1. Previously Optimization Probability Review Inventory Models Markov Decision Processes Queues

2. Agenda Simulation

3. Simulation (Ch 15) Interested in quantity X (it is random) Run simulation to get realizations of X: –X 1, X 2, X 3, …, X n Evaluate output: –look at averageE[X] ≈ AVERAGE(X 1,…,X n ) –standard deviation  [X] ≈ STDEV(X 1,…,X n ) –distribution of realizations

4. Agenda Confidence intervals –for output evaluation Generating realizations

5. Independent Case Suppose X 1, X 2, X 3, …, X n are independent, –a n = AVERAGE(X 1,…,X n ) –s n = STDEV(X 1,…,X n ) E[X] in [a n -y, a n +y] with probability 1-p –y = - NORMINV (p/2, 0, s n / n 1/2 ) –y = - NORMINV (p/2, 0, 1) s n / n 1/2 –y decrease to 50%  1/n 1/2 decrease to 50%  n 1/2 increase to 2x  n increase to 4x

6. Rare Event X = 1 with probability q, 0 otherwise –q small s n ≈ (q(1-q)) 1/2 ≈ q 1/2 > q ≈ a n for a decent error s n /n 1/2 ≈ 10% a n n ≈ (100)/q suppose y=10% a n, 1-p=90% –y= - NORMINV (p/2, 0, 1) s n / n 1/2 –1.64=- NORMINV (5%, 0, 1) –so, s n /n 1/2 = 6.1% a n –n ≈ 271/q

7. Standard Deviation X = # customers in line at lunch place at noon Data X 1,…,X 20 from last month (n=20) a n =6.2, s n =1.8 Want 90% confidence interval for  [X]:

8. Standard Deviation Want 90% confidence interval for  [X]. Var[X] = E[ (X-E[X]) 2 ] Idea: Y = (X-a n ) 2 –Get confidence interval on E[Y] ≈ Var[X] –Confidence interval on  [X]

9. Not independent a n ≈ E[X], s n ≈  [X] No good bounds on –Error –Necessary n –Generally larger Queueing example (p534): Estimating L q –relative error 10%, 1-p=90%,  =0.8 –n ≈ 240,000 arrivals needed