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Queuing Theory II

Queuing Theory II. 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.

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Queuing Theory II

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  1. Queuing Theory II

  2. 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

  3. 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

  4. 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,

  5. 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

  6. M/M/1 Queue Finite Capacity

  7. M/M/1 Queue Finite Capacity

  8. M/M/1 Queue Finite Capacity

  9. M/M/1 Queue Finite Capacity L Lq

  10. + - 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,

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. + - 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 0 1 2 3 5 6 l = 0 . 949 L = = W 2 . 025 l Little’s Revisited l l l l l l

  23. 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

  24. 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

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