The difference is that the program below is somewhat simpler as it uses only 2 branches (BRZ 09 and BR 01), whereas the program in the lecture notes uses 3 branches (BRP 05, BR 10, and BR 01). First, try to understand each instruction thoroughly and then trace the execution of each instruction.

Address Instruction

(Mnemonics)

00 IN

01 STO 25

02 BRZ 09

03 IN

04 ADD 27

05 STO 27

06 LDA 25

07 SUB 26

08 BR 01

09 LDA 27

10 OUT

11 HLT

Address Contents

25 DAT ? 2 1 0 (decremented by 1 each time the loop is executed)

26 DAT 1

27 DAT 0

Assume now that the above program will only read 3 numbers. That is, the following numbers in this order will be placed, one at a time, in the In-basket: 2, 35, and 15, where 2 is the count of numbers that follow, and 20 and 16 are the numbers that are to be added. The first column in the table on page 3 shows the order in which the instructions from the program will be executed. Trace the execution of these instructions and determine the contents of the PC before and after each instruction is executed. Also, write down in the table the contents of the In-basket; Accumulator; Memory locations 25, 26, and 27; and Out-basket after each instruction is executed. Memory location 25 controls the loop. It initially contains an unknown value (?), then 2, next 1, and finally 0. Memory location 26 always contains 1. It is used to decrease the loop count by 1. Memory location 27 is initialized with 0, and finally it stores 50, the sum of 35 and 15. The entry 00 01 in the PC column means that the PC is 00 when the instruction IN started and is changed to 01 when the instruction IN is finished.

The sequence in which instructions are executed	PC before after	In-basket	Accumulator	Memory location 25	Memory location 26	Memory location 27	Out-basket
IN	00 01	2	2	?	1	0	?
STO 25
BRZ 09
IN
ADD 27
STO 27
LDA 25
SUB 26
BR 01
STO 25
BRZ 09
IN
ADD 27
STO 27
LDA 25
SUB 26
BR 01
STO 25
BRZ 09
LDA 27
OUT
HLT