1. INDUSTRIAL ENGINEERINGLINEAR PROGRAMMING n nThe general linear programming (LP) problem is: Minimize Z = ! ! cij xij i=1j=1 Maximize Z = c1x1 + c2x2 + … + cnxn subject toSubject to: n n a11x1 + a12x2 + … + a1nxn ≤ b1 ! xij - ! x ji = bi for each node i ! N j=1 j=1 a21x1 + a22x2 + … + a2nxn ≤ b2 and h 0 ≤ xij ≤ uij for each arc (i, j)∈A am1x1 + am2x2 + … + amnxn ≤ bm x1, … , xn ≥ 0 The constraints on the nodes represent a conservation of ﬂow relationship. The ﬁrst summation represents total ﬂow out ofAn LP problem is frequently reformulated by inserting non- node i, and the second summation represents total ﬂow intonegative slack and surplus variables. Although these variables node i. The net difference generated at node i is equal to bi.usually have zero costs (depending on the application), theycan have non-zero cost coefﬁcients in the objective function. Many models, such as shortest-path, maximal-ﬂow,A slack variable is used with a "less than" inequality and assignment and transportation models can be reformulated astransforms it into an equality. For example, the inequality minimal-cost network ﬂow models.5x1 + 3x2 + 2x3 ≤ 5 could be changed to 5x1 + 3x2 + 2x3 + s1 = 5if s1 were chosen as a slack variable. The inequality STATISTICAL QUALITY CONTROL3x1 + x2 – 4x3 ≥ 10 might be transformed into3x1 + x2 – 4x3 – s2 = 10 by the addition of the surplus variable Average and Range Chartss2. Computer printouts of the results of processing an LP n A2 D3 D4usually include values for all slack and surplus variables, the 2 1.880 0 3.268dual prices, and the reduced costs for each variable. 3 1.023 0 2.574 4 0.729 0 2.282Dual Linear Program 5 0.577 0 2.114Associated with the above linear programming problem is 6 0.483 0 2.004another problem called the dual linear programming problem. 7 0.419 0.076 1.924If we take the previous problem and call it the primal 8 0.373 0.136 1.864 9 0.337 0.184 1.816problem, then in matrix form the primal and dual problems are 10 0.308 0.223 1.777respectively: Xi = an individual observation Primal DualMaximize Z = cx Minimize W = yb n = the sample size of a groupSubject to: Ax ≤ b Subject to: yA ≥ c k = the number of groups x≥0 y≥0 R = (range) the difference between the largest and smallestIt is assumed that if A is a matrix of size [m × n], then y is a observations in a sample of size n.[1 × m] vector, c is a [1 × n] vector, b is an [m × 1] vector, and X1 + X2 + f + Xn X= nx is an [n × 1] vector.Network Optimization X1 + X 2 + f + X nAssume we have a graph G(N, A) with a ﬁnite set of nodes N X= kand a ﬁnite set of arcs A. Furthermore, letN = {1, 2, … , n} R1 + R2 + f + Rkxij = ﬂow from node i to node j R= kcij = cost per unit ﬂow from i to j The R Chart formulas are:uij = capacity of arc (i, j)bi = net ﬂow generated at node i CLR = R UCLR = D4RWe wish to minimize the total cost of sending the availablesupply through the network to satisfy the given demand. The LCLR = D3Rminimal cost ﬂow model is formulated as follows: INDUSTRIAL ENGINEERING 215
2. The X Chart formulas are: Tests for Out of Control 1. A single point falls outside the (three sigma) control limits. CLX = X 2. Two out of three successive points fall on the same side of UCLX = X + A2R and more than two sigma units from the center line. LCLX = X - A2R 3. Four out of ﬁve successive points fall on the same side of and more than one sigma unit from the center line.Standard Deviation Charts 4. Eight successive points fall on the same side of the center line. n A3 B3 B4 2 2.659 0 3.267 PROCESS CAPABILITY 3 1.954 0 2.568 Actual Capability 4 1.628 0 2.266 PCRk = Cpk = min c - 5 1.427 0 2.089 n LSL USL - n m , 6 1.287 0.030 1.970 3v 3v 7 1.182 0.119 1.882 Potential Capability (i.e., Centered Process) 8 1.099 0.185 1.815 9 1.032 0.239 1.761 PCR = Cp = USL - LSL , where 10 0.975 0.284 1.716 6v μ and σ are the process mean and standard deviation, UCLX = X + A3S respectively, and LSL and USL are the lower and upper speciﬁcation limits, respectively. CLX = X LCLX = X - A3S QUEUEING MODELS UCLS = B4S Deﬁnitions Pn = probability of n units in system, CLS = S L = expected number of units in the system, LCLS = B3S Lq = expected number of units in the queue,Approximations W = expected waiting time in system,The following table and equations may be used to generate Wq = expected waiting time in queue,initial approximations of the items indicated. λ = mean arrival rate (constant), u m = effective arrival rate, n c4 d2 d3 2 0.7979 1.128 0.853 μ = mean service rate (constant), 3 0.8862 1.693 0.888 ρ = server utilization factor, and 4 0.9213 2.059 0.880 s = number of servers. 5 0.9400 2.326 0.864 6 0.9515 2.534 0.848 Kendall notation for describing a queueing system: 7 0.9594 2.704 0.833 A/B/s/M 8 0.9650 2.847 0.820 A = the arrival process, 9 0.9693 2.970 0.808 10 0.9727 3.078 0.797 B = the service time distribution, s = the number of servers, and v = R d2 t M = the total number of customers including those in service. v = S c4 t Fundamental Relationships vR = d3 v t L = λW 2 vS = v 1 - c4 , where t Lq = λWq W = Wq + 1/μv = an estimate of σ,t ρ = λ/(sμ)σR = an estimate of the standard deviation of the ranges of the samples, andσS = an estimate of the standard deviation of the standard deviations of the samples.216 INDUSTRIAL ENGINEERING
3. Single Server Models (s = 1) Calculations for P0 and Lq can be time consuming; however,Poisson Input—Exponential Service Time: M = ∞ the following table gives formulas for 1, 2, and 3 servers. P0 = 1 – λ/μ = 1 – ρ Pn = (1 – ρ)ρn = P0ρn s P0 Lq 2 L = ρ/(1 – ρ) = λ/(μ – λ) 1 1–ρ ρ /(1 – ρ) Lq = λ2/[μ (μ– λ)] 2 (1 – ρ)/(1 + ρ) 2ρ 3/(1 – ρ 2) W = 1/[μ (1 – ρ)] = 1/(μ – λ) 3 2(1 − ρ ) 9ρ 4 Wq = W – 1/μ = λ/[μ (μ – λ)] 2 + 4ρ + 3ρ 2 2 + 2ρ − ρ 2 − 3ρ 3Finite queue: M < ∞ m = m _1 - Pni u SIMULATION P0 = (1 – ρ)/(1 – ρM+1) 1. Random Variate Generation Pn = [(1 – ρ)/(1 – ρM+1)]ρn The linear congruential method of generating L = ρ/(1 – ρ) – (M + 1)ρM+1/(1 – ρM+1) pseudorandom numbers Ui between 0 and 1 is obtained using Lq = L – (1 – P0) Zn = (aZn–1+C) (mod m) where a, C, m, and Z0 are given nonnegative integers and where Ui = Zi /m. Two integers arePoisson Input—Arbitrary Service Time equal (mod m) if their remainders are the same when divided by m.Variance σ2 is known. For constant service time, σ2 = 0. P0 = 1 – ρ 2. Inverse Transform Method Lq = (λ2σ2 + ρ2)/[2 (1 – ρ)] If X is a continuous random variable with cumulative L = ρ + Lq distribution function F(x), and Ui is a random number between 0 and 1, then the value of Xi corresponding to Ui can be Wq = L q / λ calculated by solving Ui = F(xi) for xi. The solution obtained is W = Wq + 1/μ xi = F–1(Ui), where F–1 is the inverse function of F(x).Poisson Input—Erlang Service Times, σ2 = 1/(kμ2) Lq = [(1 + k)/(2k)][(λ2)/(μ (μ– λ))] F(x) = [λ2/(kμ2) + ρ2]/[2(1 – ρ)] 1 Wq = [(1 + k)/(2k)]{λ /[μ (μ – λ)]} U1 W = Wq + 1/μMultiple Server Model (s > 1) U2Poisson Input—Exponential Service Times X R V- 1 X2 0 X2 n s Ss - 1 cn S mm cn mm W 1 W n! + s ! f m pW P0 = S ! Inverse Transform Method for Continuous Random Variables Sn = 0 1 - sn W T X FORECASTING s - 1 ^ sth ^ sth = 1 > ! n! + H n s Moving Average n=0 s! ^1 - th n s ! dt - i P0 c n m t m t dt = i=1 n , where Lq = s! ^1 - th 2 t dt = forecasted demand for period t, s s+1 P0s t dt – i = actual demand for ith period preceding t, and s! ^1 - th = 2 n = number of time periods to include in the moving Pn = P0 ^m nh /n! n average. 0 # n # s Pn = P0 ^m nh / _ s!s n - si n $ s n Exponentially Weighted Moving Average Wq = Lq m dt = adt 1 + ^1 - ah dt 1, where t - t - W = Wq + 1 n t dt = forecasted demand for t, and L = Lq + m n α = smoothing constant, 0 ≤ α ≤ 1 INDUSTRIAL ENGINEERING 217
4. LINEAR REGRESSION C = T2 NLeast Squares k 2 ni SStotal = ! ! xij - C t t y = a + bx, where i=1j=1 SStreatments = ! `Ti2 ni j - C k t y - intercept: a = y - bx, t i=1 t and slope: b = Sxy Sxx, SSerror = SStotal - SStreatments Sxy = ! xi yi - _1 ni d ! xi n d ! yi n, n n n i=1 i=1 i=1 See One-Way ANOVA table later in this chapter. 2 Sxx = ! xi2 - _1 ni d ! xi n , n n RANDOMIZED BLOCK DESIGN i=1 i=1 The experimental material is divided into n randomized n = sample size, blocks. One observation is taken at random for every treatment within the same block. The total number of y = _1 ni d ! yi n, and n i=1 observations is N = nk. The total value of these observations is equal to T. The total value of observations for treatment i is Ti. x = _1 ni d ! xi n . n The total value of observations in block j is Bj. i=1 C = T2 NStandard Error of Estimate k 2 n 2 SStotal = ! ! xij - C 2 SxxSyy - Sxy i=1j=1 Se = = MSE, where Sxx ^n - 2h SSblocks = ! ` B 2 k j - C n j 2 j=1 Syy = ! yi2 - _1 ni d ! yi n n n SStreatments = ! `Ti2 nj - C k i=1 i=1 i=1Conﬁdence Interval for a SSerror = SStotal - SSblocks - SStreatments e 1 + x o MSE 2 a ! ta t See Two-Way ANOVA table later in this chapter. 2, n - 2 n Sxx 2n FACTORIAL EXPERIMENTSConﬁdence Interval for b Factors: X1, X2, …, Xn t MSE Levels of each factor: 1, 2 (sometimes these levels are b ! ta 2, n - 2 represented by the symbols – and +, respectively) Sxx r = number of observations for each experimental conditionSample Correlation Coefﬁcient (treatment), Sxy Ei = estimate of the effect of factor Xi, i = 1, 2, …, n, r= SxxSyy Eij = estimate of the effect of the interaction between factors Xi and Xj,ONE-WAY ANALYSIS OF VARIANCE (ANOVA) Y ik = average response value for all r2n – 1 observations havingGiven independent random samples of size ni from k Xi set at level k, k = 1, 2, andpopulations, then: 2 ! ! _ xij - x i k ni km Y ij = average response value for all r2n – 2 observations having i=1j=1 Xi set at level k, k = 1, 2, and Xj set at level m, m = 1, 2. 2 = ! ! _ xij - xi i + ! ni _ xi - x i or k ni k 2 Ei = Y i2 - Y i1 i=1j=1 i=1 cY ij - Y ij m - cY ij - Y ij m 22 21 12 11 SStotal = SSerror + SStreatments Eij = 2Let T be the grand total of all N = Σini observations and Tibe the total of the ni observations of the ith sample.218 INDUSTRIAL ENGINEERING
5. ANALYSIS OF VARIANCE FOR 2n FACTORIAL where SSi is the sum of squares due to factor Xi, SSij is the sumDESIGNS of squares due to the interaction of factors Xi and Xj, and so on.Main Effects The total sum of squares is equal to m r 2Let E be the estimate of the effect of a given factor, let L be SStotal = ! ! Yck - T 2the orthogonal contrast belonging to this effect. It can be c = 1k = 1 Nproved that where Yck is the kth observation taken for the cth experimental E= L condition, m = 2n, T is the grand total of all observations, and 2n - 1 N = r2n. m L = ! a^chY ^ch RELIABILITY c=1 2 If Pi is the probability that component i is functioning, a SSL = rL , where n reliability function R(P1, P2, …, Pn) represents the probability 2 that a system consisting of n components will work.m = number of experimental conditions For n independent components connected in series, (m = 2n for n factors), R _ P , P2, fPni = % Pi na(c) = –1 if the factor is set at its low level (level 1) in 1 i=1 experimental condition c,a(c) = +1 if the factor is set at its high level (level 2) in For n independent components connected in parallel, experimental condition c, R _ P , P2, fPni = 1 - % _1 - Pi i nr = number of replications for each experimental condition 1 i=1Y ^ch = average response value for experimental condition c, and LEARNING CURVESSSL = sum of squares associated with the factor. The time to do the repetition N of a task is given byInteraction Effects TN = KN s, whereConsider any group of two or more factors. K = constant, anda(c) = +1 if there is an even number (or zero) of factors in the s = ln (learning rate, as a decimal)/ln 2; or, learninggroup set at the low level (level 1) in experimental condition rate = 2s.c = 1, 2, …, m If N units are to be produced, the average time per unit isa(c) = –1 if there is an odd number of factors in the group set given by 8^ N + 0.5h^1 + sh - 0.5^1 + shBat the low level (level 1) in experimental condition K Tavg =c = 1, 2, …, m N ^1 + shIt can be proved that the interaction effect E for the factors inthe group and the corresponding sum of squares SSL can be INVENTORY MODELSdetermined as follows: For instantaneous replenishment (with constant demand rate, E= L known holding and ordering costs, and an inﬁnite stockout 2n - 1 cost), the economic order quantity is given by m L = ! a^chY ^ch EOQ = 2AD , where c=1 h 2 SSL = rLn A = cost to place one order, 2 D = number of units used per year, andSum of Squares of Random ErrorThe sum of the squares due to the random error can be h = holding cost per unit per year.computed as Under the same conditions as above with a ﬁnite SSerror = SStotal – ΣiSSi – ΣiΣjSSij – … – SS12 …n replenishment rate, the economic manufacturing quantity is given by EMQ = 2AD , where h _1 - D Ri R = the replenishment rate. INDUSTRIAL ENGINEERING 219
6. ERGONOMICSNIOSH FormulaRecommended Weight Limit (pounds)= 51(10/H)(1 – 0.0075|V – 30|)(0.82 + 1.8/D)(1 – 0.0032A)(FM)(CM)whereH = horizontal distance of the hand from the midpoint of the line joining the inner ankle bones to a point projected on the ﬂoor directly below the load center, in inchesV = vertical distance of the hands from the ﬂoor, in inchesD = vertical travel distance of the hands between the origin and destination of the lift, in inchesA = asymmetry angle, in degreesFM = frequency multiplier (see table)CM = coupling multiplier (see table) Frequency Multiplier Table ≤ 8 hr /day ≤ 2 hr /day ≤ 1 hr /day –1 F, min V< 30 in. V ≥ 30 in. V < 30 in. V ≥ 30 in. V < 30 in. V ≥ 30 in. 0.2 0.85 0.95 1.00 0.5 0.81 0.92 0.97 1 0.75 0.88 0.94 2 0.65 0.84 0.91 3 0.55 0.79 0.88 4 0.45 0.72 0.84 5 0.35 0.60 0.80 6 0.27 0.50 0.75 7 0.22 0.42 0.70 8 0.18 0.35 0.60 9 0.15 0.30 0.52 10 0.13 0.26 0.45 11 0.23 0.41 12 0.21 0.37 13 0.00 0.34 14 0.31 15 0.28220 INDUSTRIAL ENGINEERING
7. Coupling Multiplier (CM) Table (Function of Coupling of Hands to Load) Container Loose Part / Irreg. Object Optimal Design Not Comfort Grip Not Opt. Handles Not POOR GOOD or Cut-outs GOOD Flex Fingers 90 Degrees Not FAIR POOR Coupling V < 30 in. or 75 cm V ≥ 30 in. or 75 cm GOOD 1.00 FAIR 0.95 POOR 0.90Biomechanics of the Human BodyBasic Equations Hx + Fx = 0 Hy + Fy = 0 Hz + W+ Fz = 0 THxz + TWxz + TFxz = 0 W THyz + TWyz + TFyz = 0 THxy + TFxy = 0The coefﬁcient of friction μ and the angle α at which the ﬂoor is inclined determine the equations at the foot. Fx = μFzWith the slope angle α Fx = αFzcos αOf course, when motion must be considered, dynamic conditions come into play according to Newtons Second Law. Forcetransmitted with the hands is counteracted at the foot. Further, the body must also react with internal forces at all points betweenthe hand and the foot. INDUSTRIAL ENGINEERING 221
8. PERMISSIBLE NOISE EXPOSURE (OSHA) Standard Time DeterminationNoise dose (D) should not exceed 100%. ST = NT × AF where C NT = normal time, andD = 100% # ! i Ti AF = allowance factor.where Ci = time spent at speciﬁed sound pressure level, SPL, (hours) Case 1: Allowances are based on the job time. AFjob = 1 + Ajob Ti = time permitted at SPL (hours) Ajob = allowance fraction (percentage/100) based on ! Ci = 8 (hours) job time. Case 2: Allowances are based on workday.For 80 ≤ SPL ≤ 130 dBA, Ti = 2c 105 - SPL m 5 (hours) AFtime = 1/(1 – Aday)If D > 100%, noise abatement required. Aday = allowance fraction (percentage/100) based onIf 50% ≤ D ≤ 100%, hearing conservation program required. workday.Note: D = 100% is equivalent to 90 dBA time-weighted Plant Locationaverage (TWA). D = 50% equivalent to TWA of 85 dBA. The following is one formulation of a discrete plant locationHearing conservation program requires: (1) testing employee problem.hearing, (2) providing hearing protection at employee’srequest, and (3) monitoring noise exposure. Minimize m n nExposure to impulsive or impact noise should not exceed 140 z = ! ! cij yij + ! fjx jdB sound pressure level (SPL). i=1j=1 j=1 subject toFACILITY PLANNING m ! yij # mx j, j = 1, f, nEquipment Requirements i=1 n PT n ij ij ! yij = 1, Mj = ! where i = 1, f, m i = 1 Cij j=1 yij $ 0, for all i, jMj = number of machines of type j required per production period, x j = ^0, 1h, for all j, wherePij = desired production rate for product i on machine j, m = number of customers, measured in pieces per production period, n = number of possible plant sites,Tij = production time for product i on machine j, measured in hours per piece, yij = fraction or proportion of the demand of customer i which is satisﬁed by a plant located at site j; i = 1, …, m; j = 1,Cij = number of hours in the production period available for …, n, the production of product i on machine j, and xj = 1, if a plant is located at site j,n = number of products. xj = 0, otherwise,People Requirements cij = cost of supplying the entire demand of customer i from a n PT ij ij plant located at site j, and A j = ! ij , where i=1 C fj = ﬁxed cost resulting from locating a plant at site j.Aj = number of crews required for assembly operation j, Material HandlingPij = desired production rate for product i and assembly Distances between two points (x1, y1) and (x2, y2) under operation j (pieces per day), different metrics:Tij = standard time to perform operation j on product i Euclidean: (minutes per piece), D = ^ x1 - x2h + _ y1 - y2i 2 2Cij = number of minutes available per day for assembly operation j on product i, and Rectilinear (or Manhattan):n = number of products. D = x1 - x2 + y1 - y2222 INDUSTRIAL ENGINEERING
9. Chebyshev (simultaneous x and y movement): TAYLOR TOOL LIFE FORMULA D = max _ x1 - x2 , y1 - y2 i VT n = C, whereLine Balancing V = speed in surface feet per minute, Nmin = bOR # ! ti OT l T = tool life in minutes, and i C, n = constants that depend on the material and on the tool. = Theoretical minimum number of stations Idle Time/Station = CT - ST WORK SAMPLING FORMULAS Idle Time/Cycle = ! ^CT - ST h p _1 - pi 1-p D = Za 2 n and R = Za 2 pn , where Idle Time/CyclePercent Idle Time = 100, where Nactual # CT # p = proportion of observed time in an activity,CT = cycle time (time between units), D = absolute error,OT = operating time/period, R = relative error (R = D/p), andOR = output rate/period, n = sample size.ST = station time (time to complete task at each station),ti = individual task times, andN = number of stations.Job SequencingTwo Work Centers—Johnson’s Rule1. Select the job with the shortest time, from the list of jobs, and its time at each work center.2. If the shortest job time is the time at the ﬁrst work center, schedule it ﬁrst, otherwise schedule it last. Break ties arbitrarily.3. Eliminate that job from consideration.4. Repeat 1, 2, and 3 until all jobs have been scheduled.CRITICAL PATH METHOD (CPM)dij = duration of activity (i, j),CP = critical path (longest path),T = duration of project, andT = ! dij _i, j i ! CPPERT(aij, bij, cij) = (optimistic, most likely, pessimistic) durations for activity (i, j),μij = mean duration of activity (i, j),σij = standard deviation of the duration of activity (i, j),μ = project mean duration, andσ = standard deviation of project duration. aij + 4bij + cij nij = 6 cij - aij vij = 6 n = ! nij _i, j i ! CP 2 2 v = ! vij _i, j i ! CP INDUSTRIAL ENGINEERING 223
10. ONE-WAY ANOVA TABLE Degrees of Sum of Source of Variation Mean Square F Freedom Squares SStreatments MST Between Treatments k–1 SStreatments MST = k −1 MSE SS error Error N–k SSerror MSE = N −k Total N–1 SStotal TWO-WAY ANOVA TABLE Degrees of Sum of Source of Variation Mean Square F Freedom Squares SStreatments MST Between Treatments k–1 SStreatments MST = k −1 MSE SS blocks MSB Between Blocks n–1 SSblocks MSB = n −1 MSE SS error Error (k − 1 )(n − 1 ) SSerror MSE = (k − 1 )(n − 1 ) Total N–1 SStotal224 INDUSTRIAL ENGINEERING
11. PROBABILITY AND DENSITY FUNCTIONS: MEANS AND VARIANCES Variable Equation Mean VarianceBinomialCoefficient ( nx ) = x!(nn− x)! ! b(x; n , p ) = ( ) p (1 − p ) n x n− xBinomial np np(1 – p) xHyper r ( ) N −r n−x nr r (N − r )n(N − n ) h(x; n, r, N ) = x () N 2 (N − 1)Geometric (N ) n N λ x e −λPoisson f (x; λ ) = λ λ x!Geometric g(x; p) = p (1 – p)x–1 1/p (1 – p)/p2NegativeBinomial f ( y ; r, p) = ( y + r −1 r r −1 ) p (1 − p ) y r/p r (1 – p)/p2 n! xMultinomial f ( x1, …, xk) = p1x1 … pk k npi npi (1 – pi) x1! , …, xk !Uniform f(x) = 1/(b – a) (a + b)/2 (b – a)2/12 x α − 1e − x βGamma f (x ) = ; α > 0, β > 0 αβ αβ 2 β α Γ (α ) 1 −x βExponential f (x ) = e β β2 βWeibull f (x ) = α α − 1 − xα β x e β β1 α Γ[(α + 1) α ] [ ( ) ( )] β2 α Γ α +1 α − Γ2 α +1 α 2Normal f (x ) = 1 e − 2 σ ( ) 1 x−μ μ σ2 σ 2π { 2(x − a ) if a ≤ x ≤ m (b − a )(m − a ) a+b+m a 2 + b 2 + m 2 − ab − am − bmTriangular f (x ) = 3 18 2(b − x ) if m < x ≤ b (b − a )(b − m ) INDUSTRIAL ENGINEERING 225
12. HYPOTHESIS TESTING Table A. Tests on means of normal distribution—variance known. Hypothesis Test Statistic Criteria for Rejection H0: μ = μ0 H1: μ ≠ μ0 |Z0| > Zα /2 H0: μ = μ0 X − μ0 Z0 ≡ Z0 < –Z α H0: μ < μ0 σ n H0: μ = μ0 H1: μ > μ0 Z0 > Z α H0: μ1 – μ2 = γ H1: μ1 – μ2 ≠ γ |Z0| > Zα /2 X1 − X 2 − γ H0: μ1 – μ2 = γ Z0 ≡ 2 2 σ1 σ 2 Z0 <– Zα H1: μ1 – μ2 < γ + n1 n2 H0: μ1 – μ2 = γ H1: μ1 – μ2 > γ Z0 > Z α Table B. Tests on means of normal distribution—variance unknown. Hypothesis Test Statistic Criteria for Rejection H0: μ = μ0 H1: μ ≠ μ0 |t0| > tα /2, n – 1 H0: μ = μ0 X − μ0 t0 = t0 < –t α , n – 1 H1: μ < μ0 S n H0: μ = μ0 t0 > tα , n – 1 H1: μ > μ0 X1 − X 2 − γ t0 = H0: μ1 – μ2 = γ 1 1 Variances Sp + |t0| > t α /2, v H1: μ1 – μ2 ≠ γ n1 n2 equal v = n1 + n2 − 2 X1 − X 2 − γ H0: μ1 – μ2 = γ t0 = 2 2 t0 < –t α , v S1 S2 H1: μ1 – μ2 < γ Variances + unequal n1 n2 2 ) 2 2 H0: μ1 – μ2 = γ ν= ( S1 S2 + n1 n2 t0 > tα , v 2 2 H1: μ1 – μ2 > γ (S n ) + (S 2 1 1 2 2 n2 ) n1 − 1 n2 − 1 In Table B, Sp2 = [(n1 – 1)S12 + (n2 – 1)S22]/v226 INDUSTRIAL ENGINEERING
15. ERGONOMICS—HEARINGThe average shifts with age of the threshold of hearing for pure tones of persons with “normal” hearing, using a 25-year-oldgroup as a reference group.Equivalent sound-level contours used in determining the A-weighted sound level on the basis of an octave-band analysis. Thecurve at the point of the highest penetration of the noise spectrum reﬂects the A-weighted sound level. - INDUSTRIAL ENGINEERING 229
16. Estimated average trend curves for net hearing loss at 1,000, 2,000, and 4,000 Hz after continuous exposure to steady noise.Data are corrected for age, but not for temporary threshold shift. Dotted portions of curves represent extrapolation fromavailable data.Tentative upper limit of effective temperature (ET) for unimpaired mental performance as related to exposure time; data arebased on an analysis of 15 studies. Comparative curves of tolerable and marginal physiological limits are also given.Effective temperature (ET) is the dry bulb temperature at 50% relative humidity, which results in the same physiological effectas the present conditions.230 INDUSTRIAL ENGINEERING
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