This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: SOLUTION MA 206 Suggested Problems
Lesson 5 Fitting Models to EDFs NOTE: The problems below are a continuation of the Lesson 4 Suggested Problems. You are
expected to use the EDF you created for Lesson 4, Suggested Problem # 2 to complete the
following exercises. 1. Plot the function F (x) = 1 ~— 6"“ where x,a 2 0 on the same graph as the EDF. To begin
with, set a Z 1. Note that the function F(x) does not appear to match the ED]: very well. a. Increase the value of a in small increments and observe the result.
Increasing the value of a moves F (x) even farther from the EDF. b. Now, decrease the value ofa in small increments and observe the result.
Decreasing the value of (1 moves F (x) closer to the EDF. c. Using your results from above, conduct further experimentation to determine the value of
a that enables F (x) to ﬁt the EDF as closely as possible. Using a = 0.03 yields the following graph: Proportion .__‘____._______.Y_._..M.,_N MM— w.,w O 10 20 30 40 50 Time (min) d. Use Excel’s Solver to determine the best value fora. i Using Solver, we ﬁnd that the best value for a is 0033604. This has a corresponding
minimized SSE of 10.60. 2. Next, add the function G(x) = l —— 6"”)1) where x,a,b 2 O to your graph. To begin with, set
a = 1 and b = 1. It appears that the function G(x) does not fit the EDF very well either. SOLUTION a. Change the value of a to the number you determined in Problem ld (should be
approximately 0.034) and observe the result. Why does the graph of G(x) look exactly like F(x)?
When b = I, G(x) : 1 —~ (WV is mathematically equivalent to the function F(x) 2 l- 6””.
b. Adjust the value of b in small increments (up and down) and observe the results. Increasing the value of [9 moves G07) closer to the EDF. Decreasing the value of 1) moves
G(x) farther from the EDF. 0. Determine the value of b that enables G(x) to ﬁt the EDF as closely as possible. Using a Z 0.03 for F(x) and a = 0.03, b = 2.5 for G(x) yields the following graph: Uniform Prep Time 1.2' Proportion This solution is starting to look better for G(x). In part (2 (below) we will use Solver to
find the optimal value of the parameters in G(X). d. At this point, the function G(x) should provide a relatively good ﬁt to the EDF. As a
ﬁnal step, check to see if any further adjustments to the value of a (up or down) are
necessary in order to ﬁnd a better ﬁt. SOLUTION Using a = 0.03 for F (x) and a = 0.04, b :— 2.5 for 60:) yields the following graph: Uniform Prep Time 1.2' Proportion 6. Use Excel’s Solver to determine the best values for both a and b. Using Solver, the minimized SSE for G(x) occurs when a = 0.039443 and b = 2.850205. f. Why does C(x) provide a much better ﬁt to the EDF than the function F (x)? As we determined in part (a) of this problem, 0(x) is mathematically equivalent to the
function F (x) when I) = 1. However, for b > 1, G(x) changes concavity and therefore more
closely models the behavior of our EDF. F (x) is a more simplistic, less ﬂexible model. ...
View Full Document
- Spring '10