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IX
Dewey
HD28
.M414
(^
Selecting Ingot Sizes for Joint Order Processing
in
Sheet Manufacturing
Anantaram Balakrishnan
and
Srimathy Gopalakrishnan
WP# 3734-94-MSA
October 1994
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Abstract
This paper addresses a tactical plannin g problem of seiecdng standard ingot sizes lo
facilitate joini
processing of crders in a mate-to-order alunnrmm steet manafacniniig
Tne
operaiioiL
farilit}'
has traditjonallv dedicated ingots to rodi'vidBal orders, bat
is
now
considaiiig producing multiple cffders from each ingot to reduce surplus and ejq>k>ii
ecoBomi^ of scale
depends on ite
in processiag costs.
Tie
effectiveQess
a^.'silabk srapdard ingot sizes.
to decide ingot sizes given tie disiiibimon of
erf liiis
jcani processing scras^zy
We develc^ an integer programming mock:
demand for various prodacis and the order
combination constraints. The model simultaneously addresses the issue of selecting a
prespecified
The
number of standaid
sizes
objective is to minimize tie total
products.
luiiiile
deciding whicii product combinaiions lo use.
Tft'eigiii
of ingots processed to
saiisfv"
for all
We describe a dual-ascent method to generate lower bounds for the probieiii sac
heuristic psrocedures to idendiS'
actual data
good
solutions. Extensive corc5>utaiiortal results
from a leading aluminum shest manijfacturar confirm the
also illusuaie
bow tfce
nsmg
effectri-eness of the
algorithm; on average the gaps bera-'een lie upper and lower bounds are
planning.
demand
abom 4%.
nxxiel can be used to oerform sensidiirv^ analv'ses for lactica!
We
1.
Introduction
This paper addresses a
manufacturing
tactical
planning problem in a leading aluminum sheet
produces sheet and plate products to customer specifications.
facility that
The problem, motivated by technological and market
trends in the industry, deals with
selecting standard ingot sizes to produce multiple orders
produces ingots to stock
in a
few standard
sizes.
The downstream
standard ingots to produce finished coils to order.
of scale
in ingot casting
equipment
to
and
from a single
To
ingot.
The
facility
rolling mills use these
exploit the considerable
rolling costs, the facility has continuously
economies
upgraded
its
produce and process larger ingots. At the same time, customers are placing
smaller orders more frequently to reduce their raw material inventories and
move towards
just-in-time manufacturing. Consequently, the plant's traditional policy of dedicating an
ingot for each order
is
not economically viable. For instance, producing a 4,000 pound
order using a 10,000 pound ingot requires either storing the remaining 6,000 pounds of
pound
finished product in stock or recycling this metal; furthermore, processing the 10,000
ingot uses up scarce rolling mill capacity, possibly delaying other orders.
these problems, the plant
is
exploring the option of combining orders,
i.e.,
To overcome
jointly
producing multiple orders that require the same alloy and have similar specifications using
a single ingot. For a detailed discussion of the costs and benefits of order combination, see
Ventola [1991] and Balakrishnan and Brown [1992].
The choice of standard
available opportunities to
ingot sizes (widths and weights) determines the
combine
number of
and thus greatly influences the cost
orders,
effectiveness of the order combination strategy. Reviewing and revising ingot sizes
medium-term (one
to
two years) decision since adding new standard ingot
considerable lead time and investment to acquire and test
develops an optimization model to support
selects a prespecified
used
to
number of standard
new
is
sizes requires
tooling. This paf>er
this tactical ingot sizing decision. Tlie
model
ingot sizes to minimize the total weight of ingots
manufacture a representative mix of products for each alloy. Minimizing the
total
ingot weight reduces processing time and costs as well as planned scrap and "surplus"
excess production that
is
not shipped).
We
facility
Computational
(i.e.,
formulate the problem as an integer program,
and develop an optimization-based solution procedure
heuristics.
a
results using actual data
demonstrates that the solution procedure
is
that
combines dual ascent and
from an aluminum sheet manufacturing
quite effective, producing solutions that
are within
4%
solutions.
For one product family that we analyzed, the new ingot sizes chosen by
of optimality on average even without local improvement of the heuristic
-
1
this
model provide savings of hundreds of thousands of dollars compared
We
sizes.
also use the
model
to using current ingot
of ingot sizing decisions to various
to study the sensitivity
technological and operational parameters.
This paper
is
organized as follows. Section 2 presents a brief description of the sheet
manufacturing process, defines the ingot sizing problem, and describes the integer
programming formulation.
In Section 3,
we develop
compute
We also describe two stand-alone
lower bounds and generate heuristic solutions.
heuristics.
a dual ascent procedure to
Section 4 provides details of our implementation and
test
problems, and
presents computational results and sensitivity analyses.
2.
2.1
Problem Context and Formulation
Overview of sheet manufacturing process
Aluminum
sheets have a variety of uses ranging from aerospace and automotive
applications to construction and packaging.
The sheet manufacturing
facility that
we
studied produces a wide range of sheet and plate products to customer specifications.
These specifications include
The
length).
main steps
facility
alloy, width,
maintains
in the sheet
little
gauge (thickness), and
total
weight (or
total
or no uncommitted finished goods inventory.
The
manufacturing process are ingot casting, hot rolling, cold rolling,
heat treatment, and finishing
(e.g., slitting)
operations.
Ingot casting consists of melting pure and scrap
aluminum with
the required alloying
elements, and pouring and cooling the molten metal to produce solid rectangular
aluminum
ingots of the desired dimensions. This process requires separate tooling for each size, and
changing over production from one alloy and size to another entails long setup times and
wastage of metal
(e.g., to flush the furnaces).
Therefore, the ingot plant produces only a
limited set of standard sizes, primarily to stock, in large lots using a cyclic schedule.
Hot
rolling,
performed on reversing or multi-stand rolling
thickness (and elongates
it)
at
mills, reduces the ingot's
elevated temperatures. Cold rolling further reduces the
thickness to the desired finished gauge(s).
A minimum amount of cold rolling is often
necessary to ensure that the product meets tight gauge and finish tolerances, and material
specifications (e.g., temper). Several cold rolling passes might be necessary
gauge
is
small relative to the sheet's thickness
when
it
exits the hot mills.
sheet does not change during hot and cold rolling operations.
-2-
if
the finished
The width of the
Although the number of ingot sizes
is
small, the facility
and weight combinations) because the hot
different finished specifications (width, gauge,
and cold rolling operations are quite
finished gauges from the
same
flexible,
able to produce hundreds of
is
i.e.,
they can produce a wide range of
ingot, subject to width
and weight
restrictions.
Furthermore, the cold rolling process permits changing the gauge (during the final cold
rolling pass) while processing a coil, enabling the plant to jointly
produce multiple orders
with comparable (but not necessarily identical) widths and gauges in a single coil
(i.e.,
using a single ingot). Since the coil has uniform gauge until the final pass, and since the
amount of reduction
coil
is
limited.
each pass
in
To account
is
limited, the range of possible gauges in the finished
for this restriction, the plant has provided a
table" that specifies various gauge
"windows"
"gauge combination
(or ranges) for each width.
can be jointly produced from a single ingot only
if
Two products
both their gauges belong to one of these
windows.
One of the
factors contributing to the scale
"planned" scrap generated
depth of material
is
at
economies
in sheet
production
is
various stages. For instance, ingots are "scalped" (a fixed
removed from
the top and bottom surfaces) prior to hot rolling.
Similarly, fixed lengths of the coil's leading and trailing ends (called head and
are discarded after rolling,
The proportion of metal
—decreases
side trim
it
same
to
need
to
scrap)
be trimmed prior to shipping.
"planned" scrap— scalping
loss,
head and
tail
scrap,
and
i.e.,
producing multiple orders using a single ingot,
is
advantageous
permits the facility to exploit economies of scale in processing large ingots and
also reduces the
customers.
due
coil
tail
as the weight of the ingot increases.
Order combination,
because
and the sides of each
lost
the
The
amount of
surplus,
i.e.,
excess quantity produced but not shipped to
surplus might be kept in inventory to satisfy
specification.
To avoid accumulating uncommitted
some
future order for the
finished goods inventory, the
plant currently melts and reuses the surplus, incurring handling and reprocessing costs
(including melt loss).
The
the increasing quantities of
plant's decision to explore order
uncommitted metal
(i.e.,
combination was prompted by
surplus) being recycled due to the
steady decrease in the quantity per customer order.
Combining up
to three orders per ingot is apparently technically feasible (although not
actually tested in practice).
However,
the plant prefers not to
combine more than two
orders for operational reasons, especially since order combination
is
a
new and
untested
strategy.
We
say an order pair
is
feasible
if
two orders can be jointly produced using an
the
available ingot without violating any of the processing restrictions. In particular, the
following conditions must be met:
1.
2.
Alloy feasibility: Both orders require the same alloy.
Weight feasibility: The sum of the weights of the two orders must be
to the net weight (after subtracting the
less than or equal
planned scrap allowances) of an available ingot
for that alloy;
3.
Width feasibility: The width of each order must be
ingot
4.
minus the required side
Gauge
differential:
less than or equal to the
width of the
trim;
The gauges of the two products must
lie
within one of the
corresponding gauge windows.
5.
Width differential: The absolute difference
exceed a prespecified
limit
which we
operational restriction limits the
call the
widths for the two orders must not
maximum
amount of scrap
must exceed the larger of the two
we perform
in
sensitivity analysis with respect to
feasibility conditions
available standard ingot sizes.
CO.
This
(since the width of the ingot
and
coil
orders' widths). In our computational experiments,
We refer to these conditions collectively as the
and width
width differential
o).
order combination rules. The weight
imply that the feasibility of order pairs depends on the
Two
orders might require the
same
alloy
and have
;
compatible gauges and widths, and yet might not be feasible because none of the available
ingots for that alloy can
feasible, but
accommodate both
orders. Alternatively, an order pair might be
combining them might require using a very large ingot
(relative to the total
ordered weight); dedicating two smaller ingots to each order might then be more
economical. For dedicated ingots,
i.e.,
producing a single order using an ingot, the net
weight and width of that ingot must equal or exceed the order weight and width.
In the short-run, given the set of standard ingot sizes, the scheduler
combine the orders on hand and assign them
Gopalan [1992] discussed
this
to appropriate ingots.
must decide how
Ventola [1991
]
to
and
decision problem, and observed that choosing good standard
ingot sizes can considerably improve short-run performance. This paper focuses on the
medium-term
ingot sizing decision, namely, finding an effective set of standard ingot sizes
to facilitate short-run order combination.
different alloys, the ingot sizing
Since
we cannot combine
problem decomposes by
4-
alloy.
orders requiring
2.2
Problem
'^
definition
For medium-term planning, we discretize the range of widths, gauges, and weights of
orders requiring a particular alloy, and consider
all
orders within each cell as a single
we
product group or product. Using past data on actual orders,
orders or
demand (number of orders)
for each product. Ingots are specified
width, weight, and thickness (or length).
since only these
two dimensions
product pair (assuming
We
will focus
on the
this pair satisfies the
gauge and width
limits
maximum number
number of standard
investment
lot sizes,
their alloy,
width and weight
ingot's
differential constraints).
on the ingot
the plant has identified a discrete set of candidate ingot sizes
the
by
are relevant to test if an ingot can produce a particular
Equipment capacities impose upper and lower
plant also specifies the
estimate the frequency of
sizes.
from
We assume that
this range.
p of standard ingot sizes
The
to select.
ingot
Limiting
ingot sizes has several advantages including reducing the
in tooling, limiting the
number of changeovers and ensuring reasonably
large
and reducing ingot safety stocks (and hence reducing both storage space
requirements and holding costs) due to the risk pooling benefits of having greater
commonality
(see, for
example. Baker, Magazine, and Nuttle [1986]). Our model does not
by varying
explicitly incorporate these factors; instead,
p,
we can
study the sensitivity of
the overall performance to this parameter.
Given
the
demand
for each product, the parameters for the order combination rules, the
candidate ingot sizes, and the
problem (ISP)
maximum number p
of standard ingot sizes, the
mgot
requires selecting p or fewer standard ingot sizes, and deciding the
sizing
number
of ingots of each size to use for feasible product pairs (for the selected ingot sizes) or single
orders so as to meet the
demand
for all products.
The objective
planned scrap and surplus, or equivalently (since demand
is
to
minimize the
total
fixed) to
minimize the
total
is
weight of ingots used.
The ISP combines two
ingots
—
that
features
—
selecting p standard sizes,
have been addressed separately
in
and "packing" orders
in
other contexts. Several papers have
addressed the problem of selecting standard lengths to stock and the related assortment
problem
(e.g.,
Wolfson [1965], Pentico [1974], Beasley [1985], Pentico [1988]). This
problem deals with selecting standard sizes
given set of candidate sizes to meet
(e.g., lengths) for
demand
for various finished sizes.
minimize scrap and/or inventory holding and substitution
not consider order combination,
i.e., it
semi-finished items from a
assumes
that only
The
objective
costs. This class of
is
models does
one order can be produced from
each semi-finished item. For the one-dimension case, say, selecting standard lengths of
-5
to
steel rods, if
we assume
that a finished product with length
semi-finished item with length greater than or equal to
efficiently using
dynamic programming
(e.g.,
/,
/
the
can be produced from any
problem can be solved
Pentico [1974]). Vasko, Wolf, and Stott
[1987] developed and applied a set covering approach to select optimal ingot sizes for the
steel industry.
Again, their model did not incorporate the option of jointly producing
multiple orders using a single ingot.
issue of
how
The
literature
"pack" multiple orders
to optimally
on cutting stock problems considers the
in available semi-finished sizes (e.g.,
Gilmore and Gomory [1961], [1963]). For instance, given
problem
lengths, the cutting stock
rolls
of certain widths and
selects an optimal set of patterns to
Chambers and Dyson [1976] and Beasley [1985]
produce
orders.
all
present heuristic algorithms for the two-
dimensional cutting stock problem with stock size selection.
Problem formulation
2.3
and
K=
and candidate ingot
sizes.
For
order for product
For
Let
I
=
[1,
..., Ill}
pair of products
i.
i,
j
e
all
all
i
all
g
let
let d^
I,
if
denote the index sets of products
and Qj denote the demand and weight per
Wj^ denote the total weight of ingot type k. For every
< j and every
combination rules to determine
denote the set of
i
k e K,
with
I
IKI} respectively
{
the pair
feasible assignments
ingot type k e K,
we apply
the order
can be produced using ingot type
(i,j)
such that the pair of products
(i,j,k)
Let
k.
(i,j)
UK
can be
jointly produced using ingot type k. This set might include triplets such as (i,i,k) denoting
the assignment of
two orders of product
i
to
one type k
if
this triplet as (0,i,k).
(The index
represents a
dummy
any single order, and has zero demand.) Notice
feasible using ingot type k, then the pair (0,i)
denote the set of
We also include the option
a type k ingot can produce a single order of product
of dedicating ingots:
J(i)
ingot.
all
that
if,
i,
we denote
product that can be combined with
for
some j e L
the pair
(i,j) is
must also be feasible using ingot type
'i-compatible" products,
i.e., J(i)
=
or
{j: (i,j,k)
(j,i,k)
g
Let
k.
UK for
some k G K}
To
formulate the ingot sizing problem,
we
define the following
two
sets of integer
variables:
^
z.
yjj,^
For
all
f
=
1
[0
=
i,
if
ingot
type
k
^'
to
chosen as a standard
size,
and
.
,,
for all
,
^
,
k G K; and,
otherwise
number of type k
j
is
-'
\
e L define
X.jj
ingots used to produce product pair
= min
{
d,, d:
}
;
let
X^ = d
j
for all products
problem has the following integer programming formulation.
-6-
(i,j),
i.
for all (i,j,k)
The
g UK.
ingot sizing
=
Z*
[ISP]
Z
min
w^y..
(i,j,k)eUK
"^
(2.1)
'J*^
subject to:
Demand constraints:
Z
2
Z
Z
j^i
(i,j,k)eUK
+
y::.
(i,i.k)€UK"^
>
d
^
for
e I
(2.2)
y-^
<
XjjZi^
forall(i,j,k)6UK,
(2.3)
Zu
<
p,
y-j,
'J"
all
i
Forcing constraints:
Maximum # of standard sizes:
Z
k€K
and
(2.4)
"^
Integrality/nonegativity:
The
z^
=
or
yy^
>
and integer
K
for all (i,j,k)
demand
for all products. Constraints (2.2) ensure that the
The
satisfied.
first
term
in the left-hand side
ingots used to process the product-pair
these ingots generates satisfies
two
left-hand side corresponds to the
(either dedicated or joint with
products
i
ingot type
(2.5a)
€ UK.
(2.5b)
and j, with
is
(i,j,k)
(i,i) if
units of
number of
this pair is feasible for
demand
for product
assigns A.jj+5 ingots to the pair,
solution,
or product
i.
any ingot; each of
The second term
i).
The
size.
(i,j)
In this constraint, limiting the
to at
(i,j)
most
we can
\
(= min
{
dj, d:
UK.
} )
is
number of
valid since
Therefore,
if
i
two
if this
tyf)e
k
(i,j,k)
e
a solution
replace 5 of these pairs with single orders of
depending on whether dj >dj. This change retains
j
in the
forcing constraint specifies that
e UK, can be jointly produced using ingot type k only
chosen as a standard
is
ingots containing one unit each of product
implies that (0,i,k) and (0,j,k) also belong to the set
i
for every product
of this equation represents the number of
some product j ^
ingots used to produce the pair
product
k e
objective function (2.1) minimizes the total weight of ingots used to satisfy the
demand
UK
for all
1
feasibility
of the
and does not increase the objective function value. Constraint (2.4) imposes the
upper limit of p on the number of standard sizes
feasible (for instance,
dedicating this ingot
The ISP
is
we can add
is
NP-hard
for instance, Francis
we
select.
We assume that [ISP]
a "large" ingot size to the candidate
list
is
such that
feasible for every product).
since, as
we
note in the next section, the p-median problem (see,
and Mirchandani [1990])
is
-7-
a special case. Formulation [ISP] requires
prior enumeration of
feasible product pairs
all
Consequently, the number of variables
encoding the product-pair
directly
we might
can be quite large. Instead,
consider
problem
feasibility rules as constraints in the
we can
formulation. For instance,
y--^
for all candidate ingot sizes.
(i,j)
represent the problem as one of assigning products to
individual ingots, and checking the feasibility of assignments via constraints corresponding
gauge
to weight feasibility, width feasibility,
differential,
and width
differential.
Our
solution procedure exploits the special structure of [ISP].
cases
2.4 Special
Formulation [ISP] has two interesting special cases.
First,
we do
suppose
joint production, or the feasibility constraints are such that the pair (0,i)
"combination" for every product
and candidate ingot sizes as
product
i
only
The
6 UK. Note
if (0,i.k)
total surplus
we can
and assigns products
i
that, for a
which we
special case,
corresponds to the situation
An
when
ingot type k can ser\'e
given choice of standard ingot sizes, the
this
total
to the
i
same ingot type
assignment, dj(wj, -
The p-median model
minimize the
the only feasible
products as "customers"
that assigns all dj units of product
to ingot type k.
to ingots to
treat the
p-median problem.
and planned scrap generated by
"cost" of assigning product
The second
In this case,
"facilities" in a
problem has an optimal solution
k.
i.
is
not permit
assignment
selects at
q^), is the
most p
cost.
order assignment problem,
refer to as the
when
the standard ingot sizes are prespecified. or
(so [ISP] has an optimal solution with
Zj.
=
1
for all
ingots,
k € K).
We can
<p
IKI
find the optimal y-
values for this special case by solving a non-bipartite matching problem defined over the
following network G;
Figure
1
shows
(i,i) is
to
G
for a 3 product, 3 ingot example.
each product
feasible using
i.
add a single "dummy" node
we add
dj
dummy
(j,/2) for 12
i.e.,
=
1,
...,
dj.
The
We
t^
0,
we
for
Otherwise,
/
i;
if
for
(i,/)
producing
/
=
= dj+l,
we
...,
say the pair
this pair is
cost of this edge
is
(i,//;
for /7
=
1
i.
(i,j)
=
1
the cost of each of these edges
let kjj
we
(using
dj to every
node
(i,j),
denote the index of this
dj to every
is
If
in this
the weight of the best ingot for the pair
UK;
d^
more
then
dominated;
For every feasible pair
2d|.
connect every node
(i,i) is
dj.
1
produce individual orders of product
also connect each node (i,//) for /7
corresponding to product
._
i
(i,/)
to
The network contains
denote these nodes as
the ingot k with the lowest weight such that (i,j,k)G
ingot type.
I
(i,dj+l).
nodes
the available ingots) with
We
one of the available ingots, and
economical than dedicating two ingots
case,
our heuristic solution procedure.
later exploit this property in
the network
nodes corresponding
the pair
we
dummy
node
(i,/2)
the weight of the lowest
weight ingot k^j that can produce a single order of product
dummy
nodes corresponding to product
The minimum-cost matching over
assignment problem.
then
we
product
If a
node
(i,//j
assign one ingot of type
i's
dummy
nodes
in the
kj:
is
connected to a node
to the pair
The
i.
(i,j).
If
node
total cost
we
in
G by
number of
the
for all products in
increasing order of
for
/
=
1, ....,
min
demand;
2-
{d:,
reduces considerably the
others have relatively
J'.
(i,/) is
d:.)
j
to all the
e
(i,j)
J(i),
nodes
optimal matching,
connected to one of
assign an ingot of type k^j to
we can reduce
for all
j
e
J'
J'
we need
c
the
number of
J(i)\{0,i
in J(i)\{0,i
}
to connect only the
Q,12), for 12
=
},
the
must not exceed the sum of
Suppose we index the products
then, for any
j'<jJ'eJ(i)
3.
(j,l2) in the
noting that for any subset of i-compatible products
ingots assigned to produce the pairs
demands
all
of the optimal matching equals the
optimal value of the order assignment problem. Finally,
edges
I,
provides an optimal solution to the order
optimal matching, then
produce a single order of product
e
i
are connected to each other with zero-cost edges.
i
G
Finally, for each
i.
1,
...,
J
d:.
in
nodes
(i,/)
This observation
J
number of edges
in
G
if
a
few products have high demands while
low demands.
Solving the Ingot Sizing Problem
Since the ISP model
heuristics
is
NP-hard,
we
focus on developing effective optimization-based
and lower bounds. This section describes a dual ascent method to generate lower
bounds on the optimal value, and heuristic methods (including a dual-based method) to
identify feasible solutions.
These methods exploit the
features of formulation [ISP].
The gap between
sp)ecial
the lower
covering and p-median
bound and
the best heuristic
solution value provides an assessment of the quality of the heuristic solution.
3.1
Dual ascent procedure
Dual ascent refers to a broad class of heuristic strategies to generate lower bounds as
well as feasible solutions by approximately solving the linear
programming
dual. This
technique has been successfully applied to several difficult discrete optimization problems
including the plant location problem (Erienkotter [1978]), the Steiner tree problem
[1984]), set packing problems (Fisher and
(Wong
Kedia [1986]), and the uncapacitated network
design problem (Balakrishnan, Magnanti, and
Wong
[1989]).
For the ISP, the value of any feasible
LP dual
optimal value Z*. Starting with a specific set of
bound on
solution provides a lower
initial
the
dual values, the dual ascent
procedure iteratively modifies these values in a myopic fashion to monotonically increase
The procedure terminates when no
the dual objective function value.
improvement
is
The
possible.
final dual solution not
further
myopic
only provides a lower bound, but also
suggests a partial solution to the original problem.
Consider the linear programming relaxation of formulation [ISP] obtained by replacing
the integrality restrictions (2.5a)
constraints. Let Uj for all
i
e
and (2.5b) on the z and y variables with nonnegativity
\-^ for
I,
all (i,j,k)
corresponding to the demand constraints
maximum
of
all
(i,j)
a be
the dual variables
(2.2), the forcing constraints (2.3),
standard sizes constraint (2.4). Let
order pairs
e UK, and
U(k) =
UK)
€
{(i,j): (i,j,k)
and the
denote the
set
k can produce. The linear programming dual [DISP] of
that ingot
formulation [ISP] has the following form:
Zp
[DISP]
=
ZdjUj-pa
max
(3.1)
iel
subject to:
+
u,
S
X-
Note
that constraints (3.2)
V.,.
y
(ij)6U(k)
Uj
-
w--^.
-a
simplify the notation,
Uq corresponding to the
<
for all k
Uj
>
for all
i
Vjj^
>
for all
(i,j,k)
a
>
corresponding to
we assume
dummy
Our dual ascent method
nonnegative values for
all
for all (i,j,k)
wj,
e UK,
(3.2)
G K, and
(3.3)
'J*"
2u,-v,^
To
<
(3.4a)
I,
6 UK,
<
(3.4c)
have the form:
wj,.
that formulation
we
[DISP] also contains a dual variable
fix the
value of Uq
at zero.
exploits the following observations. Suppose
the v-variables, say,
(3.4b)
0.
triplets (i,i,k)
product 0, but
6
v- jj^
for all (i,j,k)
e UK.
we
We
are given
wish
to
A
"complete"
feasible in
this partial solution
V,
i.e.,
determine values for
a
and the u-variables
that are
[DISP] for the specified v-values, and maximize the dual objective function
10-
>
value. Since p
of
a
and since the objective function coefficient of
a
is
-p, the "best" value
the smallest value necessary to ensure feasibility of constraints (3.3),
is
a
=
Max
keK
(i,j).
is
must
that these values
ingots that can produce the pair
(3.5)
^ij ^ijk
(ij)eU(k)
Determining the best values of the u-variables
and exploit some properties
i.e..
not as easy; however,
satisfy.
we can
identify
Let K(iJ) denote the subset of
Since the u-values have positive coefficients
in the
maximization objective function, the best values of the u-variables must be locally
maximum,
i.e.,
we cannot
for every product
e
i
^
"tight",
if
i.e.,
at least
one constraint
(3.2) containing the variable Uj
A
+
we can
U:
is
=
must be
A
'^
U = (uj)
Uj
Otherwise,
I,
increase one u-value without decreasing another. In particular,
an optimal completion of the partial solution (V, a) then:
w.
for at least
ijk
some u-value without
increase
one j e
and k g
K(i,j).
(3.6)
violating constraints (3.2); the
solution has a higher dual objective function value. For
(w^ +
min
J(i)
all
e
i
and j g
I
new
J(i), let
(3.7)
vjjk)
k€ K(io)
Condition (3.6) implies that
(Ui
For convenience,
assume b- =
°°.
if
the pair
+
Uj)
(i,i) is
=
for
5ij
all
i
G
I
and j g
not feasible using any candidate ingot size,
Our dual ascent procedure
iteratively
heuristically selects u-values to satisfy condition (3.8).
(3.8)
J(i).
we
will
changes the v-values, and
We
first
describe
how
to select
values of the dual variables before discussing the multiplier adjustment methods and
initial
the dual-based heuristic solution.
Initial
To
dual solution
construct an
Consequently,
given v-values),
its
dual feasible solution,
a = 0. Assume,
decreasing demand.
to
initial
we
To
we
select
Vjj|^
we
for all (i,j,k)
g UK.
for convenience, that the products are indexed in order of
heuristically
maximize
consider each product
i
the dual objective function value (for the
in this
decreasing-demand sequence, and
highest possible value given the prior choice of values
particular,
=
set:
11
U:
forj=
1, ...,
i-1.
In
set Uj
u,
=
min
min
{
j<i withj
(5
e
Dual
Step
Index the products from
Set
Vjj^
a
<-
<—
Fori=l
2:
1
=
min
(3.9)
}.
•*
satisfies condition (3.8).
dual initialization method.
=
113
=
non-increasing order of
demand
dj.
e UK,
min
-
[w,^]
U;
}
keK(i,j)
[wj
if
min
i
min
{
<- min
Uj
in
III
'^^'^^''2
112
Set
to
I
j>i,jeJ(i)
^
J(»),
[w J
and 00 otherwise;
}
keK(iJ)
{|i
J,
112,1X3},
and
i;
procedure terminates, the
final
optimal value Z* of problem [ISP]. Next,
changing v-values,
5
J(i)
let
min
this
e
0.
j<i,jeJ(i)
When
j
and
0,
Ill,
Hj
Next
and
feasible
is
this
for all (ij,k)
^D^
Step
j>i with
Method:
Initialization
1:
min
•
^
•
This procedure ensures that the dual solution
The following procedure summarizes
v Ji
(i
"
J(i)
first
one
at a
Ascent procedure Phase
1
:
we
value of
^q
discuss
how
is
our
to
initial
improve
lower bound on the
this
lower bound by
time and subsequently two values simultaneously.
Changing v-values one
at a time
Increasing certain v-values might permit us to selectively increase u-values, and hence
the dual objective value.
However,
this increase
might be offset by objective value
reductions due to required changes in other variables. For instance, increasing a particular
v-value might also require increasing
a to
maintain feasibility
12
in constraints (3.3).
.
when we
Likewise,
we might have
increase one u-value,
in constraints (3.2).
maintain feasibility
Therefore,
to decrease other u-values to
we must judiciously
The
increase so as to produce a net increase in the objective function value.
our dual ascent procedure increases just one v-value
To
^^.
B =
Set of ingot types k for which constraints (3.3) are tight.
note that
if
which constraints
all triplets (i,j,k) for
and
Uj
satisfy condition (3.8)).
j-,^
correspondingly increase by
denotes the current slack
A to be
Increasing
for
one such
X-A
by
triplet
Suppose we
if
i
must also ensure
A
by
Vj:,^
So,
or u
A
units.
:.
we
j
= j)
If
k g B, then
Otherwi.se
Vj:|.
triplets
increasing
introduces a slack of
A
i
= j, we can
(i.e.,
k,
k ^ B),
a
increases the dual objective function value by
requires decreasing
increases only one v-value at a time,
remain
feasible.
we
U:.
or increasing
its
dj0
Vj:.^.
(i,j)
value by
its
j'
= j,
is tight.
must be
all
be nonnegative,
less than or equal to
which exactly one constraint
we cannot reduce
u^.
for
all
j'
13
(3.2)
u- by more than
such that
(i,j',k')
i.e., if
is
1
two
are tight, then increasing Uj
we
for
if
triplet (i,j',k')
by 0. Since Phase
if
A
we
left-hand side and
is
(i,j)
(=
But,
units.
requires decreasing Uj which implies that the original increase in Vj-^
values must
if Sj,
unchanged
is
Consider any
ignore the latter option. Also,
constraints (3.2) corresponding to the product pair
focus on product pairs
which
increase Uj by only A/2 units.)
as the candidate variable to increase. Increasing
by
must
units in the corresponding constraint (3.2)
(If
that all other constraints (3.2)
Uj
a
s^-
A\(iJ,k). Since the corresponding constraint (3.2) contains Uj in
more
consider only
corresponding to ingot type
to increase either Uj or U:.
select u
^}, and A/2
or
other constraint(s) (3.2)
units to preserve feasibility of constraints (3.3), in
in constraint (3.3)
no more than
which we can use
tight,
not currently
Therefore, increasing
Uj
case the dual objective function value decreases by pA^jjA.
e
is
the set A.
Suppose we increase v
i
some
values must be restricted by
U:
does not create new opportunities to increase either
set
next motivate
(3.2) are tight.
the constraint (3.2) corresponding to a triplet (i,j,k)
then the current
(i,j,k) in
in
define:
Set of
(assuming that these current values
we
we
our subsequent discussions,
facilitate
A =
First,
if
We
iteration.
phase
understand the implications of increasing the current value of a specific v-
first
variable, say, \
tight,
each
first
this procedure.
and describe
Let us
at
select the values to
its
unnecessary. So,
Since the u-
current value
e A\(i,j,k) for some
k'
Uj. i.e.,
e K.
For each such product j', reducing the value of
dj.0. Finally, for triplets
u-values as long as
(i,j',k')
is less
€
A\(i,j,k:),
U:.
decreases the objective function value by
increasing Uj does not require changes in other
than the current slacks in constraints (3,2) corresponding to
these triplets.
To summarize, we
consider increasing one v-^ value, and
a
correspondingly increasing
must be reduced
U:.
(i.e., if
We
k e B)
increase Uj (or
belonging to tight constraints shared with Uj (or U:) that
to preserve feasibility of constraints (3.2).
the dual objective function value, but
this value.
necessary
to preserve feasibility of constraints (3.3).
and identify other variables
U;),
if
Increasing Uj (or U:) increases
compensating changes
Furthermore, various factors limit the
maximum
in the
other variables reduce
possible change including the
slacks in constraints (3.2) that involve Uj (or U:) but are not tight, the slack in constraint
corresponding to ingot type
(3.3)
decreased.
We evaluate the
net
k,
and the current values of the u-variables that must be
change
in the
dual objective value for each triplet
A, and select the dual adjustment that produces the
adjustments produces a net increase
first
phase.
A
increase. If
dual objective function value,
formal description of the Phase
Ascent procedure Phase
Step
in the
maximum
we
terminate the
1
1:
A
=
Set of triplets
B
=
Set of ingot types k for which constraints (3.3) are tight;
Sj(^
=
slack in the constraint (3.3) corresponding to ingot k, for
For every
(i,j,k)
For /=
let
for
which constraints
A with (i,j,k')
g
A
for
(/,
m,
k')
(3.2) are tight;
any other
e
k'
K(i,j),
i,j,
M=
Calculate
0,
6
(i,j,k)
=
{m:
(/,
m,
k')
or
e
A\(i,j,k)}.
maximum allowable value of 0:
u^;
min
me M
02
=
(i'j',k')«
Ao
=
A
(w,^.
min
—
^k
A, i'orj' -
ifke
+
A
Vj.:.)^.
A
-U-.
I
B, and oo otherwise.
h
Set
<— min
{
0j,02, A3}
ifi^^j, or
14
-
A
Uj.)
;
and
all
€
none of the
procedure follows:
1
(i,j,k)
k € K.
.
<- min
{
A,
0,,02, 2
i=j-
if
)
Calculate net change in objective function value:
Z
{d, -
d^}0-Xijp3
ifke Bandi^tj,
me M
=
a..^(/)
Z d^}0-2Xijp0
{d,-
ifkG Band i=j, and
me M
Z
{d;-
d^}
ifk€ B.
me M
next
/;
next (ij,k);
Step
2:
Find the
triplet (i,j,k)
and
/
= or j with maximum increase
i
Ojj|^(0 in the dual objective
function value.
If this
maximum
increase
is
not positive,
STOP.
Else, update
<—
Uj
Uj^
+ 0;
U;
<— Uj^ -
for all
ifke B and
if
ke B
and
Return to Step
i
Ttj,
i
=j,
Phase
values,
Vjji^
before,
we
for
2,
and
we evaluate
for the
v-j^.,
consider only
which constraints
M;
a<— a+ A^j 0;
a <— a + 2X^j 0;
and.
1
Ascent Procedure Phase
In
me
2:
Increasing two v-values simultaneously
ascent opportunities by simultaneously increasing two v-
same product
triplets (i,j,k)
(3.2) are tight.
pair
and
(i,j)
(i,j,k')
but different ingot types k and
both belong to the set
that
The procedure
for increasing the
procedure, except that
now we might have
values (since increasing the two v-values by
creates a slack of
similar to the Pha.se
1
constraints). If constraint (3.3)
Otherwi.se, the slack in these
in the v-values.
is
tight for either
k or
k',
then
two constraints determines the
Other considerations that limit
15
this
As
k'.
A of triplets
two v-values
is
very
to decrease fewer u^^
units in
we must
maximum
two
tight
increase a.
allowable increase
allowable increase are the current
u^
values of the variables
We give
to
be reduced
below a formal description of
Ascent procedure Phase
Step
need
and the current slack
constraints involving u^),
u^.
that
(to preserve feasibility
of tight
in constraints that are not tight but
contain
Phase 2 procedure.
this
2:
1:
A
=
Set of triplets
B
=
Set of ingot types k for which constraints (3.3) are tight;
S|^
=
slack in the constraint (3.3) corresponding to ingot k, for
For every
For
/
(i,j,k)
=
let
i,
(i,j,k')
which constraints
(3.2) are tight;
all
k e FL
€ A,
j,
M=
Calculate
01
and
(i,j,k) for
{
m:
(/,
m, k") or (m,
maximum allowable
=
min
/,
k")
6
A\{(i,j,k), (ij,k')}
value of O:
u^;
me M
©2
=
min
(i',j',k')g
min
A3
=
(w,^.
A, i'or j'=
(S|^,S|^)/>-j:
Si^X^j
(or Sj^/A^j
)
00
Set
+
Vj.-^.
-
uj.
- U;)
;
and
/
if
k and
k'
«E
B,
if
k (or
k')
€
B and k'
(or k)
e B
,
and
otherwise.
0^
min
{
0,,02, A3
}
ifi^^^j,
<-
min
{
01,02,
^
}
ifi=j.
or
Calculate net change in objective function value:
{d;-
Z d^}0-X,jp0
ifkork'e Bandiv^j,
me M
o,jy^.(/)
=
{d/-
Z d^}0-2Xyp0
ifkork'e Band i=j, and
me M
{d,-
Z
me M
next
next
dj„}
m
if
/;
(i,j,k)
and
(i,j,k');
16
k and
k'
6 B.
.
Step
2:
Find the
triplets (i,j,k), (i,j,k')
and
/
=
i
or j with
maximum
increase Oj:,j^.(0 in the dual
objective function value.
maximum
If this
increase
is
STOP.
not positive,
Else, update
<—
u^
Uj^
<—
+ 0;
U;
Uj^
-
A
A
for all
m such that
(/,
m,
k')
g
A\(i,j,k);
^^
Vijk^^jk + Q;
if
k or
e B and
i
?t
if
kork'e B and
i
=j,
k'
Return to Step
Note
that the
triplets (i,j,k)
when we
Uj..
In
If
Phase
and
j,
a
a
<—
<—
a+
0; and,
a + lX^- 0; and.
X.^:
1
same
principle of increasing
(i',j',k')
belonging to the
increase v-. and v-:^.,
our implementation,
we
we must
two v-values applies even
A, possibly with
set
i
^
to "unrelated"
and j
i'
^j'.
In this case,
also decide whether to increase Uj or U: and u- or
consider only the special case with
=
i
i'
and j = j'.
Phase 2 changes any of the dual values, then there might be opportunities
1
We continue alternating
and further improve the dual objective function value.
between the Phase
1
and Phase 2 procedures
final dual objective function value
optimal value
Z* of
[ISP].
As we
when
until
no further improvement
the algorithm terminates
explain in the next section,
to reapply
we
is
is
a lower
The
possible.
bound on
the
also use the final dual
solution to construct a feasible solution to the ingot sizing problem.
3.2 Heuristic
procedures
This section describes three methods to select the standard ingot sizes
method, and two greedy heuristics. For each choice of standard ingot
—a dual-based
sizes,
we
solve the
order assignment problem (using a non-bipartite matching algorithm as described in
Section
3.2.1
2),
and
ISP heuristic solution.
select the best solution as the
Dual-based heuristic solution
When
the dual ascent procedure terminates,
solution using
we
try to construct a
complementary slackness conditions.
constraint (3.3) corresponding to ingot type k
is
17-
tight,
If,
in the final
primal feasible
dual solution, the
then by complementary slackness
ingot type k
is
a candidate for being included in the set of standard sizes. Hence,
if
dual ascent procedure ends with p or fewer constraints in the set (3.3) being tight,
choose the ingots corresponding
the
we
We then
to these tight constraints as the standard ingots.
determine the optimal combinations and the actual solution value by solving the nonbipartite
matching subproblem as described
(3.3) are tight at the
Section
in the
end of the dual ascent phase, we
the tight constraints as candidates for standard sizes.
that
we
If
2.
more than p constraints
in
select all the ingots corresponding to
We then
apply the greedy heuristics
describe next to select a subset of standard ingot sizes from this restricted set of
candidate ingots.
Given the
set
of candidate ingot sizes, the following two heuristics
utilization heuristic
and the greedy assignment heuristic
—
—
the ingot
select standard ingot sizes
The
a time to heuristically minimize the total weight of ingots used.
one
at
heuristics differ in the
order in which they consider the candidate sizes.
3.2.2 Ingot Utilization heuristic
This heuristic attempts to select standard ingot sizes that can satisfy a large proportion
(in
terms of weight) of the
produce the pair
(i,j),
demand
at
high utilization levels. If
the utilization of the ingot
divided by the gross weight Wj^ of ingot
combinations should be such that every ingot
specify a threshold utilization level (}% (0
By
ingot utilization.
The algorithm
<
is
parametrically varying p,
iteratively evaluates the total
two orders
chosen standard sizes and
nearly fully utilized. For this purpose,
< 100) representing
(3
use an ingot type k to
the total weight (qj+q;) of the
is
Ideally, the
k.
we
we can
the
minimum
we
acceptable
generate multiple solutions.
demand
(in
pounds)
that every
remaining
candidate ingot size can satisfy subject to the threshold utilization requirement, and selects
the size that satisfies the
maximum demand. At any
iteration, let
current set of chosen and remaining candidate ingot sizes.
assigns
demand
for various products to the
(number of orders)
for each product
product pair
(i,j)
type k ingot
p% or more,
W(k) be
sum
the
the ingot k with
(i,j)
e U(k)
as M(i,j)
=
^j:
i.e.,
i
g
I.
chosen
If
^j:
= min
U(k)=
maximum W(k)
all
{(i,j): (i,j,k)
pairs
Let
&j
{^j, &:
},
denote the residual demand
we define
all
the
volume of a
product pairs that utilize a
UK and (qj+qj) >
e U(k). The ingot
(i,j)
|3w,/100}. Let
utilization heuristic selects
as the next candidate size, assigns all the product pairs
to this ingot, updates the residual
If at
e
denote the
The algorithm progressively
(qj+q :). Let U(k) be the set of
of volumes for
candidate sizes are chosen.
sizes.
K* and R
demands, and repeats the procedure
any stage, W(k)
18
is
until
p
zero for every remaining candidate size
k G
^
and
if
the
are not completely satisfied, then
demands
P by a fixed percentage (10%
we reduce
the threshold value
our computations) and continue the algorithm.
in
Ingot utilization heuristic:
Step
^
Initialize
0:
<— set of candidate ingot sizes
K*
<-({)
Sj
=
qj
= weight
dj for all products
e
i
I.
per order of product
i.
A
Step
For every k e K,
1:
U(k,p) = Set of all feasible pairs (ij) e U(k) with (qj+q^)
B
>-^w,^.
lUU
= min
i^.
{
dj
,
dj
for
}
W(k)
Compute
all (iJ)
e U(k, p).
Z
^ij(qi
=
+
qj)
(ij)eU(k,P)
next k;
Let
If
K
= argmax { W(k): k e
k'
W(k') =
update P
0,
0.90*p, and repeat Step
<r-
Else, select ingot type
}.
as a standard ingot,
k'
i.e.,
1.
update
K* <-K*u{k'};
K<-K\{k'};
<- max
dj
If IK*I
=p
or
-
{dj
if d,
,
^jj
=
0} and
for all
i
<-
dj
g
I,
max
STOP.
{dj
-
i--
,
0} for
all (i,j)
Else, repeat Step
e U(k,P).
1.
For the chosen standard ingot sizes K*, solve the order assignment problem to
Step 2:
determine the optimal combinations.
3.2.3
Greedy Assignment heuristic
This heuristic attempts to
considers product pairs
(i,j)
select
first
in
good ingots
decreasing order of
for
total
high-volume combinations.
It
weight, and selects the best ingot
A
As
type for each pair in sequence.
before,
let dj
denote the residual demand for product
i
at
A
any intermediate
iteration.
Let
U
denote the set of feasible prtxluct pairs
A
residual
A,jj
demand
{qj+q-
sizes.
}
for both
for every pair
We select the
i
and
(i,j).
j
are positive. Define
Let
K
A,-
A
= min
(i,j)
such that the
A
{dj,d;
}
and volume M(i,j) =
denote the current set of available candidate ingot
A
pair
(i',j')
e
U
with
maximum volume
M(i',j'),
and determine the best
A
ingot
k'
G
K
for this combination,
i.e.,
the ingot for
-
19-
which the scrap and surplus
(w,^.
-
qj.
-
q:.) is
minimum among
demands
standard ingots, reduce the
for products
demand
ingots have been selected or the residual
strategy might be useful
few products
0:
when
the
demand
add ingot type
for every product
distribution
is
Initialize
^
<—
a,
=
set
=
p
zero. This heuristic
i.e.,
the
demand
for a
of candidate ingot sizes
dj for
AAA
LI
skewed,
rs
until
heuristic:
all
products
i
€
I
j
1:
to the current set of
and j', and repeat the procedure
i'
Let q denote the weight per order of product
Step
k'
exceeds other demands.
far
Greedy Assignment
Step
We
ingot types k e K.
all
set
of
X- = min
all
feasible pairs
^
with
(i,j)
i.
and ^ >
0.
A
{dj, d:
for all
}
e U.
(i,j)
M(i,j)= l,j{q,-Kij}
forall(i,j)eij.
A
(i',j')
= argmax
= argmin
{M(i,j)
(wj.
-
Select ingot type
k'
k'
{
:
-
q,.
g
i,j
q.
)
:
LI
}.
ke
R}.
as a standard ingot,
update,
i.e.,
K*<-K*u{k'};
AAA
AAA
dj.
d:.
K^K\{k'};
If IK*I
Step 2:
<—
dj.
=p
- AjY and
or
if dj
=
<—
for all
d:.-
i
e
Aj.-.
I,
.
STOP.
Else, repeat Step
1.
For the standard ingot sizes K*, solve the order assignment problem to
determine the optimal combinations.
Several variants of these heuristics are possible. For instance, in the ingot utilization
heuristic,
we might dynamically
update the residual
weight W(k) of each candidate ingot. Likewise,
candidate ingot k
is
we might
the best available ingot.
demand
in the
we compute
the total
greedy assignment method, for each
consider the total weight of
We
as
all
combinations for which ingot k
might also apply local improvement procedures to reduce
the total cost of each of these three heuristic solutions
—
the dual-ascent, ingot utilization,
or demand-based salutions. For instance, one procedure consists of considering each
current standard size in turn and evaluating the potential savings
by replacing
this size
a neighboring (either larger or smaller) size. If none of these "swaps" produces any
-20
with
savings, the procedure terminates; otherwise,
we perform
swap with maximum savings,
the
and repeat the procedure. To evaluate the savings for each swap, we solve the matching
problem
cost
is
to
the total cost with the
compute
infmte
if
the
new
set
new proposed
of ingots cannot produce
set
of standard sizes (the
products), and
all
compare
total
this cost
with the current cost.
In our computational experiments,
three heuristics
was
itself quite
we found
that the best solution
close to optimality. So,
we
produced by the
did not implement any local
improvement procedure.
Computational Results
4.
We implemented
actual orders
and tested the dual ascent procedure and heuristics using data on
from a leading aluminum sheet manufacturer.
distinct product groups,
We classified the
orders into 4
We
and separately solved the ingot sizing problem for each group.
analyzed various scenarios obtained by varying the number of candidate ingot sizes, the
number of standard
compared
sizes required,
and the
maximum
width differential.
We
also
the results with current practice and with restricted order combination options.
This section describes some implementation details and summarizes our computational
results.
Implementation details and test problems
4.1
We
implemented the dual ascent procedure and heuristics
4381 computer. To solve the order assignment subproblem,
bipartite
in
FORTRAN on
we used
an
IBM
Derigs' [1988] non-
matching algorithm.
We developed
four test problems from a database of actual orders for a popular product
family (sheet products with gauges in a specified range and requiring a particular alloy)
over a 12-month period. The database contained over 500 "small" orders
which dedicating an ingot was not considered economical).
into
all
4 natural product classes based on
their widths.
products have the same width w; approximately
divided the range of widths below
w
into
two
approximately equal number of products, and
We
first
parameter
(d)
orders for
classified these orders
For one of these classes (Problem
50%
of the orders had
3),
We
this width.
classes such that each class has
treat the
range of widths above
fourth class. Since products with large width differences (greater than the
differential
(i.e.,
cannot be combined, partitioning the data
-21
w as the
maximum
set into these four
width
product classes does not eliminate
many
feasible pairs.
Within each product
class,
we
selected appropriate intervals of widths, gauges, and weights to group the orders into
The width and weight
products.
were defined by rounding up the actual order
intervals
widths and weights to the nearest inch and thousand pounds, respectively. The gauge
intervals corresponded to non-overlapping ranges
from the table of feasible gauge
combinations (see Gopalakrishnan [1994] for additional details regarding the
test
problems).
The
facility currently
produces ingots
in 3
widths and a
of 6 standard sizes (width
total
and weight combinations). For each problem, we generated two
—a
sizes
limited set and an extended
set.
The
sets of candidate ingot
limited set has both fewer candidate widths
and fewer candidate weights for every width compared to the extended
set.
candidate weights for a particular width was chosen to reflect the relative
width. For instance,
then for this width
if
many
we might
The number of
demand
for that
products have a width close to a candidate ingot width wl,
select candidate weights at 2,500
pound
intervals in the range
of ingot weights that the ingot plant can produce; on the other hand, less popular widths
might have candidate weights
at
10,000 pound intervals. Table
1
shows
the
number of
products and the number of candidate ingot sizes for both the limited and extended sets for
each of the four problems.
For the order combination constraints, we used the actual gauge combination table
supplied by the manufacturer.
Our "standard" problems assume
differential (o of 2 inches,
we cannot combine two
than 2 inches.
compute
tail
i.e.,
We subsequently
maximum
we used
feasible combinations for every candidate ingot size based
restrictions.
i.e.,
(O.
To
first
generate
all
the
on the order combination
subject to the total weight, width, width differential, and gauge differential
The demand
for each product
corresponding product group
sizes, feasible
to
the actual values of scalping loss, head and
and trim loss used by the planners. For each problem, we
constraints,
width
orders that differ in width by more
performed sensitivity analysis with respect
the planned scrap allowances,
scrap,
a
combinations,
in
our data
we
is
set.
the
number of orders belonging
to the
Given the product demands, candidate ingot
obtain lower and upper bounds for various values of p by
applying the dual ascent procedure and heuristics. The integer programming formulation
for our largest test
problem has
1
1096 variables and
22-
1
11 66 constraints.
Framework
4.2
for experimentation
Our computational
results seek to address
two
issues:
(i)
how effective
ascent and heuristic procedures in generating bounds and solutions; and,
economic impact of changing the problem parameters.
and heuristic methods
to
In
all,
we
are the dual
(ii)
what
is
the
applied the dual ascent
over 80 problem instances.
We use the % gap between
the upper
and lower bounds, defined as (upper bound -
Low
lower bound)* 100/lower bound, to assess algorithmic effectiveness.
bound
indicate that both the dual ascent lower
perform well. For each problem instance,
we
values of
% gap
and the heuristic procedures
is tight,
select the best
among
the upper
bounds
obtained by the heuristic procedures. Recall that the ingot utilization heuristic requires a
user-specified threshold utilization value p.
our standard
test
We experimented with various values of P
for
problems to identify effective settings for subsequent experimentation.
Section 4.3 reports these results on algorithmic effectiveness and compares the
performance of the heuristic solution procedures.
Section 4.4 illustrates
how
the
ISP model and solution methodology can be used to
perform various types of sensitivity analyses. In particular,
the results to three
problem features
standard ingot sizes, and the
total scrap
and surplus when
restrictive scenarios:
if
we
limit the
for the
—
explore the sensitivity of
the set of candidate ingot sizes, the
maximum
we
we
number of
We also compare the
width differential parameter.
permit order combinations with the results for two
when we only permit dedicated
ingots
combinations to identical orders. Finally,
(i.e.,
one order per
we compare
ISP solution with the corresponding value for the current
set
ingot),
and
the objective value
of standard ingot sizes
(but with order combination permitted).
4.3 Algorithmic effectiveness
We
first
compare
the performance of various heuristics and determine an effective
range for the parameter P
in the ingot utilization heuristic
before assessing the quality of
the lower bounds.
4.3.1 Heuristic
To evaluate
performance
heuristic performance,
we
consider only problem instances with the
"standard" width differential of 2 inches and the limited set of candidate ingot sizes, but
several values for p. For the ingot utilization heuristic,
utilization levels
P ranging from 507o
to
90%. Let us
-23-
we
first
initially tried several
understand
how
this
threshold
parameter
might affect the quality of the solution. For high values of
utilize
P%
an ingot
or higher. Consequently, in the
the threshold thus permitting less effective ingot sizes.
values of
(3,
many
pairs will achieve or
However, assigning these
exceed
Our
initial
experiments confirmed
=
select
this
At the other extreme, for low
every candidate ingot.
Table 2 shows the objective values for
we
expect that the heuristic solution will
this conjecture.
For
70 or 80% was always
65,
P = 50 or 90%. So, we report
utilization heuristic, with
method might
pairs reduces the overall average ingot utilization for the final set
problems, the heuristic solution with p
solutions with
can
pairs
ingot, the algorithm lowers
this threshold for
of standard sizes. Therefore, for either extreme,
deteriorate.
the
few
demand. Subsequently, when
no longer attainable for any remaining candidate
is
relatively
initial stages,
ingots that satisfy only a small proportion of the total
threshold
(3,
results only for the
all
all
our
initial test
better than the
former three values.
four test problems for both the ingot
P = 65, 70, and 80%, and for the greedy assignment
Computation times (not reported here) for the
heuristics
heuristic.
were modest. Problems
4 with a few hundred feasible assignments required 0.3 to 4.5
CPU
1,
seconds on an
2,
and
IBM
4381, whereas Problem 3 with several thousand feasible combinations required 30 to 50
CPU
seconds.
results in
Table 2 indicate that none of the methods
Out of the 16 problem instances, the ingot
superior.
P=
The
80%
produced the best solution
and ingot
utilization heuristic with
in 5 instances
P =
70%
is
consistently
utilization heuristic with
P=
65%
and
each; the greedy assignment heuristic
produced the best solution
in 3
problem
instances each. In general, the ingot utilization heuristic performs better than the greedy
assignment method. The greedy method selects ingot sizes based on the residual volume
of a single product pair, whereas the ingot utilization heuristic considers multiple product
pairs that the ingot
can produce. Indeed, the greedy assignment heuristic might sometimes
select small ingot sizes initially,
infeasible at the
Problem
3,
end
(after
6.
certain
low-demand but "heavy-weight" products
p standard sizes are selected). This phenomenon
where the greedy
values of p less than
making
is
heuristic failed to produce a feasible set of standard sizes for
In contrast, the ingot utilization heuristic has an adaptive feature
(reducing the threshold utilization level) to avoid infeasible solutions. Finally,
computational results,
apparent for
we observed
in all
of our
that the dual-based heuristic solution often coincided
with the ingot utilization or greedy assignment solutions, but in no case was the dual-based
solution superior.
We,
therefore,
do not separately
method.
24-
report the results of this heuristic
For our subsequent experiments, we applied the ingot
utilization heuristic, for all three
values of P, as well as the greedy assignment heuristic, and selected the best heuristic
solution value as the upper bound.
4,3.2 Dual
ascent performance
Tables 3 through 6 present the results for Problems
through
1
4, respectively, for
various p values and width differentials (except Problem 3 in which
same width) with both
all
products have the
the limited and extended set of candidate ingot sizes.
The
tables
provide information on the respective problem sizes (# of feasible assignments) as well as
% of orders processed using dedicated
the
ingots in the best heuristic solution.
gaps in these tables show that the dual ascent lower bounds are
solutions are near-optimal. In almost
all
problem instances, the
the limited set of candidate sizes, the average
the four problems are 1.8%, 3.7%, 6.1%,
tight,
The low
%
and the heuristic
% gap is below 7%.
With
gap between the upper and lower bounds
and 4.7%, respectively. For the extended
set
for
of
candidate sizes, the corresponding averages are 2.7%, 3.0%, 4.9%, and 5.9%. Overall, the
procedure generates solutions that are (provably) within
In general, for a given
width differential, the
standard sizes p increases (except for Problem
differential or increasing the
%
3).
number of candidate
gap tends
set
to decrease as the
number of
However, increasing the width
ingot sizes (both of
number of feasible combinations) does not produce
Table 3 with the limited
4% on average.
which increase
the
a similar consistent effect, e.g., in
of candidate ingots, for each value of p, the gap
first
decreases
and then increases as the width differential increases.
4.4 Sensitivity analyses
We now illustrate how to use
decision support. In particular,
to three
(i)
the
to
problem
the
ISP model and methodology for planning and
we examine
the sensitivity of the objective function value
features:
number of standard
choose sizes
ingot sizes: Permitting
that are closer to the order
more standard
sizes enables the
method
combinations; consequently, the objective
function value should decrease;
(ii)
the set of candidate ingot sizes:
choice
in selecting
With more candidate
a set of standard sizes, and hence
surplus to decrease;
25
sizes, the
model has greater
we expect
the total scrap and
(iii)
the width differential:
increases the
Permitting greater differences in widths for combined orders
number of feasible
pairs, thus potentially
reducing
total
pounds of metal
processed.
Tables 2 through 6 confirm these effects. For instance. Table 2 shows that as the
number p of standard
to
2.7%
(for
Problem
ingot sizes increases
3),
from 3
assuming the standard width
To
limited set of candidate ingot sizes.
ingot sizes consider, for instance.
by 3.7% when we extend
differential (2 inches)
products
for
and
for the
assess the impact of increasing the set of candidate
Problem 3 with p =
4.
The
objective function decreases
the set of candidate ingot sizes. Finally, increasing the width
differential appears to provide considerable benefits for
improvement
Problem 2
(for
Problem
3,
Problems
width differential
is
and
1
4, but not so
not relevant since
have the same width). For instance, for Problem 4 with p =
in this class
by up
to 6, the objective value decreases
objective value decreases by
10% when we
increase the
maximum
much
all
3,
the
width differential from
2 inches to 6 inches.
The
(0
tables also
show
and extending the
the
set
the effect of increasing the
maximum
of candidate ingot sizes on the size of the problem (measured by
number of feasible assignments). Both of these changes
feasible combinations; however, increasing the
impact on problem
size.
width differential parameter
Also notice
increase the
number of candidate
that, generally, as
we extend
number of
sizes has
much
greater
the set of candidate
% of orders with dedicated ingots in the best solution tends to decrease
(since more combinations are feasible); likewise, this % decreases as the width differential
ingot sizes the
increases.
4.5 Benefits of ingot sizing
To
assess the benefits of order combination,
following two special cases
ingots, and,
for the
and order combination
(ii)
:
(i)
solved the ingot sizing problem for the
no combinations permitted,
limited combinations,
same product
we
are permitted.
i.e.,
i.e., all
orders use dedicated
only dedicated ingots or combining two orders
We also compare our proposed solution with the set of
standard sizes currently stocked by the facility.
The
results of all these
experiments are
presented in the following sections.
First,
we compare
ingot sizes.
The
the effectiveness of the
facility's
cuaent 6
sizes,
ISP solution with the current
set
of standard
although not tailored for order combination, do
permit joint processing of orders. Using the matching procedure,
26
we determined
the
optimal assignment of feasible pairs to the current set of standard ingot sizes to meet total
demand. Table 7 compares the ISP values with the
(we compute the percentage deviation
current ingot sizes
current sizes). For Problems
1
and
2, the
savings), whereas for Problems 3 and
Problem 4 with the
practice;
we
Since
products
all
ISP solution
is
4 the improvement
restricted set of candidate ingots, the
attribute part of this deterioration to the
dramatic savings.
weight of ingots processed for the
total
in
relative to the total weight for the
considerably superior (over
is
1
8%
only marginal (indeed, for
ISP solution
worse than current
is
poor quality of our heuristic solution).
Problem 3 have the same width, solving the ISP does not produce
When we
extend the set of candidate ingot sizes, the savings are
uniformly higher, highlighting the importance of using a good set of candidate ingot sizes.
The
total
savings for the product family that
we
studied
was of the order of hundreds of
10%
thousands of dollars annually, and represented almost
reduction in the total weight of
metal processed.
To
two
quantify the benefit of combining orders,
variants of
Problem 3
we
applied the ISP solution procedure to
—one without any combinations
(i.e.,
only dedicated ingots), and
one with limited combinations (within each product group only). Table 8 compares the
standard solution to these two special cases.
dramatically
for limited
variables increases
when we allow order combination, from 897 with no combinations
combinations to 9838 for the standard model. In each case,
methodology
to determine a
good
utilize the available ingots better,
that order
The number of assignment
combination reduces
set
use our solution
of standard sizes. Order combination allows us to
and reduces
total ingot
the limited-combinations model,
we
to 1441
total surplus
and scrap. The
results indicate
usage by an average of 8.3% when compared to
and by an average of
26% when compared
to the
no-
combinations model.
5.
Concluding Remarks
This paper has described the tactical planning problem of deciding standard ingot sizes
in a
make-to-order aluminum sheet manufacturing
facility.
integer program, and developed a dual ascent lower
Our computational
that are
results indicate that the
guaranteed to be within
model and solution method
4%
method
of optimality.
We modeled this problem as an
bounding procedure and
is
very effective, producing solutions
We also illustrated how to use the
to assess the sensitivity of overall
parameters such as the number of standard ingot sizes, the
27
heuristics.
performance to problem
maximum
width differential.
and the
set
effective
of candidate ingot
means
sizes.
These
results
demonstrate that order combination
is
an
to reduce total production costs.
Future work might address related algorithmic and practical issues, including
improving the two heuristic methods and exploring more sophisticated ascent methods to
increase several v- values simultaneously. In this paper
that
we
limited the
number of orders
can be combined to two, but combining more orders might become operationally
feasible in the future.
The underlying dual
ascent principles apply to this
problem as well; however, the order assignment problem
ingot sizes)
is
not as easy. Another issue to explore
is
(for a
more complex
given choice of standard
the linkage
between the medium-
term ingot sizing decisions and the short-term order combination decisions, perhaps, using
detailed simulations.
28
.
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R. T.
E
Table
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Date Due
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Lib-26-67