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Example The probability of successful start of a certain engine is ¼ and four trials are to be made. Evaluate the individual and cumulative probabilities of success in this case. n = 4, p = ¼, q = 1 ¼

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Example The probability of successful start of a certain engine is ¼ and four trials are to be made. Evaluate the individual and cumulative probabilities of success in this case. n = 4, p = ¼, q = 1 ¼ = ¾ (p+q) 4 = p 4 + 4p 3 q + 6p 2 q 2 + 4pq 3 + q 4 Number of Cumulative successes failures Individual probability probability 0 4 q 4 = (3/4) 4 = 81/256 81/ pq 3 = 4(1/4)(3/4) 3 = 108/ / p 2 q 2 = 6(1/4) 2 (3/4) 2 = 54/ / p 3 q = 4(1/4) 3 (3/4) = 12/ / p 4 = (1/4) 4 = 1/ /256 Σ = 1 Binomial Distribution Application Example: It is known that, in a certain manufacturing process, 1% of the products are defective. If the a customer purchases 200 of these products selected at random, what is the expected value and standard deviation of the number of defects? n = 50 q = 0.01 p = = 0.99 E(defects) = n.q = 200 x 0.01 = 2 σ(defects) = npq = 200 x 0.01 x 0.99 = Example: The manufacturing company has a policy of replacing, free-of-charge, all defective products that are purchased. If the product manufacturing cost is $10 per unit and each product is sold for $15, how much profit is made from a sale of 1000 products? q = 0.01 For 1000 products, n = 1000 Expected # of defects, E(defects) = n.q = 10 Therefore, 1010 products must be manufactured to sell 1000 products. Manufacturing cost = $10 x 1010 = $10,100 Income = $15 x 1000 = $15,000 Profit = $15,000 - $10,100 = $4900 Profit per unit = $4900/1000 = $4.90 Example: If the company decides to increase the manufacturing cost to $10.05 per unit in order to decrease the probability of defects to 0.1%, q = For 1000 products, n = 1000 Expected # of defects, E(defects) = n.q = 1 Therefore, 1001 products must be manufactured to sell 1000 products. Manufacturing cost = $10.05 x 1001 = $10, Income = $15 x 1000 = $15,000 Profit = $15,000 - $10, = $ Profit per unit = $ /1000 = $4.94 2 Effect of Redundancy Consider a system consisting of 4 identical components, each having a failure probability of 0.1. q = 0.1 (p = 0.9) n = 4 (p+q) 4 = p 4 + 4p 3 q + 6p 2 q 2 + 4pq 3 + q 4 System state Individual probability all components working p 4 = (0.9) 4 = working, 1 failed 4p 3 q = 4(0.9) 3 (0.1) = working, 2 failed 6p 2 q 2 = 6(0.9) 2 (0.1) 2 = working, 3 failed 4pq 3 = 4(0.9)( 0.1) 3 = all components failed q 4 = (0.1) 4 = Σ = 1 Consider 4 criteria all components required for success (no redundancy) components required for success (partial redundancy) 2 components required for success (partial redundancy) 1 component required for success (full redundancy) System reliability, R = = = System with Derated States Consider a generation plant with two 10 MW units, each having a probability of failure (forced outage rate) of 10%. q = 0.1, p = 0.9, n = 2 Binomial Distribution: Capacity Outage Probability Table: (p + q) 2 = p 2 + 2pq + q 2 Units Out Cap Out (MW) Cap In (MW) Probability Cum. Prob If the generation plant operates to supply a 15 MW load, what is the probability of load loss (system failure)? Probability of load loss = Loss of Load Probability (LOLP) = 0.19 Expected # of days of load loss = 0.19 x 365 = days/yr Loss of Load Expectation (LOLE) 3 System with Derated States If the generation plant operates to supply a 15 MW load, what is the Expected Load Loss (ELL)? Capacity Outage Probability Table: Units Out Cap Out (MW) Cap In (MW) Probability Cum. Prob Units Out Cap Out (MW) Cap In (MW) Load Loss (MW) Probability Col.4 x Col Expected Load Loss (ELL) = 1.05 MW Example A generating plant is to be designed to satisfy a constant 10 MW load. Four alternatives are being considered: a) 1 x 10 MW unit b) 2 x 10 MW units c) 3 x 5 MW units d) 4 x 3.33 MW units The probability of unit failure is assumed to be For each unit, q = forced outage rate (FOR) = unavailability, U = 0.02 p = availability, A = Capacity Outage Probability Tables units capacity (MW) Binom. Individual cum. out out in Distr. prob prob (a) 1 x 10 MW A U (b) 2 x 10 MW A AU U (c) 3 x 5 MW A A 2 U AU U (d) 4 x 3.33 MW A A 3 U A 2 U AU U LOLP = 0.01 LOLE = 365 x 0.02 = 7.3 d/yr LOLP = LOLE = 0.15 d/yr LOLP = LOLE = 0.43 d/yr LOLP = LOLE = 0.85 d/yr Expected Load Loss Capacity (MW) Load Loss Prob L i x p i Out In L i (MW) p i (MW) (a) 1 x 10 MW (b) 2 x 10 MW (c) 3 x 5 MW (d) 4 x 3.33 MW ELL = 0.2 MW = 200 kw ELL = MW = 4.00 kw ELL = MW = 5.96 kw ELL = MW = 7.89 kw 5 Comparative Analysis Effect of unit unavailability: System ELL (kw) at Different Unit FOR s 2% 4% 6%) (a) 1 x 10 MW (b) 2 x 10 MW (c) 3 x 5 MW (d) 4 x 3.33 MW System with Non-identical Components All components must be identical to apply the Binomial Distribution. If components of a system have non-identical capacities: - Units with identical capacities are grouped together - COPT is developed for each group - COPT for different groups are combined, one at a time - Final COPT for the system is used for reliability evaluation A pumping station has 2 x 20 t/hr units, each having an unavailability of 0.1, and 1 x 30 t/hr unit with an unavailability of Calculate the capacity outage probability table for this plant. COPT for 2 x 20 t/hr units: COPT for 1 x 30 t/hr unit: Units Out Cap Out (t/hr) Cap In (t/hr) Prob Units Out Cap Out (t/hr) Cap In (t/hr) Prob System with Non-identical Components Combining COPTs: 1 x 30 t/hr unit 2 x 20 t/hr units 40 / / / / / / / / / / / Each cell contains: Capacity In / probability Overall System COPT: Cap In (t/hr) Cap Out (t/hr) Prob

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