Computer Science Topics
Postgraduate level — Computer Science
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quiz Questions
Q191
An output function simplified using a 4-variable Karnaugh Map (K-Map) maps minterms at positions m(0, 2, 8, 10). Which minimized product term represents the simplified Boolean expression output?
W'X'
X'Z'
Y'Z'
W'Z'
Explanation
Minterms m(0, 2, 8, 10) represent the four corners of a 4-variable K-Map (variables W, X, Y, Z). Grouping the four corners together forms a quad of size 4. Looking at rows (00 and 10), W changes but X' remains constant. Looking at columns (00 and 10), Y changes but Z' remains constant. Thus, the grouping simplifies cleanly to X'Z'.
Q192
A 4-bit binary input sequence 1010 is processed through a digital network. What is its exact structural representation when converted cleanly into Excess-3 code?
1101
0100 0011
0011 0100
1010 0011
Explanation
The binary sequence 1010 is equivalent to decimal 10. Excess-3 code is a non-weighted self-complementing code obtained by adding decimal 3 (binary 0011) to each BCD digit of the number. For 10, the BCD digits are 1 (0001) and 0 (0000). Adding 0011 to each: 0001 + 0011 = 0100, and 0000 + 0011 = 0011. Combined sequentially, the Excess-3 configuration is 0100 0011.
Q193
A digital circuit output updates its parameters to match the Boolean expression Y = AB + A'B'. Which single physical logic gate natively tracks this operational lifecycle loop output profile?
XOR Gate
XNOR Gate
NAND Gate
NOR Gate
Explanation
The Boolean identity AB + A'B' defines the core logical operation of the Exclusive-NOR (XNOR) gate, which characteristically generates a binary '1' if and only if both binary inputs are completely identical (either both 0 or both 1).
Q194
An internal CPU cache subsystem reads a hit rate efficiency of exactly 95% with a cache access speed latency of 1 nanosecond (ns). If tracking misses require falling back to the slower main memory which displays a latency of 30 nanoseconds (ns), what is the computed Effective Memory Access Time (EMAT) for this hardware loop?
1.45 nanoseconds
2.45 nanoseconds
3.55 nanoseconds
15.50 nanoseconds
Explanation
Formula for EMAT: (Hit Rate × Cache Latency) + (Miss Rate × Main Memory Latency). Given parameters: Hit Rate = 0.95, Miss Rate = 1 - 0.95 = 0.05. Calculation: (0.95 × 1 ns) + (0.05 × 30 ns) = 0.95 ns + 1.50 ns = 2.45 nanoseconds.
Q195
If a processor architecture uses an 11-bit wide physical address bus layout natively, what is the absolute maximum number of unique memory configurations (byte-addressable storage cells) that the CPU can directly reference?
1024 locations (1 KB)
2048 locations (2 KB)
4096 locations (4 KB)
8192 locations (8 KB)
Explanation
The maximum memory reference threshold capacity is calculated mathematically as 2^n, where n is the number of bits in the address bus. Thus, 2^11 = 2048 distinct byte locations. Converting via binary scaling steps: 2048 / 1024 = exactly 2 Kilobytes (KB).
Q196
An electronic logic control block is simplified using a 4-variable Karnaugh Map (K-Map). If the minterms are mapped at positions m(5, 7, 13, 15), which of the following represents the absolute minimized product expression output?
W'X
XZ
WY'
XYZ'
Explanation
Plotting m(5, 7, 13, 15) on a 4-variable K-Map (variables W, X, Y, Z where WX represents rows and YZ tracks columns) fills the intersections of rows 01 and 11 with columns 01 and 11. This forms a single quad of size 4. Across these cells, X = 1 and Z = 1 remain perfectly constant while W and Y vary completely, minimizing cleanly to XZ.
Q197
A digital switching network utilizes a 4-bit binary code sequence 1100. What is its exact configuration profile when converted cleanly down into Gray Code format?
1110
1010
1001
1111
Explanation
To convert Binary (B₃B₂B₁B₀ = 1100) to Gray Code (G₃G₂G₁G₀): Keep the Most Significant Bit identical: G₃ = B₃ = 1. Then perform successive bitwise XOR operations: G₂ = B₃ ⊕ B₂ = 1 ⊕ 1 = 0. G₁ = B₂ ⊕ B₁ = 1 ⊕ 0 = 1. G₀ = B₁ ⊕ B₀ = 0 ⊕ 0 = 0. Combining these tracking parameters sequentially yields 1010.
Q198
Which of the following physical digital device units is functionally classified as an output terminal that uses an array of tiny piezoelectric or thermal heating nozzles to spray microscopic ink droplets directly onto paper matrices?
Dot Matrix impact wheel
Inkjet Printer
Electrostatic xerographic toner drum
Flatbed digitizer pane
Explanation
An Inkjet Printer operates natively as a non-impact raster output peripheral. It generates high-resolution textual characters and graphic configurations by systematically firing liquid ink droplets through specialized thermal or piezoelectric printhead micro-nozzles.
Q199
An internal CPU memory access sequence operates with a cache hit efficiency rate of exactly 80%. If the hardware cache speed latency is 4 nanoseconds (ns) and misses require jumping to the slower main memory which displays a latency of 50 nanoseconds (ns), what is the computed Effective Memory Access Time (EMAT)?
9.2 nanoseconds
13.2 nanoseconds
14.8 nanoseconds
43.2 nanoseconds
Explanation
Formula for Effective Access Time tracking metrics: EMAT = (Hit Rate × Cache Latency) + (Miss Rate × Main Memory Latency). Given parameters: Hit Rate = 0.80, Miss Rate = 1 - 0.80 = 0.20. Calculation: (0.80 × 4 ns) + (0.20 × 50 ns) = 3.2 ns + 10.0 ns = exactly 13.2 nanoseconds.
Q200
If a custom microchip processing architecture implements a 12-bit wide physical address bus layout natively, what is the absolute maximum memory capacity parameter (byte-addressable storage index limits) that this CPU can directly reference?
2 Kilobytes (KB)
4 Kilobytes (KB)
8 Kilobytes (KB)
16 Kilobytes (KB)
Explanation
The maximum direct reference capacity threshold is calculated mathematically as 2^n bytes, where n represents the bit width of the address bus line pipelines. Thus, 2^12 = 4096 distinct byte locations. Converting via standard binary scaling indices: 4096 / 1024 = exactly 4 Kilobytes (KB).