Question

B.31 [10] <§B.6> Instead of thinking of an adder as a device that adds: two numbers and then links the carries together, we can think of the adder as a hardware device that can add three inputs together (ai, bi, ci) and produce two outputs (s, ci + 1). When adding two numbers together, there is little we can do with this observation. When we are adding more than two operands, it is possible to reduce the cost of the carry. The idea is to form two independent sums, called S (sum bits) and C (carry bits). At the end of the process, we need to add C and S together using a normal adder. This technique of delaying carry propagation until the end of a sum of numbers is called carry save addition. The block drawing on the lower right of Figure B.14.1 (see below) shows the organization, with two levels of carry save adders connected by a single normal adder. Calculate the delays to add four 16-bit numbers using full carry-lookahead adders versus carry save with a carry-lookahead adder forming the final sum. (The time unit T in Exercise B.28 is the same)

Short answer

The total required time is 5T.

Step-by-step solution

Define the concept.

This total time is required for every operand by not considering the bit.

5T time is taken from the path of consisting the inputs of “S1”, “S2”, “S3”.

For generating all of the signals of “p” and “g”, only a single cycle is taken.

For generating the “C3” to “C0” additional two cycles are required.

And for generating the “S3” to “S0” additional two cycles are required.

The “T” is referred to as the time unit.

Determine the calculation.

The adder:

In the Ripple carry adder one bit requires 2T time.

In this Carry Look Ahead Adder,

Action	Required time
To generate Piand Gi.	1T
To carry Ci	2T
To generate Sums Si.	2T

Hence,

Required time	Signals of adder
T=0	“A3”…”A0” ,“B3”...”B0”, C0
T=1	“G3”…”G0”
T=2	“S0”
T=3	“C1”,”C2”, “C3”, “C4”
T=5	“S1” ,”S2”,”S3”

Hence total required time is 5T.