GATE2004 IT q50 - Revision history

Arjun Suresh at 11:31, 15 July 2014

2014-07-15T11:31:16Z

2014-07-13T22:45:53Z

2014-04-15T07:08:58Z

2014-04-14T14:04:31Z

2014-02-27T22:01:39Z

2014-02-27T22:01:04Z

2013-12-16T09:11:21Z

2013-12-16T09:10:41Z

2013-12-15T20:56:28Z

2013-12-15T20:54:47Z

← Older revision		Revision as of 11:31, 15 July 2014
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	[[Category: GATEIT2004]]		[[Category: GATEIT2004]]
−	[[Category: Architecture questions]]	+	[[Category: Architecture questions from GATE]]

← Older revision		Revision as of 22:45, 13 July 2014
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−	~~[[Category: Computer Architecture]]~~
	[[Category: GATEIT2004]]		[[Category: GATEIT2004]]
	[[Category: Architecture questions]]		[[Category: Architecture questions]]

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 [[Category: Computer Architecture]]
 [[Category: GATEIT2004]]
-[[Category: Previous year GATE questions]]
+[[Category: Architecture questions]]

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 '''(A) 1.155''' (B) 1.185 (C) 1.255 (D) 1.285
-===Solution===
+==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
   Speed up = Original time taken/ new time taken
   Let $x$ be the time for a fixed point operation

← Older revision		Revision as of 22:01, 27 February 2014
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−	In an enhancement of a design of a CPU, the speed of a floating point until has been increased by ~~<math>~~20%~~</math>~~ and the speed of a fixed point unit has been increased by ~~<math>~~10%~~</math>~~. What is the overall speedup achieved if the ratio of the number of floating point operations to the number of fixed point operations is ~~<math>~~2:3~~</math>~~ and the floating point operation used to take twice the time taken by the fixed point operation in the original design?	+	In an enhancement of a design of a CPU, the speed of a floating point until has been increased by 20% and the speed of a fixed point unit has been increased by 10%. What is the overall speedup achieved if the ratio of the number of floating point operations to the number of fixed point operations is 2:3 and the floating point operation used to take twice the time taken by the fixed point operation in the original design?

	'''(A) 1.155''' (B) 1.185 (C) 1.255 (D) 1.285		'''(A) 1.155''' (B) 1.185 (C) 1.255 (D) 1.285

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   Speed up = Original time taken/ new time taken
   Let $x$ be the time for a fixed point operation
-  Original time taken = (3x + 2*2x)/5 = 7x/5
+  Original time taken = (3$x$ + 2*2$x$)/5 = 7$x$/5
-  New time taken = ((3x/1.1) + (4x/1.2))/5 = 8x/1.32*5
+  New time taken = ((3$x$/1.1) + (4$x$/1.2))/5 = 8$x$/1.32*5
   So, speed up = 7*1.32/8 = 1.155