CHAPTER 1 INTRODUCTION
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TO

Q UESTIONS

1.1

Computer architecture refers to those attributes of a system visible to a programmer or, put another way, those attributes that have a direct impact on the logical execution of a program. Computer organization refers to the operational units and their interconnections that realize the architectural specifications.
Examples of architectural attributes include the instruction set, the number of bits used to represent various data types (e.g., numbers, characters), I/O mechanisms, and techniques for addressing memory. Organizational attributes include those hardware details transparent to the programmer, such as control signals; interfaces between the computer and peripherals; and the memory technology used. 1.2

Computer structure refers to the way in which the components of a computer are interrelated. Computer function refers to the operation of each individual component as part of the structure.

1.3

Data processing; data storage; data movement; and control.

1.4

Central processing unit (CPU): Controls the operation of the computer and performs its data processing functions; often simply referred to as processor.
Main memory: Stores data.
I/O: Moves data between the computer and its external environment.
System interconnection: Some mechanism that provides for communication among CPU, main memory, and I/O. A common example of system interconnection is by means of a system bus, consisting of a number of conducting wires to which all the other components attach.

1.5

Control unit: Controls the operation of the CPU and hence the computer
Arithmetic and logic unit (ALU): Performs the computer’s data processing functions Registers: Provides storage internal to the CPU
CPU interconnection: Some mechanism that provides for communication among the control unit, ALU, and registers

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CHAPTER 2 COMPUTER EVOLUTION AND
PERFORMANCE
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Q UESTIONS

2.1 In a stored program computer, programs are represented in a form suitable for storing in memory alongside the data. The computer gets its instructions by reading them from memory, and a program can be set or altered by setting the values of a portion of memory.
2.2

A main memory, which stores both data and instructions: an arithmetic and logic unit (ALU) capable of operating on binary data; a control unit, which interprets the instructions in memory and causes them to be executed; and input and output
(I/O) equipment operated by the control unit.

2.3

Gates, memory cells, and interconnections among gates and memory cells.

2.4

Moore observed that the number of transistors that could be put on a single chip was doubling every year and correctly predicted that this pace would continue into the near future.

2.5

Similar or identical instruction set: In many cases, the same set of machine instructions is supported on all members of the family. Thus, a program that executes on one machine will also execute on any other. Similar or identical operating system: The same basic operating system is available for all family members. Increasing speed: The rate of instruction execution increases in going from lower to higher family members. Increasing Number of I/O ports: In going from lower to higher family members. Increasing memory size: In going from lower to higher family members. Increasing cost: In going from lower to higher family members.

2.6

In a microprocessor, all of the components of the CPU are on a single chip.

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TO

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This program is developed in [HAYE98]. The vectors A, B, and C are each stored in 1,000 contiguous locations in memory, beginning at locations 1001, 2001, and
3001, respectively. The program begins with the left half of location 3. A counting variable N is set to 999 and decremented after each step until it reaches –1. Thus, the vectors are processed from high location to low location.
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Location Instruction
0
1
2
3L
3R
4L
4R
5L
5R
6L
6R
7L
7R
8L
8R
9L
9R
10L
10R
2.2

Comments
Constant (count N)
Constant
Constant
Transfer A(I) to AC
Compute A(I) + B(I)
Transfer sum to C(I)
Load count N
Decrement N by 1
Test N and branch to 6R if nonnegative
Halt
Update N
Increment AC by 1

a.

999
1
1000
LOAD M(2000)
ADD M(3000)
STOR M(4000)
LOAD M(0)
SUB M(1)
JUMP+ M(6, 20:39)
JUMP M(6, 0:19)
STOR M(0)
ADD M(1)
ADD M(2)
STOR M(3, 8:19)
ADD M(2)
STOR M(3, 28:39)
ADD M(2)
STOR M(4, 8:19)
JUMP M(3, 0:19)
Opcode
00000001

Modify address in 3L
Modify address in 3R
Modify address in 4L
Branch to 3L
Operand
000000000010

b. First, the CPU must make access memory to fetch the instruction. The instruction contains the address of the data we want to load. During the execute phase accesses memory to load the data value located at that address for a total of two trips to memory.
2.3

To read a value from memory, the CPU puts the address of the value it wants into the MAR. The CPU then asserts the Read control line to memory and places the address on the address bus. Memory places the contents of the memory location passed on the data bus. This data is then transferred to the MBR. To write a value to memory, the CPU puts the address of the value it wants to write into the MAR.
The CPU also places the data it wants to write into the MBR. The CPU then asserts the Write control line to memory and places the address on the address bus and the data on the data bus. Memory transfers the data on the data bus into the corresponding memory location.

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2.4
Address
08A
08B
08C

Contents
LOAD M(0FA)
STOR M(0FB)
LOAD M(0FA)
JUMP +M(08D)
LOAD –M(0FA)
STOR M(0FB)

08D
This program will store the absolute value of content at memory location 0FA into memory location 0FB.
2.5

All data paths to/from MBR are 40 bits. All data paths to/from MAR are 12 bits.
Paths to/from AC are 40 bits. Paths to/from MQ are 40 bits.

2.6 The purpose is to increase performance. When an address is presented to a memory module, there is some time delay before the read or write operation can be performed. While this is happening, an address can be presented to the other module. For a series of requests for successive words, the maximum rate is doubled. 2.7

The discrepancy can be explained by noting that other system components aside from clock speed make a big difference in overall system speed. In particular, memory systems and advances in I/O processing contribute to the performance ratio. A system is only as fast as its slowest link. In recent years, the bottlenecks have been the performance of memory modules and bus speed.

2.8

As noted in the answer to Problem 2.7, even though the Intel machine may have a faster clock speed (2.4 GHz vs. 1.2 GHz), that does not necessarily mean the system will perform faster. Different systems are not comparable on clock speed. Other factors such as the system components (memory, buses, architecture) and the instruction sets must also be taken into account. A more accurate measure is to run both systems on a benchmark. Benchmark programs exist for certain tasks, such as running office applications, performing floating-point operations, graphics operations, and so on. The systems can be compared to each other on how long they take to complete these tasks. According to Apple Computer, the G4 is comparable or better than a higher-clock speed Pentium on many benchmarks.

2.9

This representation is wasteful because to represent a single decimal digit from 0 through 9 we need to have ten tubes. If we could have an arbitrary number of these tubes ON at the same time, then those same tubes could be treated as binary bits. With ten bits, we can represent 210 patterns, or 1024 patterns. For integers, these patterns could be used to represent the numbers from 0 through 1023.

2.10 CPI = 1.55; MIPS rate = 25.8; Execution time = 3.87 ns. Source: [HWAN93]

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2.11 a.
CPIA =

∑ CPI × I = (8 ×1+ 4 × 3 + 2 × 4 + 4 × 3) ×10 i i

6

≈ 2.22

(8 + 4 + 2 + 4 ) ×10 6

Ic

f
200 ×10 6
MIPSA =
=
= 90
CPIA ×10 6 2.22 ×10 6
Ic × CPIA 18 ×10 6 × 2.2
CPU A =
=
= 0.2 s f 200 ×10 6
CPIB =

∑ CPI × I = (10 ×1+ 8 × 2 + 2 × 4 + 4 × 3) ×10 i i

(10 + 8 + 2 + 4 ) ×10 6

Ic

MIPSB =
CPU B =

6

≈ 1.92

f
200 ×10 6
=
= 104
CPIB ×10 6 1.92 ×10 6

Ic × CPIB 24 ×10 6 ×1.92
=
= 0.23 s f 200 ×10 6

b. Although machine B has a higher MIPS than machine A, it requires a longer
CPU time to execute the same set of benchmark programs.
€
2.12 a. We can express the MIPs rate as: [(MIPS rate)/106] = Ic/T. So that:
Ic = T × [(MIPS rate)/106]. The ratio of the instruction count of the RS/6000 to the VAX is [x × 18]/[12x × 1] = 1.5.
b. For the Vax, CPI = (5 MHz)/(1 MIPS) = 5.
For the RS/6000, CPI = 25/18 = 1.39.
2.13 From Equation (2.2), MIPS = Ic/(T × 106) = 100/T. The MIPS values are:
Computer A

Computer B

Computer C

Program 1

100

10

5

Program 2

0.1

1

5

Program 3

0.2

0.1

2

Program 4

1

0.125

1

Arithmetic mean Rank

Harmonic mean Rank

Computer A

25.325

1

0.25

2

Computer B

2.8

3

0.21

3

Computer C

3.26

2

2.1

1

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2.14 a. Normalized to R:
Benchmark

Processor
R

M

Z

E

1.00

1.71

3.11

F

1.00

1.19

1.19

H

1.00

0.43

0.49

I

1.00

1.11

0.60

K

1.00

2.10

2.09

Arithmetic mean 1.00

1.31

1.50

b. Normalized to M:
Benchmark

Processor
R

M

Z

E

0.59

1.00

1.82

F

0.84

1.00

1.00

H

2.32

1.00

1.13

I

0.90

1.00

0.54

K

0.48

1.00

1.00

Arithmetic mean 1.01

1.00

1.10

c. Recall that the larger the ratio, the higher the speed. Based on (a) R is the slowest machine, by a significant amount. Based on (b), M is the slowest machine, by a modest amount.
d. Normalized to R:
Benchmark

Processor
R

M

Z

E

1.00

1.71

3.11

F

1.00

1.19

1.19

H

1.00

0.43

0.49

I

1.00

1.11

0.60

K

1.00

2.10

2.09

Geometric mean 1.00

1.15

1.18

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Normalized to M:
Processor

Benchmark

R

M

Z

E

0.59

1.00

1.82

F

0.84

1.00

1.00

H

2.32

1.00

1.13

I

0.90

1.00

0.54

K

0.48

1.00

1.00

Geometric mean 0.87

1.00

1.02

Using the geometric mean, R is the slowest no matter which machine is used for normalization.
2.15 a. Normalized to X:
Processor

Benchmark

X

Y

Z

1

1

2.0

0.5

2

1

0.5

2.0

Arithmetic mean 1

1.25

1.25

Geometric mean 1

1

1

Normalized to Y:
Processor

Benchmark

X

Y

Z

1

0.5

1

0.25

2

2.0

1

4.0

Arithmetic mean 1.25

1

2.125

Geometric mean 1

1

1

Machine Y is twice as fast as machine X for benchmark 1, but half as fast for benchmark 2. Similarly machine Z is half as fast as X for benchmark 1, but twice as fast for benchmark 2. Intuitively, these three machines have equivalent performance. However, if we normalize to X and compute the arithmetic mean
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of the speed metric, we find that Y and Z are 25% faster than X. Now, if we normalize to Y and compute the arithmetic mean of the speed metric, we find that X is 25% faster than Y and Z is more than twice as fast as Y. Clearly, the arithmetic mean is worthless in this context.
b. When the geometric mean is used, the three machines are shown to have equal performance when normalized to X, and also equal performance when normalized to Y. These results are much more in line with our intuition.
2.16 a. Assuming the same instruction mix means that the additional instructions for each task should be allocated proportionally among the instruction types. So we have the following table:
Instruction Type
Arithmetic and logic
Load/store with cache hit
Branch
Memory reference with cache miss CPI
1
2
4
12

Instruction Mix
60%
18%
12%
10%

CPI = 0.6 + (2 × 0.18) + (4 × 0.12) + (12 × 0.1) = 2.64. The CPI has increased due to the increased time for memory access.
b. MIPS = 400/2.64 = 152. There is a corresponding drop in the MIPS rate.
c. The speedup factor is the ratio of the execution times. Using Equation 2.2, we calculate the execution time as T = Ic/(MIPS × 106). For the single-processor case, T1 = (2 × 106)/(178 × 106) = 11 ms. With 8 processors, each processor executes 1/8 of the 2 million instructions plus the 25,000 overhead instructions.
For this case, the execution time for each of the 8 processors is
2 ×10 6
+ 0.025 ×10 6
8
T8 =
= 1.8 ms
152 ×10 6
Therefore we have time to execute program on a single processor
11
Speedup €
=
=
= 6.11 time to execute program on N parallel processors 1.8
d. The answer to this question depends on how we interpret Amdahl's' law. There are two inefficiencies in the parallel system. First, there are additional
€
instructions added to coordinate between threads. Second, there is contention for memory access. The way that the problem is stated, none of the code is inherently serial. All of it is parallelizable, but with scheduling overhead. One could argue that the memory access conflict means that to some extent memory reference instructions are not parallelizable. But based on the information given, it is not clear how to quantify this effect in Amdahl's equation. If we assume that the fraction of code that is parallelizable is f = 1, then Amdahl's law reduces to Speedup = N =8 for this case. Thus the actual speedup is only about
75% of the theoretical speedup.

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2.17 a. Speedup = (time to access in main memory)/(time to access in cache) = T2/T1.
b. The average access time can be computed as T = H × T1 + (1 – H) × T2
Using Equation (2.8):

Speedup =

Execution time before enhancement T2
T2
1
= =
=
Execution time after enhancement
T H × T1 + (1− H )T2 1− H + H T1
(
)
T2

c. T = H × T1 + (1 – H) × (T1 + T2) = T1 + (1 – H) × T2)
This is Equation (4.2) in Chapter 4. Now,

€

Execution time before enhancement T2
T2
1
= =
=
Execution time after enhancement
T T1 + (1− H )T2 1− H + T1
(
)
T2
In this case, the denominator is larger, so that the speedup is less.

Speedup =

€

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CHAPTER 3 COMPUTER FUNCTION AND
INTERCONNECTION
A NSWERS

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Q UESTIONS

3.1

Processor-memory: Data may be transferred from processor to memory or from memory to processor. Processor-I/O: Data may be transferred to or from a peripheral device by transferring between the processor and an I/O module. Data processing: The processor may perform some arithmetic or logic operation on data. Control: An instruction may specify that the sequence of execution be altered. 3.2

Instruction address calculation (iac): Determine the address of the next instruction to be executed. Instruction fetch (if): Read instruction from its memory location into the processor. Instruction operation decoding (iod): Analyze instruction to determine type of operation to be performed and operand(s) to be used. Operand address calculation (oac): If the operation involves reference to an operand in memory or available via I/O, then determine the address of the operand. Operand fetch (of): Fetch the operand from memory or read it in from I/O. Data operation
(do): Perform the operation indicated in the instruction. Operand store (os): Write the result into memory or out to I/O.

3.3

(1) Disable all interrupts while an interrupt is being processed. (2) Define priorities for interrupts and to allow an interrupt of higher priority to cause a lower-priority interrupt handler to be interrupted.

3.4

Memory to processor: The processor reads an instruction or a unit of data from memory. Processor to memory: The processor writes a unit of data to memory. I/O to processor: The processor reads data from an I/O device via an I/O module.
Processor to I/O: The processor sends data to the I/O device. I/O to or from memory: For these two cases, an I/O module is allowed to exchange data directly with memory, without going through the processor, using direct memory access
(DMA).

3.5

With multiple buses, there are fewer devices per bus. This (1) reduces propagation delay, because each bus can be shorter, and (2) reduces bottleneck effects.

3.6

System pins: Include the clock and reset pins. Address and data pins: Include 32 lines that are time multiplexed for addresses and data. Interface control pins:
Control the timing of transactions and provide coordination among initiators and targets. Arbitration pins: Unlike the other PCI signal lines, these are not shared lines. Rather, each PCI master has its own pair of arbitration lines that connect it directly to the PCI bus arbiter. Error Reporting pins: Used to report parity and
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other errors. Interrupt Pins: These are provided for PCI devices that must generate requests for service. Cache support pins: These pins are needed to support a memory on PCI that can be cached in the processor or another device. 64-bit Bus extension pins: Include 32 lines that are time multiplexed for addresses and data and that are combined with the mandatory address/data lines to form a 64-bit address/data bus. JTAG/Boundary Scan Pins: These signal lines support testing procedures defined in IEEE Standard 1149.1.

A NSWERS

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3.1

Memory (contents in hex): 300: 3005; 301: 5940; 302: 7006
Step 1: 3005 → IR; Step 2: 3 → AC
Step 3: 5940 → IR; Step 4: 3 + 2 = 5 → AC
Step 5: 7006 → IR; Step 6: AC → Device 6

3.2

1. a. The PC contains 300, the address of the first instruction. This value is loaded in to the MAR.
b. The value in location 300 (which is the instruction with the value 1940 in hexadecimal) is loaded into the MBR, and the PC is incremented. These two steps can be done in parallel.
c. The value in the MBR is loaded into the IR.
2. a. The address portion of the IR (940) is loaded into the MAR.
b. The value in location 940 is loaded into the MBR.
c. The value in the MBR is loaded into the AC.
3. a. The value in the PC (301) is loaded in to the MAR.
b. The value in location 301 (which is the instruction with the value 5941) is loaded into the MBR, and the PC is incremented.
c. The value in the MBR is loaded into the IR.
4. a. The address portion of the IR (941) is loaded into the MAR.
b. The value in location 941 is loaded into the MBR.
c. The old value of the AC and the value of location MBR are added and the result is stored in the AC.
5. a. The value in the PC (302) is loaded in to the MAR.
b. The value in location 302 (which is the instruction with the value 2941) is loaded into the MBR, and the PC is incremented.
c. The value in the MBR is loaded into the IR.
6. a. The address portion of the IR (941) is loaded into the MAR.
b. The value in the AC is loaded into the MBR.
c. The value in the MBR is stored in location 941.

3.3

a. 224 = 16 MBytes
b. (1) If the local address bus is 32 bits, the whole address can be transferred at once and decoded in memory. However, because the data bus is only 16 bits, it will require 2 cycles to fetch a 32-bit instruction or operand.
(2) The 16 bits of the address placed on the address bus can't access the whole memory. Thus a more complex memory interface control is needed to latch the first part of the address and then the second part (because the microprocessor will end in two steps). For a 32-bit address, one may assume the first half will decode to access a "row" in memory, while the second half is sent later to access
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a "column" in memory. In addition to the two-step address operation, the microprocessor will need 2 cycles to fetch the 32 bit instruction/operand.
c. The program counter must be at least 24 bits. Typically, a 32-bit microprocessor will have a 32-bit external address bus and a 32-bit program counter, unless onchip segment registers are used that may work with a smaller program counter.
If the instruction register is to contain the whole instruction, it will have to be
32-bits long; if it will contain only the op code (called the op code register) then it will have to be 8 bits long.
3.4

In cases (a) and (b), the microprocessor will be able to access 216 = 64K bytes; the only difference is that with an 8-bit memory each access will transfer a byte, while with a 16-bit memory an access may transfer a byte or a 16-byte word. For case (c), separate input and output instructions are needed, whose execution will generate separate "I/O signals" (different from the "memory signals" generated with the execution of memory-type instructions); at a minimum, one additional output pin will be required to carry this new signal. For case (d), it can support 28 = 256 input and 28 = 256 output byte ports and the same number of input and output 16-bit ports; in either case, the distinction between an input and an output port is defined by the different signal that the executed input or output instruction generated.

3.5

Clock cycle =

1
= 125 ns
8 MHz
Bus cycle = 4 × 125 ns = 500 ns
2 bytes transferred every 500 ns; thus transfer rate = 4 MBytes/sec

Doubling the frequency may mean adopting a new chip manufacturing technology
(assuming each instructions will have the same number of clock cycles); doubling the external data bus means wider (maybe newer) on-chip data bus drivers/latches and modifications to the bus control logic. In the first case, the speed of the memory chips will also need to double (roughly) not to slow down the microprocessor; in the second case, the "wordlength" of the memory will have to double to be able to send/receive 32-bit quantities.
3.6

a. Input from the Teletype is stored in INPR. The INPR will only accept data from the Teletype when FGI=0. When data arrives, it is stored in INPR, and FGI is set to 1. The CPU periodically checks FGI. If FGI =1, the CPU transfers the contents of INPR to the AC and sets FGI to 0.
When the CPU has data to send to the Teletype, it checks FGO. If FGO = 0, the CPU must wait. If FGO = 1, the CPU transfers the contents of the AC to
OUTR and sets FGO to 0. The Teletype sets FGI to 1 after the word is printed.
b. The process described in (a) is very wasteful. The CPU, which is much faster than the Teletype, must repeatedly check FGI and FGO. If interrupts are used, the Teletype can issue an interrupt to the CPU whenever it is ready to accept or send data. The IEN register can be set by the CPU (under programmer control)

3.7

a. During a single bus cycle, the 8-bit microprocessor transfers one byte while the
16-bit microprocessor transfers two bytes. The 16-bit microprocessor has twice the data transfer rate.
b. Suppose we do 100 transfers of operands and instructions, of which 50 are one byte long and 50 are two bytes long. The 8-bit microprocessor takes 50 + (2 x
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50) = 150 bus cycles for the transfer. The 16-bit microprocessor requires 50 + 50
= 100 bus cycles. Thus, the data transfer rates differ by a factor of 1.5.
3.8

The whole point of the clock is to define event times on the bus; therefore, we wish for a bus arbitration operation to be made each clock cycle. This requires that the priority signal propagate the length of the daisy chain (Figure 3.26) in one clock period. Thus, the maximum number of masters is determined by dividing the amount of time it takes a bus master to pass through the bus priority by the clock period. 3.9

The lowest-priority device is assigned priority 16. This device must defer to all the others. However, it may transmit in any slot not reserved by the other SBI devices.

3.10 At the beginning of any slot, if none of the TR lines is asserted, only the priority 16 device may transmit. This gives it the lowest average wait time under most circumstances. Only when there is heavy demand on the bus, which means that most of the time there is at least one pending request, will the priority 16 device not have the lowest average wait time.
3.11 a. With a clocking frequency of 10 MHz, the clock period is 10–9 s = 100 ns. The length of the memory read cycle is 300 ns.
b. The Read signal begins to fall at 75 ns from the beginning of the third clock cycle (middle of the second half of T3). Thus, memory must place the data on the bus no later than 55 ns from the beginning of T3.
3.12 a. The clock period is 125 ns. Therefore, two clock cycles need to be inserted.
b. From Figure 3.19, the Read signal begins to rise early in T2. To insert two clock cycles, the Ready line can be put in low at the beginning of T2 and kept low for
250 ns.
3.13 a. A 5 MHz clock corresponds to a clock period of 200 ns. Therefore, the Write signal has a duration of 150 ns.
b. The data remain valid for 150 + 20 = 170 ns.
c. One wait state.
3.14 a. Without the wait states, the instruction takes 16 bus clock cycles. The instruction requires four memory accesses, resulting in 8 wait states. The instruction, with wait states, takes 24 clock cycles, for an increase of 50%.
b. In this case, the instruction takes 26 bus cycles without wait states and 34 bus cycles with wait states, for an increase of 33%.
3.15 a. The clock period is 125 ns. One bus read cycle takes 500 ns = 0.5 µs. If the bus cycles repeat one after another, we can achieve a data transfer rate of 2 MB/s.
b. The wait state extends the bus read cycle by 125 ns, for a total duration of 0.625 µs. The corresponding data transfer rate is 1/0.625 = 1.6 MB/s.
3.16 A bus cycle takes 0.25 µs, so a memory cycle takes 1 µs. If both operands are evenaligned, it takes 2 µs to fetch the two operands. If one is odd-aligned, the time required is 3 µs. If both are odd-aligned, the time required is 4 µs.
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3.17 Consider a mix of 100 instructions and operands. On average, they consist of 20 32bit items, 40 16-bit items, and 40 bytes. The number of bus cycles required for the
16-bit microprocessor is (2 × 20) + 40 + 40 = 120. For the 32-bit microprocessor, the number required is 100. This amounts to an improvement of 20/120 or about 17%.
3.18 The processor needs another nine clock cycles to complete the instruction. Thus, the Interrupt Acknowledge will start after 900 ns.
3.19
CLK
1

2

3

AD

Address

Data-1

C/BE#

Bus Cmd

Byte Enable

4

5

6

7

8

FRAME#

Data-2

Data-3

Byte Enable

Byte Enable

IRDY#

TRDY#

DEVSEL#
Address PhaseAddress Phase Address Phase

Address Phase
Wait State

Bus Transaction

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Wait State

Wait State

9

CHAPTER 4 CACHE MEMORY
A NSWERS

TO

Q UESTIONS

4.1

Sequential access: Memory is organized into units of data, called records. Access must be made in a specific linear sequence. Direct access: Individual blocks or records have a unique address based on physical location. Access is accomplished by direct access to reach a general vicinity plus sequential searching, counting, or waiting to reach the final location. Random access: Each addressable location in memory has a unique, physically wired-in addressing mechanism. The time to access a given location is independent of the sequence of prior accesses and is constant. 4.2

Faster access time, greater cost per bit; greater capacity, smaller cost per bit; greater capacity, slower access time.

4.3

It is possible to organize data across a memory hierarchy such that the percentage of accesses to each successively lower level is substantially less than that of the level above. Because memory references tend to cluster, the data in the higherlevel memory need not change very often to satisfy memory access requests.

4.4

In a cache system, direct mapping maps each block of main memory into only one possible cache line. Associative mapping permits each main memory block to be loaded into any line of the cache. In set-associative mapping, the cache is divided into a number of sets of cache lines; each main memory block can be mapped into any line in a particular set.

4.5

One field identifies a unique word or byte within a block of main memory. The remaining two fields specify one of the blocks of main memory. These two fields are a line field, which identifies one of the lines of the cache, and a tag field, which identifies one of the blocks that can fit into that line.

4.6

A tag field uniquely identifies a block of main memory. A word field identifies a unique word or byte within a block of main memory.

4.7

One field identifies a unique word or byte within a block of main memory. The remaining two fields specify one of the blocks of main memory. These two fields are a set field, which identifies one of the sets of the cache, and a tag field, which identifies one of the blocks that can fit into that set.

4.8

Spatial locality refers to the tendency of execution to involve a number of memory locations that are clustered. Temporal locality refers to the tendency for a processor to access memory locations that have been used recently.
-19-

4.9

Spatial locality is generally exploited by using larger cache blocks and by incorporating prefetching mechanisms (fetching items of anticipated use) into the cache control logic. Temporal locality is exploited by keeping recently used instruction and data values in cache memory and by exploiting a cache hierarchy.

A NSWERS
4.1

TO

P ROBLEMS

The cache is divided into 16 sets of 4 lines each. Therefore, 4 bits are needed to identify the set number. Main memory consists of 4K = 212 blocks. Therefore, the set plus tag lengths must be 12 bits and therefore the tag length is 8 bits. Each block contains 128 words. Therefore, 7 bits are needed to specify the word.
TAG
8

Main memory address =

SET
4

WORD
7

4.2 There are a total of 8 kbytes/16 bytes = 512 lines in the cache. Thus the cache consists of 256 sets of 2 lines each. Therefore 8 bits are needed to identify the set number. For the 64-Mbyte main memory, a 26-bit address is needed. Main memory consists of 64-Mbyte/16 bytes = 222 blocks. Therefore, the set plus tag lengths must be 22 bits, so the tag length is 14 bits and the word field length is 4 bits.
TAG
14

Main memory address =

SET
8

WORD
4

4.3
Address
a. Tag/Line/Word
b. Tag /Word
c. Tag/Set/Word
4.4

4.5

111111
11/444/1
44444/1
22/444/1

666666
66/1999/2
199999/2
CC/1999/2

BBBBBB
BB/2EEE/3
2EEEEE/3
177/EEE/3

a. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in cache: 214; size of tag: 8.
b. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in cache: 4000 hex; size of tag: 22.
c. Address length: 24; number of addressable units: 224; block size: 4; number of blocks in main memory: 222; number of lines in set: 2; number of sets: 213; number of lines in cache: 214; size of tag: 9.
Block frame size = 16 bytes = 4 doublewords
16 KBytes
= 1024
Number of block frames in cache =
16 Bytes
Number of block frames 1024
=
= 256 sets
Number of sets =
Associativity
4

-20-

20 bits

8

Tag

Set

20

8

Tag (20)

Set 0
Set 1

Decoder

4
Offset

Comp1

•
•
•

4 DWs

4

Comp2

Set
0

Comp3

Set 255

Set
1

Comp4

•
•
•
Set
255

Hit

Example: doubleword from location ABCDE8F8 is mapped onto: set 143, any line, doubleword 2:

8
A

B

C

D

E

F

(1000) (1111) (1000)
Set = 143

-21-

8

4.6

12 bits

10 bits

4.7

A 32-bit address consists of a 21-bit tag field, a 7-bit set field, and a 4-bit word field. Each set in the cache includes 3 LRU bits and four lines. Each line consists of
4 32-bit words, a valid bit, and a 21-bit tag.

4.8

a. 8 leftmost bits = tag; 5 middle bits = line number; 3 rightmost bits = byte number b. slot 3; slot 6; slot 3; slot 21
c. Bytes with addresses 0001 1010 0001 1000 through 0001 1010 0001 1111 are stored in the cache
d. 256 bytes
e. Because two items with two different memory addresses can be stored in the same place in the cache. The tag is used to distinguish between them.
-22-

4.9

a. The bits are set according to the following rules with each access to the set:
1.
2.
3.
4.
5.
6.

If the access is to L0 or L1,
If the access is to L0,
If the access is to L1,
If the access is to L2 or L3,
If the access is to L2,
If the access is to L3,

B0 ← 1.
B1 ← 1.
B1 ← 0.
B0 ← 0.
B2 ← 1.
B2 ← 0.

The replacement algorithm works as follows (Figure 4.15): When a line must be replaced, the cache will first determine whether the most recent use was from
L0 and L1 or L2 and L3. Then the cache will determine which of the pair of blocks was least recently used and mark it for replacement. When the cache is initialized or flushed all 128 sets of three LRU bits are set to zero.
b. The 80486 divides the four lines in a set into two pairs (L0, L1 and L2, L3). Bit
B0 is used to select the pair that has been least-recently used. Within each pair, one bit is used to determine which member of the pair was least-recently used.
However, the ultimate selection only approximates LRU. Consider the case in which the order of use was: L0, L2, L3, L1. The least-recently used pair is (L2,
L3) and the least-recently used member of that pair is L2, which is selected for replacement. However, the least-recently used line of all is L0. Depending on the access history, the algorithm will always pick the least-recently used entry or the second least-recently used entry.
c. The most straightforward way to implement true LRU for a four-line set is to associate a two bit counter with each line. When an access occurs, the counter for that block is set to 0; all counters with values lower than the original value for the accessed block are incremented by 1. When a miss occurs and the set is not full, a new block is brought in, its counter is set to 0 and all other counters are incremented by 1. When a miss occurs and the set is full, the block with counter value 3 is replaced; its counter is set to 0 and all other counters are incremented by 1. This approach requires a total of 8 bits.
In general, for a set of N blocks, the above approach requires 2N bits. A more efficient scheme can be designed which requires only N(N–1)/2 bits. The scheme operates as follows. Consider a matrix R with N rows and N columns, and take the upper-right triangular portion of the matrix, not counting the diagonal. For N = 4, we have the following layout:
R(1,2)

R(1,3)

R(1,4)

R(2,3)

R(2,4)
R(3,4)

When line I is referenced, row I of R(I,J) is set to 1, and column I of R(J,I) is set to 0. The LRU block is the one for which the row is entirely equal to 0 (for those bits in the row; the row may be empty) and for which the column is entirely 1
(for all the bits in the column; the column may be empty). As can be seen for N
= 4, a total of 6 bits are required.
-23-

4.10 Block size = 4 words = 2 doublewords; associativity K = 2; cache size = 4048 words; C = 1024 block frames; number of sets S = C/K = 512; main memory = 64K
× 32 bits = 256 Kbytes = 218 bytes; address = 18 bits.
.

Word bits
(6 bits)

(9)

Tag

(2) (1)

Set word select
Tag (6)

Decoder

Set 0

•
•
•

4 words
Set 0
(8 words)

Compare
0

•
•
•

Compare
1

Set 511
(8 words)

Set 511

4.11 a. Address format: Tag = 20 bits; Line = 6 bits; Word = 6 bits
Number of addressable units = 2s+w = 232 bytes; number of blocks in main memory = 2s = 226; number of lines in cache 2r = 26 = 64; size of tag = 20 bits.
b. Address format: Tag = 26 bits; Word = 6 bits
Number of addressable units = 2s+w = 232 bytes; number of blocks in main memory = 2s = 226; number of lines in cache = undetermined; size of tag = 26 bits. c. Address format: Tag = 9 bits; Set = 17 bits; Word = 6 bits
Number of addressable units = 2s+w = 232 bytes; Number of blocks in main memory = 2s = 226; Number of lines in set = k = 4; Number of sets in cache = 2d
= 217; Number of lines in cache = k × 2d =219; Size of tag = 9 bits.
4.12 a. Because the block size is 16 bytes and the word size is 1 byte, this means there are 16 words per block. We will need 4 bits to indicate which word we want out of a block. Each cache line/slot matches a memory block. That means each cache slot contains 16 bytes. If the cache is 64Kbytes then 64Kbytes/16 = 4096 cache slots. To address these 4096 cache slots, we need 12 bits (212 = 4096).
Consequently, given a 20 bit (1 MByte) main memory address:
Bits 0-3 indicate the word offset (4 bits)
Bits 4-15 indicate the cache slot (12 bits)
Bits 16-19 indicate the tag (remaining bits)
F0010 = 1111 0000 0000 0001 0000
Word offset = 0000 = 0
Slot = 0000 0000 0001 = 001
Tag = 1111 = F
01234 = 0000 0001 0010 0011 0100
Word offset = 0100 = 4
Slot = 0001 0010 0011 = 123
-24-

Tag = 0000 = 0
CABBE = 1100 1010 1011 1011 1110
Word offset = 1110 = E
Slot = 1010 1011 1011 = ABB
Tag = 1100 = C
b. We need to pick any address where the slot is the same, but the tag (and optionally, the word offset) is different. Here are two examples where the slot is 1111 1111 1111
Address 1:
Word offset = 1111
Slot = 1111 1111 1111
Tag = 0000
Address = 0FFFF
Address 2:
Word offset = 0001
Slot = 1111 1111 1111
Tag = 0011
Address = 3FFF1
c. With a fully associative cache, the cache is split up into a TAG and a
WORDOFFSET field. We no longer need to identify which slot a memory block might map to, because a block can be in any slot and we will search each cache slot in parallel. The word-offset must be 4 bits to address each individual word in the 16-word block. This leaves 16 bits leftover for the tag.
F0010
Word offset = 0h
Tag = F001h
CABBE
Word offset = Eh
Tag = CABBh
d. As computed in part a, we have 4096 cache slots. If we implement a two -way set associative cache, then it means that we put two cache slots into one set.
Our cache now holds 4096/2 = 2048 sets, where each set has two slots. To address these 2048 sets we need 11 bits (211 = 2048). Once we address a set, we will simultaneously search both cache slots to see if one has a tag that matches the target. Our 20-bit address is now broken up as follows:
Bits 0-3 indicate the word offset
Bits 4-14 indicate the cache set
Bits 15-20 indicate the tag
F0010 = 1111 0000 0000 0001 0000
Word offset = 0000 = 0
Cache Set = 000 0000 0001 = 001
Tag = 11110 = 1 1110 = 1E
CABBE = 1100 1010 1011 1011 1110
Word offset = 1110 = E
Cache Set = 010 1011 1011 = 2BB
Tag = 11001 = 1 1001 = 19
4.13 Associate a 2-bit counter with each of the four blocks in a set. Initially, arbitrarily set the four values to 0, 1, 2, and 3 respectively. When a hit occurs, the counter of the block that is referenced is set to 0. The other counters in the set with values
-25-

originally lower than the referenced counter are incremented by 1; the remaining counters are unchanged. When a miss occurs, the block in the set whose counter value is 3 is replaced and its counter set to 0. All other counters in the set are incremented by 1.
4.14 Writing back a line takes 30 + (7 × 5) = 65 ns, enough time for 2.17 single-word memory operations. If the average line that is written at least once is written more than 2.17 times, the write-back cache will be more efficient.
4.15 a. A reference to the first instruction is immediately followed by a reference to the second. b. The ten accesses to a[i] within the inner for loop which occur within a short interval of time.
4.16 Define
Ci = Average cost per bit, memory level i
Si = Size of memory level i
Ti = Time to access a word in memory level i
Hi = Probability that a word is in memory i and in no higher-level memory
Bi = Time to transfer a block of data from memory level (i + 1) to memory level i
Let cache be memory level 1; main memory, memory level 2; and so on, for a total of N levels of memory. Then
N

∑ Ci Si

Cs = i =1
N

∑ Si i=1 The derivation of Ts is more complicated. We begin with the result from probability theory that:
N

Expected Value of x =

∑ i Pr[ x = 1]

i =1

We can write:
N

Ts =

∑T i Hi

i =1

We need to realize that if a word is in M1 (cache), it is read immediately. If it is in
M2 but not M1, then a block of data is transferred from M2 to M1 and then read.
Thus:
T2 = B1 + T1
-26-

Further
T3 = B2 + T2 = B1 + B2 + T1
Generalizing:

i−1

Ti =
So

∑ Bj + T1 j =1

N i−1

Ts =

∑ ∑(

i =2 j =1

)

N

B j Hi + T1 ∑ Hi i=1 N

∑ Hi = 1

But

i =1

Finally

N i−1

Ts =

∑ ∑ (B j Hi ) + T1

i =2 j =1

4.17 Main memory consists of 512 blocks of 64 words. Cache consists of 16 sets; each set consists of 4 slots; each slot consists of 64 words. Locations 0 through 4351 in main memory occupy blocks 0 through 67. On the first fetch sequence, block 0 through
15 are read into sets 0 through 15; blocks 16 through 31 are read into sets 0 through
15; blocks 32-47 are read into sets 0 through 15; blocks 48-63 are read into sets 0 through 15; and blocks 64-67 are read into sets 0 through 3. Because each set has 4 slots, there is no replacement needed through block 63. The last 4 groups of blocks involve a replacement. On each successive pass, replacements will be required in sets 0 through 3, but all of the blocks in sets 4 through 15 remain undisturbed.
Thus, on each successive pass, 48 blocks are undisturbed, and the remaining 20 must read in.
Let T be the time to read 64 words from cache. Then 10T is the time to read 64 words from main memory. If a word is not in the cache, then it can only be ready by first transferring the word from main memory to the cache and then reading the cache. Thus the time to read a 64-word block from cache if it is missing is 11T.
We can now express the improvement factor as follows. With no cache
Fetch time = (10 passes) (68 blocks/pass) (10T/block) = 6800T
With cache
Fetch time

=
(68) (11T)
+ (9) (48) (T) + (9) (20) (11T)
=
3160T

6800T
Improvement = 3160T = 2.15
-27-

first pass other passes

4.18 a. Access 63
1 Miss
Block 3 → Slot 3
Access 64
1 Miss
Block 4 → Slot 0
Access 65-70 6 Hits
Access 15
1 Miss
Block 0 → Slot 0
Access 16
1 Miss
Block 1 → Slot 1
Access 17-31 15 Hits
Access 32
1 Miss
Block 2 → Slot 2
Access 80
1 Miss
Block 5 → Slot 1
Access 81-95 15 Hits
Access 15
1 Hit
Access 16
1 Miss
Block 1 → Slot 1
Access 17-31 15 hits
Access 32
1 Hit
Access 80
1 Miss
Block 5 → Slot 1
Access 81-95 15 hits
Access 15
1 Hit
Access 16
1 Miss
Block 1 → Slot 1
Access 17-31 15 hits
Access 32
1 Hit
Access 80
1 Miss
Block 5 → Slot 1
Access 81-95 15 hits
Access 15
1 Hit
… Pattern continues to the Tenth Loop

First Loop

Second Loop

Third Loop

Fourth Loop

For lines 63-70
2 Misses
6 Hits
First loop 15-32, 80-95
4 Misses
30 Hits
Second loop 15-32, 80-95 2 Misses
32 Hits
Third loop 15-32, 80-95
2 Misses
32 Hits
Fourth loop 15-32, 80-95 2 Misses
32 Hits
Fifth loop 15-32, 80-95
2 Misses
32 Hits
Sixth loop 15-32, 80-95
2 Misses
32 Hits
Seventh loop 15-32, 80-95 2 Misses
32 Hits
Eighth loop 15-32, 80-95 2 Misses
32 Hits
Ninth loop 15-32, 80-95
2 Misses
32 Hits
Tenth loop 15-32, 80-95
2 Misses
32 Hits
Total:
24 Misses 324 Hits
Hit Ratio = 324/348 = 0.931
b. Access 63
1 Miss
Block 3 → Set 1 Slot 2
Access 64
1 Miss
Block 4 → Set 0 Slot 0
Access 65-70 6 Hits
Access 15
1 Miss
Block 0 → Set 0 Slot 1 First Loop
Access 16
1 Miss
Block 1 → Set 1 Slot 3
Access 17-31 15 Hits
Access 32
1 Miss
Block 2 → Set 0 Slot 0
Access 80
1 Miss
Block 5 → Set 1 Slot 2
Access 81-95 15 Hits
Access 15
1 Hit
Second Loop
Access 16-31 16 Hits
Access 32
1 Hit
Access 80-95 16 Hits
… All hits for the next eight iterations
-28-

For lines 63-70
2 Misses
First loop 15-32, 80-95
4 Misses
Second loop 15-32, 80-95 0 Misses
Third loop 15-32, 80-95
0 Misses
Fourth loop 15-32, 80-95 0 Misses
Fifth loop 15-32, 80-95
0 Misses
Sixth loop 15-32, 80-95
0 Misses
Seventh loop 15-32, 80-95 0 Misses
Eighth loop 15-32, 80-95 0 Misses
Ninth loop 15-32, 80-95
0 Misses
Tenth loop 15-32, 80-95
0 Misses
Total = 6 Misses 342 Hits
Hit Ratio = 342/348 = 0.983

6 Hits
30 Hits
34 Hits
34 Hits
34 Hits
34 Hits
34 Hits
34 Hits
34 Hits
34 Hits
34 Hits

4.19 a. Cost = Cm × 8 × 106 = 8 × 103 ¢ = $80
b. Cost = Cc × 8 × 106 = 8 × 104 ¢ = $800
c. From Equation (4.1) : 1.1 × T1 = T1 + (1 – H)T2
(0.1)(100) = (1 – H)(1200)
H = 1190/1200
4.20 a. Under the initial conditions, using Equation (4.1), the average access time is
T1 + (1 - H) T2 = 1 + (0.05) T2
Under the changed conditions, the average access time is
1.5 + (0.03) T2
For improved performance, we must have
1 + (0.05) T2 > 1.5 + (0.03) T2
Solving for T2, the condition is T2 > 50
b. As the time for access when there is a cache miss become larger, it becomes more important to increase the hit ratio.
4.21 a. First, 2.5 ns are needed to determine that a cache miss occurs. Then, the required line is read into the cache. Then an additional 2.5 ns are needed to read the requested word.
Tmiss = 2.5 + 50 + (15)(5) + 2.5 = 130 ns
b. The value Tmiss from part (a) is equivalent to the quantity (T1 + T2) in Equation
(4.1). Under the initial conditions, using Equation (4.1), the average access time is Ts = H × T1 + (1 – H) × (T1 + T2) = (0.95)(2.5) + (0.05)(130) = 8.875 ns
Under the revised scheme, we have:
-29-

and

Tmiss = 2.5 + 50 + (31)(5) + 2.5 = 210 ns

Ts = H × T1 + (1 – H) × (T1 + T2) = (0.97)(2.5) + (0.03)(210) = 8.725 ns
4.22 There are three cases to consider:
Location of referenced word
In cache
Not in cache, but in main memory Not in cache or main memory

Probability
0.9
(0.1)(0.6) = 0.06

Total time for access in ns
20
60 + 20 = 80

(0.1)(0.4) = 0.04

12ms + 60 + 20 = 12,000,080

So the average access time would be:
Avg = (0.9)(20) + (0.06)(80) + (0.04)(12000080) = 480026 ns
4.23 a. Consider the execution of 100 instructions. Under write-through, this creates
200 cache references (168 read references and 32 write references). On average, the read references result in (0.03) × 168 = 5.04 read misses. For each read miss, a line of memory must be read in, generating 5.04 × 8 = 40.32 physical words of traffic. For write misses, a single word is written back, generating 32 words of traffic. Total traffic: 72.32 words. For write back, 100 instructions create 200 cache references and thus 6 cache misses. Assuming 30% of lines are dirty, on average 1.8 of these misses require a line write before a line read. Thus, total traffic is (6 + 1.8) × 8 = 62.4 words. The traffic rate:
Write through = 0.7232 byte/instruction
Write back = 0.624 bytes/instruction
b. For write-through: [(0.05) × 168 × 8] + 32 = 99.2 → 0.992 bytes/instruction
For write-back: (10 + 3) × 8 = 104 → 0.104 bytes/instruction
c. For write-through: [(0.07) × 168 × 8] + 32 = 126.08 → 0.12608 bytes/instruction
For write-back: (14 + 4.2) × 8 = 145.6 → 0.1456 bytes/instruction
d. A 5% miss rate is roughly a crossover point. At that rate, the memory traffic is about equal for the two strategies. For a lower miss rate, write-back is superior.
For a higher miss rate, write-through is superior.
4.24 a. One clock cycle equals 60 ns, so a cache access takes 120 ns and a main memory access takes 180 ns. The effective length of a memory cycle is (0.9 × 120) + (0.1 ×
180) = 126 ns.
b. The calculation is now (0.9 × 120) + (0.1 × 300) = 138 ns. Clearly the performance degrades. However, note that although the memory access time increases by 120 ns, the average access time increases by only 12 ns.
4.25 a. For a 1 MIPS processor, the average instruction takes 1000 ns to fetch and execute. On average, an instruction uses two bus cycles for a total of 600 ns, so the bus utilization is 0.6
b. For only half of the instructions must the bus be used for instruction fetch. Bus utilization is now (150 + 300)/1000 = 0.45. This reduces the waiting time for other bus requestors, such as DMA devices and other microprocessors.
-30-

4.26 a. Ta = Tc + (1 – H)Tb + W(Tm – Tc)
b. Ta = Tc + (1 – H)Tb + Wb(1 – H)Tb = Tc + (1 – H)(1 + Wb)Tb
4.27 Ta = [Tc1 + (1 – H1)Tc2] + (1 – H2)Tm
4.28 a. miss penalty = 1 + 4 = 5 clock cycles
b. miss penalty = 4 × (1 + 4 ) = 20 clock cycles
c. miss penalty = miss penalty for one word + 3 = 8 clock cycles.
4.29 The average miss penalty equals the miss penalty times the miss rate. For a line size of one word, average miss penalty = 0.032 x 5 = 0.16 clock cycles. For a line size of 4 words and the nonburst transfer, average miss penalty = 0.011 x 20 = 0.22 clock cycles. For a line size of 4 words and the burst transfer, average miss penalty
= 0.011 x 8 = 0.132 clock cycles.

-31-

CHAPTER 5 INTERNAL MEMORY
A NSWERS

TO

Q UESTIONS

5.1

They exhibit two stable (or semistable) states, which can be used to represent binary 1 and 0; they are capable of being written into (at least once), to set the state; they are capable of being read to sense the state.

5.2

(1) A memory in which individual words of memory are directly accessed through wired-in addressing logic. (2) Semiconductor main memory in which it is possible both to read data from the memory and to write new data into the memory easily and rapidly.

5.3

SRAM is used for cache memory (both on and off chip), and DRAM is used for main memory.

5.4

SRAMs generally have faster access times than DRAMs. DRAMS are less expensive and smaller than SRAMs.

5.5

A DRAM cell is essentially an analog device using a capacitor; the capacitor can store any charge value within a range; a threshold value determines whether the charge is interpreted as 1 or 0. A SRAM cell is a digital device, in which binary values are stored using traditional flip-flop logic-gate configurations.

5.6

Microprogrammed control unit memory; library subroutines for frequently wanted functions; system programs; function tables.

5.7

EPROM is read and written electrically; before a write operation, all the storage cells must be erased to the same initial state by exposure of the packaged chip to ultraviolet radiation. Erasure is performed by shining an intense ultraviolet light through a window that is designed into the memory chip. EEPROM is a readmostly memory that can be written into at any time without erasing prior contents; only the byte or bytes addressed are updated. Flash memory is intermediate between EPROM and EEPROM in both cost and functionality. Like EEPROM, flash memory uses an electrical erasing technology. An entire flash memory can be erased in one or a few seconds, which is much faster than EPROM. In addition, it is possible to erase just blocks of memory rather than an entire chip. However, flash memory does not provide byte-level erasure. Like EPROM, flash memory uses only one transistor per bit, and so achieves the high density (compared with
EEPROM) of EPROM.

5.8

A0 - A1 = address lines:. CAS = column address select:. D1 - D4 = data lines. NC: = no connect. OE: output enable. RAS = row address select:. Vcc: = voltage source.
Vss: = ground. WE: write enable.
-32-

5.9

A bit appended to an array of binary digits to make the sum of all the binary digits, including the parity bit, always odd (odd parity) or always even (even parity). 5.10 A syndrome is created by the XOR of the code in a word with a calculated version of that code. Each bit of the syndrome is 0 or 1 according to if there is or is not a match in that bit position for the two inputs. If the syndrome contains all 0s, no error has been detected. If the syndrome contains one and only one bit set to 1, then an error has occurred in one of the 4 check bits. No correction is needed. If the syndrome contains more than one bit set to 1, then the numerical value of the syndrome indicates the position of the data bit in error. This data bit is inverted for correction. 5.11 Unlike the traditional DRAM, which is asynchronous, the SDRAM exchanges data with the processor synchronized to an external clock signal and running at the full speed of the processor/memory bus without imposing wait states.

A NSWERS

TO

P ROBLEMS

5.1

The 1-bit-per-chip organization has several advantages. It requires fewer pins on the package (only one data out line); therefore, a higher density of bits can be achieved for a given size package. Also, it is somewhat more reliable because it has only one output driver. These benefits have led to the traditional use of 1-bit-perchip for RAM. In most cases, ROMs are much smaller than RAMs and it is often possible to get an entire ROM on one or two chips if a multiple-bits-per-chip organization is used. This saves on cost and is sufficient reason to adopt that organization. 5.2

In 1 ms, the time devoted to refresh is 64 × 150 ns = 9600 ns. The fraction of time devoted to memory refresh is (9.6 × 10–6 s)/10–3 s = 0.0096, which is approximately
1%.

5.3 a. Memory cycle time = 60 + 40 = 100 ns. The maximum data rate is 1 bit every 100 ns, which is 10 Mbps.
b. 320 Mbps = 40 MB/s.

-33-

5.4

Chip select

Chip select

1 Mb

1 Mb

Chip select

Chip select

1 Mb

1 Mb

Chip select

Chip select

1 Mb

1 Mb

Chip select

Chip select

1 Mb

S0 S1

1 Mb

Decoder
S2
S3

S4
S5
A22
A21

S6
S7

A20
A19
•
•
•
A0

5.5 a. The length of a clock cycle is 100 ns. Mark the beginning of T1 as time 0.Address
Enable returns to a low at 75. RAS goes active 50 ns later, or time 125. Data must become available by the DRAMs at time 300 – 60 = 240. Hence, access time must be no more than 240 – 125 = 115 ns.
b. A single wait state will increase the access time requirement to 115 + 100 = 215 ns. This can easily be met by DRAMs with access times of 150 ns.
€
5.6 a. The refresh period from row to row must be no greater than
4000/256 = 15.625 µs.
b. An 8-bit counter is needed to count 256 rows (28 = 256).
5.7 a. pulse a = write pulse b = write pulse c = write pulse d = write pulse e= write

pulse f = write pulse g = store-disable outputs pulse h = read pulse i = read pulse j = read

-34-

pulse k = read pulse l = read pulse m = read pulse n = store-disable outputs

.

5.8

b. Data is read in via pins (D3, D2, D1, D0) word 0 = 1111 (written into location 0 during pulse a) word 1 = 1110 (written into location 0 during pulse b) word 2 = 1101 (written into location 0 during pulse c) word 3 = 1100 (written into location 0 during pulse d) word 4 = 1011 (written into location 0 during pulse e) word 5 = 1010 (written into location 0 during pulse f) word 6 = random (did not write into this location 0)
c. Output leads are (O3, O2, O1, O0) pulse h: 1111 (read location 0) pulse i: 1110 (read location 1) pulse j: 1101 (read location 2) pulse k: 1100 (read location 3) pulse l: 1011 (read location 4) pulse m: 1010 (read location 5)
8192/64 = 128 chips; arranged in 8 rows by 64 columns:

AB
3

6

Ak-A10 A9-A7

A0 = L

A6-A1

Section 0 (even)
0
•
•
•

Decoder

Row 0

All zeros

Row 1

•
•
•

• • •

1
•
•
•

112

A0 = H
Section 1 (odd)

113

8

9

•
•
•

• • •

7

•
•
•

• • •

•
•
•

119

120

121

15
•
•
•

• • •

8
Rows

127

Row 7

En

8

Depends on type of processor 5.9

8

Total memory is 1 megabyte = 8 megabits. It will take 32 DRAMs to construct the memory (32 × 256 Kb = 8 Mb). The composite failure rate is 2000 × 32 = 64,000
FITS. From this, we get a MTBF = 109/64,000 = 15625 hours = 22 months.

5.10 The stored word is 001101001111, as shown in Figure 5.10. Now suppose that the only error is in C8, so that the fetched word is 001111001111. Then the received block results in the following table:
Position
Bits
Block
Codes

12
D8
0

11
D7
0

10
9
D6
D5
1
1
1010 1001

8
C8
1

7
D4
1
0111
-35-

6
D3
0

5
D2
0

4
C4
1

3
D1
1
0011

2
C2
1

1
C1
1

The check bit calculation after reception:
Position
Hamming
10
9
7
3

Code

1111
1010
1001
0111
0011
1000

XOR = syndrome

The nonzero result detects and error and indicates that the error is in bit position 8, which is check bit C8.
5.11 Data bits with value 1 are in bit positions 12, 11, 5, 4, 2, and 1:
Position 12
11
Bits
D8
D7
Block
1
1
Codes
1100 1011

10
D6
0

9
D5
0

8
C8

7
D4
0

6
D3
0

5
D2
1
0101

4
C4

3
D1
0

2
C2

1
C1

The check bits are in bit numbers 8, 4, 2, and 1.
Check bit 8 calculated by values in bit numbers: 12, 11, 10 and 9
Check bit 4 calculated by values in bit numbers: 12, 7, 6, and 5
Check bit 2 calculated by values in bit numbers: 11, 10, 7, 6 and 3
Check bit 1 calculated by values in bit numbers: 11, 9, 7, 5 and 3
Thus, the check bits are: 0 0 1 0
5.12 The Hamming Word initially calculated was: bit number:

12
0

11
0

10
1

9
1

8
0

7
1

6
0

5
0

4
1

3
1

2
1

1
1

Doing an exclusive-OR of 0111 and 1101 yields 1010 indicating an error in bit 10 of the Hamming Word. Thus, the data word read from memory was 00011001.
5.13 Need K check bits such that 1024 + K ≤ 2K – 1.
The minimum value of K that satisfies this condition is 11.

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5.14 As Table 5.2 indicates, 5 check bits are needed for an SEC code for 16-bit data words. The layout of data bits and check bits:
Bit Position
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1

Position Number
10101
10100
10011
10010
10001
10000
01111
01110
01101
01100
01011
01010
01001
01000
00111
00110
00101
00100
00011
00010
00001

Check Bits

Data Bits
M16
M15
M14
M13
M12

C16
M11
M10
M9
M8
M7
M6
M5
C8
M4
M3
M2
C4
M1
C2
C1

The equations are calculated as before, for example,
C1= M1 ⊕ M2 ⊕ M4 ⊕ M5 ⊕ M7 ⊕ M9 ⊕ M11 ⊕ M12 ⊕ M14 ⊕ M16.
For the word 0101000000111001, the code is
C16 = 1; C8 = 1; C 4 = 1; C2 = 1; C1 = 0.
If an error occurs in data bit 4:
C16 = 1 ; C8 =1; C4 = 0; C2 = 0; C1 = 1.
Comparing the two:
C16
1
1
0

C8
1
1
0

C4
1
0
1

C2
1
0
1

C1
0
1
1

The result is an error identified in bit position 7, which is data bit 4.

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CHAPTER 6 EXTERNAL MEMORY
A NSWERS

TO

Q UESTIONS

6.1

Improvement in the uniformity of the magnetic film surface to increase disk reliability. A significant reduction in overall surface defects to help reduce read/write errors. Ability to support lower fly heights (described subsequently).
Better stiffness to reduce disk dynamics. Greater ability to withstand shock and damage 6.2

The write mechanism is based on the fact that electricity flowing through a coil produces a magnetic field. Pulses are sent to the write head, and magnetic patterns are recorded on the surface below, with different patterns for positive and negative currents. An electric current in the wire induces a magnetic field across the gap, which in turn magnetizes a small area of the recording medium. Reversing the direction of the current reverses the direction of the magnetization on the recording medium.

6.3

The read head consists of a partially shielded magnetoresistive (MR) sensor. The
MR material has an electrical resistance that depends on the direction of the magnetization of the medium moving under it. By passing a current through the
MR sensor, resistance changes are detected as voltage signals.

6.4

For the constant angular velocity (CAV) system, the number of bits per track is constant. An increase in density is achieved with multiple zoned recording, in which the surface is divided into a number of zones, with zones farther from the center containing more bits than zones closer to the center.

6.5

On a magnetic disk. data is organized on the platter in a concentric set of rings, called tracks. Data are transferred to and from the disk in sectors. For a disk with multiple platters, the set of all the tracks in the same relative position on the platter is referred to as a cylinder.

6.6

512 bytes.

6.7

On a movable-head system, the time it takes to position the head at the track is known as seek time. Once the track is selected, the disk controller waits until the appropriate sector rotates to line up with the head. The time it takes for the beginning of the sector to reach the head is known as rotational delay. The sum of the seek time, if any, and the rotational delay equals the access time, which is the time it takes to get into position to read or write. Once the head is in position, the read or write operation is then performed as the sector moves under the head; this is the data transfer portion of the operation and the time for the transfer is the transfer time.
-38-

6.8

1. RAID is a set of physical disk drives viewed by the operating system as a single logical drive. 2. Data are distributed across the physical drives of an array. 3.
Redundant disk capacity is used to store parity information, which guarantees data recoverability in case of a disk failure.

6.9

0: Non-redundant 1: Mirrored; every disk has a mirror disk containing the same data. 2: Redundant via Hamming code; an error-correcting code is calculated across corresponding bits on each data disk, and the bits of the code are stored in the corresponding bit positions on multiple parity disks. 3: Bit-interleaved parity; similar to level 2 but instead of an error-correcting code, a simple parity bit is computed for the set of individual bits in the same position on all of the data disks.
4: Block-interleaved parity; a bit-by-bit parity strip is calculated across corresponding strips on each data disk, and the parity bits are stored in the corresponding strip on the parity disk. 5: Block-interleaved distributed parity; similar to level 4 but distributes the parity strips across all disks. 6: Blockinterleaved dual distributed parity; two different parity calculations are carried out and stored in separate blocks on different disks.

6.10 The disk is divided into strips; these strips may be physical blocks, sectors, or some other unit. The strips are mapped round robin to consecutive array members. A set of logically consecutive strips that maps exactly one strip to each array member is referred to as a stripe.
6.11 For RAID level 1, redundancy is achieved by having two identical copies of all data. For higher levels, redundancy is achieved by the use of error-correcting codes. 6.12 In a parallel access array, all member disks participate in the execution of every
I/O request. Typically, the spindles of the individual drives are synchronized so that each disk head is in the same position on each disk at any given time. In an independent access array, each member disk operates independently, so that separate I/O requests can be satisfied in parallel.
6.13 For the constant angular velocity (CAV) system, the number of bits per track is constant. At a constant linear velocity (CLV), the disk rotates more slowly for accesses near the outer edge than for those near the center. Thus, the capacity of a track and the rotational delay both increase for positions nearer the outer edge of the disk.
6.14 1. Bits are packed more closely on a DVD. The spacing between loops of a spiral on a CD is 1.6 µm and the minimum distance between pits along the spiral is 0.834 µm. The DVD uses a laser with shorter wavelength and achieves a loop spacing of
0.74 µm and a minimum distance between pits of 0.4 µm. The result of these two improvements is about a seven-fold increase in capacity, to about 4.7 GB. 2. The
DVD employs a second layer of pits and lands on top of the first layer A dual-layer
DVD has a semireflective layer on top of the reflective layer, and by adjusting focus, the lasers in DVD drives can read each layer separately. This technique almost doubles the capacity of the disk, to about 8.5 GB. The lower reflectivity of the second layer limits its storage capacity so that a full doubling is not achieved.
-39-

3. The DVD-ROM can be two sided whereas data is recorded on only one side of a
CD. This brings total capacity up to 17 GB.
6.15 The typical recording technique used in serial tapes is referred to as serpentine recording. In this technique, when data are being recorded, the first set of bits is recorded along the whole length of the tape. When the end of the tape is reached, the heads are repositioned to record a new track, and the tape is again recorded on its whole length, this time in the opposite direction. That process continues, back and forth, until the tape is full.

A NSWERS
6.1

TO

P ROBLEMS

It will be useful to keep the following representation of the N tracks of a disk in mind: 0

1

•••

j–1

•••

N–j

•••

N–2 N–1

a. Let us use the notation Ps [j/t] = Pr [seek of length j when head is currently positioned over track t]. Recognize that each of the N tracks is equally likely to be requested. Therefore the unconditional probability of selecting any particular track is 1/N. We can then state:

Ps[j /t] =

1
N

if

t ≤ j – 1 OR t ≥ N – j

Ps[j /t] =

2
N

if

j–1