opm ASSGMNT 5.docx - Pg 209 15 The Money Pit a Lower Specification Calculation Upper Specification Calculation 13.066 5.00 0.64 3 4.21 25.00 13.066 0.94

opm ASSGMNT 5.docx - Pg 209 15 The Money Pit a Lower...

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Pg. 209 15. The Money Pit a Lower Specification Calculation   13.066 5.00 0.64 3 4.21 Upper Specification Calculation   25.00 13.066 0.94 3 4.21 min 0.64, 0.94 0.64 pk C 25 5 0.79 6 4.21 p C b Because p C and pk C have values less than 1, the process is not capable of meeting specifications c The variability of the process must be greatly reduced. Also, the process should be better centered between the specification limits. Pg. 259 Q5. Service rate µ = 60/3 min. per customer = 20 customers/hour. a. Average utilization, ρ=λ / s µ =50/ [3(20)]= 0.8333. b. = [ 1 + 2.5+3.125 + 15.622 ] 1 = 0.0449 c. = [(0.0049)(50/20)^3(0.8333)] ÷ [3! (1-0.8333)] =3.5063 d.
= 3.5063/50 = 0.0701 hours or 4.2 minutes e. L = λ W = λ ( W q + ( 1 / µ ) ) = 50 (0.0701+1/ 20 ) = 6.05 or 6 customers Pg 281 1 Bill’s Barbershop a 10 + 8 + (15+10)/2 + 9 = 39.5 minutes b Step B1 is the bottleneck, it can only handle 6 customers per hour while the rest of the steps can handle 7.5, 10 (60/10 +60/15), and 6.67 customers per hour. c This process is limited by step B1, therefore the entire process can only serve 6 customers per hour. 2 Barbara’s Boutique a 3 [the bottleneck is step T4 at 18 minutes – 3.33 customers per hour or 3 cust/hr] b Step T6 at 22 minutes limits Type B to 60/22 = 2.73 customers/hr or 2 cust/hr. c 3.33(.3) + 2.73(.7) = 2.91 customers on average With an arrival rate greater than 5 customers per hour into the process, then type A

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• Fall '17
• Krejewski
• Distance, Euclidean space