Computer Science Topics
Postgraduate level — Computer Science
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quiz Questions
Q161
A specialized computer storage channel exhibits an internal bus width layout of exactly 32 bits. If the hardware execution clock updates natively at a frequency of 400 MHz, what is the maximum theoretical data transfer throughput speed achievable across this system?
800 MB/s
1200 MB/s
1600 MB/s
3200 MB/s
Explanation
To calculate throughput: Bus Width in Bytes = 32 bits / 8 = 4 Bytes. Throughput = Bus Width (in Bytes) × Clock Frequency = 4 Bytes × 400 MHz = 1600 MB/s (or 1.6 GB/s).
Q162
If a basic processor architecture uses a 16-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?
16,384 locations (16 KB)
32,768 locations (32 KB)
65,536 locations (64 KB)
1,048,576 locations (1 MB)
Explanation
The maximum memory addressing capacity is calculated mathematically as 2^n, where n is the number of bits in the address bus. Thus, 2^16 = 65,536 distinct byte locations, which is exactly equivalent to 64 KB (65,536 / 1024).
Q163
A digital document file tracking index updates its configuration profiles to exactly 4 Megabytes (MB). What is the exact mathematical size of this file when converted cleanly down into binary data bits inside standard computing matrices?
4,194,304 bits
8,388,608 bits
33,554,432 bits
34,359,738,368 bits
Explanation
Conversion path: 4 MB × 1024 KB/MB = 4,096 KB. 4,096 KB × 1024 Bytes/KB = 4,194,304 Bytes. 4,194,304 Bytes × 8 bits/Byte = 33,554,432 bits.
Q164
An internal CPU cache subsystem reads a hit rate efficiency of exactly 90% with a cache access speed latency of 2 nanoseconds (ns). If tracking misses require falling back to the slower main memory which displays a latency of 20 nanoseconds (ns), what is the computed Effective Memory Access Time (EMAT) for this hardware loop?
2.2 nanoseconds
3.8 nanoseconds
4.2 nanoseconds
11.0 nanoseconds
Explanation
Formula for EMAT: (Hit Rate × Cache Latency) + (Miss Rate × Main Memory Latency). Given Hit Rate = 0.90, Miss Rate = 1 - 0.90 = 0.10. EMAT = (0.90 × 2 ns) + (0.10 × 20 ns) = 1.8 ns + 2.0 ns = 3.8 nanoseconds.
Q165
In computer systems running binary logic representations, what is the exact decimal value computed by performing a bitwise AND operation between the binary sequences 1100 and 1010?
6
8
10
14
Explanation
Perform the bitwise AND: 1 1 0 0 AND 1 0 1 0 ------- 1 0 0 0 Converting the binary result 1000 to decimal: (1 × 2³) + (0 × 2²) + (0 × 2¹) + (0 × 2⁰) = 8.
Q166
A secondary storage magnetic disk drive tracks data layout configurations across exactly 8 platter surfaces natively. If each surface contains 1024 tracking cylinders, and each track maps precisely 64 sectors holding 512 bytes of data per block, what is the absolute storage capacity parameter of this hardware system in Megabytes (MB)?
128 MB
256 MB
512 MB
1024 MB (1 GB)
Explanation
To calculate the absolute capacity: Total Storage = Surfaces × Cylinders × Sectors × Bytes per Sector = 8 × 1024 × 64 × 512 = 268,435,456 Bytes. Converting cleanly to Megabytes based on binary base-2 metrics: 268,435,456 / (1024 × 1024) = exactly 256 MB.
Q167
If a specialized microprocessor core pipeline is natively configured with a 32-bit physical address bus layout, what is the absolute maximum memory capacity range (byte-addressable space mapping thresholds) that this processing chip can directly manage?
64 Kilobytes (KB)
16 Megabytes (MB)
4 Gigabytes (GB)
64 Gigabytes (GB)
Explanation
The maximum direct reference threshold capacity is calculated mathematically as 2^n bytes, where n represents the bit width of the address bus line pipelines. Thus, 2^32 = 4,294,967,296 Bytes. Converting via binary scaling steps: 4,294,967,296 / (1024 × 1024 × 1024) = exactly 4 Gigabytes (GB).
Q168
An active system processing matrix shifts an application software storage layout allocation to exactly 2 Megabytes (MB). What is the exact mathematical length of this code buffer when converted down into individual binary data bits?
2,097,152 bits
8,388,608 bits
16,777,216 bits
17,179,869,184 bits
Explanation
Conversion workflow layout: 2 MB × 1024 KB/MB = 2048 KB. 2048 KB × 1024 Bytes/KB = 2,097,152 Bytes. 2,097,152 Bytes × 8 bits/Byte = exactly 16,777,216 bits.
Q169
A basic hardware chip memory access tracking loop performs with a cache hit efficiency index of exactly 80%. The internal cache memory access speed is 5 nanoseconds (ns), while tracking misses require jumping to the main memory, which exhibits an access speed latency of 50 nanoseconds (ns). What is the calculated Effective Memory Access Time (EMAT) for this loop execution layout?
9 nanoseconds
14 nanoseconds
15 nanoseconds
41 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 × 5 ns) + (0.20 × 50 ns) = 4 ns + 10 ns = exactly 14 nanoseconds.
Q170
In computer binary machine logic representations, what is the exact decimal value output generated by executing a bitwise OR operation between the 4-bit sequences 1010 and 0110?
10
12
14
15
Explanation
Perform the bitwise OR tracking path: 1 0 1 0 OR 0 1 1 0 ------- 1 1 1 0 Converting the binary matrix output 1110 back down to decimal: (1 × 2³) + (1 × 2²) + (1 × 2¹) + (0 × 2⁰) = 8 + 4 + 2 + 0 = exactly 14.