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Industrial Engineering

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Jan 9, 2024

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3-7 The airlines industry, customer requirements are: Price of ticket, convenience of schedule, delay in arrival, lost baggage, and in-flight service, among others. Based on customer survey, priority ratings may be applied to the above requirements. Technical descriptors are: Select cities to serve based on competition and demand, type and size of fleet, baggage identification barcoding and handling procedures, training of in-flight attendants and provision of desirable meals. The relationship matrix should be completed through allocation of 31 relative scores to the degree to which each technical descriptor meets each customer requirement. 3-30 As major errors happen in Typographical and proofreading these areas should be targeted and the $18000 should be allotted to these areas accordingly. 3-31 According to the given data, with an increase in disposable income, life insurance coverage increases non-linearly.

35 40 45 50 55 60 65 70 75 80 0 20 40 60 80 100 120 140 160 4-2 a) The appropriate distribution for modeling the number of customers that arrive in a 2-hour period will be Poisson distribution. b) The stated assumptions not hold true in situations where the rate of arrival is different for two or more different time slots of 2 hours. c) We will have to select a two hour period and observe the number of customers who enter the store during that period and then Based on data collected we can estimate the mean number of arrivals and calculate the probability distribution. d) We have to obtain data on population size, average income, number of similar stores in proximity, etc. for the new location. Based on locations, with similar characteristics, where there are existing stores, estimate the mean number of arrivals in a given time period. 4-8 P(X = 0)=5c₀+20c₄/25c₄=(4850/12650)=0.3833 P(x=1)=5c₁+20c₃/25c₄=(5700/12650)=0.4505 E(X) = 4(5)/25 = 0.8 Variance (X) =(4*5)/25*(20/25)(25-4)/(25-1)=14/25=0.560 Standard deviation(X)=√0.560=0.748

4-21 Using Cumulative Binomial probabilities a) P(x ≥3)=1-p(x ≤2)=1-0.889=0.111 b) P(x ≤5)=0.999 c) P(l < X < 5) = P(X < 5) - P(X < 0) = 0.999 - 0.282 = 0.717 d) Expected number of sensors that will malfunction = 12(0.10) = 1.2 e) Standard deviation(X) = √12(0.10)(0.90) = 1.039 4-30 a) μ = 0.98, σ = 0.02, specification limits are 1.0 + 0.04 mm. z1=(0.96-1)/0.02=-2.00 z2=(1.04-1)/0.02=2.00 Proportion of conforming washers = 1 - (0.1587 + 0.0013) = 0.84. Daily cost of scrap = 10000 (0.1587)(0.15) = $238.05. Daily cost of rework = 10000 (0.0013)(0.10) = $ 1.30. Total daily cost of rework and scrap = $239.35. b) μ =1.0, σ = 0.02. z1=(0.96-1)/0.02=-2.00 z2=(1.04-1)/0.02=2.00 Proportion of rework = 0.0228; proportion of scrap = 0.0228. Total daily cost of scrap and rework = 10000(0.0228)(0.15 + 0.10) = $57 c) μ = 1.0, σ = 0.015 z1=(0.96-1)/0.015=-2.67 z2=(1.04-1)/0.015=2.67 Proportion of scrap = 0.0038; proportion of rework = 0.0038. Total daily cost of scrap and rework = 10000(0.0038)(0.15 + 0.10) = $9.50.

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Percentage decrease in the total daily cost of rework and scrap compared to part a) = (239.35 - 9.50)7239.35 = 0.9603 = 96.03%. 4-45 Let μ , = mean time lost before implementation of OSHA program, and μ 2 = mean time lost after implementation of OSHA program. It is given that θ \ = 02 = 3.5 hours, ni = 40, X, = 45, n2 = 45, and J 2 =39 . a) 90% confidence interval for (u1-u2):((45-39) ±z.05√(3.5)*2/45-+3.5)*2/45) =6 ±(1.645)(0.7606)=6 ±1.251=(4.749,7251). b) H0:u1-u2 ≤0,ha:u1-u2>0. The test statistic is Z0=(45-39)/ √(3.5)*2/45-+3.5)*2/45)=7.889 The critical value of z is 1.282, with the rejection region of the null hypothesis being z0 > 1.282. Since z0 = 7.889 > 1.282, we reject H0 and conclude that implementation of the OSHA program has reduced the mean employee lost time.

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