Inventories with Multiple Supply Sources and Networks of Queues with Overflow Bypasses Jing-Sheng...

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Inventories with Multiple Supply Sourcesand Networks of Queues with Overflow Bypasses

Jing-Sheng SongPaul Zipkin

Duke University

WLU, 25 April, 2008

Why multiple sources?

Extensive global supply chainsprone to disruptions

Shipment tracking technology (GPS, RFID)allows us to use more information

Review: Minner 2003

Two sources, constant leadtimesMoinzadeh & Schmidt 1989

One itemPoisson demand with rate No scale economiesStockouts are backlogged (for now)Two supply sources

with different (constant) leadtimesT1 + T2 and T2

Demand

T1 T2Stock

State variables

IN = net inventoryNk = units in transit in node k, k

= 1,2IP2 = inventory position = IN +

N2

IP1 = inventory position = IP2 + N1N1 N2

IN

IP2IP1

Heuristic policy

Policy variables sk , s1 s2

Maintain IPk sk

So, a standard base-stock policy generates orders.

An order uses the fast source, when too many units are too far away.

Equivalent rule

Start with (and keep) IP1 = s1

So, every demand generates an order

Define u1 = s1 – s2

Route units to keep N1 u1

Network of queueswith overflow bypass

N1 N2Poisson

N1 u1

SolutionDefine

p(n1,n2) = Pr{ N1 = n1 , N2 = n2 } (steady state)

k(n) = exp(-Tk) (Tk)n / n!

N(u1) = feasible states = { (n1,n2) : n1 u1 }

G(u1) = normalizing constant

Then,p(n1,n2) = 1(n1)2(n2) / G(u1) , (n1,n2) N(u1)

IN = s1 – ( N1 + N2 )

From here on, it’s easy to evaluate a policy and optimize

Lost salesMinor adjustment: Only N and G change

Two sources, stochastic leadtimes

Expand each node to a Jackson sub-network

N1 N2Poisson

N1 u1

Solution

Reinterpret notationNk = total number in sub-network k, k = 1,2

k(n) = equilibrium distribution of Nk for sub-network k alone, with Poisson input

Then, the solution is the samep(n1,n2) = 1(n1)2(n2) / G(u1) , (n1,n2) N(u1)

IN = s1 – ( N1 + N2 )

LiteratureJackson 1963, Lam 1977, van Dijk 1989,

Serfozo 1989

Example

M/M/1 = 8

ConstantT2 = 1/2

N1 u1

stockoutprobability

1.E-07

1.E-06

1.E-05

1.E-04

1.E-03

1.E-02

1.E-01

1.E+005 10 15 20

One source

Two sourcesu1 = 8

s1

Exogenous leadtimes and other demand processes

Idea: A queue with many other inputsOur orders have negligible effectsTk(t) = virtual leadtime for sub-

network kTime required for a unit arriving at

tThis is a given stochastic process

Tk = virtual leadtime in equilibrium

Analysis

DenoteD(s,t] = demands in (s,t]D(s,t] = slow orders in (s,t]

Then,IN(t+T2(t)) = IP2(t) – D(t,t+T2(t)]

IP2(t+T1(t)) = IP1(t) – D(t,t+T1(t)]

Combine these and take the limitIN = s1 – ( N1 + D2 )

Here, N1 is the content of a loss system in equilibrium,and D2 = D(0,T2]

The solution has the same form as before

Markov-modulated demand

Suppose X is a Markov chainWhile X = x, demand is Poisson with rate

(x)

The analysis above goes throughIN = s1 – ( N1 + D2 )

Now we need the joint distribution of (N1,X)

D2 = D(0,T2], conditional on X(0)

All this can be done numerically

Batch orders

Special case – equal batch sizes qReorder points rk

Use (r,q) policy, to maintain IPk > rk

This works just like the non-Poisson demand systemIN = s1 – ( N1 + D2 )

s1 = r1 + q

X = IP1

N1/q is a the content of a loss system with non-Poisson (Erlang-q) input

More sources

Demand

T1Stock

T2 T3

Heuristic policy

K = number of sourcesPolicy variables sk , s1 … sK

Maintain IPk sk

EquivalentDefine uk = s1 – sk+1

Route units to keep N1 + … + Nk uk

Network of queues

Poisson

N1 u1

N1 + N2 u2

N1 N2 N3

Solution

n = (n1,…,nK)

u = (u1,…,uK)

p(n) = 1(n1)…K(nK) / G(u) , n N(u)

IN = s1 – ( N1 + … + NK )

Discussion

This is quite a flexible modeling approach

What does it not cover?Independent sources

The sources here are on a line, in effect

Asia –> west coast –> mid-westBut what if the fast source is on the east coast?

Dynamic expediting

Optimal policy?