engm 732 queuing applications. motivation idea: we want to minimize the total cost of a queuing...
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ENGM 732ENGM 732
Queuing Queuing ApplicationsApplications
MotivationMotivation
Idea: We want to minimize the total cost of a queuing system
Let SC = cost of serviceWC = cost of waiting TC = total cost of system
min E[TC] = E[SC] + E[WC]
MotivationMotivation
E[TC] = E[SC] + E[WC]
E[TC]
E[SC]
E[WC]
Service Level
Cost
ExampleExample
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 ?
Example (rate Example (rate diagrams) diagrams)
0 31 2 10
8/20 8/20 8/20 7/20 1/20
1/2 1/2 1/2 1/2 1/2
0 31 2 10
8/20 8/20 8/20 7/20 1/20
1/2 1 1 1 1
M/M/1 Queue
M/M/2 Queue
Example (rate Example (rate diagrams) diagrams)
0 31 2 10
8/20 8/20 8/20 7/20 1/20
1/2 1/2 1/2 1/2 1/2
M/M/1 Queue
P C Pn n 0
C Cnn n
n n
n
nn
1 2 0
1 1
11
...
...
Example (rate Example (rate diagrams) diagrams)
0 31 2 10
8/20 8/20 8/20 7/20 1/20
1/2 1/2 1/2 1/2 1/2
/ n Pn' Pn
0 1 0.271 8/10 1 0.8 0.217 8/10 2 0.64 0.173 8/10 3 0.512 0.139 7/10 4 0.3584 0.097 6/10 5 0.215 0.058 5/10 6 0.108 0.029 4/10 7 0.043 0.012 3/10 8 0.013 0.003 2/10 9 0.003 0.001 1/10 10 0.000 0.000
Sum = 3.692
M/M/1
Waiting Costs ( g(N) Waiting Costs ( g(N) form )form )
The current rate at which costs are being incurred is determined primarily by the current state N.
g Nn
n n( )
, , ,
( ) , , ,...,
RST0 0 1 2
400 2 3 4 10
Waiting CostsWaiting Costs
For g(n) linear; g(n) = CwnPn
E WC E g N
g n Pnn
[ ] [ ( ) ]
( )
0
E WC g n P C nP
C nP
C L
nn
w nn
w nn
w
[ ] ( )
0 0
0
Example 2Example 2A 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.
Computer Service Rate Lease Cost
MBI computer ( = 30) $5,000 / dayCRAB computer (= 25) $3,750 / day
Example 2Example 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
wherew = wait time for a customer
Waiting Costs ( h(w) Waiting Costs ( h(w) model )model )
Since customers arrive per day
E h w for customer wait
h w f w dww
[ ( )]
( ) ( )
zexpected cost
0
E WC E h w
h w f w dww
[ ] [ ( ) ]
( ) ( )
z
0
Waiting Costs ( h(w) Waiting Costs ( h(w) model )model )
Recall, for an M/M/1 queue, the distribution of the wait time is given by
f w eww( ) ( ) ( )
E WC h w f w dw
w w e dw
w
w
[ ] ( ) ( )
( )( ) ( )
zz
0
2
020 500 400
Example 2 (rate Example 2 (rate diagram)diagram)
0 31 2 10
20 20 20 20 20
25 25 25 25 25
0 31 2 10
20 20 20 20 20
30 30 30 30 30
MBI Comp.
CRAB Comp.
MBI Computer (MBI Computer ( – – = = 10)10)
zE WC w w e dw
we dw w e dw
w e dw w e dw
w
w w
w w
[ ] ( )
( ) ( )
( ) ( )
,( )
,( )
$ ,
z zz z
20 500 400 10
20 500 10 20 400 10
20 500 10 20 400 10
100 0002
1080 000
3
101160
2 10
10 2 10
2 1 10 3 1 10
2 3
CRAB Computer (CRAB Computer ( – – = = 5)5)
zE WC w w e dw
w e dw w e dw
w
w w
[ ] ( )
( ) ( )
,( )
,( )
,
$ ,
z z
20 500 400 5
20 500 5 20 400 5
50 0002
540 000
3
52 000 640
2 640
2 5
2 1 5 3 1 5
2 3
Expected Total CostExpected Total Cost
E WCMBI
CRAB[ ]
,
,
1160
2 640
E TC
MBI
CRAB
[ ], ,
, ,
,
,
1160 5 000
2 640 3 750
6 160
6 390
Decision ModelsDecision Models
Unknown s
Let Cs = cost per server per unit time
Obj: Find ss.t.min E[TC] = sCs + E[WC]
Example (Repair Example (Repair Model)Model)
min E[TC] = sCs + E[WC]
s sCs E[WC] E[TC]
1 280 280 5612 560 48 6083 840 0 840
Decision ModelsDecision Models
Unknown & s
Let f() = cost per server per unit time A = set of feasible
Obj: Find , ss.t.min E[TC] = sf() + E[WC]
ExampleExampleFor MBI = 30
CRAB = 25
f ( ), ,
, ,
5 000 30
3 750 25
E TC f E WC[ ] ( ) [ ]
, ,
, ,
6 160 30
6 390 25
Decision ModelsDecision 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
Decision ModelsDecision Models
Unknown & s
Let Cs = marginal cost of server / unit time
Cf = fixed cost of service / facility – unit time
p = mean arrival rate for population
n = no. service facilities = p/
Decision ModelsDecision Models
Unknown & s
Cost / facility = fixed + marginal cost of service + expected waiting cost
+ travel time cost
= Cf + Cs +E[WC] + CtE[T]
Decision ModelsDecision Models
Unknown & s
Cost / facility = Cf + Cs +E[WC] + CtE[T]
Min E[TC] = n{ Cf + Cs +E[WC] + CtE[T] }
Example Example
Alternativesone tool crib at location 2two cribs at locations 1 & 3three cribs at locations 1, 2, & 3
1 2
3
Example Example
Each mechanic is assigned to nearest crib. Walking rate = 3 mph
1 2
3
E T
alt
alt
alt
[ ]
. ,
. ,
. ,
0 04 1
0 278 2
0 02 3
Example Example
Fixed cost / crib = $16 / hr (Cf)
Marginal cost / crib = $20 / hr (Cs)
Travel cost = $48 / hr (Ct)
p = 120 / hr. = 120 / hr (1 crib)
1 2
3
Example Example 1 2
3
E TC n s E WC C E T
n s E WCn
E T
t[ ] { [ ] [ ]}
{ [ ] ( ) [ ]}
16 20
16 20120
48
E WC C Lw[ ]
E TC n s Ln
E T[ ] { ( ) [ ]} 16 20 48120
48
But,
Example Example 1 2
3
E TC n s Ln
E T[ ] { ( ) [ ]} 16 20 48120
48
Consider 1 facility, 2 servers ( M/M/2 )
P0 = 0.333Lq = 0.333L = Lq + / = 1.333
Example Example 1 2
3
P0 = 0.333Lq = 0.333L = Lq + / = 1.333
E TC L E T[ ] { ( ) ( ) [ ]}
( . ) ( )( . )
.
1 16 20 2 48 120 48
16 40 48 1333 120 48 0 04
350 40
Example Example 1 2
3
n s L E[T] Cf + Css E[WC] CtE[T] E[TC]
1 120 1 M 0.04 36 M 230.4 M1 120 2 1.333 0.04 56 64.00 230.4 350.41 120 3 1.044 0.04 76 50.11 230.4 356.52 60 1 1.000 0.0278 36 48.00 80.0 328.02 60 2 0.534 0.0278 56 25.63 80.0 323.33 40 1 0.500 0.02 36 24.00 38.4 295.23 40 2 0.344 0.02 56 16.51 38.4 332.7