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ENGM 732 Queuing Applications
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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
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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]
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Motivation E[TC] = E[SC] + E[WC] E[TC] E[SC] E[WC] Service Level Cost
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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 ( =0.05). Service time on a failed machine is also exponential ( =0.5). Should the firm have 1 or 2 repairmen ?
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2 031210 8/20 8/20 8/20 7/20 1/20 1/2 1 1 1 1 M/M/1 Queue M/M/2 Queue
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2 M/M/1 Queue PCP nn 0 CC n nn nn n n n 120 11 1 1......
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1 1 1 1
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Waiting Costs ( g(N) form ) The current rate at which costs are being incurred is determined primarily by the current state N. gN n nn (),,, (),,,..., R S T 0012 40023410
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2
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Example (rate diagrams) 031210 8/20 8/20 8/20 7/20 1/20 1/2 1/2 1/2 1/2 1/2 031210 8/20 8/20 8/20 7/20 1/20 1/2 1 1 1 1
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Waiting Costs ( g(N) form ) The current rate at which costs are being incurred is determined primarily by the current state N.
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Waiting Costs For g(n) linear; g(n) = C w nP n EWCEgN gnP n n [][()] () 0
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Waiting Costs For g(n) linear; g(n) = C w nP n EWCEgN gnP n n [][()] () 0 E gnPCnP C CL n n wn n wn n w []() 00 0
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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 =20). Service time is exponential with service rate dependent upon the type of computer used. Service rates and lease costs are shown below. ComputerService RateLease Cost MBI computer ( = 30) $5,000 / day CRAB computer ( = 25)$3,750 / day
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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 + 400w 2 where w = wait time for a customer
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Waiting Costs ( h(w) model ) Ehwforcustomerwait hwfwdw w [()] ()() z expectedcost 0
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Waiting Costs ( h(w) model ) Since customers arrive per day Ehwforcustomerwait hwfwdw w [()] ()() z expectedcost 0 EWCEhw hwfwdw w [][()] ()() z 0
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Waiting Costs ( h(w) model ) Recall, for an M/M/1 queue, the distribution of the wait time is given by fwe w w ()() () EWChwfwdw wwe w w []()() ()() () z z 0 2 0 20500400
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Example 2 (rate diagram) 031210 20 20 20 20 20 25 25 25 25 25 031210 20 20 20 20 20 30 30 30 30 30 MBI Comp. CRAB Comp.
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MBI Computer ( – = 10) EWCwwedw w []() z 2050040010 2
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MBI Computer ( – = 10) EWCwwedw wedwwe w ww []() ()() z zz 2050040010 20500102040010 2 2
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MBI Computer ( – = 10) z EWCwwedw wedwwe we we w ww ww []() ()() ()() zz zz 2050040010 20500102040010 20500102040010 2 2 21 31
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MBI Computer ( – = 10) z
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CRAB Computer ( – = 5) z EWCwwedw w []() 205004005 25
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CRAB Computer ( – = 5) z EWCwwedw we we w ww []() ()() zz 205004005 205005204005 25 215315
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CRAB Computer ( – = 5) z EWCwwedw we we w ww []() ()(), (), () zz 205004005 205005204005 50000 2 5 40000 3 5 25 215315 23
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CRAB Computer ( – = 5) z EWCwwedw we we w ww []() ()(), (), (), $, zz 205004005 205005204005 50000 2 5 40000 3 5 2 640 2 25 215315 23
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Expected Total Cost EWC MBI CRAB [],, 1160 2640 ETC MBI CRAB [],,,,,, 11605000 26403750 6160 6390
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Decision Models Unknown s Let C s = cost per server per unit time Obj: Find s s.t. min E[TC] = sC s + E[WC]
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Example (Repair Model) min E[TC] = sC s + E[WC] ssCsE[WC]E[TC] 1280 280 561 2560 48 608 3840 0 840
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Decision Models Unknown & s Let f( ) = cost per server per unit time A = set of feasible Obj: Find , s s.t. min E[TC] = sf( ) + E[WC]
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Example For MBI = 30 CRAB = 25 f(),,,, 500030 375025 ETCfEWC[]()[],,,,,, 5000116030 3750264025
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Example For MBI = 30 CRAB = 25 f(),,,, 500030 375025 ETCfEWC[]()[] ,,,, 616030 639025
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Decision Models Unknown & 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
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Decision Models Unknown & s Let C s = marginal cost of server / unit time C f = fixed cost of service / facility – unit time p = mean arrival rate for population n = no. service facilities = p /
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Decision Models Unknown & s Cost / facility = fixed + marginal cost of service + expected waiting cost + travel time cost = C f + C s +E[WC] + C t E[T]
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Decision Models Unknown & s Cost / facility = C f + C s +E[WC] + C t E[T] Min E[TC] = n{ C f + C s +E[WC] + C t E[T] }
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Example Alternatives one tool crib at location 2 two cribs at locations 1 & 3 three cribs at locations 1, 2, & 3 12 3
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Example Each mechanic is assigned to nearest crib. Walking rate = 3 mph 12 3 ET alt [].,.,., 0041 02782 0023
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Example Fixed cost / crib = $16 / hr (C f ) Marginal cost / crib= $20 / hr (C s ) Travel cost = $48 / hr (C t ) p = 120 / hr. = 120 / hr (1 crib) 12 3
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Example 12 3 ETCnsEWCCET nsE n ET t []{[][]} {[]()[]} 1620 1620 120 48
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Example 12 3 ETCnsEWCCET nsE n ET t []{[][]} {[]()[]} 1620 1620 120 48 EWCCL w [] ETCnsL n ET[]{()[]} 162048 120 48 But,
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Example 12 3 ETCnsL n ET[]{()[]} 162048 120 48 Consider 1 facility, 2 servers ( M/M/2 ) P 0 = 0.333 L q = 0.333 L = L q + / = 1.333
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Example 12 3 P 0 = 0.333 L q = 0.333 L = L q + / = 1.333 ETCLET[]{()()[]} (.)()(.). 1162024812048 164048133312048004 35040
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Example 12 3
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