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Records Management Program <[log in to unmask]>
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From:
Glen Sanderson <[log in to unmask]>
Date:
Wed, 24 Feb 2016 12:50:49 +0000
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Records Management Program <[log in to unmask]>
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For those of you that want to understand the calculations behind the question here you go. (thanks to PS for supplying the calculation and ?)

Since 65,536 is 2 16, we will need 16 bits for each pixel. We then compute the amount of video memory required: 
1.	? bytes = number of pixels * bytes per pixel 
2.	number of pixels = 800 * 600 = 480,000 pixels 
3.	bytes per pixel = 16 bits / pixel / ( 8 bits / byte ) = 2 bytes / pixel 
4.	bytes of video RAM = 480,000 pixels * 2 bytes / pixel = 960,000 bytes 
We can again use proportionalities to do additional problems: 
•	For 1024 by 768 16 bit resolution, we need 
960,000 bytes * ( 1024 * 768 pixels / ( 800 * 600 pixels))
= 1,572,864 bytes / ( 2 20 bytes / MB )
= 1.5 MB 
•	a Truecolor (24 bits per pixel) digital camera image which is 2048 by 1536 pixels requires 
960,000 bytes * 24 bits / pixel / ( 16 bits / pixel) * 2048 * 1536 pixels / ( 800 * 600 pixels)
= 9,437,184 bytes / ( 2 20 bytes / MB )
= 9 MB


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