Chapter 7 Integer Programming Models

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Chapter 7 Integer Programming Models

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Chapter 7Integer Programming Models

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In manufacturing, products are often indivisible, so a production plan that calls for fractional output is not acceptable.

There are also many situations that require logical decisions of the form yes/no, go/no go, assign / don't assign.

Designers faced with selecting from a finite set of alternatives, schedulers seeking the optimal sequence of activities, or transportation planners searching for minimum cost vehicle routes all face discrete decision problems.

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SITE SELECTION EXAMPLE

A manufacturer is planning to construct new buildings at four local sites designated 1,2, 3, and 4. At each site, there are three possible building designs labeled A, B, and C. There is also the option of not using a site.

The problem is to select the optimal combination of building sites and building designs.

Preliminary studies have determined the required investment and net annual income for each of the 12 options.

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• The company has an investment budget of $100 million ($100M). The goal is to maximize total annual income without exceeding the investment budget .

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Let I = {A, B, C},

J= (1,2,3,4) be the set of site options.

Jj Iiyij

and for otherwise 0

j siteat used is i design if 1Let

Also, denote by pij the annual net income and by aij the i

nvestment required for the design/site combination i,j.

As a first try, you propose the following model for finding the maximum net annual income.

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JjIiy

z

ij

ijIi Jj

ij

ijIi Jj

ij

ya

yp

,},1,0{

100 subject to

Maximize

the optimal solution is

yA1= yA3= yB3=yB4=yC1=1

with all other values of yij equal to zero and z = 40. O

f the available budget, $99M is used.

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You seem to have omitted some of the logic of the problem, because two designs are built on the same site—that is, Al and Cl, and also A3 and B3, are all in the solution.

In addition, your supervisor now realizes that you were not alerted to several other logical restrictions imposed by the owners and architects—i.e.,

(a) site 2 must have a building; (b) design A can be used at sites 1,2, and 3 only if it is also selected for site 4; (c) at most two of the designs may be included in the plans.

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The following additional constraints are needed to guarantee a feasible solution.

Site 2 must have a building:

There can be at most one building at each of the other sites:

12

Ii

iy

1,3,4 for 1

jIi

ijy

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Design A can be used at sites 1,2, and 3 only if it is also selected for site 4:

To formulate the constraints associated with design selection, three new binary variables are introduced.

Let Wi =

A4A3A2A1 3yyyy

CB,A,for otherwise0

used is design if1

ii

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At most two designs may be used:

WA + WB + WC 2

Finally, the yij and wi variables must be tied together:

the optimal solution is

yA1 = yA4 = yB2 = yB3 = wA =wB = 1

with all other variables equal to zero and z = 37.

All the budget is spent.

CB,A, for 44

1

iwyj

iij

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Logical Constraints

Bound Constraints: yi=0 or 1 for i=1,2,…,n

The n decisions are mutually exclusive: y1 + y2 +…+yn 1

At most k in the subset may be chosen: y1 + y2 +…+yn k

At least k in the subset must be chosen: y1 + y2 +…+yn k

Exactly k must be chosen: y1 + y2 +…+yn = k

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Implication Constraints

Let w be a binary variable that corresponds to a decision implied by one or more related decisions.

• Decision w is implied if any one of the other n decision variables has a value of l:

y1 + y2 +…+yn nw

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• Decision w is implied if all of the other decision variables are 1:

y1 + y2 +…+yn n-1+w

• Decision w is implied if at least k of the other n decision variables are 1:

y1 + y2 +…+yn (k-1)+[n-(k-1)]w

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Binary Variable Implied by a Real Variable

Consider the binary variable y representing the decision whether or not to build a facility and the real variable x representing the number of products produced by the facility. A logical constraint that restricts the variable x to be 0 unless y is 1 is

where u is an upper bound on x.

uyx

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Maximize z =

subject to bi ,i=1, . . .m

xj 0 and integer, j = 1,...,p, and xj > 0, j = p+ l,...,n

n

jjj xc

1

n

pjjij

p

jjij xaxa

11

Model Structure

GENERAL CONSIDERATIONS

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The first p variables are restricted to integer values, whereas the remaining n - p variables can assume any nonnegative real values.

When all the variables are constrained to the values 0 and 1, this restriction allows us to represent binary decisions such as yes/no and gives rise to what is called a binary or 0-1 programming problem.

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Simple Transformations

Although it is somewhat inefficient from a computational point of view, it is possible to use only binary variables in IP models.

A simple transformation replaces a bounded general integer variable with a weighted sum of several binary variables.

Let x be an integer variable bounded by zero from below and by u from above.

Let t be the smallest integer such mat 2t+1 > u.

Converting a General IP to a 0-1 Model

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x= y1 +2 y2 +4y3…+2tyt+1=

yj= 0 or 1, j =1 ,..., t + l

where 2t u < 2t+1

t

jj

j uy0

12

For example, if x is bounded by 15, then t = 3 (because 24 > 15 23).

The transformation is x = y1 + 2y2 + 4y3 + 8y4.

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Discrete Decision Values and Functions

Suppose that x can take on only one of a finite set of values—i.e., x D = {d1, d2,..., dr}. We can model this situation as

x= d1y1 +d2 y2 + d3y3…+ dryr

y1 + y2 +…+ yr= l

yj=0 or l , j=1,...,r

Ex: x {5, 8, 13, 20}

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Complemented Variables

For some algorithms, it is necessary to have all positive signs in the objective function. When only binary variables appear in the model, replacing the corresponding variable with its complement can reverse a negative coefficient. The complement of binary variable xj is 1- xj which is also binary.

Ex. Z=x1-4x2+5x3

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Fixed Charge Model

SYSTEM DESIGN WITH FIXED CHARGES

In many situations, undertaking an activity means that a fixed charge or setup cost is incurred in addition to the variable cost associated with the level of the activity.

In the telecommunications network design problem, there is a tradeoff between construction costs and operating costs for a given demand. In a manufacturing problem, there is usually a fixed charge for setting up a machine and a variable cost for each item produced. The same is true for building and operating most facilities.

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To construct a model for the network design problem, let xk be

the flow on proposed link k A'. The total cost for link k can be modeled by the following nonlinear concave function.

0 when0

0 when

k

kkkkkk x

xxcf)(xh

where fk is the fixed cost coefficient and ck is the variable cost

coefficient.

LP formulations cannot handle this kind of function directly.

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Akk

yk

allfor otherwise0

built islink if1Let

When hk(xk) is to be minimized and fk > 0, the following transfor

mation allows us to formulate the objective as a linear function.

hk(xk)=fkyk+ckxk

xk 0 , yk = 0 or 1

xk uyk (u is the upper bound of xk)

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KO(i) is the set of arcs originating at node i and KT(i) is the

set of arcs terminating at node i, we have

C1:

C2:

C3:

C4:

'

z MinimizeAk Ak

kkkk xcyf

)( )(

, Subject toiO iTKk Kk

ikk TDSibxx

',0 Akyux kkk

',0 Akux kk ',1or0 Akyk

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All of the data for the model are included in Figure 7.1.

Constraint C2 represents the implication that if xk is gr

eater than zero, yk must be 1.

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FACILITY LOCATION PROBLEM

A logistics company wants to set up a distribution network in a new region of the country. There are five possible locations for warehouses and five customer locations that use the commodities supplied by the warehouses.

Table 7.2 displays the data defining the problem, including unit shipping costs between potential warehouse sites and customers, fixed and variable costs for constructing the warehouses, customer demands, and warehouse capacities.

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The goal is to select warehouse sites and sizes and to establish a shipping pattern between warehouses and customers that minimizes total shipping costs as well as the amortized cost of construction .All demands must be met and the capacity at each facility must not be exceeded.

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Let us say that m potential sites for warehouses have been identified and that the locations and demands of the n customers are known. Let dj be the demand for customer j. The shippin

g cost between each potential warehouse site i and each customer j has been estimated as cij The cost of establishing a ware

house at location i consists of a fixed cost fi and a variable cost

vi per unit of warehouse capacity.

The maximum capacity at warehouse site i is ui

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Define the following variables.

zi = size of warehouse at location i

xij = amount of product shipped from warehouse i to customer j

i

iyi siteat locatednot is ea warehous if 0

siteat located is ea warehous if1

The mathematical programming model is as follows.

All demands must be met:

Supply must not be exceeded:

m

i

m

i

m

i

n

jijijiiii xczvyfz

1 1 1 1

Minimize

njdxm

ijij , . . . 1,,

1

, . . . ,mizxn

iiij 1

1

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Shipping from a location implies that the warehouse has been built:

Nonnegative shipments:

Nonnegative size:

Integrality:

m, . . . , i yuz iii 1,

n, . . . , m, j , . . . , i xij 11,0

m, . . . , i zi 1,0

m, . . . , i yi 1,1or0

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Uncapacitated Facility Location Problem

It is assumed that there is no limit on the size of the warehouses. Although the same mathematical programming model applies with ui , set at an arbitrarily large value, we present a sli

ghtly different formulation that allows for a more efficient solution technique.

Let

xij = proportion of demand j satisfied by warehouse i

otherwise 0

built is warehouseif 1 iyi

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Because capacity is unlimited, it can be shown that it is optimal to meet the demand of each customer from a single warehouse. As such, the unit transportation cost and the variable facility cost can be combined with the demand to obtain the cost coefficient associated with the new variable xij. This is

the cost of supplying the entire demand of customer j from warehouse i.

jijiij dcvc )(

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The new model is as follows.

Minimize

All demands must be met:

m

i

m

i

n

jijijii xcyfz

1 1 1

, . . . ,n j xm

iij 1,1

1

Shipping from a location implies that the warehouse has been built

Simple bounds:

Integrality: yi = 0 or 1, i = 1, . . . , m

1

, 1n

ij ij

x ny i , . . . ,m

n, . . . , m ; j , . . . , i xij 11,10

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For the second constraint, it is necessary to multiply the yi v

ariable on the RHS by n to allow for the extreme case in which all customers are serviced by warehouse i.

In an alternative formulation, these m implication constraints representing the potential warehouse sites are replaced by mn implication constraints each representing the relation between an individual transportation link and a site:

xij yi , i=1,...,m ; j=1,....n

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Although this is inefficient from a modeling point of view because we have increased the number of constraints by a factor of n, IP algorithms may now be able to find solutions more quickly.

This increase in efficiency results from the fact that if we relax the integrality requirement on all yi, the feasible region as

sociated with the expanded model is much tighter than that of the original.

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Covering Problem

Consider a microelectronics company that would like to manufacture six new products. Initial cost estimates indicate that the equipment needed to make any of the products is very expensive, and that to make each product individually would probably require too much investment. It is possible, however, to produce composite devices that through different inter connections can perform the functions of two or more of the products.

In fact, a very complex device can be constructed that would have the same functionality as all six products, but it would require the use of unproven technology and so has been ruled out.

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After studying the design and manufacturing issues, the company's engineers have come up with 14 options, each performing a subset of functions, for management to consider. The first six options are the products themselves.

To define the problem mathematically, let cj be the equipment c

ost for device j and let the column vector Aj represent the set of

functions that the device performs. For instance, the vector A14

= (1, 0,0,1,1,0)T indicates that device 14 can be used for products 1,4, and 5.

The problem is to find the set of devices with the minimum total equipment cost that can perform the functions of all six products.

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For convenience, Figure 7.6 shows the 14 vectors A 1, A2,..., A14 arrayed in a 6×14 matrix in which each

row corresponds to a particular function.

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Assume that there are, in general, n devices or options, and let

For x the model is

Minimize cx

subject to Ax e

xj = 0 or l, j =1,..., n

where e = (1...., 1)T is an n–dimensional vector of 1's.

1 if device is selected for manufacture

0 otherwisej

jx

,n

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We must select devices to cover all functions. The ith row of A identifies the set of devices that includes the ith function. The corresponding constraint ensures that at least one of the devices selected will perform the function.

In the optimal solution for this example, x5= x7= x13= 1, and

all other variables are zero, with cost z = 62. Each function is covered by this solution, and function 2 is covered twice.

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Partitioning Model

If we now add the restriction that each function must be included once and only once in the selected devices , the partitioning problem results. The general formulation is as follows.

Minimize cx

subject to Ax = e

xj = 0 or l, j =1,..., n

For this problem, the subsets cannot overlap, and thus the solution obtained for the covering model is not feasible because function 2 is covered twice. Solving this problem, we find that x1=x3= x5= x13=

1 and that all other variables are zero, with cost z = 63.

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Traveling Salesman Problem

Starting at an arbitrary location, we would like to find a tour that visits each node exactly once and returns to the original location. The goal is to minimize the sum of the lengths of the selected arcs.

Let n be the number of nodes in the network and let cij be

the length of the arc passing from i to j. When it is infeasible to go from i to j either the coefficient cij is set to an arbitraril

y large number or the corresponding variable is omitted from the formulation.

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The matrix shown in Figure 7-7 gives the distance between each pair of nodes. The dashes along the diagonal rule out self-loops. Also, the data imply an asymmetric problem structure, because the distance from node i to node j is generally not equal to the distance from node j to node i.

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A subtour is a sequence of unique nodes that starts and ends at the same location but visits only a subset of the n nodes. A solution containing subtours is not feasible.

Define the following decision variables:

,n, j iji

xij

1 allfor otherwise0

selected is to from arc theif1

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When i =j, xij does not exist so it is not included in the model.

We now give the mathematical programming formulation of the asymmetric TSP.

Tour length: Minimize z =

Cl: Exactly one successor for each node:

C2: Exactly one predecessor for each node:

C3: Subtour elimination:

C4: Integrality. xij = 0 or l, i≠j = l , . . ., n

n

i

n

jijij xc

1 1

n , . . ., i xn

jij 1,1

1

n , . . ., jxn

iij 1 ,1

1

22,,1 nSNSSxSi Sj

ij

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Constraint C3 prevents formation of subtours or cycles of size less than n. It is based on the observation that any subtour constructed from the nodes in a subset S must have exactly arcs.

A major difficulty in solving a TSP is dealing with the subtour elimination constraints, of which there are 2n–1 – n –1.

The second difficulty in solving a TSP is the loss of the integrality property commonly associated with network flow problems. The subtour elimination constraints do not admit a total unimodular structure.

S

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Returning now to the six-node example, if one relaxes the subtour elimination constraints, the assignment problem (AP) results, which automatically has integer solutions. Solving the AP relaxation, we obtain the solution {(1, 4), (4, 2), (2, 1), (3, 5), (5, 6), (6,3)}. The corresponding objective function value z = 54 provides a lower bound on the length of me optimal TSP tour. This solution is not feasible, because it contains two subtours:(1,4)→(4,2)→(2,1) and (3,5)→(5,6)→(6,3).

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To eliminate the first subtour associated with S = {1, 2, 4}, we add the constraint x12 + x21 + x14 + x41 + x24+ x42 2 to the ≦relaxed formulation. This constraint indirectly eliminates the second subtour associated with the subset S = (3, 5,6) as well as the subtour in the opposite direction: (1,2)→(2,4)→(4,1).

Solving the new problem, we obtain the solution {(1,4), (4,3), (3,5), (5,6), (6,2), (2,1)}, with z = 63. This solution contains no subtours and so is optimal (see Figure 7.9).

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The idea of solving a relaxed version of the problem and then adding constraints sequentially to remove subtours is the strategy followed by the most sophisticated algorithms. The basic approach is simply to try to solve the full model using implicit enumeration (branch and bound).

The combination of implicit enumeration and the sequential addition of constraints is called branch and cut.

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Homework

Solve the following TSP with a branch and cut method.

- 17 10 15 17

18 - 6 10 20

12 5 - 14 19

12 11 15 - 7

16 21 18 6 -

1 2 3 4 5

1 2 3 4 5

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Directed Minimal Spanning Tree Problem

Given a directed graph G = (N, A) with node set N and arc set A, a tree is a subgraph G' = (N, A') that has no cycles and is connected (contains a directed path from the root node to all other nodes in N ). The directed minimal spanning tree (MST) problem is to find a tree rooted at, say, node 1 with a directed path to every other node. The goal is to minimize the sum of the arc lengths used in the tree.

The model for this problem is similar to that of the TSP except now one or more arcs may leave node 1; however, there is no requirement that any arcs leave nodes 2,..., n.

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Mathematical programming formulation is

Tree length: Minimize z =

At least one arc must leave root node 1:

One predecessor for all other nodes :

TSP subtour elimination constraints :

Integrality. xij = 0 or l, i≠j = l , . . ., n

n

i

n

jijij xc

1 1

n

jjx

11 1

1

1, 2n

iji

x j , . . ., n

22,,1 nSNSSxSi Sj

ij

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Again we solve the relaxation of the problem obtained by removing the subtour elimination constraints. The solution to this problem is {(1,4), (3,6), (3,5), (6,2), (6,3)}, with z = 31.

It contains the cycle (6,3)→(3,6),so we add the constraint x63+

x36 1 and resolve.

At the next two iterations, we add

x56+x65 1 and x35+ x53+ x56+ x65+ x63+ x36 2, respectively, t

o eliminate the corresponding cycles.

The solution at this step is {(1,6), (6,2), (6,3). (6,5), (2,4)}, with z = 42, and is optimal .

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Shortest Path Tree Problem

Given a directed graph, we wish to find a tree rooted at, say, node 1 with a path leading to every other node. The goal is to minimize the sum of the individual path lengths.

To develop a mathematical programming formulation for the problem, we must define the variables in a way that allows the arcs to be used multiple times. Accordingly,

let xij = number of paths using the arc from i to j

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The model for the shortest path tree problem is as follows.

Length of the n–1 paths :

Minimize z =

Cl: At node 1, supply = n –1:

C2: Conservation of flow:

C3: Nonnegativity: xij 0 , i≠j = l , . . ., n

n

i

n

jijij xc

1 1

n

jj nx

11 1

n , . . ., i xxn

j

n

jjiij 2,1

1 1

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Constraint Cl requires that exactly n –1 units of flow leave the root node 1, thus establishing n – 1 paths. For every other node, the amount of flow entering that node must be one unit greater than the amount leaving. This is guaranteed by Constraint C2.

Note that we have not required the variables to have integer values in Constraint C3. Since this is a pure network flow problem, the constraint matrix is totally unimodular. The solution will automatically be integer valued.

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The Cutting Stock Problem

A paper company sells rolls of paper of fixed length in five standard widths; 5,8,12,15, and 17 feet. Its manufacturing process produces 25-foot-wide rolls only, so all orders must be cut from stock of this size.

Demands for the 5-, 8-, 12-, 15-, and 17-foot rolls are 40,35,30,25, and 20, respectively.

The problem is to cut the manufactured rolls in a fashion that minimizes the total number required.

Manufactured rolls can be cut in 11 different patterns, as shown in Figure 7.14. Some patterns result in excess paper (shown in black) that must be discarded. .

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Simplify the situation somewhat, no pattern is included hat has an excess as great as the smallest standard width (5 feet). Also, each roll can be cut in only one pattern.

Because each pattern is cut from a 25-foot-wide roll, we can assign a unit cost to each pattern. The goal is to select an integral number of rolls to be cut in each pattern so that the total cost is minimized and the demand satisfied.

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To formulate the model, let xj = number of rolls cut in pattern j

1 2 3 4 5 6 7 8 9 10 11

2 4 6 7 9 10 11

Minimize

subject to

2 2 3 5 40

z x x x x x x x x x x x

x x x x x x x

1 3 6 8 9 10

(Demend for 5-foot rolls)

3 2 35

x x x x x x

5 6 7


2 x x x

3 4

30


x x

1 2

25


x x

20

(De

mend for 17-foot rolls)

0 and integer, 1,....,11jx j

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Three optimal solutions are shown in Table 7.3, with the total number of rolls cut being equal to 64 in each case. Only the second solution satisfies the demand exactly.

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Production Scheduling

You are given the demands for two products A and B over a 5-day period. Both products are manufactured on the same machine, but on any given day the machine can be used for only one product because of the extensive changeover time.

Job is to establish a production schedule for the machine that shows which product, and the amount of that product, that should be manufactured on each day. The schedule must ensure that all demands are met without shortages.

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The machine is set up for production each morning. Let dAt a

nd dBt be the respective product demands, with t ranging from 1

to 5. The corresponding setup costs are $100 for product A and $50 for product B.

For each of the 5 days, you are to determine if product A, product B, or neither is to be manufactured.

When product A is being manufactured, the machine can produce at most 10 units per day, when product B is being manufactured, the machine can produce at most 20 units per day. On any given day, you can produce more than is needed and place the excess in inventory.

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The cost of storing product A is $2 unit per day , the cost of storing product B is $3 per unit per day. There are 10 units of each product in inventory.

At the end of the 5–day planning period, inventories are to be zero. For convenience, assume that all additions to and withdrawals from inventory occur at the beginning of each day.

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Definition of Variables

Real variables:

At = quantity of product A produced on day t

Bt = quantity of product B produced on day t

It = inventory of product A at the end of day t

Jt= inventory of product B at the end of day t

Binary variables:

xt = decision to produce product A on day t

yt = decision to produce product B on day t

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Objective Function

Minimize z =

5

1

)3250100(t

tttt JIyx

Constraints

Conservation of inventory (t=l,...,5):

Cl:

Initial and final inventories:

C2: I0=10 , J0=10 , I5=0 , J5=0

Atttt dIIA 1

Btttt dJJB 1

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Capacity implications (t = 1,..., 5):

C3:

At most one product on any day (t = 1,.... 5):

C4:

Integrality and nonnegativity (t = 1,..., 5):

C5:

tt xA 10

20t tB y

1 tt yx

0,,,;1or 0, tttttt JIBAyx

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Production Scheduling with Continued Setup

We now modify the preceding example so that the setup for a particular product can be carried over from one day to the next. In this situation, if the machine is set up for product A on day t at a cost of $ 100, it is available to manufacture product A on any number of consecutive days with no additional setup cost.

This feature can be easily accommodated in the existing model by noting that the expression

has the value 1 only if xt= 1 and xt–1=0. This will occur wheneve

r production of product A starts on day t.

1(1 )t tx x

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The model can be modified by inserting this factor (and a similar one for product B) into the objective function, yielding

]32)1(50)1(100[ z Minimize 1

5

11 tttt

ttt JIyyxx

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Unfortunately, the objective now has nonlinear terms that are simple in appearance but are not allowed in linear–integer programming models. An acceptable alternative is to introduce two binary variables ut and vt to replace the multiplicative terms in the o

bjective function.

Let

We must now add constraints that determine the values of ut,

and vt in terms of xt and yt

otherwise0

day on Aproduct on begins production if1

tut

otherwise0

day on product B on begins production if1

tvt

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Constraints defining production initiation (t = 1,.... 5):

C6:

The new objective function is

which now has a linear form.

ttt uxx 1)1( 1

ttt vyy 1)1( 1

5

1

]3250100[ Minimizet

tttt JIvuz

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The primary reason why this transformation works is that we are minimizing cost and so there is no incentive for ut or vt to be posi

tive unless there is a desire to begin manufacturing the respective products on day t.

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Days-Off Scheduling

The most common example is the (5,7)-cyclic staffing problem, in which each employee works 5 days per week given two consecutive days off. To formulate the model for this problem,

let xj = number of employees assigned to days-off pattern j

cj = weekly cost of days-off pattern j per employee

ri = number of employees required on day i

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In the objective function, the coefficient ci can account for such f

actors as premium pay for weekend assignments, different pay rates for different labor categories, assuming a more elaborate model that included, say, full-time and part-time employees.

7

1

z Minimizej

jj xc

70

1

7

1

where7 1,..., integer and0

...1,Subject to

xxjx

irxxx

j

iiij

j

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or

Minimize z = cx

subject to

integer and 0x

rx

0011111

1001111

1100111

1110011

1111001

1111100

0111110

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Each column of the A matrix represents a feasible days-off pattern.

Although the A matrix is not totally unimodular, it does have a special structure called the circular property. In particular, a 0-1 vector is said to be circular if its 1's occur consecutively, where the first and last entries are considered to be adjacent. Veinott and Wagner [1962] recognized that a related ILP with consecutive 1's could be transformed into network flow problems. Building on that work, Bartholdi, Orlin, and Ratliff [1980] transformed the (k, m)-cyclic staffing ILP into a bounded series of network flow problems.

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In addition, they provided an alternative solution technique to the original problem based on LP rounding. The steps of this technique are as follows.

1. Ignoring the integrality restrictions in the days-off ILP, solve the LP relaxation to obtain the solution ,…, If these values are all integer, this is the optimal solution to the ILP; if not, go to Step 2.

1x 7x

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Form two linear programs LP1 and LP2 from the relaxation in Step 1 by adding, respectively, the constraints x1+ x2+…..+ x7 = and x1+

x2+…..+ x7 = where is the smallest

integer greater than or equal to y and is the largest integer less than or equal to y. LP1 is always feasible, but it is possible that LP2 is infeasible. In either case, it can be shown that if an optimal solution exists, then an integral optimal solution exists, so the better of the two solves the days-off ILP.

2.

71 ..... xx 71 ..... xx y

y

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Homework

Solve the previous days-off scheduling example with

[c1, c2, c3, c4, c5, c6, c7]=[8, 10, 13, 15, 14, 12, 9] and

[r1, r2, r3, r4, r5, r6, r7]=[30, 60, 65, 70, 60, 55, 40].

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Assembly Line Balancing

A wide range of products are assembled on fixed-paced flow lines consisting of a collection of workstations. One or more operations are performed at each station, with some restrictions on the order, these are called precedence relations. There is also a limit on the amount time a product can spend at any particular workstation. This value is known as the cycle time and is assumed to be fixed.

Consider an example of a product whose manufacture requires five operations. The decisions involve allocating each operation to a particular workstation so that the number of stations is minimized. Table 7.4 gives the precedence relations and the time needed to complete each operation.

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Logically, operation i is either performed or not performed at station j.

Suppose that the maximum time available at each workstation is 12 minutes. Along with the data in Table 7.4, this implies that at most four stations are needed. Hence, we have the following four constraints.

otherwise0

stationat done is operation if 1Let

jixij

5

1

.,4 . . 1,,12i

iji jxp

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The total time taken for all operations assigned to a station must be no more than 12 minutes.

Expanding these inequalities yields

1256756 5141312111 xxxxx

1256756 5242322212 xxxxx

1256756 5343332313 xxxxx

1256756 5444342414 xxxxx

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By saying that operation 3 must be done before operation 4, we mean that the former must be performed either at the same workstation as the latter or at a prior workstation. In mathematical terms, operation i is done at or before station k if and is done after station k if . If then operation 4 cannot be done at station k unless operation 3 has been done, because x4k = 1 only if . The precedence relations

must hold at all stations, so

k

j ijx1

1

k

j ijx1

0 kj jk xx 1 34

k

j jx1 3 1

.,4 . . 1,,1 34

kxxk

j jk

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If neither operation is done at or prior to station k, this expression holds trivially (i.e.,0 0); it also holds if both operations are done at station k (i.e.,1 1).

One set of constraints is required for each precedence relation For job 5, the constraints are

4,...,1

145

125

k

xx

xx

k

jjk

k

jjk

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To ensure that each operation is performed once and only once, we require

The objective is to find the minimum number of workstations needed to assemble the product. This can be achieved by assigning a smaller "cost" to operation i performed at station 1 than for operation i performed at station 2, and so on. By minimizing these costs, we force each operation to be allocated to the earliest possible workstation. One way to specify the objective function coefficients to obtain the desired result is to let cij=j for all i.

4

1

5,....,1,1j

ij ix

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Consequently we have

Preceding equations together with the nonnegativity and integrality conditions on the variables

make up the integer program for this problem. Note that we do not need to impose an upper bound of 1 on each xij, because the last co

nstraint does this implicitly.

5

1

5

143

5

1

5

121 432 Minimize

i iii

i iii xxxxz

jixij and allfor integer and0

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5

1

5

143

5

1

5

121 432 Minimize

i iii

i iii xxxxz

jixij and allfor integer and0

s.t.

5

1

.,4 . . 1,,12i

iji jxp

.,4 . . 1,,1 34

kxxk

j jk

4,...,1

145

125

k

xx

xx

k

jjk

k

jjk

4

1

5,....,1,1j

ij ix

Chapter 7 Integer Programming Models

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Transcript of Chapter 7 Integer Programming Models