28.02.2013 Views

fundamentals of engineering supplied-reference handbook - Ventech!

fundamentals of engineering supplied-reference handbook - Ventech!

fundamentals of engineering supplied-reference handbook - Ventech!

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Poisson Input—Arbitrary Service Time<br />

Variance σ 2 is known. For constant service time,<br />

σ 2 = 0.<br />

P0 = 1 – ρ<br />

Lq = (λ 2 σ 2 + ρ 2 )/[2 (1 – ρ)]<br />

L = ρ + Lq<br />

Wq = Lq / λ<br />

W = Wq + 1/µ<br />

Poisson Input—Erlang Service Times, σ 2 = 1/(kµ 2 )<br />

Lq = [(1 + k)/(2k)][(λ 2 )/(µ (µ– λ))]<br />

= [λ 2 /(kµ 2 ) + ρ 2 ]/[2(1 – ρ)]<br />

Wq = [(1 + k)/(2k)]{λ /[µ (µ – λ)]}<br />

W = Wq + 1/µ<br />

Multiple Server Model (s > 1)<br />

Poisson Input—Exponential Service Times<br />

⎡ n s<br />

⎛λ⎞ ⎛λ⎞ ⎤<br />

⎢<br />

⎛ ⎞<br />

s−1<br />

⎜ ⎟ ⎜ ⎟ ⎥<br />

⎢ ⎝µ⎠ ⎝µ⎠ ⎜ 1 ⎟⎥<br />

P0<br />

= ⎢∑+ ⎜ ⎟<br />

n=<br />

0 n! s!<br />

λ ⎥<br />

⎢ ⎜1−⎟⎥ ⎢ ⎝ sµ⎠<br />

⎣<br />

⎥<br />

⎦<br />

L<br />

q<br />

s−1<br />

n s<br />

( sρ) ( sρ)<br />

∑ n! s!1<br />

( −ρ)<br />

⎡ ⎤<br />

= 1 ⎢ + ⎥<br />

⎢ ⎥<br />

⎣n= 0<br />

⎦<br />

⎛λ⎞ P0<br />

⎜<br />

⎝µ⎠<br />

⎟<br />

=<br />

s!1<br />

Ps<br />

=<br />

s!1<br />

s<br />

( −ρ)<br />

( −ρ)<br />

ρ<br />

2<br />

s s+<br />

1<br />

0 ρ<br />

2<br />

Pn = P0 (λ/µ) n /n! 0 ≤ n ≤ s<br />

Pn = P0 (λ/µ) n /(s! s n – s ) n ≥ s<br />

Wq = Lq/λ<br />

W = Wq + 1/µ<br />

L = Lq + λ/µ<br />

Calculations for P0 and Lq can be time consuming; however,<br />

the following table gives formulas for 1, 2, and 3 servers.<br />

−1<br />

s P0 Lq<br />

1 1 – ρ ρ 2 /(1 – ρ)<br />

2 (1 – ρ)/(1 + ρ) 2ρ 3 /(1 – ρ 2 )<br />

3 2(<br />

1−<br />

ρ)<br />

2<br />

2 + 4ρ<br />

+ 3ρ<br />

4<br />

9ρ<br />

2 3<br />

2 + 2ρ<br />

− ρ − 3ρ<br />

191<br />

INDUSTRIAL ENGINEERING (continued)<br />

SIMULATION<br />

1. Random Variate Generation<br />

The linear congruential method <strong>of</strong> generating pseudorandom<br />

numbers Ui between 0 and 1 is obtained using<br />

Zn = (aZn+C) (mod m) where a, C, m, and Zo are given nonnegative<br />

integers and where Ui = Zi /m. Two integers are<br />

equal (mod m) if their remainders are the same when<br />

divided by m.<br />

2. Inverse Transform Method<br />

If X is a continuous random variable with cumulative<br />

distribution function F(x), and Ui is a random number<br />

between 0 and 1, then the value <strong>of</strong> Xi corresponding to Ui<br />

can be calculated by solving Ui = F(xi) for xi. The solution<br />

obtained is xi = F –1 (Ui), where F –1 is the inverse function <strong>of</strong><br />

F(x).<br />

U1 U1<br />

F(x)<br />

1<br />

U2<br />

U2<br />

X2 X1<br />

X2<br />

Inverse Transform Method for Continuous Random Variables<br />

FORECASTING<br />

Moving Average<br />

n<br />

X2 0<br />

d<br />

d<br />

n<br />

ˆ<br />

∑ t i<br />

i<br />

t = =<br />

−<br />

1 , where<br />

d ˆ<br />

t = forecasted demand for period t,<br />

dt–i = actual demand for ith period preceding t, and<br />

n = number <strong>of</strong> time periods to include in the moving<br />

average.<br />

Exponentially Weighted Moving Average<br />

dt ˆ = αdt–1 + (1 – α) dt 1<br />

ˆ , where<br />

dt ˆ = forecasted demand for t, and<br />

α = smoothing constant, 0≤α≤1<br />

−<br />

X X

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!