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Design and Uncertainty Discussion Points Mean Variance Standard Deviation Probabilities Uniform Distribution Normal Distribution Computational Form for Variance and Standard Deviation Poisson Distribution Manufacturing Scheduling and Uncertainty Mean ( µ ) Given a set of N numbers, X1 , X2 , ... , XN , the mean, or average, is calculated by summing the numbers and dividing by N. N ∑ Xi i=1 µ = --------N For example, for the numbers 3, 4, 5 µ = (3+4+5)/3 = 12 / 3 = 4 Variance ( σ2 ) The variance is a measure of the spread of a distribution of numbers. It is computed as the average squared deviation of each number from the mean of the distribution. N ∑ ( X i - µ)2 i=1 σ = --------------N 2 For the numbers 3, 4, 5 σ2 = ( (3 – 4)2 + (4 – 4)2 + (5 – 4)2 ) / 3 = (1 + 0 + 1) / 3 = 0.667 Standard Deviation ( σ ) The standard deviation of a distribution of numbers is the square root of the variance. For the numbers 3, 4, 5 σ2 = 0.667 σ = 0.816 Another Example For the numbers 2, 4, 6 µ = (2+4+6)/3 = 12 / 3 = 4 σ2 = ( (2 – 4)2 + (4 – 4)2 + (6 – 4)2 ) / 3 = (4 + 0 + 4) / 3 = 2.667 σ = 1.633 Coin Toss Possible Outcomes (Head) (Tail) Each outcome has a probability of 1/2. Single Standard Die Possible Outcomes (1) (2) (3) (4) (5) (6) Each outcome has a probability of 1/6. These are both Uniform Distributions. Pair of Standard Dice Possible Outcomes (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Frequency Distribution 2 3 4 5 6 7 8 9 10 11 12 - 1 2 3 4 5 6 5 4 3 2 1 This is a Normal Distribution. Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 The 68-95-99.7 Rule All normal density curves satisfy the following property, which is often referred to as the Empirical Rule. 68% of the observations fall within 1 standard deviation of the mean, that is between - and + . 95% of the observations fall within 2 standard deviations of the mean, that is between - 2 and + 2. 99.7% of the observations fall within 3 standard deviations of the mean, that is between - 3 and + 3. For a pair of Standard 6-Sided Dice: =7 2 = 5.833 = 2.415 - to + = 4.585 - 9.415 24 / 36 = 66.666% - 2 to + 2 = 2.17 - 11.83 34 / 36 = 94.444% - 3 to + 3 = -0.245 - 14.245 36 / 36 = 100.0% Computation Form for Variance Computational form for the variance in a population. All summations are assumed to be from i equals 1 to N. ∑ (X - µ)2 σ2 = --------------N ∑ (X2 - 2 X µ + µ2) = ---------------------------N ∑ X2 - 2 µ ∑ X + ∑ µ2 = -------- ----------- -----N N N ∑ X2 - (2 µ) µ + µ2 = -------N ∑ X2 - µ2 σ2 = -------N The Poisson Distribution The Poisson distribution is used to model the number of random occurrences of some phenomenon in a specified unit of space or time. For example, The number of phone calls in a 30-minute period. The number of defective parts produced during an eight hour shift. The number emergency assistance calls received in an hour. For a Poisson random variable, the probability that X is some value x is given by the formula: x e- P (X = x) = -------x! x = 0, 1, . . . where: : mean number of occurrences in a specified interval (time period). x : number of occurrences of interest. e : Base of the natural logarithmic function ln ( 2.71828 ) Example On an average day in January, a furnace repair person receives 4 calls for emergency furnace repair service. Find the probability that this person will receive 5 calls today. (i.e., = 4, X = 5) 45 e-4 P (X = 5) = -------- = 0.1563 5! Find the probability that this person will receive fewer than 4 calls today. (i.e., = 4, X < 4) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335 Problem Statement The plant manager of a toy manufacturing plant has been given an order for 10,000 new toys with a shipping date 10 weeks from now. He currently has 8 machines that produce 100 toys per week per machine, assuming one shift of 40 hours per week. He needs to know if the machines need to run overtime to meet the order and if so how much overtime is necessary. The goal is to minimize overtime. Calculate how many toys the 8 machines will produce in 10 weeks working one 40 hour shift. Do the machines need to run additional shifts (each shift for each machine is 40 hours per week)? Write a formula for calculating the number of weeks with an extra shift to meet this order in 10 weeks. Now, in actuality, the machines have been known to produce on the average 2 defective toys in a week (one shift) per machine. Also, the machines periodically require maintenance. This occurs on the average of 1 machine down for maintenance per week (one shift), with the maintenance requiring 1 hour. With these uncertainties calculate the necessary extra weeks of overtime to meet the order. Given: Ttotal = 10,000 toys 8 machines 10 weeks production time 100 toys/machine . shift-week Determine R, the rate of production for all 8 machines for one shiftweek R = (8 machines * 100 toys/ machine . shift-week) = 800 toys/shift-week Determine Tp, the production for 10 one shift-weeks Tp = 800 toys/shift-week * 10 shift-weeks Tp = 8000 toys But we need 10000, so we are 2000 short. How many shift-weeks to add (as overtime) to production? 2000 toys * 1 shift-week/800 toys = 2.5 shift-weeks of overtime Write a formula for calculating the number of weeks with an extra shift to meet this order in 10 weeks. S (shift-weeks) = T (toys) / R (toys/shift-week) Dealing with uncertainty The number of defective toys produced are the number of toys lost each shift-week. Average 2 defective toys in one shift-week per machine. Use Poisson Distribution as model. λ =2 P(X = 0) = 0.1353 P(X = 1) = 0.2707 P(X = 2) = 0.2707 P(X = 3) = 0.1804 P(X = 4) = 0.0902 ---------------------P(X = 5) = 0.0361 Let L = toys lost/shift-week, based on above probability, choose X = 4 L = 8 machines * 4 toys lost/machine . shift-week L = 32 toys lost/shift-week Tlost = L * 12.5 shift-weeks Tlost = 32 toys lost/shift-week * 12.5 shift weeks = 400 toys lost Substitute into formula: Shift-weeks = 400 toys / 800 toys/shift-week = 0.5 shift-weeks Dealing with uncertainty, Part 2 Average 1 machine down for maintenance in one shift-week with maintenance requiring 1 hour to complete. Use Poisson Distribution as model. λ =1 P(X = 0) = 0.3679 P(X = 1) = 0.3679 P(X = 2) = 0.1839 P(X = 3) = 0.0613 ----------------------P(X = 4) = 0.0153 P(X = 5) = 0.0031 Let H = hours lost to maintenance, choose X = 3 machines H = 3 machines * 1 hr/machine . shift-week = 3 hrs/shift-week 3 hr/shift-week * 13.0 shift-weeks = 39.0 hours 39.0 hours * 2.5 toys/hour = 97.5 toys lost Substitute into formula: Shift-weeks = 97.5 toys / 800 toys/ shift-weeks = 0.1219 Putting it all together: Total production time = 12.5 shift-weeks + 0.5 shift-weeks + 0.1219 shift-weeks Total production time = 13.1219 shift-weeks Any time over 10 shift-weeks is overtime. Therefore the plant manager will have to schedule 4 shift-weeks of overtime during the 10 week run of production on this new toy.