240 likes | 374 Views
This document explores the dynamics of an M/M/1 queuing system with finite capacity, specifically focusing on a workstation that receives parts from a conveyor. The system has a buffer capacity allowing for six parts (five in addition to the one being processed). Utilizing a Poisson arrival process at a rate of one part per minute and an exponential service time with a mean of 45 seconds, we derive critical metrics such as average waiting time (W) and system capacity (L). This analysis contributes valuable insights into optimizing queuing systems in operational environments.
E N D
l m - l = n p ( ) ( ) n m m l l 2 = L = L m - l q l m - l ( ) 1 l = W = W m - l q l m - l ( ) Model
l + m = m + l p p p p 0 2 1 1 l = m p p 0 1 l = m p p - N 1 N M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N State Balance Eq. 0 1 N
N å = p 1 n = 0 n l N å = n 1 p ( ) 0 m = 0 n M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N Now,
M/M/1 Queue Finite Capacity • Workstation receives parts form a conveyor. Station has buffer capacity for 5 parts in addition to the 1 part to work on (N=6). Parts arrive in accordance with a Poisson process with rate of 1 / min. Service time is exp. with mean = 45 sec. (m = 4/3). 0 1 2 3 5 6
+ - 1 N N 1 x å = l n N x å = n - 1 p ( ) 1 x = 0 0 n m = 0 n M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N Recall,
l + - N 1 1 ( ) m = 1 p l l N 0 å - = n 1 ( ) 1 p ( ) m 0 m = 0 n M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N
l - 1 ( ) m = p l 0 + - N 1 1 ( ) m l + - N 1 1 ( ) m = 1 p l 0 - 1 ( ) m M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N
l - 1 ( ) m = p l 0 + - N 1 1 ( ) m l - 1 ( ) l m = n p ( ) l n m + - N 1 1 ( ) m M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N
l - 1 ( ) l m N N å å = = n L np n ( ) l n m + - 1 N 1 ( ) = = 0 0 n n m l - 1 ( ) l m N å = n n ( ) l m + - 1 N 1 ( ) = 0 n m M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N
l l + l + - + 1 N N [ 1 N ( ) ( N 1 )( ) ] m m = L l + m - l - 1 N ( )[ 1 ( ) ] m M/M/1 Queue Finite Capacity 0 1 2 3 N-1 N Miracle 37 b
l = ( ) 0 . 75 m M/M/1 Queue Finite Capacity • Workstation receives parts form a conveyor. Station has buffer capacity for 5 parts in addition to the 1 part to work on (N=6). Parts arrive in accordance with a Poisson process with rate of 1 / min. Service time is exp. with mean = 45 sec. (m = 4/3). 0 1 2 3 5 6
+ - 7 6 1 [ 1 6 ( 0 . 75 ) ( 7 )( 0 . 75 ) ] = = L 1 . 92 l - - 7 ( 1 . 33 1 ) [ 1 ( 0 . 75 ) ] = ( ) 0 . 75 m M/M/1 Queue Finite Capacity • l = 1 N = 6 • m = 4/3 = 1.33 0 1 2 3 5 6
0 1 2 3 5 6 L = = W 1 . 92 l q = = W 1 . 21 q l 1 = + = W W 1 . 96 q m = l = L W * 1 . 96 ??? Little’s Revisted L
0 1 2 3 5 6 ¥ å l = l p n n = 0 n M / M / 1 ¥ ¥ å å l = l = l = l p p n n = = 0 0 n n Little’s Revisited l l l l l l
0 1 2 3 5 6 ¥ å l = l p n n = 0 n M / M / 1 / 6 l = l + l + l + l + l + l p p p p p p 0 1 2 3 4 5 = l + + + + + ( p p p p p p ) 0 1 2 3 4 5 = l - ( 1 p ) 6 Little’s Revisited l l l l l l
0 1 2 3 5 6 l = - 1 ( 1 p ) 6 = - 1 ( 1 . 051 ) = 0 . 949 Little’s Revisited l l l l l l
0 1 2 3 5 6 l = 0 . 949 L = = W 2 . 025 l Little’s Revisited l l l l l l
0 1 2 3 5 6 l = 0 . 949 = W 2 . 025 1 = - W W q m = - 2 . 025 0 . 75 = 1 . 275 Little’s Revisited l l l l l l
0 1 2 3 5 6 l = 0 . 949 = W 2 . 025 = W 1 . 275 q = l L W q q = . 949 ( 1 . 275 ) = 1 . 210 Little’s Revisited l l l l l l