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fundamentals of engineering supplied-reference handbook - Ventech!

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Standard Deviation Charts<br />

n A3 B3 B4<br />

2<br />

3<br />

4<br />

5<br />

6<br />

7<br />

8<br />

9<br />

10<br />

2.659<br />

1.954<br />

1.628<br />

1.427<br />

1.287<br />

1.182<br />

1.099<br />

1.032<br />

0.975<br />

0<br />

0<br />

0<br />

0<br />

0.030<br />

0.119<br />

0.185<br />

0.239<br />

0.284<br />

3.267<br />

2.568<br />

2.266<br />

2.089<br />

1.970<br />

1.882<br />

1.815<br />

1.761<br />

1.716<br />

UCLX<br />

= X + A3S<br />

CLX<br />

= X<br />

LCL X = X − A3S<br />

UCLS<br />

= B4<br />

S<br />

CLS<br />

= S<br />

LCLS<br />

= B3S<br />

Approximations<br />

The following table and equations may be used to generate<br />

initial approximations <strong>of</strong> the items indicated.<br />

n c4 d2 d3<br />

2<br />

3<br />

4<br />

5<br />

6<br />

7<br />

8<br />

9<br />

10<br />

0.7979<br />

0.8862<br />

0.9213<br />

0.9400<br />

0.9515<br />

0.9594<br />

0.9650<br />

0.9693<br />

0.9727<br />

σ= ˆ R/d<br />

σ= ˆ S/c<br />

3<br />

2<br />

4<br />

σ = d σˆ<br />

R<br />

2<br />

4<br />

1.128<br />

1.693<br />

2.059<br />

2.326<br />

2.534<br />

2.704<br />

2.847<br />

2.970<br />

3.078<br />

σ =σˆ1− c , where<br />

s<br />

ˆσ = an estimate <strong>of</strong> σ,<br />

0.853<br />

0.888<br />

0.880<br />

0.864<br />

0.848<br />

0.833<br />

0.820<br />

0.808<br />

0.797<br />

σR = an estimate <strong>of</strong> the standard deviation <strong>of</strong> the ranges<br />

<strong>of</strong> the samples, and<br />

σS = an estimate <strong>of</strong> the standard deviation <strong>of</strong> the standard<br />

deviations <strong>of</strong> the samples.<br />

Tests for Out <strong>of</strong> Control<br />

1. A single point falls outside the (three sigma) control<br />

limits.<br />

2. Two out <strong>of</strong> three successive points fall on the same side<br />

<strong>of</strong> and more than two sigma units from the center line.<br />

3. Four out <strong>of</strong> five successive points fall on the same side<br />

<strong>of</strong> and more than one sigma unit from the center line.<br />

4. Eight successive points fall on the same side <strong>of</strong> the<br />

center line.<br />

190<br />

PROCESS CAPABILITY<br />

C<br />

pk<br />

⎛µ −LSL USL −µ ⎞<br />

= min ⎜ , ⎟ , where<br />

⎝ 3σ 3σ<br />

⎠<br />

INDUSTRIAL ENGINEERING (continued)<br />

µ and σ are the process mean and standard deviation,<br />

respectively, and LSL and USL are the lower and upper<br />

specification limits, respectively.<br />

QUEUEING MODELS<br />

Definitions<br />

Pn = probability <strong>of</strong> n units in system,<br />

L = expected number <strong>of</strong> units in the system,<br />

Lq = expected number <strong>of</strong> units in the queue,<br />

W = expected waiting time in system,<br />

Wq = expected waiting time in queue,<br />

λ = mean arrival rate (constant),<br />

λ ~ = effective arrival rate,<br />

µ = mean service rate (constant),<br />

ρ = server utilization factor, and<br />

s = number <strong>of</strong> servers.<br />

Kendall notation for describing a queueing system:<br />

A / B / s / M<br />

A = the arrival process,<br />

B = the service time distribution,<br />

s = the number <strong>of</strong> servers, and<br />

M = the total number <strong>of</strong> customers including those in<br />

service.<br />

Fundamental Relationships<br />

L = λW<br />

Lq = λWq<br />

W = Wq + 1/µ<br />

ρ = λ /(sµ)<br />

Single Server Models (s = 1)<br />

Poisson Input—Exponential Service Time: M = ∞<br />

P0 = 1 – λ/µ = 1 – ρ<br />

Pn = (1 – ρ)ρ n = P0ρ n<br />

L = ρ/(1 – ρ) = λ/(µ – λ)<br />

Lq = λ 2 /[µ (µ– λ)]<br />

W = 1/[µ (1 – ρ)] = 1/(µ – λ)<br />

Wq = W – 1/µ = λ/[µ (µ – λ)]<br />

Finite queue: M < ∞<br />

P0 = (1 – ρ)/(1 – ρ M+1 )<br />

Pn = [(1 – ρ)/(1 – ρ M+1 )]ρ n<br />

L = ρ/(1 – ρ) – (M + 1)ρ M+1 /(1 – ρ M+1 ~<br />

λ = λ(<br />

1−<br />

Pn<br />

)<br />

)<br />

Lq = L – (1 – P0)

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