Another specialty valve widely used in hydronic systems provides the ability to isolate, flow balance, and even drain individual
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Answer:
a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Explanation:
From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.
Part 1
For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time
The reduction on this case is
And since the new total time would be given by
Part 2
For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time
The original time for INT operations is calculated as:
For this part the only time that was changed is assumed the INT operations so then:
And then:
And we can quantify the decrease using the relative change:
of reduction
Part 3
A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.
Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + ..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 +
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - ) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - )
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = ..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 =
solve it we get
e = 2%