hw6_sol

# hw6_sol - HW6 Solution 3.a Please see the attached Arena...

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HW6 Solution 3.a) Please see the attached Arena model. Results for initial 8 runs… "Project:","Project Schedule Network Figure 6.17" "User:","Kurt Palmer" "Data item:","Record Project Time" "Run date:",10/17/2010 "Options:",YDT ,1 Time,Observation 184.6621681299,184.6621681299 378.66521324171,178.66521324171 578.57526125112,178.57526125112 779.91884966079,179.91884966079 971.53866924382,171.53866924382 1177.4325038682,177.43250386822 1382.8460453036,182.84604530358 1568.4632702679,168.4632702679 -1,-1 From the Arena simulation results, we get 177.76 5.41 x s = = . 2.5 d = , let 0.05 α = , so 2 1.96 z α = ( ) ( ) 2 2 /2 2 2 5.41 1.96 17.99 2.5 s z n d α × = = = Rounding up, use n = 18. 3.b) Simulate the project with sample size 18. Results for final run… "Project:","Project Schedule Network Figure 6.17" "User:","Kurt Palmer" "Data item:","Record Project Time" "Run date:",10/17/2010 "Options:",YDT ,1 Time,Observation 184.6621681299,184.6621681299 378.66521324171,178.66521324171 578.57526125112,178.57526125112 779.91884966079,179.91884966079 971.53866924382,171.53866924382 1177.4325038682,177.43250386822 1382.8460453036,182.84604530358 1568.4632702679,168.4632702679 1782.4501017307,182.45010173066 1967.5036817257,167.50368172565

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2175.0553762607,175.05537626068 2370.6004293299,170.6004293299 2581.8671423714,181.86714237142 2778.3190221618,178.31902216181 2970.1690128635,170.1690128635 3184.3408464676,184.34084646759 3387.2954114095,187.29541140952 3574.6630283898,174.66302838981 -1,-1 From the Arena simulation results, we get 177.46 5.98 x s = = Assume that the project time is normally distributed, i.e. 2 (177.46, 5.98 ) X N ( ) ( ) ( ) ( ) 165 170 165 170 165 177.46 170 177.46 5.98 5.98 2.08 1.25 1.25 2.08 0.1056 0.0188 0.0868 x X x P X P s s P Z P Z μ σ - - - = - - = = - ≤ - = Φ - - Φ - = - = A 95% confidence interval for the mean project time [ ] 2, 2, 0.025,17 , for 18, 2.11 5.98 5.98 177.46 2.11 ,177.46 2.11 18 18 174.5,180.4 s s x t x t n n n t α ν α ν μ μ μ - + = = - + 3.c) ( ) 45000 500 ( ( ) 120) 45000 500 (177.46 120) 16270 E financial result E project time = - × - = - × - = Under the terms of this contract, the company would expect a profit of \$16,270.

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Homework 6 Solution 7.a) See the Arena model flowchart to show that the warm-up period is approximately 2000 minutes. The 6000 minutes of steady-state operation were divided into eight 750-minute batches. The resulting batch means for waiting time (Wq), number waiting (Lq), and instantaneous utilization (rho) are shown below, along with appropriate confidence intervals. MTB > print c1-c3 Data Display Row Wq_FIFO Lq_FIFO rho_FIFO 1 7.7902 1.9079 0.7496 2 6.6538 1.6686 0.7081 3 14.5631 3.5107 0.8012 4 5.7604 1.4481 0.7452 5 10.2121 2.5780 0.8157 6 13.4255 3.5763 0.8741 7 8.9203 2.3428 0.7113 8 4.9131 1.0991 0.6611 MTB > Onet 'Wq_FIFO'-'rho_FIFO'.
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