Consider the following pseudo assembly code for computing c $=$ a $+$ b $.$ Assume that a$,$ b$,$ and c are assigned to consecutive memory "words" $($memory is generally addressed byte by byte and assume that a word is $4$ bytes$)$ and address for $"$a$"$ is $0$x$0000$ec$00,$ Also, we have a $=24$ b $=110$ and c $=10$ at the starting time. Assume that the first instruction of the code is stored in $0$x$0000$b$138.$ Also, each instruction has the opcode in the first byte $($most significant byte$)$ and the remaining $3$ bytes specify the corresponding address. The opcode for store is $5,$ load is $4,$ and add is $6.$

$0$x$0000$b$138$ load a

$0$x$0000$b$13$c add b

$13810$x$0000$b$140$

store c

a$.$ Show the memory addresses and contents for all the instructions and data involved. HR$-$Use the format as follows to express your answer $($but the following is not the answer$).$

For all data, use hexadecimal representatio

Addresses

contents

$0$x$00002104$

$0$x$00000001$

$0$x$00002108$

$0$x$00000002$

$44$ b$.$ Write the micro instructions for the code segment. Assume that current PC $($program counter$)$ is $0$x$0000$b$138.$ For each micro$-$instruction, also indicate the data is transferred $($if it is a memory access$).$ For all data, use the hexadecimal representation. The following are the first two micro$-$instructions and the data transferred. Complete the rest. $[2+2+2=6$ Marks$]$

$(80$s that

$1381$

Micro$-$instructions

Data

PC$-$MAR

$0$x$0000$b$128$

HRM$-$MBR

$0$x$0200$ec$00$