The production inventory problem
Post on 03-Jan-2016
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The production inventory problem
• What is the expected inventory level?
• What is expected backorder level?
• What is the expected total cost?
• What is the optimal base-stock level?
I: level of finished goods inventoryB: number of backorders (backorder level)IO: inventory on order.
Three basic processes
Under a base-stock policy, the arrival of each customer order triggers the placement of an order with the production system
s = I + IO – B
s = E[I] + E[IO] – E[B]
Three basic processes
I and B cannot be positive at the same time
I = max(0, s - IO) = (s – IO)+
E[I] = E[(s – IO)+]
B = max(0, IO - s) = (IO - s)+
E[B] = E [(IO - s)+]
Three basic processes
The production system behaves like an M/M/1 queue, with IO corresponding to the number of customers in the system.
Pr( ) (1 ), where /
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E IO
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E B E IO s
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Expected backorder level
1
[ ] ( ) [ ]1 1
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E I s E IO E B s
Expected inventory level
1 1
( ) : expected cost (holding cost + backorder cost)
( ) [ ] [ ] ( )1 1 1
s s
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z s hE I bE B h s b
Expected cost
Optimal base-stock level
2 2 1 1
2 1 2 1
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( 1) ( ) ( 1 ) ( ) 01 1 1 1 1 1
(1 ) ( ) 01 1 1 1
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Optimal base-stock level
1
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ln ln* 1
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lnIf we ignore the integrality of *, then * .
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Queueing Theory
The study of queues – why they form, how they can be evaluated, and how they can be optimized.
Building blocks – arrival process and a service process.
Arrival process – individually/in groups, independent/correlated, single source/multiple sources, infinite/finite population, limited/unlimited capacity.
Service process – single/multiple servers, single/multiple stages, individually/in groups, independent/correlated, service discipline (FCFS/priority).
Some characteristics of arrival and service processes
GX/GY/k/N
A Common notation
G: distribution of inter-arrival times
X: distribution of arrival batch (group)
size
G: distribution of service times
Y: distribution of service batch size
k: number of servers
N: maximum number of
customers allowed
Common examples
M/M/1M/G/1 M/M/k M/M/1/NMX/M/1GI/M/1M/M/k/k
Fundamental quantities
L: expected number of customers in the system, L =E(n).
LQ: expected number of customers waiting in queue.
W: expected time a customer spends in the system.
WQ: expected time a customer spends waiting in queue
E[S]: expected time customer spends in service.
: customer arrival rate, = limt ∞ N(t)/t, where N(t) is the number of arrivals up to time t.
Fundamental relationships
L = LQ + Ns
W = WQ + E(S)
L = W
LQ = WQ
Ns = E(S)
The relationship L = W is often referred to as Little’s law.
t1 t2 t3 t4 t5 t6 t7
T
Number in
system3
2
1
A heuristic proof
L = [1(t2-t1)+2(t3-t2)+1(t4-t3)+2(t5-t4)+3(t6-t5)+2(t7-t6)+1(T-t7)]/T
= (area under curve)/T
= (T+t7+t6-t5-t4+t3-t2-t1)/T
W = [(t3-t1)+(t6-t2)+(t7-t4)+(T-t5)]/4
= (T+t7+t6-t5-t4+t3-t2-t1)/4
= (area under curve)/N(T)
L = (area under curve)/T, W = (area under curve)/N(T)LT=WN(T) L=WN(T)/TSince as T∞, N(T)/T , L= W as T∞.
A similar heuristic proof can be used to show LQ = WQ and Ns = E(S).
L = (area under curve)/T, W = (area under curve)/N(T)LT=WN(T) L=WN(T)/TSince as T∞, N(T)/T , L= W as T∞.
A similar heuristic proof can be used to show LQ = WQ and Ns = E(S).
01 1 1
0
( )
( 1) 1
( ) /
is the utilization of the server (fraction of time server is busy).
Q
Q n n nn n n
L L E S
L L np n p p P
P E S
For a single server queue:
Case 1Customers arrive at regular & constant intervalsService times are constantArrival rate < service rate (< )
WQ = 0
W = WQ + E(S) = E(S)
L = W = E(S) WQ = WQ = 0
TH (output rate) =
Why do queues form?
Case 1Customers arrive at regular & constant intervalsService times are constantArrival rate < service rate (< )
WQ = 0
W = WQ + E(S) = E(S)
L = W = E(S) WQ = WQ = 0
TH (output rate) =
Why do queues form?
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