beykent üni̇versi̇tesi̇ - Hasan Hüseyin BALIK

ISTANBUL AREL UNIVERSITY
Name, Surname :
Number :
Course Code : CEN208 & MTB204
Course Name : Computer Organization & Computer Architechture
Exam : Quiz ⊗ Midterm Final
Date :03.05.2013
Please make sure to write your name and student number on each paper that you have used
Question Number 1 2 3 Total
Mark
Note: Exam duration is 75 (seventy five) minutes only. No extra paper is allowed
QUESTIONS
1. True or False (25p)
T/F
a. F Computer organization refers to attributes of a system visible to the programmer
b. T Changes in technology not only influence organization but also result in the introduction of more powerful
and more complex architectures
c. T The hierarchical nature of complex systems is essential to both their design and their description
d. F A vacuum tube is a solid-state device made from silicon
e. F System software was introduced in the third generation of computers.
f. T Designers wrestle with the challenge of balancing processor performance with that of main memory and
other computer components.
g. F An I/O module cannot exchange data directly with the processor.
h. T In general, the more devices attached to the bus, the greater the bus length and hence the greater the
propagation delay.
i. T Because all devices on a synchronous bus are tied to a fixed clock rate, the system cannot take advantage
of advances in device performance
j. F Cache is not a form of internal memory.
k. F The L1 cache is slower than the L3 cache
l. T To achieve greatest performance the memory must be able to keep up with the processor.
m. T The two traditional forms of RAM used in computers are DRAM and SRAM.
n. T All DRAMs require a refresh operation.
o. F The advantage of RAM is that the data or program is permanently in main memory and need never be
loaded from a secondary storage device.
p. F During a read or write operation, the head rotates while the platter beneath it stays stationary.
q. T A removable disk can be removed and replaced with another disk.
r. T RAID is a set of physical disk drives viewed by the operating system as a single logical drive.
s. F A set of I/O modules is a key element of a computer system.
t. T A disadvantage of memory-mapped I/O is that valuable memory address space is used up.
u. F The rotating interrupt mode allows the processor to inhibit interrupts from certain devices.
v. T Privileged instructions are certain instructions that are designated special and can be executed only by the
monitor.
w. F Uniprogramming is the central theme of modern operating systems.
x. F An interrupt is not a hardware-generated signal to the processor.
y T The Pentium II includes hardware for both segmentation and paging.
2. Short Answer (25p)
a. The basic functions that a computer can perform are: data processing, data movement, control, and data storage.
b. When data are received from or delivered to a device that is directly connected to the computer, the process is
known as I/O.
c. Often referred to as processor the CPU controls the operation of the computer and performs its data processing
functions.
d. At the most fundamental level, the speed of a processor is dictated by the pulse frequency produced by the clock,
measured in cycles per second, or Hertz.

e. Amdahls law deals with the potential speedup of a program using multiple processors compared to a single
processor.
f. The Intel 8080 was the world’s first general-purpose microprocessor.
g. A MBR register contains the data to be written into memory or receives the data read from memory.
h. A(n) I/O interrupt is generated by an I/O controller to signal normal completion of an operation, request service
from the processor, or to signal a variety of error conditions.
i. Bus lines can be separated into two generic types: dedicated and multiplexed.
j. The most commonly used physical types of memory are: semiconductor memory, magnetic surface memory
(used for disk and tape), and optical and magneto-optical.
k. External, nonvolatile memory is referred to as secondary or auxiliary memory.
l. One byte equals 8 bits.
m. The simplest of the error-correcting codes is the Hamming code.
n. Three common forms of read-mostly memory are: EPROM, EEPROM, and flash memory.
o. A dynamic RAM is made with cells that store data as charge on capacitors.
p. The floppy disk is a small, flexible platter and the least expensive type of disk.
q. The RAID strategy employs multiple disk drives and distributes data in such a way as to enable simultaneous
access to data from multiple drives, thereby improving I/O performance and allowing easier incremental increases
in capacity.
r. Data is organized on the platter in a concentric set of rings called tracs.
s. In DMA mode the I/O module and main memory exchange data directly, without processor involvement.
t. Interface to the processor and memory via the system bus or central switch and interface to one or more
peripheral devices by tailored data links are two major functions of an I/O Module.
u. When the processor, main memory, and I/O share a common bus, two modes of addressing are possible: memory
mapped and isolated.
v. The Operating System is a program that manages the computer’s resources, provides services for programmers,
and schedules the execution of other programs.
w. The Kernel , or nucleus, contains the most frequently used functions in the OS.
x. Early computer systems presented two main problems: setup time and schedulling.
y. Because a process executes only in main memory, that memory is referred to as real memory.
3. Consider two different machines, with two different instruction sets, both of which have a clock rate of 200 MHz.
The following measurements are recorded on the two machines running a given set of benchmark programs:
Instruction Type Instruction
Count
(millions)
Cycles per
Instruction
Machine A
Arithmetic and logic
Load and store
Branch
Others
Machine B
Arithmetic and logic
Load and store
Branch
Others
8
4
2
4
10
8
2
4
a. Determine the effective CPI, MIPS rate, and execution time for each machine (30p)
b. Comment on the results. (20p)
1
3
4
3
1
2
4
3

a.
CPI
CPI
A
MIPS
CPU
B
CPU
A
MIPS
B
=
A
=
B
∑
f
=
CPI × 10
I
=
∑
f
=
CPI × 10
I
=
c
c
CPI × I
I
× CPI
f
I
c
A
CPI × I
c
B
i
i
× CPI
f
A
B
i
6
( 8×
1+
4×
3 + 2×
4 + 4×
3)
6
( 8 + 4 + 2 + 4)
× 10
200×
10
=
2.22×
10
6
18×
10 × 2.2
=
= 0.2 s
6
200×
10
i
6
=
=
( 10×
1+
8×
2 + 2×
4 + 4×
3)
6
( 10 + 8 + 2 + 4)
× 10
200×
10
=
1.92×
10
= 90
= 104
6
24×
10 × 1.92
=
= 0.23s
6
200×
10
6
6
6
6
× 10
6
× 10
≈ 2.22
6
≈1.92
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.