1. Flynn’s Taxonomy
• The purpose of parallel processing is to speed up the computer
process and also increase its throughput, so the hardware will also
increase for this purpose which will make the system costly.
• Michael Flynn proposed a classification for computer architectures
based on the number of instruction steams and data streams (Flynn’s
Taxonomy).
• The instruction stream is defined as the sequence of instructions
performed by the processing unit or fetched from memory.
• The data stream is defined as the operations performed on the data
in the processor.
• A stream simply means a sequence of items (data or instructions).
Flynn’s Taxonomy:
• SISD: Single instruction single data (executing 1 instruction or data
at a time)
– Classical von Neumann architecture
Isha Padhy,CSE Dept, CBIT

2. • SIMD: Single instruction multiple data
There are no. of processing elements: no. of registers but control unit is
same. All the processing elements are executing same instruction. Ex image
processing , add 2 images: 2 matrix m*n, every matrix point will represent
a intensity value in digitized form. same operation(add) to be performed on
all the elements of the matrix.
• MISD: Multiple instructions single data
- Non existent, just listed for completeness.
- Theoretical true because multiple operation on same data is hardly possible.
• MIMD: Multiple instructions multiple data
- Most common and general parallel machine
- Leads to a type of computing called distributed computing. no. of
independent computers , each computer is executing diff instruction ,each
instruction working on diff data set. If different instructions belong to same
task then each computer has to interact with each other and that can be
done LAN, closely n/w.
Isha Padhy,CSE Dept, CBIT

3. SISD SIMD
MISD MIMDIsha Padhy,CSE Dept, CBIT

4. Non- pipeline process
Pipeline process
- Pipelining is a technique of decomposing a sequential process into sub- processes, with
each sub-process being executed in a special dedicated segment that operates concurrently with
all other segments. It is an implementation technique whereby multiple instructions are executed
in an overlapped manner. Min aim is to increase the throughput of the processor.
- This technique is efficient for applications that are going to repeat the same task on
different set of data.
Isha Padhy,CSE Dept, CBIT

5. Non- pipeline process Pipeline process
Isha Padhy,CSE Dept, CBIT

6. General considerations: 4 segment pipeline
• The operands pass through all the segments in a fixed
sequence.
• Task T is divided into sub task:t1,t2,t3, t4.
• Assume there are 4 computing block S1,S2,S3,S4 is capable
of executing S1(t1),S2(t2),S3(t3),S4(t4), each taking same
amount of time to execute (τ)
• So for completion of 4 subtask or task T in case of single
instruction is 4 τ. If n no. of task, then completion time is 4n
τ.
• If we have identical jobs to be done, pipeline concept can be
used to increase the throughput.
• Once the pipeline is full, when every processing element
has a job to execute, now at every τ time we can get an
output. In single inst after every 4 τ we will get an output.
Isha Padhy,CSE Dept, CBIT

7. Isha Padhy,CSE Dept, CBIT

8. • if there are K stages in the pipeline, where every stage
takes τ time to execute, and n jobs to be executed,
- In a non-pipeline architecture: amount of time taken is
nkτ.
- In a pipeline architecture: (n+(k-1)) τ
Ex. No. of stages=4(k)
no. of sub task= 6(n), so 9 clock cycles.
- In general all stages will not take equal amount of time:
τ=max{τi}1
K +τl , τi :time taken by the ith stage, K: K
no. of stages, τl : latching delay
τ : time taken to operate, depends on which stage is taking
max time.
• 2 areas where pipeline organization is applicable:
arithmetic pipeline, instruction pipeline.
Isha Padhy,CSE Dept, CBIT

9. Arithmetic Pipeline
• Units are found in high speed computers for floating point operations.
• Ex. 2 floating point number we r going to add,
• a,b are mantissas p,q are exponents
• Steps to execute:
1. Compare the exponents
2. Align the mantissa
3. Add or subtract mantissa
4. Normalize the result.
• A=a*2p 0.5*103 equalize the power values:
0.005*105
B=b*2q 0.3*105 0.3*105
0.305*105
• Always after point, 0 values are removed so depending on power of 10 is
reduced. ex.003*104=.3*102
Isha Padhy,CSE Dept, CBIT

10. Pipeline for floating addition
1. 1st latch will take fraction and exponent part: a , p.
2nd latch will take b , q
2. Compare the exponents 10 power(p,q)-> result, depending on the result
we have to normalize the fraction part(to same exponents ex 105 )
3. Fraction selector: input a,b , result from comparator, then make right
shift(.5->.005) depending between difference between p,q.
4. |p-q| will denote how many bits shift will take place.
5. No. of latches should be equal both sides( exponent and fraction) other
wise time delay will lead to data loss(we will get fraction and no
exponent or vice versa).
6. From exp comparator we get max exponent.
7. Leading 0 counter: gives no. of bits left shift (re -normalization of result.)
8. Now depending on leading 0’s we have to give left shift.
9. Exp subtractor: no. of left shift will be the no. that will be deducted from
exponent.(ex.003*104=.3*102 ) No. of left shift= subtract that factor from
exponent.
10. Final result will be d*2s.
Isha Padhy,CSE Dept, CBIT

11. Nt: ex. Addition of 2 linear array of floating numbers, pipeline can be used.
Latches are required only when the computation speed is different in different pipeline
stages. This type is called execution pipeline.
s d
Isha Padhy,CSE Dept, CBIT

12. Instruction Pipeline
• An instruction pipeline reads consecutive instructions from memory while previous
instructions are getting executed in other segments.
• Dividing the instruction execution across several stages ,require each stage access
only a subset of CPU’s resources(ALU, Registers, Bus etc). Ideally(may change) a
new instruction can be issued in every cycle.
• Any instruction can be done in following steps:-
- Fetch the instruction
- Decode the instruction
- Calculate the effective address
- Fetch the operands from memory
- Execute the instruction
- Store the result in the proper place
• Sometimes there will be difficulty in pipelining instructions because all the
segments do not require same amount of time, some segments are skipped(ex in
register transfers effective address calc is not required),2 segments may require to
read memory same time etc.
Isha Padhy,CSE Dept, CBIT

13. 4 segment instruction pipeline
• We assume all segments take same time.
• Step 2,3 are combined, steps 5,6 can be combined and
executed. We assume that we use processor registers to
hold the result.
• Fig 1. When an instruction is executed in segment 4,
another instruction is busy fetching operand from
memory in segment3.
• When the instruction is program control type(a branch
out of normal sequence), the pending operations in the
last 2 segments are completed and all the instructions in
instruction buffer is deleted. Then a new address is
entered to PC.
Isha Padhy,CSE Dept, CBIT

14. Isha Padhy,CSE Dept, CBIT

15. Timing of instruction pipeline
Isha Padhy,CSE Dept, CBIT
-We assume that the
processor has separate
instruction and data
memories so FI and FO
can be done at same
time.
- No branch instruction,
each segment operates
on different instructions.
-When DA of 3rd
instruction was decoded
and found that it was a
branch instruction, FI of
4th instruction is halted
till the branch instruction
is executed. If branch is
taken new instruction is
send to step 7 or else
normal 4th step is
executed.
FI: segment 1 that fetches the instruction.
DA: segment 2 that decodes the instruction and
calculates the effective address.
FO: segment 3 that fetches the operands.
EX: segment 4 that executes the instruction.

16. Problems with instruction pipeline (Pipeline Hazards).
• A pipeline hazard occurs when the instruction pipeline
deviates at some phases, some operational conditions
that do not permit the continued execution. The major
pipeline hazards are described below:
• Resource hazard (Resource conflict).
• Data hazard (data dependency).
• Branch hazard (branch difficulties).
1. Resource hazard
It is caused by the access to shared memory by 2
segments at the same time. This can be resolved by
using separate instruction and data memories.
Isha Padhy,CSE Dept, CBIT

17. • 2. Data Hazard
It arises when instructions depend on the result of previous instruction but
the previous instruction is not available yet. Ex. An instruction in FO
segment may need to fetch an operand which is produced by previous
instruction. The most common techniques used to resolve data hazard are:
(a) Hardware interlock - a hardware interlock is a circuit that detects
instructions whose source operands are destinations of instructions farther
up in the pipeline. It then inserts enough number of clock cycles to delays
the execution of such instructions.
(b) Operand forwarding - This method uses a special hardware to detect
conflicts in instruction execution and then avoid it by routing the data
through special path between pipeline segments.
for example, instead of transferring an ALU result into a destination result,
the hardware checks the destination operand, and if it is needed in next
instruction, it passes the result directly into ALU input, bypassing the
register.
c) Delayed load - It is software solution where the compiler is designed in
such a way that it can detect the conflicts; re-order the instructions to delay
the loading of conflicting data by inserting no operation instruction.
Isha Padhy,CSE Dept, CBIT

18. • 3. Branch hazard
The major obstruction that a instruction pipeline is branch instructions. Such
instructions can be conditional(depend on the condition, if yes branch to target
inst address or next inst in sequence will be executed) or unconditional(PC is
assigned the target instruction address).Computers use hardware technique to
minimize this hazard.
(a) Multiple streaming - It is an approach which replicates the initial portions of the
pipeline and allows the pipeline to fetch both instructions(target instruction of
branch and instruction next to branch), making use of two streams (branches). If
condition is successful then target is selected or the next is selected.
(b) Branch target buffer (BTB)- BTB is a memory present in fetch segment of an
instruction. Each entry consists of address of previously executed branch
instruction and their target instruction. After decoding the instruction as branch
instruction it refers its own BTB, if any information of the required branch
instruction is available then use it or pipeline shifts to a new instruction stream
and stores the target instruction in BTB. Thus branch instructions previously used
are readily available.
Isha Padhy,CSE Dept, CBIT

19. (c) Loop buffer - A variation of BTB. A loop buffer is a small, very-high-
speed memory maintained by the instruction fetch stage of the pipeline.
When a loop is detected in the program , complete instruction set is kept in
this memory area along with any branches present in the loop. And this
loop can be executed without disturbing the memory until loop breaks.
• (d) Branch prediction - uses additional logic to predict the outcomes of a
(conditional) branch before it is executed.
• (e) Delayed branch - This technique is employed in most RISC
processors. In this technique, compiler detects the branch instructions and
re-arranges the instructions by inserting no-operation instruction after
branch instructions to avoid pipeline hazards.
Isha Padhy,CSE Dept, CBIT

20. RISC PIPELINE
• In RISC, all the instruction formats are of same
length so decoding of instruction and register
selection will occur at same time. Every operand is
present in registers so no calculation of effective
address.
• The pipeline can be done in 2 or 3 segments. So the
steps can be:
- Fetching instruction from memory.
- Executing operation in ALU.
- Storing the result in destination register.
• For register transfers , only 2 instructions are present
load, store.
• To prevent conflicts between a memory access to
fetch an instruction and to load /store operand,
there are 2 separate memories ,one for instruction
and one for data.
• RISC has ability to complete an instruction in 1 clock
cycles so pipeline segments can be achieved easily,
but CISC will have many segments with the longest
segment taking 2 or 3 clock cycles.
Isha Padhy, CSE Dept, CBIT
-Fetch the instruction
-Decode the instruction
-Calculate the effective
address
-Fetch the operands from
memory
-Execute the instruction
-Store the result in the
proper place

21. • RISC instruction formats: 32 bit, 31 instructions.
• Addressing modes: register addressing, immediate addressing, relative to PC
addressing for branch instructions.
• 138 registers: 8 windows of 32 registers each(10 global, 10 local, 6 low
overlapping, 6 high overlapping)
• Instruction format: 32 bit: 8(7bits operation+ 1 b status bit depending on ALU
oper), 5(5b to specify 1 out of 32 reg), 13 bit=0(low order of S2 specifies another
source reg),13th bit=1(13 b operand), Y(relative address to jump), COND.
• 31 instructions: Data manipulation, transfer, program control.
• All instructions have 3 operands: 2nd operand can be a register, immediate
operand,3rd register can be specified as R0(all bits 0).
• LDL (R22)#150,R5 //R5<-M[R22]+150
LDL(R22)#0, R5 //R5<-M[R22]
Isha Padhy,CSE Dept, CBIT
Opcode Rd Rs 0 Not Used S2
Opcode Rd Rs 1 S2
Opcode COND Y
8 5 5 1 8 5
8 5 5 1 13
8 5 19
Register Mode
Register imm Mode
PC Relative Mode

22. 3 segment Instruction pipeline
Isha Padhy,CSE Dept, CBIT
The instruction cycle is divided into 3 segments:-
- (I) Instruction fetch: fetches the instruction from program
memory and decoded, the registers needed for execution are also
selected.
- (A) ALU performs the operation depending on the decoded
instruction. There are 3 operations that can be done by ALU.
a. Data manipulation instruction.
b. Evaluate the effective address for a load or store
instruction(register indirect method)
c. Calculates the branch address for a program control instruction.
PC<-PC+Y
- (E) directs the output of the ALU to any 1 destination(destination
register/data memory/ branch address to PC)

23. Delayed Load
• LOAD : R1<- M[address 1]
LOAD: R2<-M[address 2]
ADD: R3<-R1+R2
STORE: M[address 3]<- R3
• For the above instructions to be pipeline: A segment in cycle 4 is using the
data from R2, but the value of R2 is not a correct value as the data is not
transferred from memory at that time. So to solve this problem we can
insert a no- operation instruction between the 2 conflicting instruction such
that a clock cycle is wasted and thus use of data loaded from memory is
delayed.
Isha Padhy,CSE Dept, CBIT

24. Vector processing
Isha Padhy,CSE Dept, CBIT
In computing, a vector processor or array processor is a CPU that implements an
instruction set containing instructions that operate on 1- dim arrays of data called
vectors, compared to scalar processors, whose instructions operate on single data
items.
Vector processors have high-level operations that work on linear arrays of numbers:
"vectors“
=> Each result independent of previous result
=> long pipeline, compiler ensures no dependencies
=> high clock rate

25. • Five basic types of vector operations:
1. V <- V Example: Complement all elements
2. S <- V Examples: Min, Max, Sum
3. V <- V x V Examples: Vector addition, multiplication,
division
4. V <- V x S Examples: Multiply or add a scalar to a
vector
5. S <- V x V Example: Calculate an element of a matrix
• Application areas where vector processing is used:
Long range weather fore-casting, biological modeling, car
crash simulation, speech, image processing etc.
Isha Padhy,CSE Dept, CBIT

26. Vector Operations
Isha Padhy,CSE Dept, CBIT
• V=[V1, V2,…..Vn] row vector of length n.
• The element Vi of vector V is written as V[I], I is index refers
to memory address or register where the number is stored.
• int a[10], b[10]; Initialize i=0
int c[10]; Read a[I], Read b[I]
…. for (i = 0; i < n; i++) Store C[I]=a[I]+b[I]
c[i] = a[i] + b[i]; Increment I=I+1
//pointing to next location
If I<10, continue.
// Vector processors have the ability to remove overhead of
instruction fetch and execution in loop.

27. • Instruction can be written as
c(1:10)=A(1:10)+B(1:10)
• Any vector instruction has:
• Addition is done with pipelined floating point
adder in same process as scalar.
Isha Padhy,CSE Dept, CBIT

28. Isha Padhy,CSE Dept, CBIT
Complexity =27
operations
N*N mul requires, n3
multiply- add
operations. Ex
C11=((c11+a11b11)+a
12b21)+a13b31. so
9*3=27 operations.

29. Isha Padhy,CSE Dept, CBIT
• Every inner product consists of sum of k- products
C= A1B1+A2B2+….AkBk
• The multiplier and adder have 4 segments pipeline.To each segment 1 mul
is allowed in clock cycle, so in 1st 4 cycles: A1B1, A2B2, A3B3, A4B4 are
finished and mean time the outputs of adder are all 0.
• For next 4 cycles 0 is added to 1st 4 multiplications . So in next 8 cycles ,
A1B1 to A4B4 are in adder and A5B5 to A8B8 are in multiplier segment.At
beginning of 9th cycle the output of adder A1B1 is added to A5B5, then
A2b2+A6B6….
• So we get 4 sections:
A1B1+A5B5+A9B9+..
+A2B2+A6B6+…
+A3B3+A7B7+…
+A4B4+….

30. Memory Interleaving
Isha Padhy,CSE Dept, CBIT
- For pipelining and vector processing, memory can be required to access
simultaneously by 2 sources.
- The memory can be partitioned into a no. of modules connected to common
address bus and data bus.
- A memory module consists of memory array, its own registers. The least 2 sig bits in
the address can be used to distinguish between the 4 modules and the rest bits for
exact location.

31. Array Processors
• Suitable for scientific computations involving two dimensional matrices
• An array processor has single instruction for mathematical operations, for example,
SUM a, b, N, M
a ─ an array base address , b ─ another vector base address, N─ the number of elements in
the column vector, M ─ the number of elements in row vector
• 2 types:
- Attached array processor:
1. In this an auxiliary processor is attached to a general purpose computer. It enhances the
performance of the host computer in specific numerical computation tasks.
2. This attachment is done by parallel processing i.e. it contains one or more pipelined
floating point adders and multipliers.
3. Can be programmed by user.
4. The array processor is connected through an i/p-o/p controller to the computer. Data is
received from main memory through high speed bus.
Isha Padhy,CSE Dept, CBIT

32. SIMD Array Processor Ex ILLIAC IV
Isha Padhy,CSE Dept, CBIT

33. Isha Padhy,CSE Dept, CBIT

34. Why use array processor
Isha Padhy,CSE Dept, CBIT

35. Computer arithmetic
• Addition , subtraction, multiplication and
division of :-
1. fixed-point(total,fraction) binary data in
signed magnitude representation.
2. Fixed point binary data in signed 2’s
complement representation
3. Floating point binary data
4. Binary coded decimal data(BCD)
Isha Padhy,CSE Dept, CBIT

36. Addition and subtraction
• Negative fixed point binary numbers can be represented as: signed magnitude, signed- 1’s
complement, signed 2’s complement.
• This representation of negative numbers are used to represent number in registers before and
after arithmetic operation.
• Addition(subtraction) algorithms:
- When the signs of A and B are identical(different), add(subtract) the 2 magnitudes and attach
the sign of A to the result,
- when different (identical), compare the magnitudes and subtract the smaller from the larger,
choose the sign of the result to be same as A if A>B, or complement the sign of A if A<B. If
A=B , then subtract B from A make the result sign +.
- Hardware Implementation: 2 registers(Ra,Rb) to hold 2 operands, 2 flip-flops(As, Bs) to store
the signs, parallel adder(A+B), a comparator circuit to compare (A>B, B>A, A=B) , 2 parallel
subtractor circuits (A-B,B-A). Sign can be formed from an X-OR gate with As,Bs as inputs.
This can be implemented as:
- Subtraction can be done as A+2’s comp of B.
- E flip-flop determines the relative magnitude of A,B
- AVF holds the overflow bit when A,B are added.
- M –signal=0,the B is transferred to adder, input carry is 0, so the sum=A+B.
- M=1, 1’s complement of B(complementer)+A+1(input carry)= A-B
Isha Padhy,CSE Dept, CBIT

37. Addition and subtraction with signed-
magnitude
Isha Padhy,CSE Dept, CBIT

38. Multiplication Algorithms
• Multiplication: If the multiplier is a 1, then copy the multiplicand ,if 0 then
all the bits are 0. the numbers copied down in successive lines are shifted to
left by 1 position from the previous number, finally all the numbers are
added.
• Steps:
1. Initially the multiplicand is in B, multiplier in Q with their signs in Bs,Qs.
2. Registers A, E are cleared(A<-0,E<-0), sequence counter= number equal
to number of bits of the multiplier(SC<-n-1). We assume n-bit
word(operand) is transferred from memory, where magnitude=n-1 bits, 1
bit for sign.
3. Low order bit of Qn is tested, if 1, the multiplicand is added to A(0
initially).if 0 then nothing is done. Register EAQ is shifted right.
Sequence counter is decremented by 1 and new value checked, if not 0
then process is repeated.
4. The partial product formed in A is shifted into Q slowly and ultimately
the low order bits are in Q and MSB bits of the result in A.
Isha Padhy,CSE Dept, CBIT

39. Isha Padhy,CSE Dept, CBIT
10111
*10011
10111
10111
00000
.
.
10111
110110101

40. Isha Padhy,CSE Dept, CBIT

41. Booth’s Algorithm
Isha Padhy,CSE Dept, CBIT
-Algorithm is used for
multiplying binary
integers in signed 2’s
compl.
- mp:2p
- multiply multiplicand
M by 14(23+1-
21)[multiplier]
- Algorithm also
requires checking
multiplier bits and
shifting of partial
product, but prior to
shifting following
steps may be done:

42. Isha Padhy,CSE Dept, CBIT

43. Ex of booth algorithm
• 6*2 ,6:0110 2:0010
6 in BR(multiplicand register)
2 in QR(multiplier register)
Isha Padhy,CSE Dept, CBIT
Q0 Qn+1 Action
0 0 RS
1 1 RS
1 0 A<-A-BR
0 1 A<-A+BR
BR A QR Qn+1 SC
0110 0000 0010 0 100
00 shift 0000 0001 0 011
10 subtract
shift
0000-0110
1010
1101
0001
0000
0
1
010
01 Add,
shift
1101+0110
0011
0001
0000
1000
1
0
001
00 shift 0000 1100 0 000
Ex. -9(10111)*-
13(10011)

44. 2 bit by 2 bit array multiplier
Isha Padhy,CSE Dept, CBIT
- Why AND gate: a0=1 b0=1 a0b0
will be 1 only if a0*b0=1, otherwise
0.
- For j multiplier bits, k multiplicand
bits, we need j*k AND gates and (j-
1) k bits adders to produce product
of (j+k) bits.

45. 4 bit by 3 bit array multiplier
Isha Padhy,CSE Dept, CBIT

46. Division
• 1000)1001010(0001
001
1000
10
101
1010
- 1000
10
Isha Padhy,CSE Dept, CBIT
-Shift divisor right and compare it with
current dividend
-If divisor is larger, shift 0 as the next bit of
the quotient
-If divisor is smaller, subtract to get new
dividend and shift 1 as the next bit of the
quotient

47. Division Algorithm
Isha Padhy,CSE Dept, CBIT
Steps:
- Divisor is in B, double length dividend
is stored in A and Q.
-Dividend is shifted to left and divisor
is subtracted by adding 2’s
complement value.
- E gives the relative value of B and A.
-If E=1(A>=B), I is inserted in Qn, shift
left, subtract B(2’s comp of B+1)
-If E=0, Qn=0, add B to restore the
value of B, shift left + subtract B.

48. Divide Overflow
• Condition for overflow:
1. High order half bits of dividend is greater than or equal to divisor.
2. Division by 0.
• Overflow condition can be checked by a flip flop called DVF.
• Handling of overflow:
- Check the overflow condition after every divide operation and if
DVF is set then there will be branch to a subroutine which will
take measures to rescale the data.
- Divide stop: stop the operation of computer.
- Interrupt request is sent .
- Stop the program and send to user an error message explaining
the reason why program is stopped.
Isha Padhy,CSE Dept, CBIT

49. Isha Padhy,CSE Dept, CBIT

50. Explanation of flow-chart
• Dividend is kept in AQ, divisor in B
• Sign bit of A=As, B=Bs, As XOR Bs= Qs(Both A and B are of same
magnitude or not, If yes Qs=0,or Qs=1)
• EA:A+B’+1 //A-B to check whether A>B(Dividend>divisor)
• E=1, DVF<-1, Dividend > divisor, so hardware will not allow, DVF
condition prevails.
E=0, A<B, DVF<-0, so we move on to next step.
For both above steps, B is added(A+B) to get back the value of
A(previous step it was deducted(A-B) to check whether A>B)
Step 1. E=0: 1. shl EAQ
• E=0, EA<-A+B’+1(ADD B 2’s complement)
• E=1, A<-A+B’+1
• E=1, Qn=1
• E=0, EA<- A+B
Isha Padhy,CSE Dept, CBIT
SC<-SC+1
Continue the process from step-1

51. Floating –point arithmetic operations
• Floating point number register consists of 2 parts:
mantissa m and exponent e --- m*re
• A floating point number that has a 0 in the MSB
of the mantissa is said to have an underflow. To
normalize the number it is necessary to shift the
mantissa to the left and decrement the exponent
until a nonzero digit appears in the 1st position.
Ex .00345*105= .345*103
• Register configuration: same as for fixed point
operation. The difference lies in the way
exponents are handled.
Isha Padhy,CSE Dept, CBIT

52. Registers for floating point arithmetic
operations
• 3 registers :BR,AC,QR
• All have 2 parts: mantissa(upper case
letter), exponent(lower case letter)
• Assumed that each floating point
number has a mantissa in signed
magnitude and a biased exponent(bias is
no. that is added to exponent. Ex exp
that ranges from -50 to 49, we consider
00 to 99 as +ve number. Positive
exponents range from 99 to 50 and 49 to
00 as negative exponent)
• Comparator: used to compare the +ve
exponents to adjust mantissa part for
normalization.
• As:sign bit of A, Qs: sign bit of Q, Bs: sign
bit of B., A1: 1 if the number is to
normalized.
• 2 parallel adder, A: mantissa Q+B, E:carry
bit
a: exponent b+q
Isha Padhy,CSE Dept, CBIT

53. Addition and subtraction
Isha Padhy,CSE Dept, CBIT

54. Multiplication of floating point numbers
Isha Padhy,CSE Dept, CBIT

55. Division of floating point numbers
Isha Padhy,CSE Dept, CBIT

56. BCD ADD
- To add 2 BCD decimal numbers, max value for a operand allowed is 9. So 2
nos addition will be 9+9+1(carry)=19.The adder will form addition in form
of binary and result will be within 19.
- The binary representation should be converted to BCD form, we see till
1001 the corresponding BCD is equal but after that if we add binary
6(0110) to the binary sum converts correctly to BCD form.
- The ckt detects the necessry condition to know which binary nos. are to
converted to BCD. Correction is required when
1. K bit =1
2. For 1010 to 1111, k8 bit =1
3. K8 bit of 1001 and 1000 is also 1, so k8 with K4 or K2 bit =1
- We can get the boolean function
C= K+Z8Z4+Z8Z2
- When C=1, add 0110 to binary sum and provide carry to next stage.
Isha Padhy,CSE Dept, CBIT

57. Isha Padhy,CSE Dept, CBIT

58. Isha Padhy,CSE Dept, CBIT