Video memory sizing: How much video RAM is required to support a resolution of 1024 × 768 at 24-bit true color (no overlays or extra buffers)?
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A1MB
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B4MB
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C8MB
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D512KB
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ENone of the above
Answer
Correct Answer: 4MB
Explanation
Introduction / Context:Sizing video memory for a given resolution and color depth is a classic calculation. Even though modern GPUs have abundant VRAM, understanding the framebuffer requirement remains useful for embedded systems, thin clients, and legacy hardware.
Given Data / Assumptions:
- Resolution: 1024 × 768 pixels.
- Color depth: 24 bits per pixel (true color), which equals 3 bytes per pixel.
- Only a single framebuffer is considered (no double buffering, Z-buffer, or textures).
Concept / Approach:
Video memory required for a single framebuffer equals pixel_count * bytes_per_pixel. Compute pixel_count, multiply by 3 bytes, then convert bytes to megabytes. Choose the smallest offered VRAM size that meets or exceeds this requirement from the options provided.
Step-by-Step Solution:
Compute pixels: 1024 * 768 = 786,432 pixels.Bytes per pixel at 24-bit: 3 bytes.Required bytes: 786,432 * 3 = 2,359,296 bytes.Convert to MB (1 MB = 1,048,576 bytes): 2,359,296 / 1,048,576 ≈ 2.25 MB.Select the next standard size ≥ 2.25 MB from options: 4MB.Verification / Alternative check:
Older graphics cards offering 2 MB could not do 1024 × 768 at 24-bit; cards with 4 MB were commonly specified for this mode, confirming the selection.
Why Other Options Are Wrong:
- 512KB / 1MB: Insufficient for the 2.25 MB requirement.
- 8MB: Works but is not the minimum required from the set; the question expects the adequate choice.
- None of the above: Incorrect because 4MB meets the need.
Common Pitfalls:
Confusing bits with bytes; forgetting overhead for additional buffers (which would further increase the requirement); assuming decimal megabytes instead of binary in legacy specs.
Final Answer:
4MB