A n ali sis K eluar an Simul asi Ter m inat ing

11.4 A n ali sis K eluar an Simul asi Ter m inat ing

² Use independent replicat ions, i.e. t he simulat ion is repeat ed a t ot al of R t imes, each run using a di¤erent random number st ream and independent ly chosen init ial condit ions.

² Let Y ri be t he it h observat ions wit hin replicat ion r , for i = 1; 2; : : : n r and r = 1; 2; :::; R:

² For a … xed r , Y r1 ;Y r2 ; : : : is an aut o correlat ed sequence. For di¤erent replicat ions r 6 = s; Y ri and Y sj are st at ist ically independent .

² De…ne a sample mean

Y ri ; r = 1; 2; : : : ; R

n r i=1

² There are Rsamples, so Rsample means, t he overall point est imat e is

b 1 µ= X b µ r

R r=1

11.4.1 Est i m asi I nt er val unt uk Jum l ah R epl i kasi yang t et ap

² The sample mean can be calculat ed as above ² The sample variance

1 X 2 µ r ¡b µ

¾ b (b µ) = R r=1 R¡1

² When t he out put dat a are of t he form f Y r (t ); 0 · t· T E g for r = 1; 2; :::; R

Y r (t )dt ; r = 1; 2; : : : ; R

b Á=

² The sample variance is essent ially a measure of error. I n st at ist ics, we use it s square root as t he standard er ror .

² Example 12.10 on page 446 (T he Able-Baker carhop problem, cont in- ued) : calculat e t he con…dence int erval for Able’s ut ilizat ion and average wait ing t ime.

² Example 12.11 on page 447: t he more runs, t he more accurat e t he result is.

11.4.2 Est i m asi I nt er val dengan P resi si Ter t ent u

² Previous sect ion discusses for a given set of replicat ions t o calcualt e t he con… dence int erval and error. Somet imes we need t o do t he inverse, given a level of error and con…dence, how many r eplicat ions are needed?

² The half-lengt h (h.i.) of a 100(1 ¡ ®)% con…dence int erval for a mean µ, based on t he t dist r ibut ion, is

h:i: = t ®=2;R¡ 1 ¾(b b µ) (¤) p

where b ¾(b µ) = S= R , S is t he sample st andard deviat ion, R is t he number of replicat ions.

² A ssume an error crit erion is speci… ed wit h a con…dence level 1 ¡ ®, it

is desired t hat a su¢ cient ly large sample size R be t aken such t hat P(bjb µ ¡ µj < ²) ¸ 1 ¡ ®

² Since we have t he relat ion (* ), t he desired t he err or cont rol condit ion can be writ t en as

t ®=2;R¡ 1 S 0

h:i : =

² Solve t he above relat ion, we have ² Solve t he above relat ion, we have

z ®=2 S 0 R¸ ²

For R ¸ 50; t ®=2;R ¡ 1 ¼z ®=2 t he inequalit y wit h st andard normal dist ri- but ion holds.

This says we need t o run t hat many (R) replicat ions t o sat isfy t he error requir ement .

² The t rue value of is in t he following range wit h probabilit y of 100(1 ¡

b ®=2;R ¡ 1 S

®=2;R¡ 1 S

·µ·b µ+

² Example 12.12 on page 449