Computer Science Topics
Postgraduate level — Computer Science
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quiz Questions
Q211
An electronics engineer uses a 4-variable Karnaugh Map (K-Map) with variables (W, X, Y, Z) to minimize a Boolean function. If the minterms are mapped at positions m(0, 2, 5, 7, 8, 10, 13, 15), what is the most simplified sum-of-products (SOP) expression output?
W'X + YZ'
X'Z' + XZ
W'Z + WZ'
X'Y' + XY
Explanation
Plotting m(0, 2, 5, 7, 8, 10, 13, 15) on a 4-variable K-Map yields two distinct quad groups of size 4. The first quad is formed by the four corners m(0, 2, 8, 10), which simplifies cleanly to X'Z'. The second quad is formed by the center block m(5, 7, 13, 15), which simplifies cleanly to XZ. Combining these two product terms yields the minimized expression X'Z' + XZ (or X ⊙ Z).
Q212
Which of the following character data tracking codes represents an unweighted unit-distance metric system where any two consecutive numerical integers differ structurally by exactly one binary bit state shift?
Standard 8-4-2-1 BCD framework
Gray Code system
EBCDIC cluster alphanumeric layout
Excess-3 self-complementing layout
Explanation
Gray Code is classified as a cyclic, unweighted unit-distance code because transitioning from one sequential value to the next updates exactly one single binary bit. This minimizes transient switching errors in digital systems natively.
Q213
A digital circuit processes data using an 8-bit Byte register. If the system memory allocation updates to track exactly 3.5 Megabytes (MB), what is the mathematically equivalent buffer size when scaled cleanly down into individual Bytes inside standard binary processing setups?
3,500,000 Bytes
3,670,016 Bytes
29,360,128 Bytes
3,750,112 Bytes
Explanation
In standardized binary database systems using the base-2 scale layout, 1 Megabyte (MB) is mathematically equal to 1024 Kilobytes (KB), and 1 Kilobyte equals 1024 Bytes. Conversion tracking: 3.5 MB × 1024 KB/MB = 3584 KB. Then, 3584 KB × 1024 Bytes/KB = exactly 3,670,016 Bytes.
Q214
An internal CPU memory access sequence registers a cache hit efficiency rate of exactly 70%. If the microchip cache hardware speed latency is 3 nanoseconds (ns) and misses force jumping to main memory which registers a latency of 50 nanoseconds (ns), what is the computed Effective Memory Access Time (EMAT) for this hardware pipeline?
11.1 nanoseconds
17.1 nanoseconds
21.3 nanoseconds
37.1 nanoseconds
Explanation
Formula for Effective Memory Access Time tracking metrics: EMAT = (Hit Rate × Cache Latency) + (Miss Rate × Main Memory Latency). Given parameters: Hit Rate = 0.70, Miss Rate = 1 - 0.70 = 0.30. Calculation parameters: (0.70 × 3 ns) + (0.30 × 50 ns) = 2.1 ns + 15.0 ns = exactly 17.1 nanoseconds.
Q215
Which computing generation is historically categorized by the baseline implementation of transistors replacing high heat-generating vacuum tube infrastructures natively?
First Generation computing framework
Second Generation architecture loop
Third Generation silicon array
Fourth Generation micro framework
Explanation
The Second Generation of computers (spanning roughly the mid-1950s to mid-1960s layer) is defined technologically by discrete solid-state transistors replacing bulky first-generation vacuum tube relays, maximizing structural processing efficiency.
Q216
An output function simplified using a 4-variable Karnaugh Map (K-Map) with variables (A, B, C, D) contains minterms at positions m(0, 1, 2, 3, 4, 5, 6, 7). Which core minimized product expression represents the simplified Boolean output?
A'
B'
C'D'
A'B'
Explanation
Plotting m(0, 1, 2, 3, 4, 5, 6, 7) on a 4-variable K-Map (where AB tracks rows and CD tracks columns) fills the first two rows completely. This forms a single massive octet of size 8. Across these rows, variables B, C, and D change completely through all variations, while A remains constantly equal to 0, minimizing cleanly to A'.
Q217
A basic computer processor handles numeric data logic layouts natively. What is the decimal integer equivalent value represented by the octal number configuration profile (57)₈?
45
47
57
63
Explanation
To convert an octal number to its decimal equivalent, calculate positional weights using base-8 tracking rows: (5 × 8¹) + (7 × 8⁰) = 40 + 7 = 47.
Q218
Which fundamental basic computer peripheral device is functionally categorized as an alphanumeric impact input matrix that converts mechanical keystroke metrics cleanly into system binary data streams?
Keyboard
Optical Mouse tracker
Flatbed transparency scanner
Dot Matrix impact wheel
Explanation
A computer Keyboard operates natively as an alphanumeric text input device. Pressing a key closes a mechanical or electronic circuit switch, allowing the internal micro-controller matrix to transmit unique scan code bit streams safely to system buffer lines.
Q219
An internal CPU memory storage loop operates with a cache hit efficiency rate of exactly 75%. If the hardware cache speed latency is 5 nanoseconds (ns) and misses force jumping to main memory which registers an access speed latency of 60 nanoseconds (ns), what is the computed Effective Memory Access Time (EMAT) for this hardware loop?
11.25 nanoseconds
18.75 nanoseconds
23.75 nanoseconds
46.25 nanoseconds
Explanation
Formula for Effective Access Time tracking metrics: EMAT = (Hit Rate × Cache Latency) + (Miss Rate × Main Memory Latency). Given parameters: Hit Rate = 0.75, Miss Rate = 1 - 0.75 = 0.25. Calculation metrics setup: (0.75 × 5 ns) + (0.25 × 60 ns) = 3.75 ns + 15.00 ns = exactly 18.75 nanoseconds.
Q220
Which type of computer memory layout acts as permanent non-volatile storage natively holding the foundational code routine required to initiate the Power-On Self-Test (POST) firmware configurations during bootstrap?
Dynamic RAM (DRAM)
Read-Only Memory (ROM)
SRAM Cache space
Virtual Memory buffers
Explanation
Read-Only Memory (ROM) is a non-volatile memory layout that permanently stores firmware systems such as the BIOS natively inside silicon motherboard chips, preserving binary state arrays even when external power rails are inactive.