very easy visualization methods projectPLEASE MAKE SURE TO FOLLOW THE INSTRUCTIONS TO THE LETTER, AND PLEASE WRITE IN AN INTERNATIONAL STUDENT STYLE OF WRITING.these are the guidlines:AT6 is a single, extended synoptic, summative assessment delivered over the course of three weeks, in weeks 11, 12 and 13. Portfolio Project assess your visualization methods learning skills from all TLAs and ATs starting with week 1 through week 10, and to put that into action, which demonstrates your summative work towards meeting this course four intended learning outcomes. You will utilize their visualization methods knowledge and skills they developed in this course to use applicable visualization methods software applications and web-based research skills to develop their project and not limited to MS PowerPoint and MS Word documents. The product of this AT from course topics, problem solving, and the use of software applications are aligned with this course four CILOs. You will also create an ePortfolio in Blackboard. This is the step you ONLY DO ONCE. You can add more materials to it later on.TaskFor this task, you will answer questions through a Blackboard Assignment. The output is a report, submitted through blackboard, that describes the entirety of the your work include the formulation of a question, data planning and collection, processing, analysis, interpretation, data visualization using multiple software applications and drawing of appropriate conclusions. You will submit individual portfolio projects as files that get uploaded into blackboard.Process1. To start the assignment, click on this link: Portfolio Project2. Attach your assignment3. When done, click submit4. Read Creating and Building your ePortfolio in Blackboard guide.5. Create an ePortfolio in Blackboard. This is the step you ONLY DO ONCE.Estimated Time on task: 22 HoursAssessmentThe Portfolio project is to be marked using the CILO 1, CILO 2, CILO 3 and CILO 4 rubrics. All CILO criteria apply, no criteria other than these may be imposed. All rubric criteria are weighted equally. This is a summative assessment. No feedback on AT6 may be given to you before final submission for marking.Deadline11pm 3 December 2020Visualization Methods Portfolio Project.pdf.doc file

Adasd Instructions with hexadecimal code(Machine code):First assign address to each instructions startinf from 200(Given as starting address 200)
To generate the machine code(hex), get the opcode for the instruction from the manual. For example, the opcode for LOAD instruction is 1.
Then check the operand/label associated with the instruction. For example, LOAD Base, Base is the label associated with the instruction.
Get the address of the label associated. For example, get the address of the Base label(the address assigned for the label at first step), ie 209
Now by follow the instruction format of the instruction and form machine code as shown below:

Opcode
Address/Condition

1
209

ie machine code for the LOAD Base instruction is 1209

Address
Label
Instructions
Machine Code

200
Begin,
LOAD Base
1209

201
 
ADD Offs
3209

202
Loop,
SUBT One
420B

203
 
STORE Addr
220C

204
 
SKIPCOND 800
8800

205
 
JUMP Done
9208

206
 
JUMPI Addr
C20C

207
 
CLEAR
A000

208
Done,
HALT
7000

209
Base,
HEX 200
0200

20A
Offs,
Dec-09
0009

20B
One,
HEX 001
0001

20C
Addr,
HEX 000
0000

Symbol Table:It contains all the symbols defined in the program and the corresponding address. In the given program, 7 labels are used and the corresponding addresses are given below and is called symbol table which is generated during the assembly operation of the code.

Symbol
Address

Addr
20C

Base
209

Begin
200

Done
208

Loop
202

Offs
20A

One
20B

Accumulator always have a positive value and hence the SKIPCOND 800 goes to JUMPI instruction and this will enter into an infinite loop and hence the value of the Accumulator cannot be determined.  

 Code with hexadecimal equivalent:

Address
Label
Instructions
Machine Code

100
If,
LOAD X
110C

101
 
SUBT Y
410D

102
 
SKIPCOND 400
8400

103
 
JUMP Else
9108

104
Then,
LOAD X
110C

105
 
ADD X
310C

106
 
STORE X
210C

107
 
JUMP Endif
910B

108
Else,
LOAD X
110C

109
 
SUBT X
410C

10A
 
STORE Y
210D

10B
Endif,
HALT
7000

10C
X,
DEC 12
000C

10D
Y,
DEC 20
0014

Symbol
Address

If
100

Else
108

Endif
10B

Then
104

X
10C

Y
10D

Accumulator content is 0000 since the instructions at line 108 and 109, load the value X into the register A and subtract the same number from it.  

If,                            LOAD X                                 /Load X                                SUBT five                             /Subtract 5, Compare X to 5 store in AC                                SKIPCOND 800                   /if AC > 0 (X > 5), skip next instruction                                JUMP Endif                         /jump to Endif if X is not grater than 5Then,                    LOAD X                                 /reload X so it can be doubled                                ADD X                                   /Double X                                ADD X                                   /Tripe X                                STORE Y                                /Y = X + X + X                                CLEAR                                   /move 0 into AC                                STORE X                               /set X to 0Endif                      LOAD Y                                                 /load Y into AC                                SUBT one                            / subtract 1 from Y                                STORE Y                                /Y = Y – 1                                HALT                                      /terminate programX,                            DEC ?Y,                            DEC ?one,                       DEC 1                                     /use as constantfive,                       DEC 5                                     /use as constant

ORG 100LOAD oneSTORE X                                                               /initializing XLoop,                     LOAD X                                 /Loading Loop constant                                SUBT TEN                            /comparing x to 10                                SKIPCOND 000                   /if x < 10 continue loop                                JUMP EndLoop                 /if x not less thank 10 go to EndLoop                                LOAD X                                 / begin inside loop statement                                ADD one                              /ADD 1 to X for X = X + 1                                STORE X                               /store new value in X                                JUMP LoopEndLoop,             HALT                                      /terminate loop and programX,                            DEC 0                                     /Declaring X storageone,                       DEC 1                                     /constant as 1ten,                        DEC 10                                  / loop constant

INPUT                                      /Read a number n. where n stores into ACJNS FindSum                           /Call FindSum by passing nOUTPUT                                  /Print the return value(sum)HALT                                       /Exit the programFindSum, HEX 000                  /Store the return addressSTORE n                                 /Save the passed value into nLOOP, SKIPCOND 800           /if n is 0, go to ReturnJUMP Return   /Do sum=sum+nLOAD sum                               /AC=sumADD n                                      /AC=sum+nSTORE sum                             /sum=AC/DO n=n-1LOAD n                                    /AC=nSUBT ONE                              /AC=n-1STORE n                                 /n=ACJUMP LOOP                            /Go to LOOPReturn, LOAD sum                  /Send sum as return valueJUMPI FindSum                       /Return from FindSum/Declare necessary variablesn, DEC 0sum, DEC 0ONE, DEC 1
MARIE’s STORE:MAR <= PCIR <= M[MAR]MAR <= XMBR <= ACM[MAR] <= MBR

1
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0

0
1
0
0
0
0
0
1
1
1
0
0
0
1
0
0
0
0

0
0
1
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0

0
0
0
1
0
0
0
0
1
1
1
0
0
0
0
0
0
0

0
0
0
0
1
0
0
0
0
0
0
1
1
0
1
0
0
1

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