We know that f(x) = 48x° +42.x° + 3.5x – 16x°. 61г — 16.52? | We have also been given the initial bounds of -1 and 4 with an error tolerance of 0.08 so we start with checking the values of f(x) at x = -1 and 4 as the lower bound and the upper bound respectively. To carry out the bisection method we form the following table where the iterations are repeated until Ɛ or the error tolerance is less than or equal to a value of 0.08. b f(a) f(b) c=(a+b)/2 f(c) Ɛ = b-a a -1 4 -187 -29562 1.5 -16.6875 -1 1.5 -187 -16.6875 0.25 0.3046875 2.5 0.304688 -16.6875 0.304688 0.24823 0.25 1.5 0.875 0.24822998 1.25 0.25 0.875 0.5625 0.312483788 0.625 0.304688 0.312484 0.40625 0.312412 0.312484 0.484375 0.25 0.5625 0.312411889 0.3125 0.40625 0.5625 0.31249994 0.15625 0.484375 0.5625 0.3125 0.312484 0.523438 0.312499696 0.078125 From these calculations at the error tolerance of .078 < 0.08 we get to know that the maximization of the above function will happen approximately between 0.48 and 0.56 with the approximate value of 0.3125.
We know that f(x) = 48x° +42.x° + 3.5x – 16x°. 61г — 16.52? | We have also been given the initial bounds of -1 and 4 with an error tolerance of 0.08 so we start with checking the values of f(x) at x = -1 and 4 as the lower bound and the upper bound respectively. To carry out the bisection method we form the following table where the iterations are repeated until Ɛ or the error tolerance is less than or equal to a value of 0.08. b f(a) f(b) c=(a+b)/2 f(c) Ɛ = b-a a -1 4 -187 -29562 1.5 -16.6875 -1 1.5 -187 -16.6875 0.25 0.3046875 2.5 0.304688 -16.6875 0.304688 0.24823 0.25 1.5 0.875 0.24822998 1.25 0.25 0.875 0.5625 0.312483788 0.625 0.304688 0.312484 0.40625 0.312412 0.312484 0.484375 0.25 0.5625 0.312411889 0.3125 0.40625 0.5625 0.31249994 0.15625 0.484375 0.5625 0.3125 0.312484 0.523438 0.312499696 0.078125 From these calculations at the error tolerance of .078 < 0.08 we get to know that the maximization of the above function will happen approximately between 0.48 and 0.56 with the approximate value of 0.3125.
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter12: Queueing Models
Section12.3: The Exponential Distribution
Problem 3P
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Please see screenshot. I need the formulas to achieve this result, please.
Transcribed Image Text:We know that f(x) :
482 + 42.x + 3.5x – 162® – 614 – 16.5æ?
We have also been given the initial bounds of -1 and 4 with an error tolerance of 0.08 so we start with
checking the values of f(x) at x = -1 and 4 as the lower bound and the upper bound respectively.
To carry out the bisection method we form the following table where the iterations are repeated until Ɛ or
the error tolerance is less than or equal to a value of 0.08.
f(a)
f(b)
|c=(a+b)/2|
f(c)
E = b-a
a
-1
4
-187
-29562
1.5
-16.6875
-1
1.5
-187
-16.6875
0.25
0.3046875
2.5
0.25
1.5
0.304688 -16.6875
0.875
0.24822998
1.25
0.25
0.875
0.304688 0.24823
0.5625
0.312483788
0.625
0.25
0.5625
0.304688 0.312484 0.40625
0.312411889
0.3125
0.40625
0.5625
0.312412 0.312484 0.484375
0.31249994
0.15625
0.484375
0.5625
0.3125
0.312484 0.523438
0.312499696
0.078125
From these calculations at the error tolerance of .078 < 0.08 we get to know that the maximization of the
above function will happen approximately between 0.48 and 0.56 with the approximate value of 0.3125.
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