6 B = Problem 3. Add the phigits A (from Problem ) and B (from Problem 2). F 6 F 5 F 4 F 3 F 2 F F 0. F - F -2 F -3 F -4 F -5 F -6 A = B = A+B= Problem 4. Transform the sum, A+B, computed in Problem 3 into minimal Phi base representation. In the left column, write the rule number being used for each step of the transformation. F 6 F 5 F 4 F 3 F 2 F F 0. F - F -2 F -3 F -4 F -5 F -6 A+B= R# R# R# Problem 5. In the answer to Problem 4, substitute expressions for F n in terms of n and f. Reduce the expression to the fewest number of terms. Φ 6 + Φ + Φ -2 + Φ -6 = 3+8ϕ + +ϕ + -ϕ + 5-8ϕ = ϕ = 20

24 Number Systems and Data Representation 24 Continued Fraction Answer Problem. Evaluate the first six terms. Show all your steps = = = = X = + = + + = K The 6-term continued fraction approximation to the golden ratio is.625. By comparison, the ten digit approximation of the golden ratio is Compute as follows: () Enter (2) x + (3) Repeat step (2) four times

25 Number Systems and Data Representation 25 Complement Arithmetic Answers Problem. 8 Problem Problem 3. 20C 6 Problem D 6 Problem Problem Fast Two s Complement Answers Problem. Consider the binary number a. The right-most -bit is in bit position 2. b. Leave bit positions 2,, and 0 alone. c. The two s complement of this binary number is Problem 2. Consider the binary number a. The right-most -bit is in bit position 4. b. Leave bit positions 4, 3, 2,, and 0 alone. c. The two s complement of this binary number is

26 Binary Multiplication Answer Column Position Number X 0 Y 0 0 Carry Column Column 0 Column 2 0 Column Column 4 0 Sum Check the answer by converting to decimal. Place Value Sum The answer is 638. Hexadecimal Multiplication Answer Problem. Multiply X = FADH times EB 6. Sum Carry 2 Carry term A 6 8 Carry term 2 D 8 B X F A D times Y E B 5 E F 2 C 6 Answer E 6 3 C F Answer = E63CFH = 953,055 0.

27 Number Systems and Data Representation 27 Fixed Point Binary Division Answers Problem. Divide by The answer is. 2 Prime Number Answers Problem. a. Show that 7 is a prime number using base 2. b. Show that 7 is a prime number using base 3. c. Compare the results. a. Demonstrate 7 is a prime number using base 2. Divide by R 3 = R 0 =2 R 3 = R 3 = R 0 = R 0 Divide by 7 R 3 = 0 R 0 =0

28 Number Systems and Data Representation 28 The first integer that evenly divides 2 is 2. Therefore, 2 = 7 0 is a prime number. b. Demonstrate that 7 is a prime number using base 3. Divide by R 3 = R 3 = 2 0 R 0 = R 0 =2 R 3 = R 0 = Divide by R 3 = 0 R 0 =0 The first integer that divides 2 3 evenly is 2 3. Therefore, 2 3 = 7 0 is a prime number. c. In both problems, the equivalent of 7 0 is a prime number. Scientific Notation Conversion Answers Problem x 0 34 J/s is Plank s constant. Problem x 0 8 m/s is the speed of light in a vacuum. Scientific Notation Multiplication Answers Problem. Approximate the distance of one light-year: ( x 0 8 m/s ) x (3.47 x 0 7 s/yr ) x ( yr) = x 0 5 m = x 0 2 km Problem 2. Naively estimate the kinetic energy of the earth revolving around the Sun, assuming a circular orbit (ignoring Kepler): (/2) x (5.98 x 0 24 kg ) x ( 2.99 x 0 4 m/s ) 2 = 2.67 x 0 33 J

29 Number Systems and Data Representation 29 Floating Point Conversion Answers Problem. IEEE Single Precision Floating Point S Characteristic Mantissa Mantissa (Continued) Procedure: Step : Let M = (IEEE Single Precision Floating Point characteristic) (IEEE Single Precision Floating Point Excess). M = = 2. Note: The IEEE format uses implicit normalization. The radix point is immediately to the left of the IEEE mantissa most significant bit. Step 2: Insert the prefix bit pattern 000 immediately to the left of the radix point Step 3: Let MM = M MOD 4. This is the remainder obtained by dividing M by 4. MM = 2 MOD 4 = 2. Step 4: Compute the number of bits (J) to right-shift the mantissa and prefix bit pattern. If MM = 0, then J = 4. If MM < 0, then J = MM. MM is the absolute value of MM. If MM > 0, then J = 4 MM. J = 4 2 = 2. The number of bits of precision in the mantissa lost is J = 2 =. Step 5: Copy the shifted bit pattern into the IBM Short HFP mantissa. Step 6: Compute the IBM HFP characteristic, without excess K = M = =.5 = 4 = 4 4. Step 7: Add the IBM HFP excess to K and store it in the HFP characteristic field. K + 64 = 65 = 4 6 = Step 8: Copy the sign bit from the IEEE Single Precision FP word to the sign bit of the IBM HFP word. IBM Hexadecimal Short Floating Point S Characteristic Mantissa

30 Number Systems and Data Representation 30 Mantissa (Continued) Problem 2. IEEE Single Precision Floating Point S Characteristic Mantissa Mantissa (Continued) Procedure: Step : Let M = (IEEE Single Precision Floating Point characteristic) (IEEE Single Precision Floating Point Excess). M = = 5. Note: The IEEE format uses implicit normalization. The radix point is immediately to the left of the IEEE mantissa most significant bit. Step 2: Insert the prefix bit pattern 000 immediately to the left of the radix point Step 3: Let MM = M MOD 4. This is the remainder obtained by dividing M by 4. MM = 5 MOD 4 =. Step 4: Compute the number of bits (J) to right-shift the mantissa and prefix bit pattern. If MM = 0, then J = 4 If MM < 0, then J = MM. MM is the absolute value of MM. If MM > 0, then J = 4 MM = 4 = 3. The number of bits of precision in the mantissa lost is J = 3 = 2. Step 5: Copy the shifted bit pattern into the IBM Short HFP mantissa. Step 6: Compute the IBM HFP characteristic, without excess K = M = = 2.25 = 2 4 = 4 4. Step 7: Add the IBM HFP excess to K and store it in the HFP characteristic field. K + 64 = = 66 = 42 6 = Step 8: Copy the sign bit from the IEEE Single Precision FP word to the sign bit of the IBM HFP word. IBM Hexadecimal Short Floating Point S Characteristic Mantissa

31 Number Systems and Data Representation 3 Mantissa (Continued) Problem 3. IEEE Single Precision Floating Point S Characteristic Mantissa Mantissa (Continued) Procedure: Step : Let M = (IEEE Single Precision Floating Point characteristic) (IEEE Single Precision Floating Point Excess). M = = - 2. Note: The IEEE format uses implicit normalization. The radix point is immediately to the left of the IEEE mantissa most significant bit. Step 2: Insert the prefix bit pattern 000 immediately to the left of the radix point Step 3: Let MM = M MOD 4. This is the remainder obtained by dividing M by 4. MM = -2 MOD 4 = -2. Step 4: Compute the number of bits (J) to right-shift the mantissa and prefix bit pattern. If MM = 0, then J = 4 If MM < 0, then J = MM. J = -2 = 2. If MM > 0, then J = 4 MM. The number of bits of precision in the mantissa lost is J = 2 =. Step 5: Copy the shifted bit pattern into the IBM Short HFP mantissa. Step 6: Compute the IBM HFP characteristic, without excess K = M = = 0.5 = 0 4 = 4 4. Step 7: Add the IBM HFP excess to K and store it in the HFP characteristic field. K + 64 = = 64 = 40 6 = Step 8: Copy the sign bit from the IEEE Single Precision FP word to the sign bit of the IBM HFP word. IBM Hexadecimal Short Floating Point S Characteristic Mantissa

32 Number Systems and Data Representation 32 Mantissa (Continued) Problem 4. IBM Hexadecimal Extended Floating Point High Order Word S High Order Characteristic High Order Mantissa High Order Mantissa (Continued ) High Order Mantissa (Continued 2) High Order Mantissa (Continued 3) Low Order Word S Low Order Characteristic Low Order Mantissa Low Order Mantissa (Continued ) Low Order Mantissa (Continued 2) Low Order Mantissa (Continued 3) Procedure: Step : Subtract the IBM Hexadecimal Floating Point Excess from the high order characteristic. Step 2: Let K = the numerical value of the high order characteristic. K = = 0

33 Number Systems and Data Representation 33 Step 3: Step 4: Let N = 4 K. Each positive increment of the IBM Hexadecimal Floating Point characteristic represents a shift of the radix by 4 bit positions to the left to perform the normalization. N = 4 x 0 = 40. Let J = the number of bit positions from the left end of the mantissa occupied by the first one-bit. J = 4. Step 5: Let P = N J = 40 4 = 36. Step 6: Add the IEEE Double Extended Precision Floating Point Excess to P. M = P + XS IEEE DEP FP = = 649. Step 7: Convert M to binary. 649 = = Step 8: Step 9: Step 0: Step : Record the result in the characteristic field of the IEEE Double Extended Precision Floating Point word. The most significant bit of the mantissa of the IBM Hexadecimal Floating Point becomes the implicit bit in the IEEE Double Extended Precision Floating Point word, and therefore does not explicitly get recorded. Beginning with the bit immediately to the right of the most significant bit, copy the remaining bits from the IBM Hexadecimal Floating Point mantissa into the IEEE Double Extended Precision Floating Point mantissa. Stop after the IEEE Double Extended Precision Floating Point mantissa is filled. Note that 44 to 47 least-significant-bits of precision in the mantissa were lost in the conversion. The number of bits of precision lost is 48 J = 48 4 = 44. IEEE Double Extended Precision Floating Point S Characteristic Mantissa Mantissa (Continuation ) Mantissa (Continuation 2) Mantissa (Continuation 3)

34 Number Systems and Data Representation 34 Problem 5. Convert the following IEEE Double Extended Precision Floating Point word to an IBM Hexadecimal Extended Floating Point word. IEEE Double Extended Precision Floating Point S Characteristic Mantissa Mantissa (Continuation ) Mantissa (Continuation 2) Mantissa (Continuation 3) Procedure: Step : Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Step 8: Step 9: Set all IBM Hexadecimal Extended Floating Point bits to zero. Copy the IEEE Double Extended Precision Floating Point sign bit into the IBM Hexadecimal Extended Floating Point High Order sign bit and Low Order sign bit. Evaluate the IEEE Double Extended Precision Floating Point Excess from the characteristic. 3F4A 6 = (3 x 6 3 ) + ( F x 6 2 ) + ( 4 x 6 ) + (A x 6 0 ) = = Subtract the IEEE Double Extended Precision Floating Point Excess from the characteristic. P = = -8. Compute the IBM Hexadecimal Floating Point Characteristic without the IBM HFP Excess. K = P = = = 45 4 = 4 4 Compute the IBM Hexadecimal Floating Point Characteristic with the IBM HFP Excess. N = K + 64 = = 9 = 3 6 = Write N into the IBM HFP High Order Characteristic. Append the implicit most significant bit to the left of the radix point for the IEEE DEP FP mantissa. Let MM = P MOD 4. This is the remainder obtained by dividing P by 4. MM = -8 MOD 4 = -.

35 Number Systems and Data Representation 35 Step 0: Step : Step 2: Step 3: Compute the number of places to shift the appended mantissa to obtain the IBM HFP mantissa. If MM = 0, then J = 4 If MM > 0, then J = 4 MM If MM < 0, then J = MM where the vertical bars identify the absolute value function. J = - = Beginning with the most significant bit of the shifted mantissa, copy the remaining bits from the IEEE Double Extended Precision Floating Point mantissa into the IBM Hexadecimal Floating Point Mantissa. Compute the IBM HFP Low Order Characteristic. Q = N 4 = 9 4 = 5. Write Q into the IBM HFP Low Order Characteristic. IBM Hexadecimal Extended Floating Point High Order Word S High Order Characteristic High Order Mantissa High Order Mantissa (Continued ) High Order Mantissa (Continued 2) High Order Mantissa (Continued 3) Low Order Word S Low Order Characteristic Low Order Mantissa Low Order Mantissa (Continued )

36 Number Systems and Data Representation 36 Low Order Mantissa (Continued 2) Low Order Mantissa (Continued 3) Bits 8 through are all default values of zero. Problem 6. IEEE Double Extended Precision Floating Point S Characteristic Mantissa Mantissa (Continuation ) Mantissa (Continuation 2) Mantissa (Continuation 3) Procedure: Step : Step 2: Step 3: Step 4: Set all IBM Hexadecimal Extended Floating Point bits to zero. Copy the IEEE Double Extended Precision Floating Point sign bit into the IBM Hexadecimal Extended Floating Point High Order sign bit and Low Order sign bit. Evaluate the IEEE Double Extended Precision Floating Point Excess from the characteristic. 3F0B 6 = (3 x 6 3 ) + ( F x 6 2 ) + ( 0 x 6 ) + (B x 6 0 ) = = 639. Subtract the IEEE Double Extended Precision Floating Point Excess from the characteristic. P = = -244.

37 Number Systems and Data Representation 37 Step 5: Step 6: Step 7: Step 8: Step 9: Step 0: Step : Step 2: Compute the IBM Hexadecimal Floating Point Characteristic without the IBM HFP Excess. K = P = = 60 4 = 4 4 Compute the IBM Hexadecimal Floating Point Characteristic with the IBM HFP Excess. N = K + 64 = = 4. Write N into the IBM HFP High Order Characteristic. Append the implicit most significant bit to the left of the radix point for the IEEE DEP FP mantissa. Let MM = P MOD 4. This is the remainder obtained by dividing P by 4. MM = -244 MOD 4 = 0. Compute the number of places to shift the appended mantissa to obtain the IBM HFP mantissa. If MM = 0, then J = 4 If MM > 0, then J = 4 MM If MM < 0, then J = MM where the vertical bars identify the absolute value function. J = 4 Beginning with the most significant bit of the shifted mantissa, copy the remaining bits from the IEEE Double Extended Precision Floating Point mantissa into the IBM Hexadecimal Floating Point High Order Mantissa. Compute the IBM HFP Low Order Characteristic. Q = N 4 = 4 4 = -0. We cannot proceed further because 0 is not a possible value for an IBM HFP Characteristic. Note that we can approximate the IEEE word by using an IBM Hexadecimal Long Floating Point word which does not need to use a low order characteristic.

38 Number Systems and Data Representation 38 IBM Hexadecimal Extended Floating Point High Order Word S High Order Characteristic High Order Mantissa High Order Mantissa (Continued ) High Order Mantissa (Continued 2) High Order Mantissa (Continued 3) Low Order Word S Low Order Characteristic Low Order Mantissa Low Order Mantissa (Continued ) Low Order Mantissa (Continued 2) Low Order Mantissa (Continued 3)

39 Number Systems and Data Representation 39 Problem 7. Convert the following IEEE Double Extended Precision Floating Point word to an IBM Hexadecimal Extended Floating Point word. IEEE Double Extended Precision Floating Point S Characteristic Mantissa Mantissa (Continuation ) Mantissa (Continuation 2) Mantissa (Continuation 3) Procedure: Step : Set all IBM Hexadecimal Extended Floating Point bits to zero. Step 2: Copy the IEEE Double Extended Precision Floating Point sign bit into the IBM Hexadecimal Extended Floating Point High Order sign bit and Low Order sign bit. Step 3: Evaluate the IEEE Double Extended Precision Floating Point Excess from the characteristic. 3B96 6 = (3 x 6 3 ) + ( B x 6 2 ) + ( 9 x 6 ) + (6 x 6 0 ) = = Step 4: Subtract the IEEE Double Extended Precision Floating Point Excess from the characteristic. P = = -29. Step 5: Compute the IBM Hexadecimal Floating Point Characteristic without the IBM HFP Excess. K = P = = = = 4 4 We cannot proceed further because 282 is not a possible value for an IBM HFP Characteristic.

40 Number Systems and Data Representation 40 IBM Hexadecimal Extended Floating Point High Order Word S High Order Characteristic High Order Mantissa High Order Mantissa (Continued ) High Order Mantissa (Continued 2) High Order Mantissa (Continued 3) Low Order Word S Low Order Characteristic Low Order Mantissa Low Order Mantissa (Continued ) Low Order Mantissa (Continued 2) Low Order Mantissa (Continued 3)