Queuing Applications

1. Queuing Applications

2. Motivation Idea: We want to minimize the total cost of a queuing system Let SC = cost of service WC = cost of waiting TC = total cost of system min E[TC] = E[SC] + E[WC]

3. Cost E[TC] E[SC] E[WC] Service Level Motivation E[TC] = E[SC] + E[WC]

4. Example Suppose we have 10 CNC machines, 8 of which are required to meet the production quota. If more than 2 machines are down, the estimated lost profit is $400 per day per additional machine down. Each server costs $280 per day. Time to failure is exponential (l=0.05). Service time on a failed machine is also exponential (m=0.5). Should the firm have 1 or 2 repairmen ?

5. 8/20 8/20 8/20 7/20 1/20 8/20 8/20 8/20 7/20 1/20 0 0 1 1 2 2 3 3 10 10 1/2 1/2 1/2 1/2 1/2 1/2 1 1 1 1 Example (rate diagrams) M/M/1 Queue M/M/2 Queue

6. 8/20 8/20 8/20 7/20 1/20 0 1 2 3 10 1/2 1/2 1/2 1/2 1/2 l l l l . . . = = - - - 1 2 0 1 n n n C C - 1 n n m m m m . . . - 1 1 n n n = P C P 0 n n Example (rate diagrams) M/M/1 Queue

7. 8/20 8/20 8/20 7/20 1/20 0 1 2 3 10 1/2 1/2 1/2 1/2 1/2 Example (rate diagrams)

8. = 0 , n 0 , 1 , 2 = g ( N ) - = 400 ( n 2 ) , n 3 , 4 , . . . , 10 Waiting Costs ( g(N) form ) The current rate at which costs are being incurred is determined primarily by the current state N.

9. = E [ WC ] E [ g ( N ) ] ¥ å = g ( n ) P n = 0 n ¥ ¥ å å = = E [ WC ] g ( n ) P C nP n w n = = n 0 n 0 ¥ å = C nP w n = n 0 = C L w Waiting Costs For g(n) linear; g(n) = CwnPn

10. Example 2 A University is considering two different computer systems for purchase. An average of 20 major jobs are submitted per day (exp with rate l=20). Service time is exponential with service rate dependent upon the type of computer used. Service rates and lease costs are shown below. Computer Service Rate Lease Cost MBI computer (m = 30) $5,000 / day CRAB computer (m = 25) $3,750 / day

11. Example 2 Scientists estimate a delay in research costs at $500 / day. In addition, due to a break in continuity, an additional component is given for fractional days. h(w) = 500w + 400w2 where w = wait time for a customer

12. = E [ h ( w )] expected cost for customer wait ¥ = h ( w ) f ( w ) dw w 0 = l E [ WC ] E [ h ( w ) ] ¥ = l h ( w ) f ( w ) dw w 0 Waiting Costs ( h(w) model ) • Since l customers arrive per day ʃ ʃ

13. - m - l = m - l ( ) w f ( w ) ( ) e w ¥ = l E [ WC ] h ( w ) f ( w ) dw w 0 ¥ - m - l = + m - l 2 ( ) w 20 ( 500 w 400 w )( ) e dw 0 Waiting Costs ( h(w) model ) • Recall, for an M/M/1 queue, the distribution of the wait time is given by ʃ ʃ

14. 20 20 20 20 20 20 20 20 20 20 0 0 1 1 2 2 3 3 10 10 25 25 25 25 25 30 30 30 30 30 Example 2 (rate diagram) MBI Comp. CRAB Comp.

15. - = + 2 10 w E [ WC ] 20 ( 500 w 400 w ) 10 e dw - - = + 10 2 10 w w 20 ( 500 ) 10 we dw 20 ( 400 ) 10 w e dw - - - - = + 2 1 10 3 1 10 w w 20 ( 500 ) 10 w e dw 20 ( 400 ) 10 w e dw G G ( 2 ) ( 3 ) = + 100 , 000 80 , 000 2 3 10 10 = $ 1 , 160 MBI Computer (m – l = 10) ʃ ʃ ʃ ʃ ʃ

16. - = + 2 5 w E [ WC ] 20 ( 500 w 400 w ) 5 e dw - - - - = + 2 1 5 3 1 5 w w 20 ( 500 ) 5 w e dw 20 ( 400 ) 5 w e dw G G ( 2 ) ( 3 ) = + 50 , 000 40 , 000 2 3 5 5 = + 2 , 000 640 = $ 2 , 640 CRAB Computer (m – l = 5) ʃ ʃ ʃ

17. , 160 MBI = E [ WC ] 2 , 640 CRAB + 1 , 160 5 , 000 = E [ TC ] + 2 , 640 3 , 750 6 , 160 MBI = 6 , 390 CRAB Expected Total Cost 1

18. Decision Models Unknown s Let Cs = cost per server per unit time Obj: Find s s.t. min E[TC] = sCs + E[WC]

19. Example (Repair Model) min E[TC] = sCs + E[WC] s sCs E[WC] E[TC] 1 280 280 561 2 560 48 608 3 840 0 840

20. Decision Models Unknown m & s Let f(m) = cost per server per unit time A = set of feasible m Obj: Find m, s s.t. min E[TC] = sf(m) + E[WC]

21. m = 5 , 000 , 30 m = f ( ) m = 3 , 750 , 25 = m + E [ TC ] f ( ) E [ WC ] m = 6 , 160 , 30 = m = 6 , 390 , 25 Example For MBI m = 30 CRAB m = 25

22. Decision Models Unknown l & s Choose both the number of servers and the number of service facilities Ex: What proportion of a population should be assigned to each service facility # restrooms in office building # storage facilities

23. Decision Models Unknown l & s Let Cs = marginal cost of server / unit time Cf = fixed cost of service / facility – unit time lp = mean arrival rate for population n = no. service facilities = lp/l

24. Decision Models Unknown l & s Cost / facility = fixed + marginal cost of service + expected waiting cost + travel time cost = Cf + Cs +E[WC] + lCtE[T]

25. Decision Models Unknown l & s Cost / facility = Cf + Cs +E[WC] + lCtE[T] Min E[TC] = n{ Cf + Cs +E[WC] + lCtE[T] }

26. 3 1 2 Example Alternatives one tool crib at location 2 two cribs at locations 1 & 3 three cribs at locations 1, 2, & 3

27. 3 1 2 0 . 04 , alt 1 = E [ T ] 0 . 0278 , alt 2 0 . 02 , alt 3 Example Each mechanic is assigned to nearest crib. Walking rate = 3 mph

28. 3 1 2 Example Fixed cost / crib = $16 / hr (Cf) Marginal cost / crib = $20 / hr (Cs) Travel cost = $48 / hr (Ct) lp = 120 / hr. m = 120 / hr (1 crib)

29. 3 = + + + l E [ TC ] n { 16 20 s E [ WC ] C E [ T ] } t 1 2 120 = + + + n { 16 20 s E [ WC ] ( 48 ) E [ T ] } n = E [ WC ] C L w 120 = + + + E [ TC ] n { 16 20 s 48 L ( 48 ) E [ T ] } n Example But,

30. 3 1 2 120 = + + + E [ TC ] n { 16 20 s 48 L ( 48 ) E [ T ] } n Example Consider 1 facility, 2 servers ( M/M/2 ) P0 = 0.333 Lq = 0.333 L = Lq + l/m = 1.333

31. 3 1 2 = + + + E [ TC ] 1 { 16 20 ( 2 ) 48 L 120 ( 48 ) E [ T ] } = + + + 16 40 48 ( 1 . 333 ) 120 ( 48 )( 0 . 04 ) = 350 . 40 Example P0 = 0.333 Lq = 0.333 L = Lq + l/m = 1.333

32. 3 1 2 Example