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CS501 – Advance Computer Architecture

In CS501 Advance Computer Architecture we have you covered with Digitized Past Papers From Fall of Mid Term and Final Term.

NOTE: Tab/Click on Preparation Tab to take the MCQ’s Tests.

FINAL TERM
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CS501_Final_Subj_Ref_Moaaz                          View     Download

MID TERM

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CS501_All_Current_Subjec                              View     Download
CS501_Mid_Subj_Mega_File                              View     Download
POSTED DATE:30-01-2019

Solution Of Question No.01:
In one second transfer rate = 9600 bytes
Equelant to 9600/4 = 2400 DMA cycles as 4 bytes are transferred in every cycle
Earlier in 1 second 2000000 instructions were executed.
Now in 1 second (2000000-2400) = 1997600 instructions are executed

Solution Of Question No.02:

8086 Interrupts:

CategorySub_Category
Non-MaskableInvalid opcode
MouseFile Deletion
BreakpointArithmetic overflows
Stack OverflowTrap
Divide by zero
Mode Switching

CS501– Practice Quiz 1

CS501– Practice Quiz 2

CS501– Practice Quiz 3

2 Comments

  1. AYESHA MUMTAZ

    1. Define PROM? (2 Marks)
    Answer:- (Page 356)
    The PROM stands for Programmable Read only Memory. It is also nonvolatile and may be written into only
    once. For PROM, the writing process is performed electrically in the field. PROMs provide flexibility and
    convenience.
    2. How we refer the register to the RTL? Give an example? (2 Marks)
    Answer:- (Page 66)
    Specifying Registers
    The format used to specify registers is
    Register Name
    For example, IR means bits numbered 31 to 0 of a 32-bit register named “IR”
    (Instruction Register).
    3. What is the use of modem? (2 Marks)
    Answer:- (Page 391)
    To interconnect different computers by using twisted pair copper wire, an interface is used which is called
    modem. Modem stands for modulation/demodulation. Modems are very useful to utilize the telephone network
    (i.e. 4 KHz bandwidth) for data and voice transmission.
    4. What is the advantage of RAID level 0? (2 Marks)
    Answer:- (Page 330)
    • The user and system data are distributed across all the disks in the array.
    • Notable advantage over the use of a single large disk.

  2. AYESHA MUMTAZ

    Q1-Write two lines on connection oriented communication? 2
    Answer:- (Page 394)
    In this method, same path is always taken for the transfer of messages.
    It reserves the bandwidth until the transfer is complete. So no other server could use that path until it becomes
    free.
    Q2-What is the advantage of direct cache memory? 2
    Answer:- (Page 361)
    Advantage is Simplicity.
    4
    Q3-Write the drawbacks of DMA? 2
    Answer:- (Page 315)
    The disadvantage however, would be that an additional DMA controller would be required, that could make
    the system a bit more complex and expensive. Generally, the DMA requests have priority over all other bus
    activities including interrupts. No interrupts may be recognized during a DMA cycle.
    Q4-What is the difference between selection channel and multiple channels? 3
    Answer:- (Page 320)
    Selector Channel
    It is the DMA controller that can do block transfers for several devices but only one at a time.
    Multiplexer Channel
    It is the DMA controller that can do block transfers for several devices at once.
    Q5-What is the advantage of linker in assembly language program? 3
    Answer:- (Page 26)
    When developing large programs, different people working at the same time can develop separate modules of functionality. These modules can then be „linked‟ to form a single module that can be loaded and executed. The modularity of programs, that the linking step in assembly language makes possible, provides the same convenience as it does in higher-level languages; namely abstraction and separation of concerns. Once the functionality of a module has been verified for correctness, it can be re-used in any number of other modules.
    The programmer can focus on other parts of the program. This is the so-called “modular” approach, or the “topdown” approach.

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