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24 CHAPTER 2 / COMPUTER EVOLUTION AND PERFORMANCE

Table 2.1 The IAS Instruction Set

Instruction Symbolic
Type Opcode Representation Description
00001010 LOAD MQ Transfer contents of register MQ to the accumulator AC
00001001 LOAD MQ,M(X) Transfer contents of memory location X to MQ
00100001 STOR M(X) Transfer contents of accumulator to memory location X
Data transfer 00000001 LOAD M(X) Transfer M(X) to the accumulator
00000010 LOAD M(X)- Transfer M(X) to the accumulator-
00000011 LOAD |M(X)| Transfer absolute value of M(X) to the accumulator
00000100 LOAD |M(X)|- Transfer |M(X)| to the accumulator-
Unconditional 00001101 JUMP M(X,0:19) Take next instruction from left half of M(X)
branch 00001110 JUMP M(X,20:39) Take next instruction from right half of M(X)
00001111 JUMP+ M(X,0:19) If number in the accumulator is nonnegative, take next in-
Conditional struction from left half of M(X)
branch 00010000 JUMP+ M(X,20:39) If number in the accumulator is nonnegative, take next
instruction from right half of M(X)
00000101 ADD M(X) Add M(X) to AC; put the result in AC
00000111 ADD |M(X)| Add |M(X)| to AC; put the result in AC
00000110 SUB M(X) Subtract M(X) from AC; put the result in AC
00001000 SUB |M(X)| Subtract |M(X)| from AC; put the remainder in AC
Arithmetic 00001011 MUL M(X) Multiply M(X) by MQ; put most significant bits of result
in AC, put least significant bits in MQ
00001100 DIV M(X) Divide AC by M(X); put the quotient in MQ and the
remainder in AC
00010100 LSH Multiply accumulator by 2; i.e., shift left one bit position
00010101 RSH Divide accumulator by 2; i.e., shift right one position
00010010 STOR M(X,8:19) Replace left address field at M(X) by 12 rightmost bits
Address of AC
modify 00010011 STOR M(X,28:39) Replace right address field at M(X) by 12 rightmost
bits of AC

Table 2.1 presents instructions in a symbolic, easy-to-read form.Actually, each
instruction must conform to the format of Figure 2.2b. The opcode portion (first
8 bits) specifies which of the 21 instructions is to be executed. The address portion
(remaining 12 bits) specifies which of the 1000 memory locations is to be involved in
the execution of the instruction.
Figure 2.4 shows several examples of instruction execution by the control unit.
Note that each operation requires several steps. Some of these are quite elaborate.
The multiplication operation requires 39 suboperations, one for each bit position ex-
cept that of the sign bit.

COMMERCIAL COMPUTERS The 1950s saw the birth of the computer industry with
two companies, Sperry and IBM, dominating the marketplace.