Operations Management Assignment

7. The kinetic graph was correct as my hypothesis. My potential graph did not start in the same position as my data graph.

5) Graph the equation you wrote in step four superimposed over the original data. Comment on how well or how poorly the equation fits the data.

In order for me to find the appropriate model I need to evaluate them by using ln(Y) and Ŷ calculations. I analyzed the Linear: Ŷ= 10.4446 + .755 R2= .9719 , squared: Ŷ= 11.31 + .5951(x1) + .0062(x12) R2= .9745 , cubic: Ŷ= 12.0508 + .3627 (x1) + .0264(x12) - .00051 (x13) R2= .9751 , log log: ln(Y)= exp(2.0403 + ln(x1) + (.0597/2)^2 R2=.9072 , logarithmic: Ŷ = 2.1503 + ln(x1) R2= .8523 , and exponential: ln(Y) = exp(2.4924 + .0383(x1) + (.0375/2)^2 R2= .9634. After analyzing the R2’s I narrowed it down to Cubic. Cubic has the highest R2 and is the most promising of the models. Based on my results from (b) and (c) also regards to how I achieved Mileage Cubic to be the best

1. This exercise involves the Auto data set studied in the lab. Make sure that the missing values have been removed from the data

6. The smallest interval was 10OC, i.e. the difference between 20OC and 30OC. Yes this was a suitable interval because by moving from 20OC to 30OC there was a difference in the resistance of _____ Ohms. It was also possible to see a pattern in the results using intervals of 10OC. However to create a more accurate line of best fit I could have used intervals of 5OC. If I had the resistance value of 15OC would have been ______ Ohms (read this off your graph).

(Mrs Bolding I can’t really talk about accuracy and precision because I don’t now how to create a graph from these results…I tried to dicuss about error and unexpected results in the discussion… that’s about it, so

167 2) was converted to a linear form by the logarithmic transformation of the dependent variable. Then,

Sampling 4 pieces of precision-cut wire (to be used in computer assembly) every hour for the past 24 hours has produced the following results:

D | 12 | 7.11 | E | 10 | 1.00 | F | 6 | 0.44 | G | 8 | 7.11 | H | 6 | 0.44 | I | 7 | 2.78 | J | 4 | 0.11 | K | 4 | 0.44 | Total Expected time 76 weeks Activity | ES | EF | LS | LF | Slack | Critical |

1) Expected Return on Stock Market E(Rstock market) = [(0.20 * -0.10) + (0.35 * 0.10) + (0.30 * 0.15) + (0.15 * 0.25)

(2) > # Find 1st derivative; > D(f); (3) > # Find turning/stationary point in the interval [1,2], 1st derivative expression = 0, 10 significant figures!; > evalf(fsolve(3*cos((1/4)*x^4)*x^3-3*sin((3/4)*x)^3*cos((3/4)*x)= 0,x=1..2),10); 1.562756908 (4) > # Find 2nd derivative at x= 1.562756908; 10 significant figures!; > evalf(D[1$2](f)(1.562756908),10); (5) > # Remember to unassign variables/restart; > restart; > # Question 3; (same process as above) > f:=x-> 2^x + 2*cos(x); (6) > # Find 1st derivative; > D(f); (7) > # Find turning/stationary point

The following is a graphic representation of the new function found and the given data. The new function found is much more accurate when comparing to the linear function found earlier.

This paper intends to define operations management and analyze an ethics decision made by operations managers in the workplace or in a known organization.

a. What is the value of the slope parameter for the price of the good? Does it have the correct algebraic sign?

From equation 6 I obtained the linear equation 7, model to be used in the study.