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CS620 Assignment 1 Solution and Discussion

CS620 - Modelling and Simulation

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Re: CS620 Assignment 1 Solution and Discussion

Assignment No. 1
Semester: Spring 2020
Total Marks: 10 Due Data: May 30, 2020
Instructions:
Make sure that you upload the solution file before due date. No assignment will be acceptable through e-mail after the due date.
Formatting Guidelines
• Use the font style “Times New Roman” and font size “12”.
• Compose your document in MS-Word, any file created in any other format will not be accepted and marked zero.
• Use black and blue font colors only.
Solution Guidelines
• To solve this assignment, you should have good command over Topics.
• This is not a group assignment, it is an individual assignment so be careful and avoid copying others’ work
• Give the answer according to question only and avoid irrelevant details.
Please note that your assignment will not be graded if:
• It is submitted after due date
• The file you uploaded does not open
• The file you uploaded is copied from someone else
• It is in some other format instead of .doc
• Cheating or copying of assignment from internet, ppt lessons or from each other is strictly prohibited. The cheated or copied assignment will be marked ‘Zero’.
• Answers should be précised and clear with no useless material.

Assignment Topic
Q1. How modeling and simulation help in the construction of biological systems? What are the applications of biological modeling and simulation? (5)
Q2: Find at least 5 simulation tools for biological data? Briefly explain about their working and figure out which biological data these tool process, what are their outputs and what parameters and features they required to process the data? (5)
For any query about the assignment, contact at [email protected]
Note: Submit your solution through your LMS account. Solutions submitted through email will not be acceptable.
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The monopolist has no supply curve because:
a406ee45-d2a9-4aa8-af1a-2da4ca41e3af-image.png
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ANSWER’S:
His prophet hood is universal:

EXPLANATION:
Hazrat Muhammad (PBUH) is the last Prophet means that his Prophet Hood is universal. He was the best example for the the world .His Prophet hood is sufficient for all humanity up to the end of this world. He is not the prophet of his age but he is the Prophet of every age.

 “In God’s messenger you have indeed a good example for everyone who looks forward with hope to God and the Last Day …”
 "Say: Obey Allah and obey the Messenger, … If you obey him, you shall be on the right guidance”
He is seal of prophet hood:
EXPLANATION:
Hazrat Muhammad (s.a.w) is the last prophet of Allah Almighty, Quran is last truth book. The Holy Prophet is seal of Prophet Hood .Allah called the Holy Prophet is Khata-mun-nabyeen. No one Prophet will come after the Prophet (s.a.w).
 “Muhammad is not the father of any of your men,1 but is the Messenger of Allah and the seal of the prophets”
Deen is completed upon him:
EXPLANATION:
As you know that the Holy Prophet (s.a.w) is the last divine Prophet of Allah .No one is comes after the Mustafa (s.a.w).
Deen is completed upon him. If we want to be a successful person in our both lives (In the world and at the Judgment day)we ought to obey the Holy Prophet(s.a.w)’s firmaans and AHADEES .Because of the deen –e- islam is the Deen of Allah and Mustafa (s.a.w).

 “SAY, ALLAH IS ONE /ALONE AND MUHAMMAD (S.A.W) IS THE LAST DIVINE PROPHET OF ALLAH ALLMIGHTY”

It is known by necessity that the Deen of Islam is complete, as Allah, the Exalted, says:
 “This day, I have perfected your Deen for you, completed My Favor upon you, and have chosen for you Islam as your Deen.” (Sarah Al-Ma’idah,) 5:3
A position of praise and glory will be granted to him on the Day of Resurrection:
EXPLANATION:

 ON the day of resurrection a position of praise and glory will be granted to Mustafa (s.a.w).
 Allah creates the world just for the praise of Mustafa (s.a.w),
 On the day of Judgement HE will be the cause of intercession.

MUHAMMAD (S.A.W) SAYS:
 “The banner of praise will be in my hand on the Day of Resurrection”,

 All of creation will praise Muhammad for that status on the Day of Judgment! … so that He will start to judge between His slaves, and no one will be granted this except Muhammad (peace and blessings of Allah be upon him.

 “This is what the Beneficent (Allah) had promised, and the Messengers did speak the truth.”
.
 “And that the Hour (of Resurrection) is coming, there is no doubt therein; and that Allah will raise up those in the graves.” (Qur’an, 22:1-7
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@zareen said in CS620 Assignment 1 Solution and Discussion:

Question 2: How can do make a discrete event simulation in excel? 15.

I. Introduction to Discrete-Event System Simulation. Chapter 1 Introduction to Simulation … Simulation models designed for training make learning possible without the … The second rule says that simulation should not be used if the problem can … 2. 9. 1 1. I. 15. 15. 4. Table 2.5 Chronological Ordering of Events. Event Type.

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Solution # 1
Instruction Processor Hexadecimal Code Behavioral RTL JPL R5, [26] Falcon - A 10000 101 00011010, 1000 0101 0001 1010, 851A (R[5] ≥ 0): PC ← PC+ (26-PC); STS R7, R2 (100) Falcon - E 00101 111 010, 000000000000001100100, 0010 1111 0100 0000, 0000 0000 0110 0100, 2F40 0064 M[R[2]+100] ← R [7] STACC R4, 36 Modified EAGLE 10111 100 00100100, 1011 1100 0010 0100, BC24 M[R[4] + ([email protected]<7>)©C] ← ACC C represents the constant 36 DIV R2 EAGLE 10000 010, 1000 0010, 82 R[0] ← R[0]/R[2] R[2] ← R[0]%R[2] SHIFTL R5, R2, 7 FALCON - A 01100 101 010 00111, 0110 0101 0100 0111, 6547 R[5]<15…0> ← R [2]<(15- N)…0>©([email protected]) N represents constant 7
Solution # 2
Instruction Value of Destination Operand Data Bus <15…0> Address Bus (15…0> LDACC B 55CEh 55CEh 3320h SUB R1 55B9h N/A N/A LDACC C 2015h 2015h AB0Eh ADD R2 45E0h N/A N/A STACC A 45E0h 45E0h AB10h
Table 3 Data Bus and Address Bus Contents for Modified Eagle

Calculation Steps (Instruction-By-Instruction)

• LDACC B
LDACC stands for load accumulator. In LDACC, the destination operand is accumulator and source operand is the memory location labelled as B. The memory label B points to the memory address 3320h. When this instruction is executed, the value stored at memory address 3320h will be read and loaded in Accumulator register. This address 3320h will be copied into Address Bus which will then read its contents from memory and load the contents at data bus. The operand size in Modified EAGLE is 2-byte. Therefore, the values stored at addresses 3320h and 3321h will be loaded at data bus. These values are CEh and 55h respectively. Since, Modified EAGLE employs Little endian notation hence, the 2-bye value will be read as 55CEh. The value of data bus will also be 55CEh and this will be loaded into destination operand Accumulator.
• SUB R1
SUB R1 means to subtract the value of source operand register R1 which is 0015h, from the destination operand Accumulator (ACC) register which contains 55CEh. The result of subtraction will be stored back into ACC. After subtraction, the value stored in ACC will be 55B9h. Because SUB is not a memory instruction, we are not concerned with the contents of Data Bus or Address Bus because the values we need to execute the instruction are already available in registers. Hence, Data Bus and Address Bus values will be labelled as N/A.
• LDACC C
When this instruction is executed, the value stored at memory address labelled with C is read and loaded in Accumulator register ACC. In this case, the address of the C is AB0Eh which is also the value of address bus. The operands in Modified EAGLE are 2-byte values. The contents at addresses AB0Eh and AB0Fh will be copied into data bus which will then be loaded into ACC register. These contents are 15h and 20h. Due to Little endian notation, the 2-bye value will be 2015h. The value of data bus will be 2015h and same will be loaded in destination register ACC.
• ADDR2
When ADD is executed, the value of register R2 is added to ACC register. Hence, after the execution, the ACC register will hold 45E0h. AS usual, ADD is not a memory instruction, so we are not concerned with the values of Data Bus and Address Bus and both are labelled as N/A.
• STACC A
STACC stands for Store ACC. There is one destination operand which is a memory label A. When the instruction is executed, the value of Accumulator register ACC is stored at the memory address labelled by A. The destination memory address will be AB10h. The value of ACC is 45E0h will be stored as address AB10h. However, due to Little-Endian notation, the address will be stored as E0h at memory location AB10h and then 45h at memory location AB11h.

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CS620 Assignment 1 Solution and Discussion

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