Category: Computer Organization and Architecture

Memory Hierarchy | Memory Hierarchy Diagram

Memory Hierarchy-

 

  • Memory hierarchy is the hierarchy of memory and storage devices found in a computer system.
  • It ranges from the slowest but high capacity auxiliary memory to the fastest but low capacity cache memory.

 

Need-

 

There is a trade-off among the three key characteristics of memory namely-

  • Cost
  • Capacity
  • Access time

Memory hierarchy is employed to balance this trade-off.

 

Memory Hierarchy Diagram-

 

 

Level-0:

 

  • At level-0, registers are present which are contained inside the CPU.
  • Since they are present inside the CPU, they have least access time.
  • They are most expensive and therefore smallest in size (in KB).
  • Registers are implemented using Flip-Flops.

 

Level-1:

 

  • At level-1, Cache Memory is present.
  • It stores the segments of program that are frequently accessed by the processor.
  • It is expensive and therefore smaller in size (in MB).
  • Cache memory is implemented using static RAM.

 

Level-2:

 

  • At level-2, main memory is present.
  • It can communicate directly with the CPU and with auxiliary memory devices through an I/O processor.
  • It is less expensive than cache memory and therefore larger in size (in few GB).
  • Main memory is implemented using dynamic RAM.

 

Level-3:

 

  • At level-3, secondary storage devices like Magnetic Disk are present.
  • They are used as back up storage.
  • They are cheaper than main memory and therefore much larger in size (in few TB).

 

Level-4:

 

  • At level-4, tertiary storage devices like magnetic tape are present.
  • They are used to store removable files.
  • They are cheapest and largest in size (1-20 TB).

 

Observations-

 

The following observations can be made when going down in the memory hierarchy-

  • Cost / bit decreases
  • Frequency of access decreases
  • Capacity increases
  • Access time increases

 

Goals of Memory Hierarchy-

 

The goals of memory hierarchy are-

  • To obtain the highest possible average access speed
  • To minimize the total cost of the entire memory system

 

To gain better understanding about Memory Hierarchy-

Watch this Video Lecture

 

Next Article- Memory Organization | Simultaneous Vs Hierarchical

 

Get more notes and other study material of Computer Organization and Architecture.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Addressing Modes | Practice Problems | COA

Addressing Modes in Computer Architecture-

 

Before you go through this article, make sure that you have gone through the previous article on Addressing Modes.

 

We have discussed-

  • Addressing modes are the different ways of specifying the location of an operand.
  • Various kinds of addressing modes and their applications.

 

 

Also Read- Syntax Of Addressing Modes

 

In this article, we will discuss practice problems based on addressing modes.

 

PRACTICE PROBLEMS BASED ON ADDRESSING MODES-

 

Problem-01:

 

The most appropriate matching for the following pairs is-

 

Column-1:

X: Indirect addressing

Y: Immediate addressing

Z: Auto decrement addressing

 

Column-2:

1. Loops

2. Pointers

3. Constants

 

  1. X-3, Y-2, Z-1
  2. X-1, Y-3, Z-2
  3. X-2, Y-3, Z-1
  4. X-3, Y-1, Z-2

 

Solution-

 

Option (C) is correct.

 

Problem-02:

 

In the absolute addressing mode,

  1. The operand is inside the instruction
  2. The address of the operand is inside the instruction
  3. The register containing the address of the operand is specified inside the instruction
  4. The location of the operand is implicit

 

Solution-

 

Option (B) is correct.

 

Problem-03:

 

Which of the following addressing modes are suitable for program relocation at run time?

  1. Absolute addressing
  2. Base addressing
  3. Relative addressing
  4. Indirect addressing

 

  1. 1 and 4
  2. 1 and 2
  3. 2 and 3
  4. 1, 2 and 4

 

Solution-

 

Option (C) is correct.

 

Problem-04:

 

What is the most appropriate match for the items in the first column with the items in the second column-

 

Column-1:

X: Indirect addressing

Y: Indexed addressing

Z: Base register addressing

 

Column-2:

1. Array implementation

2. Writing relocatable code

3. Passing array as parameter

 

  1. X-3, Y-1, Z-2
  2. X-2, Y-3, Z-1
  3. X-3, Y-2, Z-1
  4. X-1, Y-3, Z-2

 

Solution-

 

Option (A) is correct.

 

Problem-05:

 

Which of the following addressing modes permits relocation without any change whatsoever in the code?

  1. Indirect addressing
  2. Indexed addressing
  3. Base register addressing
  4. PC relative addressing

 

Solution-

 

Option (C) is correct.

 

Problem-06:

 

Consider a three word machine instruction-

ADD A[R0], @B

The first operand (destination) “A[R0]” uses indexed addressing mode with R0 as the index register. The second operand operand (source) “@B” uses indirect addressing mode. A and B are memory addresses residing at the second and the third words, respectively. The first word of the instruction specifies the opcode, the index register designation and the source and destination addressing modes. During execution of ADD instruction, the two operands are added and stored in the destination (first operand).

The number of memory cycles needed during the execution cycle of the instruction is-

  1. 3
  2. 4
  3. 5
  4. 6

 

Solution-

 

For the first operand,

  • It uses indexed addressing mode.
  • Thus, one memory cycle will be needed to fetch the operand.

 

For the second operand,

  • It uses indirect addressing mode.
  • Thus, two memory cycles will be needed to fetch the operand.

 

After fetching the two operands,

  • The operands will be added and result is stored back in the memory.
  • Thus, one memory cycle will be needed to store the result.

 

Total number of memory cycles needed

= 1 + 2 + 1

= 4

 

Thus, Option (B) is correct.

To watch video solution, click here.

 

Problem-07:

 

Consider a hypothetical processor with an instruction of type LW R1, 20(R2), which during execution reads a 32-bit word from memory and stores it in a 32-bit register R1. The effective address of the memory location is obtained by the addition of a constant 20 and the contents of register R2. Which of the following best reflects the addressing mode implemented by this instruction for operand in memory?

  1. Immediate Addressing
  2. Register Addressing
  3. Register Indirect Scaled Addressing
  4. Base Indexed Addressing

 

Solution-

 

Clearly, the instruction uses base indexed addressing mode.

Thus, Option (D) is correct.

 

Problem-08:

 

The memory locations 1000, 1001 and 1020 have data values 18, 1 and 16 respectively before the following program is executed.

 

MOVI Rs, 1 Move immediate
LOAD Rd, 1000(Rs) Load from memory
ADDI Rd, 1000 Add immediate
STOREI 0(Rd), 20 Store immediate

 

Which of the statements below is TRUE after the program is executed?

  1. Memory location 1000 has value 20
  2. Memory location 1020 has value 20
  3. Memory location 1021 has value 20
  4. Memory location 1001 has value 20

 

Solution-

 

Before the execution of program, the memory is-

 

 

Now, let us execute the program instructions one by one-

 

Instruction-01: MOVI Rs, 1

 

  • This instruction uses immediate addressing mode.
  • The instruction is interpreted as Rs ← 1.
  • Thus, value = 1 is moved to the register Rs.

 

Instruction-02: LOAD Rd, 1000(Rs)

 

  • This instruction uses displacement addressing mode.
  • The instruction is interpreted as Rd ← [1000 + [Rs]].
  • Value of the operand = [1000 + [Rs]] = [1000 + 1] = [1001] = 1.
  • Thus, value = 1 is moved to the register Rd.

 

Instruction-03: ADDI Rd, 1000

 

  • This instruction uses immediate addressing mode.
  • The instruction is interpreted as Rd ← [Rd] + 1000.
  • Value of the operand = [Rd] + 1000 = 1 + 1000 = 1001.
  • Thus, value = 1001 is moved to the register Rd.

 

Instruction-04: STOREI 0(Rd), 20

 

  • This instruction uses displacement addressing mode.
  • The instruction is interpreted as 0 + [Rd] ← 20.
  • Value of the destination address = 0 + [Rd] = 0 + 1001 = 1001.
  • Thus, value = 20 is moved to the memory location 1001.

 

Thus,

  • After the program execution is completed, memory location 1001 has value 20.
  • Option (D) is correct.

 

To watch video solution, click here.

 

Problem-09:

 

Consider the following memory values and a one-address machine with an accumulator, what values do the following instructions load into accumulator?

  • Word 20 contains 40
  • Word 30 contains 50
  • Word 40 contains 60
  • Word 50 contains 70

Instructions are-

  1. Load immediate 20
  2. Load direct 20
  3. Load indirect 20
  4. Load immediate 30
  5. Load direct 30
  6. Load indirect 30

 

Solution-

 

Instruction-01: Load immediate 20

 

  • This instruction uses immediate addressing mode.
  • The instruction is interpreted as Accumulator ← 20.
  • Thus, value 20 is loaded into the accumulator.

 

Instruction-02: Load direct 20

 

  • This instruction uses direct addressing mode.
  • The instruction is interpreted as Accumulator ← [20].
  • It is given that word 20 contains 40.
  • Thus, value 40 is loaded into the accumulator

 

Instruction-03: Load indirect 20

 

  • This instruction uses indirect addressing mode.
  • The instruction is interpreted as Accumulator ← [[20]].
  • It is given that word 20 contains 40 and word 40 contains 60.
  • Thus, value 60 is loaded into the accumulator.

 

Instruction-04: Load immediate 30

 

  • This instruction uses immediate addressing mode.
  • The instruction is interpreted as Accumulator ← 30.
  • Thus, value 30 is loaded into the accumulator.

 

Instruction-02: Load direct 30

 

  • This instruction uses direct addressing mode.
  • The instruction is interpreted as Accumulator ← [30].
  • It is given that word 30 contains 50.
  • Thus, value 50 is loaded into the accumulator

 

Instruction-03: Load indirect 30

 

  • This instruction uses indirect addressing mode.
  • The instruction is interpreted as Accumulator ← [[30]].
  • It is given that word 30 contains 50 and word 50 contains 70.
  • Thus, value 70 is loaded into the accumulator.

 

To watch video solution, click here.

 

Next Article- System Bus in Computer Architecture

 

Get more notes and other study material of Computer Organization and Architecture.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Addressing Modes in Computer Architecture

Addressing Modes in Computer Architecture-

 

Before you go through this article, make sure that you have gone through the previous article on Addressing Modes.

 

We have discussed-

  • Addressing modes are the different ways of specifying the location of an operand.
  • Various kinds of addressing modes and their applications.

 

 

In this article, we will discuss syntax used for addressing modes.

 

Syntax Of Addressing Modes-

 

  • The computer instructions may use different addressing modes.
  • Different addressing modes use different syntax for their representation.

 

The syntax used for addressing modes are discussed below-

 

1. Immediate Addressing Mode-

 

Syntax-

# Expression

 

Interpretation-

Operand = Expression

 

Examples-

 

  • Load R1, #1000 is interpreted as R1 ← 1000
  • ADD R2, #3 is interpreted as R2 ← [R2] + 3

 

2. Direct Addressing Mode-

 

Syntax-

Constant

 

Interpretation-

 

Effective address of the operand = Constant

Operand = Content of the memory location ‘Constant’

Operand = [Constant]

 

Examples-

 

  • Load R1, 1000 is interpreted as R1 ← [1000]
  • ADD R2, 3 is interpreted as R2 ← [R2] + [3]

 

3. Register Direct Addressing Mode-

 

Syntax-

Rn or [Rn]

 

Interpretation-

 

Operand = Content of the register Rn

Operand = [Rn]

 

Examples-

 

  • Load R1, R2 is interpreted as R1 ← [R2]
  • ADD R1, R2 is interpreted as R1 ← [R1] + [R2]

 

4. Indirect Addressing Mode-

 

Syntax-

@Expression or @(Expression) or (Expression)

 

Interpretation-

 

Effective address of the operand = Content of the memory location ‘expression’

Operand = [[ Expression]]

 

Examples-

 

  • Load R1, @1000 is interpreted as R1 ← [[1000]]
  • ADD R1, @(1000) is interpreted as R1 ← [R1] + [[1000]]
  • ADD R1, (1000) is interpreted as R1 ← [R1] + [[1000]]

 

5. Register Indirect Addressing Mode-

 

Syntax-

@Rn or @(Rn) or (Rn)

 

Interpretation-

 

Effective address of the operand = Content of the register Rn

Operand = [[ Rn ]]

 

Examples-

 

  • Load R1, @R2 is interpreted as R1 ← [[R1]]
  • ADD R1, @(R2) is interpreted as R1 ← [R1] + [[R2]]
  • ADD R1, (R2) is interpreted as R1 ← [R1] + [[R2]]

 

6. Displacement Addressing Mode-

 

This addressing mode may be-

  • Relative addressing mode
  • Index addressing mode
  • Base register addressing mode

 

Syntax-

disp (Rn)

 

Interpretation-

 

Effective address of the operand = disp + Content of the register Rn

Operand = [disp + [ Rn ]]

 

Examples-

 

  • Load R1, 100(R2) is interpreted as R1 ← [100 + [R1]]
  • ADD R1, 100(R2) is interpreted as R1 ← [R1] + [100 + [R2]]

 

7. Auto-Increment Addressing Mode-

 

Syntax-

(Rn)+

 

Interpretation-

 

Effective address of the operand = Content of the register Rn

Operand = [[ Rn ]]

After fetching the operand, increment [Rn] by step size ‘d’

 

Examples-

 

  • Load R1, (R2)+ is interpreted as R1 ← [[R2]] followed by R2 ← [R2] + d
  • ADD R1, (R2)+ is interpreted as R1 ← [R1] + [[R2]] followed by R2 ← [R2] + d

 

8. Auto-Decrement Addressing Mode-

 

Syntax-

-(Rn)

 

Interpretation-

 

Before fetching the operand, decrement [Rn] by step size ‘d’

Then, Effective address of the operand = Content of the register Rn

Operand = [[ Rn ]]

 

Examples-

 

  • Load R1, -(R2) is interpreted as R2 ← [R2] – d followed by R1 ← [[R2]]
  • ADD R1, -(R2) is interpreted as R2 ← [R2] – d followed by R1 ← [R1] + [[R2]]

 

To gain better understanding about Syntax of Addressing Modes,

Watch this Video Lecture

 

Next Article- Practice Problems On Addressing Modes

 

Get more notes and other study material of Computer Organization and Architecture.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Magnetic Disk | Practice Problems | COA

Magnetic Disk in Computer Architecture-

 

Before you go through this article, make sure that you have gone through the previous article on Magnetic Disk.

 

We have discussed-

  • Magnetic disk is divided into platters which are further divided into tracks and sectors.
  • Read / write head is a mechanical arm that is used to write to and read from the disk.
  • Various important formulas.

 

In this article, we will discuss practice problems based on magnetic disk.

 

PRACTICE PROBLEMS BASED ON MAGNETIC DISK-

 

Problem-01:

 

Consider a disk pack with the following specifications- 16 surfaces, 128 tracks per surface, 256 sectors per track and 512 bytes per sector.

Answer the following questions-

  1. What is the capacity of disk pack?
  2. What is the number of bits required to address the sector?
  3. If the format overhead is 32 bytes per sector, what is the formatted disk space?
  4. If the format overhead is 64 bytes per sector, how much amount of memory is lost due to formatting?
  5. If the diameter of innermost track is 21 cm, what is the maximum recording density?
  6. If the diameter of innermost track is 21 cm with 2 KB/cm, what is the capacity of one track?
  7. If the disk is rotating at 3600 RPM, what is the data transfer rate?
  8. If the disk system has rotational speed of 3000 RPM, what is the average access time with a seek time of 11.5 msec?

 

Solution-

 

Given-

  • Number of surfaces = 16
  • Number of tracks per surface = 128
  • Number of sectors per track = 256
  • Number of bytes per sector = 512 bytes

 

Part-01: Capacity of Disk Pack-

 

Capacity of disk pack

= Total number of surfaces x Number of tracks per surface x Number of sectors per track x Number of bytes per sector

= 16 x 128 x 256 x 512 bytes

= 228 bytes

= 256 MB

 

Part-02: Number of Bits Required To Address Sector-

 

Total number of sectors

= Total number of surfaces x Number of tracks per surface x Number of sectors per track

= 16 x 128 x 256 sectors

= 219 sectors

Thus, Number of bits required to address the sector = 19 bits

 

Part-03: Formatted Disk Space-

 

Formatting overhead

= Total number of sectors x overhead per sector

= 219 x 32 bytes

= 219 x 25 bytes

= 224 bytes

= 16 MB

 

Now, Formatted disk space

= Total disk space – Formatting overhead

= 256 MB – 16 MB

= 240 MB

 

Part-04: Formatting Overhead-

 

Amount of memory lost due to formatting

= Formatting overhead

= Total number of sectors x Overhead per sector

= 219 x 64 bytes

= 219 x 26 bytes

= 225 bytes

= 32 MB

 

Part-05: Maximum Recording Density-

 

Storage capacity of a track

= Number of sectors per track x Number of bytes per sector

= 256 x 512 bytes

= 28 x 29 bytes

= 217 bytes

= 128 KB

 

Circumference of innermost track

= 2 x π x radius

= π x diameter

= 3.14 x 21 cm

= 65.94 cm

 

Now, Maximum recording density

= Recording density of innermost track

= Capacity of a track / Circumference of innermost track

= 128 KB / 65.94 cm

= 1.94 KB/cm

 

Part-06: Capacity Of Track-

 

Circumference of innermost track

= 2 x π x radius

= π x diameter

= 3.14 x 21 cm

= 65.94 cm

 

Capacity of a track

= Storage density of the innermost track x Circumference of the innermost track

= 2 KB/cm x 65.94 cm

= 131.88 KB

≅ 132 KB

 

Part-07: Data Transfer Rate-

 

Number of rotations in one second

= (3600 / 60) rotations/sec

= 60 rotations/sec

 

Now, Data transfer rate

= Number of heads x Capacity of one track x Number of rotations in one second

= 16 x (256 x 512 bytes) x 60

= 24 x 28 x 29 x 60 bytes/sec

= 60 x 221 bytes/sec

= 120 MBps

 

Part-08: Average Access Time-

 

Time taken for one full rotation

= (60 / 3000) sec

= (1 / 50) sec

= 0.02 sec

= 20 msec

 

Average rotational delay

= 1/2 x Time taken for one full rotation

= 1/2 x 20 msec

= 10 msec

 

Now, average access time

= Average seek time + Average rotational delay + Other factors

= 11.5 msec + 10 msec + 0

= 21.5 msec

 

Problem-02:

 

What is the average access time for transferring 512 bytes of data with the following specifications-

  • Average seek time = 5 msec
  • Disk rotation = 6000 RPM
  • Data rate = 40 KB/sec
  • Controller overhead = 0.1 msec

 

Solution-

 

Given-

  • Average seek time = 5 msec
  • Disk rotation = 6000 RPM
  • Data rate = 40 KB/sec
  • Controller overhead = 0.1 msec

 

Time Taken For One Full Rotation-

 

Time taken for one full rotation

= (60 / 6000) sec

= (1 / 100) sec

= 0.01 sec

= 10 msec

 

Average Rotational Delay-

 

Average rotational delay

= 1/2 x Time taken for one full rotation

= 1/2 x 10 msec

= 5 msec

 

Transfer Time-

 

Transfer time

= (512 bytes / 40 KB) sec

= 0.0125 sec

= 12.5 msec

 

Average Access Time-

 

Average access time

= Average seek time + Average rotational delay + Transfer time + Controller overhead + Queuing delay

= 5 msec + 5 msec + 12.5 msec + 0.1 msec + 0

= 22.6 msec

 

Problem-03:

 

A certain moving arm disk storage with one head has the following specifications-

  • Number of tracks per surface = 200
  • Disk rotation speed = 2400 RPM
  • Track storage capacity = 62500 bits
  • Average latency = P msec
  • Data transfer rate = Q bits/sec

What is the value of P and Q?

 

Solution-

 

Given-

  • Number of tracks per surface = 200
  • Disk rotation speed = 2400 RPM
  • Track storage capacity = 62500 bits

 

Time Taken For One Full Rotation-

 

Time taken for one full rotation

= (60 / 2400) sec

= (1 / 40) sec

= 0.025 sec

= 25 msec

 

Average Latency-

 

Average latency or Average rotational latency

= 1/2 x Time taken for one full rotation

= 1/2 x 25 msec

= 12.5 msec

 

Data Transfer Rate-

 

Data transfer rate

= Number of heads x Capacity of one track x Number of rotations in one second

= 1 x 62500 bits x (2400 / 60)

= 2500000 bits/sec

= 2.5 x 106 bits/sec

 

Thus, P = 12.5 and Q = 2.5 x 106

 

Problem-04:

 

A disk pack has 19 surfaces and storage area on each surface has an outer diameter of 33 cm and inner diameter of 22 cm. The maximum recording storage density on any track is 200 bits/cm and minimum spacing between tracks is 0.25 mm. Calculate the capacity of disk pack.

 

Solution-

 

Given-

  • Number of surfaces = 19
  • Outer diameter = 33 cm
  • Inner diameter = 22 cm
  • Maximum recording density = 200 bits/cm
  • Inter track gap = 0.25 mm

 

Number Of Tracks On Each Surface-

 

Number of tracks on each surface

= (Outer radius – Inner radius) / Inter track gap

= (16.5 cm – 11 cm) / 0.25 mm

= 5.5 cm / 0.25 mm

= 55 mm / 0.25 mm

= 220 tracks

 

Capacity Of Each Track-

 

Capacity of each track

= Maximum recording density x Circumference of innermost track

= 200 bits/cm x (3.14 x 22 cm)

= 200 x 69.08 bits

= 13816 bits

= 1727 bytes

 

Capacity Of Disk Pack-

 

Capacity of disk pack

= Total number of surfaces x Number of tracks per surface x Capacity of one track

= 19 x 220 x 1727 bytes

= 7218860 bytes

= 6.88 MB

 

Problem-05:

 

Consider a typical disk that rotates at 15000 RPM and has a transfer rate of 50 x 106 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller’s transfer time is 10 times the disk transfer time. What is the average time (in milliseconds) to read or write a 512 byte sector of the disk?

 

Solution-

 

Given-

  • Rotation speed of the disk = 15000 RPM
  • Transfer rate = 50 x 106 bytes/sec
  • Average seek time = 2 x Average rotational delay
  • Controller’s transfer time = 10 x Disk transfer time

 

Time Taken For One Full Rotation-

 

Time taken for one full rotation

= (60 / 15000) sec

= 0.004 sec

= 4 msec

 

Average Rotational Delay-

 

Average rotational delay

= 1/2 x Time taken for one full rotation

= 1/2 x 4 msec

= 2 msec

 

Average Seek Time-

 

Average seek time

= 2 x Average rotational delay

= 2 x 2 msec

= 4 msec

 

Disk Transfer Time-

 

Disk transfer time

= Time taken to read or write 512 bytes

= 512 bytes / (50 x 106 bytes/sec)

= 10.24 x 10-6 sec

= 0.01024 msec

 

Controller’s Transfer Time-

 

Controller’s transfer time

= 10 x Disk transfer time

= 10 x 0.01024 msec

= 0.1024 msec

 

Average Time To Read Or Write 512 Bytes-

 

Average time to read or write 512 bytes

= Average seek time + Average rotational delay + Disk transfer time + Controller’s transfer time + Queuing delay

= 4 msec + 2 msec + 0.01024 msec + 0.1024 msec + 0

= 6.11 msec

 

Problem-06:

 

A hard disk system has the following parameters-

  • Number of tracks = 500
  • Number of sectors per track = 100
  • Number of bytes per sector = 500
  • Time taken by the head to move from one track to another adjacent track = 1 msec
  • Rotation speed = 600 RPM

What is the average time taken for transferring 250 bytes from the disk?

 

Solution-

 

Given-

  • Number of tracks = 500
  • Number of sectors per track = 100
  • Number of bytes per sector = 500
  • Time taken by the head to move from one track to another adjacent track = 1 msec
  • Rotation speed = 600 RPM

 

Average Seek Time-

 

Average seek time

= (Time taken by the head to move from track-1 to track-1 + Time taken by the head to move from track-1 to track-500) / 2

= (0 + 499 x 1 msec) / 2

= 249.5 msec

 

Time Taken For One Full Rotation-

 

Time taken for one full rotation

= (60 / 600) sec

= 0.1 sec

= 100 msec

 

Average Rotational Delay-

 

Average rotational delay

= 1/2 x Time taken for one full rotation

= 1/2 x 100 msec

= 50 msec

 

Capacity Of One Track-

 

Capacity of one track

= Number of sectors per track x Number of bytes per sector

= 100 x 500 bytes

= 50000 bytes

 

Data Transfer Rate-

 

Data transfer rate

= Number of heads x Capacity of one track x Number of rotations in one second

= 1 x 50000 bytes x (600 / 60)

= 50000 x 10 bytes/sec

= 5 x 105 bytes/sec

 

Transfer Time-

 

Transfer time

= (250 bytes / 5 x 105 bytes) sec

= 50 x 10-5 sec

= 0.5 msec

 

Average Time Taken To Transfer 250 Bytes-

 

Average time taken to transfer 250 bytes

= Average seek time + Average rotational delay + Transfer time + Controller overhead + Queuing delay

= 249.5 msec + 50 msec + 0.5 msec + 0 + 0

= 300 msec

 

Problem-07:

 

A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders.

The address of a sector is given as a triple (c, h, s) where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0,0,0), the 1st sector as (0,0,1) and so on.

 

Part-01:

 

The address <400, 16, 29> corresponds to sector number-

  1. 505035
  2. 505036
  3. 505037
  4. 505038

 

Part-02:

 

The address of 1039 sector is-

  1. <0, 15, 31>
  2. <0, 16, 30>
  3. <0, 16, 31>
  4. <0, 17, 31>

 

Solution-

 

Know this Concept?

 

In general, when counting of items is started from 0, then-

  • For any item-n, number ‘n’ specifies the number of items that must be crossed in order to reach that item.

 

Example-

 

If counting is started from 0, then-

  • To reach cylinder-5, the number of cylinders that must be crossed = 5 cylinders
  • To reach surface-5, the number of surfaces that must be crossed = 5 surfaces
  • To reach sector-5, the number of sectors that must be crossed = 5 sectors

 

To solve this question, we assume there is only one track on each surface.

 

Part-01:

 

We have to calculate the sector number for the address <400, 16, 29>

 

Step-01:

 

To reach our desired cylinder, we have to cross 400 cylinders.

Total number of sectors that are crossed in 400 cylinders

= Number of cylinders x Number of surfaces per cylinder x Number of tracks per surface x Number of sectors per track

= 400 x (10 x 2) x 1 x 63

= 504000

 

Now, after crossing 400 cylinders (cylinder-0 to cylinder-399), we are at cylinder-400.

 

Step-02:

 

To reach our desired surface, we have to cross 16 surfaces.

Total number of sectors that are crossed in 16 surfaces

= Number of surfaces x Number of tracks per surface x Number of sectors per track

= 16 x 1 x 63

= 1008

 

Now, after crossing 16 surfaces (surface-0 to surface-15) in cylinder-400, we are at surface-16.

 

Step-03:

 

To reach our desired sector, we have to cross 29 sectors.

Now, after crossing 29 sectors on surface-16 of cylinder-400, we are at sector-29.

 

Thus

Total number of sectors that are crossed

= 504000 + 1008 + 29

= 505037

 

Thus,

  • After crossing 505037 sectors, we are at sector-505037.
  • So, required address of the sector is 505037.
  • Option (C) is correct.

 

Part-02:

 

We have to find the address of the sector-2039.

Let us check all the options one by one.

 

Option-A:

 

For the address <0, 15, 31>, the sector number is-

Sector number = 0 + (15 x 1 x 63) + 31 = 976

 

Option-B:

 

For the address <0, 16, 30>, the sector number is-

Sector number = 0 + (16 x 1 x 63) + 30 = 1038

 

Option-C:

 

For the address <0, 16, 31>, the sector number is-

Sector number = 0 + (16 x 1 x 63) + 31 = 1039

 

Option-D:

 

For the address <0, 17, 31>, the sector number is-

Sector number = 0 + (17 x 1 x 63) + 31 = 1102

 

Thus, Option (C) is correct.

 

Next Article- Addressing Modes

 

Get more notes and other study material of Computer Organization and Architecture.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Magnetic Disk in Computer Architecture

Magnetic Disk in Computer Architecture-

 

In computer architecture,

  • Magnetic disk is a storage device that is used to write, rewrite and access data.
  • It uses a magnetization process.

 

Architecture-

 

  • The entire disk is divided into platters.
  • Each platter consists of concentric circles called as tracks.
  • These tracks are further divided into sectors which are the smallest divisions in the disk.

 

 

  • A cylinder is formed by combining the tracks at a given radius of a disk pack.

 

 

  • There exists a mechanical arm called as Read / Write head.
  • It is used to read from and write to the disk.
  • Head has to reach at a particular track and then wait for the rotation of the platter.
  • The rotation causes the required sector of the track to come under the head.
  • Each platter has 2 surfaces- top and bottom and both the surfaces are used to store the data.
  • Each surface has its own read / write head.

 

 

Disk Performance Parameters-

 

The time taken by the disk to complete an I/O request is called as disk service time or disk access time.

Components that contribute to the service time are-

 

 

  1. Seek time
  2. Rotational latency
  3. Data transfer rate
  4. Controller overhead
  5. Queuing delay

 

1. Seek Time-

 

  • The time taken by the read / write head to reach the desired track is called as seek time.
  • It is the component which contributes the largest percentage of the disk service time.
  • The lower the seek time, the faster the I/O operation.

 

Specifications

Seek time specifications include-

  1. Full stroke
  2. Average
  3. Track to Track

 

1. Full Stroke-

 

  • It is the time taken by the read / write head to move across the entire width of the disk from the innermost track to the outermost track

 

2. Average-

 

  • It is the average time taken by the read / write head to move from one random track to another.

 

Average seek time = 1 / 3 x Full stroke

 

3. Track to Track-

 

  • It is the time taken by the read-write head to move between the adjacent tracks.

 

2. Rotational Latency-

 

  • The time taken by the desired sector to come under the read / write head is called as rotational latency.
  • It depends on the rotation speed of the spindle.

 

Average rotational latency = 1 / 2 x Time taken for full rotation

 

3. Data Transfer Rate-

 

  • The amount of data that passes under the read / write head in a given amount of time is called as data transfer rate.
  • The time taken to transfer the data is called as transfer time.

 

It depends on the following factors-

  1. Number of bytes to be transferred
  2. Rotation speed of the disk
  3. Density of the track
  4. Speed of the electronics that connects the disk to the computer

 

4. Controller Overhead-

 

  • The overhead imposed by the disk controller is called as controller overhead.
  • Disk controller is a device that manages the disk.

 

5. Queuing Delay-

 

  • The time spent waiting for the disk to become free is called as queuing delay.

 

NOTE-

 

All the tracks of a disk have the same storage capacity.

 

Storage Density-

 

  • All the tracks of a disk have the same storage capacity.
  • This is because each track has different storage density.
  • Storage density decreases as we from one track to another track away from the center.

 

Thus,

  • Innermost track has maximum storage density.
  • Outermost track has minimum storage density.

 

Important Formulas-

 

1. Disk Access Time-

 

Disk access time is calculated as-

 

Disk access time

= Seek time + Rotational delay + Transfer time + Controller overhead + Queuing delay

 

2. Average Disk Access Time-

 

Average disk access time is calculated as-

 

Average disk access time

= Average seek time + Average rotational delay + Transfer time + Controller overhead + Queuing delay

 

3. Average Seek Time-

 

Average seek time is calculated as-

 

Average seek time

= 1 / 3 x Time taken for one full stroke

 

Alternatively,

If time taken by the head to move from one track to adjacent track = t units and there are total k tracks, then-

Average seek time

= { Time taken to move from track 1 to track 1 + Time taken to move from track 1 to last track } / 2

= { 0 + (k-1)t } / 2

= (k-1)t / 2

 

4. Average Rotational Latency-

 

Average rotational latency is calculated as-

 

Average rotational latency

= 1 / 2 x Time taken for one full rotation

 

Average rotational latency may also be referred as-

  • Average rotational delay
  • Average latency
  • Average delay

 

5. Capacity Of Disk Pack-

 

Capacity of a disk pack is calculated as-

 

Capacity of a disk pack

= Total number of surfaces x Number of tracks per surface x Number of sectors per track x Storage capacity of one sector

 

6. Formatting Overhead-

 

Formatting overhead is calculated as-

 

Formatting overhead

= Number of sectors x Overhead per sector

 

7. Formatted Disk Space-

 

Formatted disk space also called as usable disk space is the disk space excluding formatting overhead.

It is calculated as-

 

Formatted disk space

= Total disk space or capacity – Formatting overhead

 

8. Recording Density Or Storage Density-

 

Recording density or Storage density is calculated as-

 

Storage density of a track

= Capacity of the track / Circumference of the track

 

From here, we can infer-

Storage density of a track ∝ 1 / Circumference of the track

 

9. Track Capacity-

 

Capacity of a track is calculated as-

 

Capacity of a track

= Recording density of the track x Circumference of the track

 

10. Data Transfer Rate-

 

Data transfer rate is calculated as-

 

Data transfer rate

= Number of heads x Bytes that can be read in one full rotation x Number of rotations in one second

OR

Data transfer rate

= Number of heads x Capacity of one track x Number of rotations in one second

 

11. Tracks Per Surface-

 

Total number of tracks per surface is calculated as-

 

Total number of tracks per surface

= (Outer radius – Inner radius) / Inter track gap

 

Points to Remember-

 

  • The entire disk space is not usable for storage because some space is wasted in formatting.
  • When rotational latency is not given, use average rotational latency for solving numerical problems.
  • When seek time is not given, use average seek time for solving numerical problems.
  • It is wrong to say that as we move from one track to another away from the center, the capacity increases.
  • All the tracks have same storage capacity.

 

To gain better understanding about magnetic disk-

Watch this Video Lecture

 

Next Article- Practice Problems On Magnetic Disk

 

Get more notes and other study material of Computer Organization and Architecture.

Watch video lectures by visiting our YouTube channel LearnVidFun.