Tracing the characteristic curve of a quadratic black box

Tracing the Characteristic Curve of a Quadratic Black Box Jie Sun Department of Industrial Engineering and Management Sciences, The Technological Institute, North western University, Evanston, Illinois 60208

A quadratic black box is a two-port network with each arc incurring a quadratic cost. This paper studies the properties of the minimum cost function of a quadratic black box with respect to an external flow input. An algorithm is presented with a numerical example for determining the subdifferential mapping of the minimum cost function. The algorithm can be applied to the piecewise quadratic case without change. If all data satisfy the commensurability condition, the algorithm terminates in a finite number of steps.

1. INTRODUCTION

Given a network consisting of arcs with separable convex quadratic costs and any two distinct nodes of the network, say node 1 and node 2, consider the following parametric network quadratic program:

(minimize F ( x ) = 4 ( x j ) j = 1

subject to Ax = b + (6 , - t,O, . . . ,0)7 (QP) I

I c- s x s c+

wheref,(xi) = &pjxy + q j x j ( p i 2 0), A E R""" is the incidence matrix of the network, b € R", x , c- and c+ € R". 5 is a parameter, representing an inflow and an outflow at nodes 1 and 2, respectively. The vector inequality is understood coordi- natewise. c; could be --cO and c: could be +-cO. The optimal value +(6) of (QP) is then a function of 5. In this paper we discuss the properties of +(5) and develop an algorithm which can determine the entire subdifferential mapping of +(t), namely all points ( 5 , ~ ) in R2 satisfying q E [+-(t),++(t)], where +- and ++ are the left and right derivatives of 4. Since the function +(€J can be obtained through its subdifferential by integration [7], the algorithm actually offers an explicit expression of +(E).

NETWORKS, Vol. 19 (1989) 637-650 0 1989 John Wiley & Sons, Inc. CCC 0028-3045l89/060637-0 1 4$04.00

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This type of parametric programming problem arises from various practical situa- tions. For instance, suppose that the network is an electrical circuit. Let x j represent the electrical current on arc j and let b = 0. If we set c[ = - m, c,’ = + m, qj = 0 and pi = R j , then the a rc j stands for a resistor with resistance R j . If c; = 0 instead, then we have a resistor R j with a diode which rejects negative current. If c j = c; = c j , then this arc can be thought as a generator which offers constant current c j . On the other hand, c; = -m, c,+ = + m, pi = 0 and 4 j = V j represent a battery with constant voltage V j . If we think of this network as a two-port “black box” with endpoints 1 and 2, then the subdifferential to be determined is the voltage-current characteristic of the black box [7, sections 81 and 8Nl. In addition to electrical networks, a number of models in operations research can be formulated as finding the function 4(5) or the like; e.g., the sensitivity analysis models which have applications in the pipeline industry, capital budgeting, farm decision making, etc. For brief references to those applications, see [ 13 and 121. Recently, due to interest in parallel computing, decomposition techniques for network programming have been a focal point of research (e.g., see related papers in [ 5 ] ) . One of the problems in this area is to determine the property of a “lumped” network, which in the simplest case means to determine the function +(5) for a black box.

Mathematically, a quadratic black box consists of a network G with m nodes, n arcs, and two of the nodes, nodes 1 and 2, being specified as the input and output ports. Each arc j of G is associated with a convex quadratic cost function. The cost function of the black box is defined as the function $(5) in problem (QP). It will be shown in Section 2 that 4(() is convex and is piecewise quadratic; i.e., there exist

that - m C g < C1 < ’ ’ ‘ < Ck 6 P I , . . . , p k , 4 1 , . . . , q k and r l , . . . , r k such

+ m, +pP15’ + q , t + r l ,

445) = * . . 1 + m,

if 5 < co if co 6 5 6 c1

$p&’ f q k [ + r k , if C k - 1 5 6 c k

if 5 > c k .

In some applications we want to regard G as a single arc with input node 1 and output node 2 and to consider a black box whose arcs are also black boxes. Therefore, to generalize, in this paper we discuss the case of (QP) in which eachf,(xj) is convex and piecewise quadratic like $([). Of course, all arguments and the algorithm will apply to the convex quadratic case as well.

Several algorithms have been published for solving the problem (QP) or more general single parameter problems; e.g., Kurata [3], Markowitz [4] and Van de Panne [6 ] . The algorithm in this paper differs from them in the following aspects. First, the algorithm can handle the piecewise quadratic case; second, it determines the subdifferential of 4(5) instead of +([) itself. This is advantageous when the algorithm is used as a subroutine in an optimization process where the information on derivatives is of more concern. Third, the algorithm takes advantage of the network structure when applied to black boxes and can be easily extended to solve general single- parameter piecewise quadratic programs (see remark in Section 3, for a general dis- cussion of piecewise quadratic programs, see [9]).

QUADRATIC BLACK BOX 639

Let us call the graph of the subdifferential of any one-dimensional convex function its characteristic curve, or c-curve for short. Due to the equivalence of mini- mization problems and equilibrium problems in optimization, the determination of the subdifferential of +([) can also be stated as an equilibrium problem as follows.

Given the c-curve rj of function f, for each j , find all points ((,q) such that there exist vectors x , u, and v, satisfying Ax = b + (6, -(,O, . . . , O ) T , ~ = -ATu, q = u2 - uI and (xj,vj) E r, Vj . ( u , and u2 are the first two components of u . )

An elegant proof of this equivalence can be found in Rockafellar’s book [7, section 8N]. The problem of determining the entire c-curve of I$( [ ) may be called “tracing” the c-curve of the black box. Rockafellar [7] has developed an algorithm which can trace the c-curve of +([) for linear and piecewise linear black boxes. This paper deals with quadratic and piecewise quadratic cases. Section 2 discusses some properties of function +((). The algorithm for tracing the c-curve of the piecewise quadratic black box is presented and justified in Section 3. A numerical example is given in Section 4.

2. SOME PROPERTIES OF PIECEWISE QUADRATIC BLACK BOXES

Generally, an n-dimensional function @ ( x ) is said to be piecewise quadratic if its domain {xl@(x) < +m} is a nonempty union of convex polyhedra on each of which @(n) is quadratic. It is easy to see that any one-dimensional convex piecewise quadratic function will necessarily assume the aforementioned form of I$((). The importance of the piecewise quadratic function in connection with the black box is indicated by the following proposition.

Proposition 2.1. Let

1 fj(x,)lAx = b + y

where A E RmX”, y and b E R”, x E R”, each f , is convex piecewise quadratic. Suppose that @(y) is finite for some yo. Then @(y) is a convex and piecewise quadratic function of y.

Proof. See [9].

Corollary 2.2. If for some piecewise quadratic function of 5.

the function +(() is finite, then +([) is a convex and

Proof. Under the linear transformation y = (1, - 1 ,0, . . . ,O)’< the convexity and piecewise-quadraticness are invariant (see [9] and Part I of [S]). 8

Since +([) is convex and piecewise quadratic, its c-curve is a brokenline that goes from “southwest” to “northeast” on R2 [7, section 8C]. The following proposition and corollary show how ++(() and +-(c) can be determined.

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Proposition 2.3. Let @’(y,d) be the directional derivative of @ at y along d, where @(y) is defined as in Proposition 2.1. Suppose the function @(y) is finite at y. Let U ( y ) be the dual optimal solution set (i.e., the Lagrange multipliers) of the problem

min { f,(x,)(Ax = b + y J - 1

Then

cP’(y,d) = sup { - d * u} . uEu(Yi

Proof. See [9].

Corollary 2.4. If for some Eo the function +(t) is finite, then

+ ‘ ( E ) = sup {u2 - ul} and +-([) = - sup {u, - uz}. U E W E ; ) u E W 6 )

where U(k) is the set of dual optimal solutions to problem (QP).

Proof. Let d = (1, - 1,0, . . . ,O)T. Observe that for +(€J = @(Ed) one gets +‘(t) = @‘([d,d) and +-([) = -@’(Ed, -4. Thus by Proposition 2 . 3 one has

+’([) = @’([d,d) = sup { - d * u} U€U(E;)

and

+-([) = -@(Ed, - d ) = - sup { - ( - d ) * u} = - sup {d - u}. rn uEU(E;) IdEU(S)

According to the theory of monotropic programming, if x is an optimal solution to the problem (QP), then

U(6) = {u E R”1v = -ATu such that f ; (x j ) S v, s f,”(x,) V j }

(See the equilibrium theorem of 11D in [7]). Hence knowing x, the intervals u;(xj), f: ( x j ) ] can be computed and c$’(c) can be obtained by solving the linear program in Corollary 2.4.

From the viewpoint of network programming, a dual feasible solution u in problem (QP) can be interpreted as a “price” or “potential” vector associated with the node set, and the vector v (= -ATu) is the “price difference” or “tension” associated with the arc set. Corollary 2.4 says that + ‘ ( E ) is the maximal difference between the potentials at node 2 and node 1 over the set U ( Q . This accounts for solving the so-called max tension problem for which there exist efficient network algorithms. For instance, because of the max tension min path theorem (see section 6C of [ 7 ] ) this task can be reduced to finding a min path between nodes 1 and 2. Similarly +-(Q is the negative max tension from node 2 to node 1 subject to the same feasible set U ( 6 ) . In summary, given a point (E,q) on r, we can trace r upward to ( ( , + ‘ ( E ) ) and downward to (<,c$-(()) by solving a pair of network max tension problems.


We will now consider how to tracer northeastwise from (c,++(c)). For ( x J , v J ) E rJ let us define

p; = lim f l ( y j ) '7 1. "j

p; = lim f l ( y j ) Y j t xj

whenever these limits exist. Define for each j the function5 as follows:

if vj = . f - (xj) < f ,+(x j ) ; (2) +m for y j > 0

if h-(xJ) < vj < f,+(x,). I for yJ = 0 otherwise (4)

Then the following piecewise quadratic program is well-defined k

(SQ): minimize c f , ( y , ) subject to Ay = [ I , - I,O, . . . ,0lT

because all p ' s that appear in the definition of 5 are necessarily finite numbers. Moreover, ifJ is quadratic, thenf, must be quadratic, too.

Proposition 2.5. Let the c-curve of f , be rJ Vj . Then there exists E > 0 such that ( x J , v J ) E rJ and ( y J , w J ) E rJ imply (xJ + 6yJ,vJ + 6 w,) E rJ Vj VO s 6 s E.

From the definition off,, it is easy to see that, as two sets in R2, the curve ( x J , v J ) + rJ coincides with the curve rJ at least in a neighborhood of ( x J , v J ) , so there is an E~ > 0 such that (xJ + 6yJ,vJ + 6 w,) E rJ VO S 6 S E ~ . The minJ{eJ} is

Corollary 2.6. Let x and u be an optimal solution and a Lagrange multiplier vector to (QP), respectively. Let v be -ATu. Construct problem (SQ) according to ( x , v ) . Then there is an E > 0 such that for all 6 E [O,E], (6 + 8,u2 - u1 + 6 ( z 2 - z , ) ) E r, where r is the c-curve of + and z is the Lagrange multiplier vector to (SQ).

By equilibrium theorem of monotropic programming [7, section 1 1 D] , w = -ATz satisfies ( y J , w J ) E rJ , where y is any optimal solution to (SQ) that is constructed according to ( x , v ) . Proposition 2.5 says that there is an E > 0 such that for any 0 =Z 6 S E we have (x , + 6yJ,vJ + 6 w J ) E T J . We obviously have A(n + Sy) = b + (5 + 6)(1,-1,0, . . . ,O)= and v + 6w = -AT(u + 6 z ) . By the equilibrium statement of the black box problem mentioned in Section 1, (6 + 6,

I - 1

Proof.

then the needed E in the proposition. rn

Proof.

u2 - u1 + 6 ( z 2 - z , ) ) then has to be in r. rn

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Remark. If specifically u2 - u1 provides the max tension from node 1 to node 2, then ([ + S,++([) + S (z2 - z,)) E r VO G 13 < E in view of Corollary 2.4. Thus r is traced in [e,( + E]. In addition, in this case (SQ) must have an optimal solution, which is shown in the next proposition.

Proposition 2.7. If the problem (SQ) is derived from such ( x , u ) that u2 - u1 is equal to + + ( c ) , then (SQ) has an optimal solution.

Proof.

Painted Index Theorem [7, section lOG].

Let us first state an important theorem about linear systems.

Arbitrarily assign one of the four colors-white, black, green and red-to each index in the index set J = { 1, . . . ,n}. A vector x E R” is called primal compatible with the colors if xJ 0 V j white, xJ S 0 Vj black and xJ = 0 Vj red. A vector v E R“ is called dual compatible with the colors if vJ < 0 Vj white, vJ 3 0 Vj black and v, = 0 Vj green. Let M be a k by n matrix. Then for any particular white or black index j, E J, one and only one of the following is true:

1) There is a primal compatible vector x in the null space of M with xJo # 0. 2) There is a dual compatible vector v in the range space of MT with vJo # 0. We now proceed to prove Proposition 2.7. Note that all r, pass through the origin,

so the zero vector is a dual feasible solution to ( S Q ) . In view of the duality of monotropic programming, we only need to verify that a primal feasible solution exists. Consider the linear system Ay - dy,+ I = 0, where the indices 1, . . . ,n are assigned white, black, green, and red, respectively in accordance with cases (1)-(4) in the definition of function%. The index n + 1 is assigned white. Let d = [ l , - 1,0, . . . ,0IT. There can’t exist a dual compatible vector [@,@“+[IT in the range space of [A, -dIT with W n + , < 0, for otherwise there would be a vector z E R” such that - ATz = @ and d * z = < 0, and because of the color compatibility of @ we would have vJ + &GJ E rJ 0’ = 1, . . . ,n) for a small E > 0. Hence u + EZ E V ( [ ) and con- sequently -d * ( u + EZ) s ++ ([) according to Corollary 2.4. This is a contradiction since -d * ( u + EZ) = - d - u - d . EZ = ++([) - > +,‘(Q. Because of nonexistence of the dual compatible vector, the Painted Index Theorem says that there is a primal compatible vector [ j&+ with yn+ > 0 satisfying Ay - djjn+ = 0. Multiplying by a positive constant- if necessary, we may assume j n + I = 1. Then A9 = d , and it is obvious that nofJ(jJ) could be infinity because of the color com-

The following proposition will play a role in finite termination of Algorithm 3.1.

Proposition 2.8. Suppose that all data (p ’ s , q’s, r’s, c’s, b, 6 and the matrix A ) in problem (QP) are rational. Then (QP) has a rational optimal solution whenever it has an optimal solution. The same is true for problem (SQ).

patibility of j . Thus is a primal feasible solution to ( S Q ) . H

Proof. This can be derived from the Kuhn-Tucker conditions.

3. THE ALGORITHM

Suppose that the function +([) is finite for some 5. Then for any primal and dual optimal solutions ( x , u ) to (QP) with v = -ATu, we have (xj,vj) E Ti Vj and


(5,u2 - u,) E r. Starting from this point on r we can trace r in the interval [t, + 03)

(northeastwise) as follows. Algorithm 3.1.

We set the “span” of arc j as v;- ( x i ) , f:: ( x i ) ] ; find the max tension from node 1 to node 2 subject to vj E K-’ (x i ) ,&+ ( x j ) ] V j . Let the max tension be 7. Then ++(t) = 7 and the vertical line segment between (5,q) and (5,q) belong to r. If q = 03, stop; the northeastern trace terminates. Otherwise let ( 5 , ~ ) = (5,q) and go to step 2.

Step 2. Let (xi,vj) be the current point on r j . Construct problem (SQ) as in Section 2. Let a pair of its optimal solutions be (y,w), where w = -ATz and z is the Lagrange multiplier in problem (SQ). Calculate

E = max{e((xj + E Y ~ , V ~ + EW/) E rj, fo r j = 1, . . . ,n}

If E = 03, stop, the half line (5 + E,q + E ( Z ~ - zl)le 2 0) is the rightmost piece of r. Otherwise 0 < ii < a, the line segment between (5,q) and (5 + E,q + E(z2 - zl)) C r. Replace (5,q) by (5 + E,q + E(z2 - 2 , ) ) and go to step 1.

If we want to trace the c-curve of @(t) = infx{Ey= dj (x j ) (Ax = b + sd} where A is a general matrix and d is a general vector, all we have to do is to change the right hand side of the constraints in (SQ) into d, and to solve the linear program in Corollary 2.4 directly.

The initial inclusion ( x j , v j ) E Tj is preserved both in step 1 and step 2, because of the setting of spans in step 1 and Proposition 2.5. The trace of r in step 1 can be seen from Corollary 2.4. The trace in step 2 is due to Corollary 2.6; the solvability of (SQ) in step 2 comes from Proposition 2.7. Moreover, by Corollary 2.2, + ( E ) is convex piecewise quadratic. Thus once an infinite piece of its c-curve is found, it must be the rightmost one. This clarifies the stop criteria in steps 1 and 2.

Since 71 could be equal to q in step 1, step 1 could end up with no moving at all. If there is an infinite consecutive sequence of step 1 and step 2 with such step 1, then the algorithm can not finitely terminate. To avoid this, the following assumption is enough. Commensurability Assumption. All the data including to are rational. The optimal solutions obtained in solving (QP) and (SQ) are rational and if we solve the same (SQ) twice, the solutions are same, too.

Step I .

-

Remark.

JustiJication of Algorithm 3.1.

Note that the consistency of the assumption is due to Proposition 2.8.

The Finite Termination of the Algorithm. assumption holds. In step 2 we have the initial rational x and v. Then

Suppose that the commensurability

- E = max{e/(x, + E Y ~ , V ~ + E W ~ ) E Ti, f o r j = 1, . . . ,n}

is rational because for eachj,xj,yj,vj,wi and all the bounds of variables are all rational. Moreover, the possible E’s in different iterations form a finite set, because we only have a finite number of different (SQ)’s; therefore, all possible xj,yj,vi,wj and Z in Algorithm 3.1 will have a common denominator. On the other hand, we already showed that E > 0 in step 2, so after finitely many iterations, the 5 must reach the rightmost piece of r and the algorithm will halt due to E = 03 or fl = a.

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node 3 node 6

node 3 node 4

node 5 node 6

E

FIG. I .

The analogous version of Algorithm 3.1 for southwestern trace can be obtained by a similar way. The differences are that in step 1 we find the negative max tension from node 2 to node 1 and that in step 2 we solve

k

(SQ): minimize 2 f-(yj) subject to Ay = [ - 1,1,0, . . . ,0lT j = 1

and replace 6 + F by 6 - E.

4. A COMPUTATIONAL EXAMPLE

Figure 1 depicts a black box with nodes 1 and 2 being the two ports. For simplicity we set b = 0 and assume that allh’s are quadratic and all variables have lower bound 0 and upper bound +m. The coefficients p , and q, of the cost functionf,(x,) = ip, xf + qJxJ are listed below:

j 1 2 3 4 5 6 7 8 9 1 0 ~ ~ 1 1 5 3 5 3 5 4 4 6 qJ 3 5 3 4 4 6 6 6 2 2

Obviously (0,O) is a point of every rJ and hence is also on r. Starting with this point, we first trace r southwestwise.

Zrerution 1. The following spans are introduced for finding the max tension from node 2 to node 1.

arc 1 2 3 4 5

span (-m,3] (-m,51 (-m,31 (-X,41 (-W,41

arc 6 7 8 9 10

span (-m,6] (--03,61 (-m,61 (-X,21 (-a,21

The max tension from node 2 to node 1 is found to be +a. Thus the c-curve r of the black box goes downward from (0,O) to (0, - m). The southwestern trace terminates. We now begin the northeastern trace.


Iteration 2 . The same spans are used but this time the max tension from node 1 to node 2 is sought. The optimal value is 9 and r therefore goes from (0,O) to (0,9). We denote this fact by r + (0,9). In the same time, we have

rl .+ (0~3) r2 -+ (0,3) r3 .+ (0~0) r4 + (0,4) rs .+ (0,4) r6 + (0~4) r7 -+ (0~4) rx .+ (o,o) r9 .+ (02) rlo .+ (0,2).

We then set up the following quadratic program and solve it.

minimize 0 . 5 ~ : + 1 . 5 ~ 2 + 2 . 5 ~ : + 2y; + 3Yfo

subject to Y l + Y2 = 1

- Y9 - YlO = - 1

- yl - y3 + y4 + y6 = 0

- Y4 - Y5 + yx + y9 = 0

- Y2 + Y3 + y5 + y7 = 0

- y6 - y7 - y8 + yl0 = 0

Y I 3 09Y2 = OJzi = OJ’4 5 o,y, 2 0 ,

y6 =z o,y7 = o,y8 = Oyy9 o,y,O 0 The dual optimal solution w is obtained by w = -ATz, where z is the Lagrange multiplier vector of this quadratic program. The pair of primal and dual solutions is

a r c 1 2 3 4 5 6 7 8 9 1 0

y c 1 0 0 1 0 0 0 0 1 0

w, 1 4 - 3 3 0 7 4 4 4 0

The maximal E that keeps (x , + e y , , ~ , + ew,) E r, for all j is 2/7. So r +

(0 + 2/7,9 + 2/7(z2 - z I ) ) = (2/7,79/7) and I?, -+ (x, + EY,,V, + EW,) gives

Iteration 3 . The spans are

arc 1 2 3 4 5

arc 6 7 8 9 10

span [-w,6] [-m,61 [-m,61 [T, 71 [-00,21

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The rnax tension from node 1 to node 2 is 79/7. Therefore r + (2/7,79/7) (There is no “jump” at (2/7,79/7).) However, there are changes in vjs during the execution of the max tension algorithm:

(Since the max tension remains the same, these changes are not necessary.) The following quadratic program is solved:

minimize 0 . 5 ~ : + 0 . 5 ~ : + 1 . 5 ~ : + 1 . 5 ~ 2 + 2y; + 3yf0

subject to Y1 + Y2 = 1

- Y9 - Y l O = - 1

- YI - Y3 + Y4 + Y6 = 0

- Y4 - Y5 + ys + y9 = 0

- Y2 + y3 + Y5 + y7 = 0

- y6 - y7 - y8 + yl0 = 0

yz 3 o,y, = o,y5 = OJ’6 3 o,y7 = o,y8 = 0,Yio 2 0 The solution is

The maximal available E = 96/301. Hence

Zreration 4 . The spans are

arc I 2 3 4 5

155 155 232 232 [-m,5] [-m,3] - - [ 43 9 4 3 1


arc 6 7 8 9 10

276 276 166 166 122 122 [-m,6] [-m,6] - - [ 437 4 3 1 [Z%]

The max tension from node 1 to node 2 remains the same. This time we deliberately do not make any changes in v j ’s and hence need solve

minimize 0 . 5 ~ : + 0 . 5 ~ : + 1.5~2 + 2 . 5 ~ : + 1 . 5 ~ : + 2 . 5 ~ : 2y; + 3y:,

subject to Y1 + y2 = I

- Y9 - Y I O = -1

- y1 - y3 + y4 + y6 = 0

- y4 - y 5 + YS + y9 = 0

- Y 2 + y3 + y5 + y7 = 0

- y6 - y7 - yS f y10 = 0

Y 2 a OJzi = 0,)’s a o,y7 = 0 , Y s = 0

The solution is

arc 1 2 3 4 5 6 7 8 9 10

108 43 0 50 43 58 0 0 93 58 151 151 151 151 151 151 151

yj - - - - - - -

108 43 65 150 215 174 239 24 372 348 151 151 151 151 151 151 151 151 151 151

wj - - - - - - - - __ __

The maximal available E = 6342110277. Hence

r9 +(202 1286) rl0 ( 9 0 - - 1018) 239’ 239 239’ 239

and

r-+ F””) 239’ 239 ‘

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U C

span

U C

span

1

239’ 239

6

1704 1704 239 ’ 239 --

2

239’

7

3

8

143

4

1, 6 1 r116

5

1

9 10

1286 12861 [ 1018 1018 239 ’ 239 239 ’ 239

The max tension from node 1 to node 2 remains the same. Using the old points in Ti’s we have to solve

minimize 0 . 5 ~ : + 0 . 5 ~ : + 1.5~: + 2 .5~ : + 1 .5~2 + 2.5~: + 2y; + 3yf0

subject to Yl + Y 2 = 1

- Y9 - Y l O = -1

- Y1 - y3 f y4 + Y6 = 0

- Y4 - ys + ys + y9 = 0

- Y 2 + y3 + ys + y7 = 0

- Y6 - y7 - y8 + yl0 = 0

Y3 = o,y, 3 0,ys = 0

The solution is

arc 1 2 3 4 5 6 7 8 9 10

445 311 325 239 756 565 0 - - yi 1320 1320 1320 1320 1320 1320 1320 1320

0 - - - - 770 550

770 550 220 1335 1555 975 1195 360 3024 3384 1320 1320 1320 1320 1320 1320 1320 1320 1320 1320 - - - - - - - -- - __

The maximal available E = 5922/239. Hence

3410 4070 2310 3410

1323 7495 1425 5595 987 6225 220’ 220

r 8 + (O9-?E) r9+ (=.=) 3308 13672 ria+ (%.=) 2412 14912


250.00 - r

200.00 - -

150.00 --

100.00

50.00

0.00 i t

-50.00 10.00 100.00

FIG. 2.

250.00 r

200.00

150.00

100.00

50.00

0.00 i t

-50.00 9 10.00

FIG. 2.

and


arc 1 2 3 4

100.00

5

6835 6835 7495 7495

arc 6 7 8 9 10

13672 136721 [ 14912, 149121 220 ’ 220 220 220

[-cC,6] - - - - 5595 5595 6255 6255 span - - __ - [ 220 ’ 2201 [ 220 ’ 220 I [ Again the max tension from node 1 to node 2 remains the same. Using the old points in rj’s we solve

minimize 0 . 5 ~ : + 0 . 5 ~ : + 2.5~: + 1.5~: + 2.5~:

+ 1.5~: + 2.5~: + 2y< + 3 ~ : ~

subject to Y1 + Y2 = 1

- Y9 - Y l O = - 1

- y l - y 3 + y 4 + y 6 = 0

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a r c 1 2 3 4 5 6 7 8 9 10

4788 3572 0 --

yJ 8360836083608360836083608360 8360 8360 4730 3630 220 2855 1933 2095 1477

4730 3630 1100 8565 9665 6285 7385 2280 19152 21432 , . - - - - - - - . - - -- 8360 8360 8360 8360 8360 8360 8360 8360 8360 8360

The maximal available E = + m. Hence we finish the northeastern trace with the slope of the rightmost piece being z2 - z , = 32447/8360. The c-curve of the black box is graphed in Figure 2.

ACKNOWLEDGMENT

The author wishes to express his appreciation to Professor R. T. Rockafellar for valuable discussions in this research.

This work was partially supported by the National Science Foundation under grant ECS-872 1709.

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[2] T. Gal, Postoptimal Analysis, Parametric Programming and Related Topics, McGraw-Hill,

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[4] H. M. Markowitz, The optimization of a quadratic function subject to linear constraints.

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Received January 1988 Accepted November 1988

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Tracing the characteristic curve of a quadratic black box