The production inventory problem

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 /

[ ]1

nIO n

E IO

0

0

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(1 )

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Expected backorder level

1

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Expected inventory level

1 1

( ) : expected cost (holding cost + backorder cost)

( ) [ ] [ ] ( )1 1 1

s s

z s

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 ) ( ) 01 1 1 1

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z s z s h s b h s b

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Optimal base-stock level

1

ln1

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ln ln* 1

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lnIf we ignore the integrality of *, then * .

ln

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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?