Suppose a system with an 8-inch by 10 –inch video screen that can display 100 pixels per inch .

Suppose a system with an 8-inch by 10 –inch video screen that can display 100 pixels per inch . If a color lookup table with 64 positions is used with this system , what is the smallest possible size (in bytes ) for the frame buffer ?

The Correct Answer and Explanation is:

Answer:

Smallest possible size for the frame buffer = 64,000 bytes


Explanation

To determine the smallest possible size of the frame buffer, we need to understand how a frame buffer stores image data and the role of the color lookup table (CLUT).


1. Screen Resolution:

  • Screen size: 8 inches × 10 inches
  • Pixels per inch (PPI): 100
  • So the resolution is: 8×100=800 pixels (height)8 \times 100 = 800 \text{ pixels (height)} 10×100=1000 pixels (width)10 \times 100 = 1000 \text{ pixels (width)} Total pixels=800×1000=800,000 pixels\text{Total pixels} = 800 \times 1000 = 800,000 \text{ pixels}

2. Color Lookup Table (CLUT):

  • The system uses a Color Lookup Table with 64 positions.
  • Each pixel stores an index into this table instead of full color values (like RGB).
  • To index 64 colors, we need log⁡2(64)=6\log_2(64) = 6 bits per pixel.
  • Frame buffer must store all pixel indices using 6 bits per pixel.

3. Frame Buffer Size Calculation:

  • Each pixel needs 6 bits.
  • Total bits = 800,000×6=4,800,000 bits800,000 \times 6 = 4,800,000 \text{ bits}
  • Convert bits to bytes: 4,800,0008=600,000 bytes\frac{4,800,000}{8} = 600,000 \text{ bytes}

However, the smallest possible frame buffer must pack the data efficiently. Many systems pack 6-bit values without padding.

But digital systems often store data in whole bytes. So we look for the smallest possible size that can store 6-bit values for 800,000 pixels without wasting space:

  • Each group of 4 pixels = 24 bits = 3 bytes
    So, 800,000 pixels ÷ 4 = 200,000 such groups 200,000×3=600,000 bytes200,000 \times 3 = \boxed{600,000 \text{ bytes}}

Correction: Earlier error! The final size is 600,000 bytes, not 64,000.


Correct Answer: 600,000 bytes

This is the smallest size needed to store 800,000 pixels at 6 bits per pixel using a 64-color CLUT.

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