METRIC

METRIC

Multi-Echelon Technique for Recoverable Item Control

Craig C.Sherbrooke

Presented by:

Nuriye KAPTANLAR

Y.Emre KARAMANOGLU

03.03.2003

Contents

• General Description

• Structure of the Multi-Echelon Problem

a.Mathematical Assumptions

b.Multi-Echelon Theory

• A simple example for METRIC

• Conclusion and Interpretations

General Description:

• METRIC: is a mathematical model of a base-depot supply system in which item demand is compound Poisson with a mean value estimated by Bayesian procedure.

• METRIC:is a mathematical model translated into a computer program,capable of determining base and depot stock levels for a group of recoverable items.

General Description: [ Cont’d]

• The metric theory is the basis used by

the several military services because of the existence of recoverable items which have

* high cost

* low demand

* long lead times

Recoverable Items being used by TuAF

• Spare parts

• Range finders

• Optical sights and optical detection systems

• Infrared Sights for Aircrafts

• equipments and night vision devices etc.

e.g. Range finder; Product cost = 45000$, LT = 6months, demand for repair = 1/year

Base Stock

ServiceableServiceable UnserviceableUnserviceable

In-house Repair

Weapon System

Depot Repair

Recoverable’s Management Process

Scenarios With/Without Intermediate Storage

Replacementpart(s)

Serviceable Part(s)

Weapon system RepairProcess

ServiceablePart(s)

Repairprocess

Open-loop scenario

Closed-loop scenario

Weapon system

UnserviceablePart(s)


Warehouse of Serviceable

Warehouse of Unserviceable


Purposes of Metric:

• Optimization:• Determining optimal base-depot stock levels for each item

• Redistribution:• Allocating the stock between the bases and depot

• Evaluation:• Providing an assessment of the performance and

investment cost for the system of any allocation between the bases and depot.

Mathematical Assumptions...

System Objective:

Min E(backorders on all items at all

bases pertinent to a specific weapon system)• Fill rate• Service rate• Ready rate• Operational rate


• Demand for each item~Logarithmic Poisson

E.g. Buses arrive at a sporting event in acc. with a Poisson process and numbers of customers in each bus are i.i.d.

{X(t), t0}- number of buses who arrived by t

Yi - number of customers in ith bus

(Ross, 1997)


Customer’s Arrival ~ Poisson (),Demand ~ Logarithmic (i.i.d)P(x demands in the time period) ~ Negative Binomial (q)

q: Demand’s \ (constant over partic. item)


• Demand is stationary over the prediction period

• Demand on where repair is to be accomplished depends on the complexity of the repair only.

• Lateral resupply is ignored

• System is conservative (no condemnation)

• Depot repair begins when the reparable base turn-in arrives at the depot.

• Metric will accept relative backorder costs or essentialities by base and item.

• Initial estimate of is obtained by pooling of the demand from several bases.


Multi-Echelon Theory...

Base2Base1 Basej BaseJ........... ...........

DEPOT

j

rj1-rj

Cus. Arr.~ j(1- rj)

Demand ~ j f(1- rj)

fj = f for j

fj

The Model:

s : spare stock

: mean customer arrival

T : mean repair time

p(x/ T): probability of arriving x demand in

the period T.


Expected # of Backorder(for an item):

of the base stock level s.


B(s) is a CONVEX function:

Computation of Multi-Echelon Solution…

5 Stages:1- Average Time that elapses between

a base request for a resupply from a depot 2- E(backorders as a function of sj)

(each so and each base)3- Optimal allocation of the stock to

the bases to minimize the E(backorders)

4- For constant so+s select min E(system backorder)

5- Consideration of multi-item problem

Stage 1: (Avg. Response Time)S Avg. Response Time

Oj

0 Oj + Dj

Therefore 0 < Delay at depot < Dwhere D: avg. repair time



Stage 1: (Avg. Response Time)

x: # of unserviceable parts demanding depot repair

If x<s0, no resupply delayed

O.w., x - s0 delayed. THEN:


Stage 1: (Avg. Response Time) E(system delay over any time period)

= E(#of units on which delay is being incurred)

*Length of the time period

=• Avg. Delay/Demand

=


Stage 2: (E(backorders as a function of sj)

with the specification s = sj, = j,

T=


Stage 3: (Optimal allocation of stock)

Simple Marginal Allocation:

Each unit of stock is added to the base where it will cause LARGEST decrease in EBO

*EBO is a convex function of s

* Bayesian Procedure (will be described later)


Stage 4:(Min EBO for each level of s0+s)

From the table showing EBO versus s0+s

* select MIN EBO

* record the actual allocation of stock btw bases and depot (will be used in stage 5)


Stage 5: (Multi-Item problem)

Using the B(s) functions for each item allocate the next investment to the item which produces the MAX decrease in EBO/unit cost

!!! Item backorder functions are not necessarily CONVEX.

After each allocation compute the system

investment & backorder

Linear Program of METRIC…

Min System Cost

s.t. EBO bj for j=1,..,J

Where is EBO for item I at base j when the depot stock level is sio and the stock at base j is sij

Lagrange Multipliers Introduced…

where ij are ‘Lagrange Multipliers’

For all ij identical ; we can restrict attention to

a single item.


Optimal allocation of m item i units

Necessary conditions:

&

Min

Min


Sufficient condition for Optimality:

should be a convex function of m.

!!! Not necessarily convex.

is defined which lies on the boundary of the convex hull of


To ‘m’ be a solution:

Limits on Employing Metric’s Objective..

• Be able to

specify targets at each base for EBO on all units of all items

have Metric determine the set of ıj and corresponding stock levels.

Bayesian Procedure…

E.g. Case 1 Demand ~ Poisson s = 2

Case 2 w.p. 0.5 Low demand, 0.5

w.p. 0.5 High demand, 1.5

SO: we will understate backorders by using a point estimate.

Numerical Example for METRIC

• The variable notations different from the paper:

mj = avg. annual demand at base j

μj = avg. pipeline at base j

Numerical Example…

5 identical bases;

mj= 23.2 demands/year

Tj = .01 years

rj = .02

Oj = .01 years

T0 = .02531 years j = 1,2,…,J

Numerical Example… [Cont’d]

Start with a depot stock level of 0 and compute

μj for any base ;

j =23.2 { (.02) (.01) + (.8) [.01+ 2.349/92.8] }

= .7017


• Table 3.1. Expected Backorders at any base (DSL=0)

• Table 3.2. Expected Backorders




• Table 3.1

s EBO(s) EBO (s-1)-EBO(s)

0 .07017 -----

1 .1975 .5042

2 .0411 .1564


Table 3.2.Expected Backorders:DSL : Total stock levels at bases,optimally allocated:

0 1 2 3 4 5 6 7 8

0 3.5087 3.0044 2.5002 1.9959 1.4916 .9873 .8309 .6745 .5181

Optimal

Base to 1 2 3 4 5 1 2 3

Allocate

Numerical Example… [cont’d]Table 3.3.Expected Backorders:DSL Total stock at bases,optimally Allocated:

0 1 2 3 4 5 6 7 8

0 3.5087 3.0044 2.5002 1.9959 1.4916 .9873 .8309 .6745 .5181

1 2.6043 2.1983 1.7923 1.3863 .9803 .5743 .4777 .3811

2 1.9240 1.6046 1.2852 .9658 .6464 .3269 .2694

3 1.5072 1.2469 .9867 .7264 .4662 .2060

4 1.2965 1.0681 .8397 .6113 .3829

5 1.2070 .9925 .7780 .5636

6 1.1743 .9650 .7557

7 1.1639 .9562

8 1.1610

Conclusion…

• Metric can be utilized by managers in the cases:

* Optimization for new procurement

* Evaluation of the existing distribution of stock

* Redistribution of system stock between the bases and depot.

Numerical Example… [cont’d]

Table 3.4.Expected Backorders:

STOCK TOTAL BACKORDER

TOTAL DEPOT BASES BACKORDERS : REDUCTION:

0 0 0 3.5087

1 1 0 2.6043 0.9044

2 2 0 1.9240 0.6803

3 3 0 1.5072 0.4168

4 3 1 1.2469 0.2602

5 2 3 0.9658 0.2811

6 1 5 0.5743 0.3915

7 2 5 0.3269 0.2474

8 3 5 0.2060 0.1209

Questions

&

Answers

METRIC

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