A numerical method for solving minimax problems

A NUMERICAL METHOD FOR SOLVING MINIMAX PROBLEMS N. I. GRACHEV and Yu. G. EVTUSI-IENKO Moscow (Received 21 April 1970)

A TRIBAL method for solving ~nirn~ and maximin problems is described. The results of numerical computations are quoted. 1. Formulation of the problem A very wide class of operational research problems reduces to finding a max-min and min-max. Such problems may be stated thus: Let K(x, y) be a twice continuously d~ferentiable function for all elements x and y belonging to compact sets X and Y respectively (2 E X c E,, y E Y c: E,, where E 1 is i-dimensional Euclidean space). l-et Q = x x Y; I‘,, I’,, I‘* are the boundaries of the sets X, Y and G? respectively. To find the min-max of K:(x, v): (1.1)

1, = min. max K(z, y). IfeY

XEX

This problem satisfying

solved

we find, for all

y cz Y

the function

XEX

and then find the set iV C Y such that, given any y* E A? JV(~2,

z4 =

minWW,

UEY

Y).

The set N and function .x”(y) will be termed the synthesis of the min-max problem. The necessary and sufficient condition for .Z(IJ) and N to be a synthesis of a *2X v~/M. Mat. mat. Fir

11, 2, 375-384,

1971.

97

N. I. Grachev and Yu. G. Ey~ush~nko

98

min-max

problem

y. E N, y E

Y, J:E X

llo)GK(l(!l), Y). K(x/ Y)GKK(fb), Y), K(.~(!/o),

(1.3)

Similarly,

we can pose the problem of finding the max-min

I2 =

t 1.4)

is that, given any

max rni~~~~,~)~ SEX ltEY

Using the condition

min K(z, y) = K(z, g(z)). On the set X, the function that, given any x* E M H(z,,

g(z.))

The synthesis

g(x),

=

and the set M C X is then found, such

is determined,

max K(z, g(s)). =ZX

of (1.4) is the combination

Q (x) .

M and

Conditions

similar to

(1.3) hold for (1.4): whatever the Ir’(z, y) > The functions the solutions,

Z(y)

and

therefore,

The function

K(?

*K(zo, y”(G))

y”(s) may be many-valued. account

3

KC? y”(x) 1.

When speaking of the properties

will be taken of all the branches of

K(x, y) has a saddle-point

I, = K(f(y*), In general,

y”(r))7

if 5, E Mand

y*) = 12 = K(s*,

of

,2?(y), g(x).

yl EZ A?exist such that

y”(G)).

12 < I,.

In certain practical problems, a complete solution of problem (1 .l) or (1.4) is unneces~ry. For instance, in the case of the min-max problem it is sometimes sufficient to find just one point

y, E A’ and just one value

5, E

{Z(I/,)}

such that

K(x., y.) = max K(x, y*) = 1,. XEY The pair x *, Y *, will be termed a solution It is difficult

to apply directly

of the min-max

the rule given above

problem.

for finding the synthesis

of

problem (1.1) or (1.41, since this involves seeking global extrema. Since numerical methods for finding such extrema have only been developed to a limited extent, it becomes a question of constructing a mesh covering R during the computational process. In the simplest case, when X and Y are intervals on a straight line, and, say, problem (1.1) is to be solved, each interval is divided into n parts; then the function K(x, y) has to be evaluated at rz2 base-points in order to find Z(y) ,after which the function (P(Y) = R(f(IJ)* Y) 1s . evaluated at n points. Numerical solution of the problem by

Numerical method for solving minimax problems

99

this method is obviously hardly worth while; and the same applies to problem (1.4). It has sometimes been suggested fl, 21 that the min-max problem might be solved numerically by utilizing the theorem on the directional differentiability of the function cp(Y)=t max K (x, II)

.

%-SE

The main dif~culty here is to find the values of V(Y) for a sufficientIy large set of values of the argument y. When describing our own method, we first’take the case, in Sections 2 and 3, when

A method for obtaining the synthesis of (1.1) under special constraints on the function K(x, y) is described in Section 2. In Section 3 a numerical method for solving min-max problems in the case m = n = 1 is described, while the case n > 1, m = 1 is treated in Section 4. 2.

Synthesis of a min-max problem

Consider the problem (2.9

1%= min

MtlXK(Xi,

y),

i=

1,2,.,.,

VEX

where {.7:< = .z (Y)) is a system of functions defined in Y, taking values in X, and satisfying the equation (U)

(X(Y) - 0) (X(Y) - b)K(X(Y)t

Y) = 0

and the condition (Z.8)

(x(y)

-

n)‘(z(y)

-

b)zKx(~(Y),

y) <

0.

As in Section 1, a function P,V(Y) in general many-valued, is found for the problem (2.1) in Y, and a set N, c: Y from the conditions (2.4)

R(Z*(y).y)=

(2.5)

K(I* (y*). Y*) = miuK(~*(Y),Y)(~Y,~~N,). l/EY

max R(G (Y) ) Y) ,

By virtue of the assumptions that arc made regarding X(x, y), for every YiiE 1’ the function 9 (2) = fi’(.r.. Y,,) is continuous, differei~tiable with respect to x, and reaches a global maximum in some set .i: c X.All the elements of the sets EandP, (Yo) are identical. For, all the elements of .i’ E 5’ are either the same as points of r, or the ame as interior points .I*E S ‘\ I’.,, a’t which

N. I. Grachev and Yu. G. Evtushenko

100

i.e., all the elements of X are roots of (2.2), and (2.3) holds when z E E, Hence 5~ {xi(&) f and f, (yo) \Z = 0 by virtue of (2.4). It can easily be shown that the synthesis 5 as the synthesis f. (y), N, af problem (2.1).

(9), N

of problem (1 .l) is the same

For, let Z(y), Nbe the synthesis of (1.1). Using (l-2), the function Z(y), is found, globally maximizing K(x, y,,) with a fixed y. CGY in X. But then, the set f (yo) is the same as i.e. f, (yp), whatever the 21 E (f(y0) }, X2= {K (Ye) 1 K(r,,y,)=

K&y,)=

rnax~~~,~~). x=:x

Since y. is arbitrary, the functions f(y) that N must be identical with N,.

and Z, (y) are identical for any y E Y, so

The reduction of problem (1.1) to the simpler problem (2.1) enables the structure of the synthesis of min-max problems to be investigated, and provides a constructive method for fiiding the synthesis. Properties of min-max problems are considered in [3]. Only those which have a direct bearing on the numerical method will be mentioned here. If G! contains no points (x0, yo) such that .&(X0, g(o) = lila(&, I/o) = JL%(~o, &Jo)= 0, the equation (2.6)

Kr(X, I/)=

0

will be said to have no singularities in !LZ. Assume that, at the point (x0, yo), &(X0, yo)=

0,

Kr(~o, l/o)+0

and that 52 contains no singular points. Using the implicit function theorem, Eq. (‘2.6) is found to have a unique solution x&y) in the neighbourhood of (x,,, yO), such that (2.7)

K$&(y,),

9,) = 0,

dxi

_

89

In some neighboxhood

= _ KW(zi(y); y) h;,(s,(yf:

of (x0. yO), where

fiir(20, $40)= 0,

~W(XO,Yo) # 0,

g) *

101

Numerical method for solving minimax problems

there exists a unique solution yi (x) such that

(2.8)

Kx(20,Yi(20))

=

Kcc(4YW Kwtz7Yitz))

dyi

O7

z

=

-’

’

Equations (2.7) and (2.8) are best combined by writing them in a parametric form. Taking t as the independent variable, we get

$L.= (2.9)

dyi

-

at

Yi),

Kw(&

Xi

= - Ku (xi, yi)

(0)

yi

9

=

GO,

(0) = YiO*

On solving the Cauchy problem for (2.9) with t 2 0, t < 0, the solutions xi (f), yi (t) are obtained, Equations (2.9) are integrated to the point where xi(t) E X and yi(t) E Y. In the phase plane (x, y), the system (2.9) may have singularities of the centre and saddle types. Hence, in a finite interval ItI < 00 the integral curves either cut the boundary of S2, or are periodic functions of t. At points (xi, ur), where K,,(si, yi) = 0, the graph of xi(v) will have a horizontal tangent (otherwise we should have K,(z,, yi) = 0, which contradicts the fact that there are no. singularities in Q). At points (x2, y2), where Krr(z2, yz) = 0, the graph of xib) has a vertical tangent; given a small enough neighbourhood of these points, xrG) is two-valued and Kxx changes sign. Discarding the xi (v) for which K, > 0, a function xi (Jo) single-valued in Y is obtained. If new conditions, not lying on the integral curves of (2.9) already obtained, are taken as the initial Cauchy conditions for (2.9), a new branch X~(JJ)will be obtained. If there are no singularities in a, no two of the solutions xi(y) and xi(~) of th. ,ystem (2.9) will intersect. On passing to the solution of of Eq. (2.2), z(y) = a, and z(y) z b must be added to the set of single-valued Z(y) can be found, and integral branches of (2.9). On then using (2.4), the function from condition (23, the set N determined. The graph of Z(y) consists either of points of the set Y X F_ or of pieces or points of the branches Xi(j). The function K (f ( y) , y) is continuously dependent on y. The numerical method for finding the synthesis of (1 .l) is as follows. 1. We “inspect” Q, (2.10)

and find the points (xi, _vi)where

Kx (xi, yt) = 0,

and then evaluate fo(y)

i=

I,2 ,...,

k,

and X0(y) from the conditions

102

N. I. Grachev and Yu. G. Bvtushenko

fo(!/)== K(G(Y), ?I)==max[K(it,

on I’Q, the set of values y* E Y, for which

2. If (2.10) has no solutions mi~~*(~) lJEY

v), K(D, !:)I.

= ~~(~~),

and the function Z,,(y) represent the synthesis of problem (1 .l). If (2.10) has solutions, we integrate the system (2.9) with initial data satisfying (2.10). The functions pi,

and Z(y)

are obtained

f,(Y)

simultaneously,

Y) = max B-t(Y),

= K(5i(Y),

As a result, it is found that f,(y) The synthesis

problem

of the Cauchy problem

uniform

mesh is constructed

(z,

~~),.representing

0 <

Here,

8 <

= Z(y).

with the conditions

K(zi(.!I),

Y)].

The set N is found from condition

(2.5).

is thus reduced to finding the roots of (2.10) and repeated

solution

found at the base-points

in accordance

for (2.9). When realizing the method numerically,

in Y and, instead of finding

p(y)

the values of

a

2

are of this mesh. Instead of the roots of (2.10), we take the points

base-points

1 is a quantity

of a mesh covering FQ, for which

characterizing

the accuracy of the solution.

To ensure

high accuracy when solving (2.10), either a reasonably fine mesh must be used for covering I’a, or a coarse mesh may be used in conjunction with local methods for finding the roots. The integration

step for (2.9) depends on the method employed,

chosen in such a way that the error of integration The proposed has no singularities (1)

lis(X(1))

in R, and at least one of the following

there are no interior

Condition(2)

two conditions

is satisfied:

points (x0, yu) of Sz at which, simultaneously, ?#of = 0;

the system (2.9) has no periodic solutions y(t))

is not greater than e.

scheme may be used for solving problem (1 .I) provided that (2.6)

KX(Z0, yo) = KY(% (2)

and has to be

lying inside Sz and such that

= 0. is weaker than(l)

but is more difficult

The max-min problem (1.4) may be solved similarly. in this case the problem

to cheek. Instead of (1.4), we consider

Numerical method for solving minimax problems

where y<(X) are the solutions (!/(x) satisfying

-

103

of the equation

c) (Y(S)

-

WG@,

Y(X))

= 01

the condition (Y(X) -

C)“(Y(Z)

-

~)ZK/,y(5V Y(Z))

2

0.

Instead of system (2.9), we have

The initial Cauchy data for (2.11) are the points (Zj, yj) E r~, at which K,(xj,

Yj) =

j = 1,2, . . . ) s.

0,

As an example of the approach I, =

described,

min max sip 2n (Ic OGY
consider the problem y) .

The system (2.9) is (2.12)

$-=$-=4n2sin2n(Z--y).

There are four points on RL where K,(ri,

ill,=

(l/4, O),

M, =

yZ) = 0 and Kxx(zi, yi) < Onamely,

(‘I& I),

M, =

(I: S/4), M, =

(0, 3/d).

The system (2.12) is integrated twice, first taking the coordinates of the point Mr as the initial data, and then the coordinates of Mz . Two branches are obtained:

x2(y) = -“/c + y.

x1(y) = ‘I&+ Y, Successive approximations C(y)=

0<

Xl(Y)7 _

Y

d

3/1,

“14 < y < 1;

{ X0(Y),

Z(Y)7

to the function

Fz(Y)

=

Z(y)

are .F~(.Y) E

i%(y) =

0< X*(Y)7 {x2(y), “I& <

1;

fl (Y),

0d

Xz(Y),

“I& <

1

Yd y <

“ii, 1;

y < 3/&, y <

1.

The set N is the same as Y, and II = 1.

3. Search for the solution of problem ( I .I) Though the method descibed in the last section is simple from the point of view of numerical realization, it demands a special investigation of the function K(x, y). The method will therefore be modified, and a numerical method for solving problem (1.1) devised in which it is only required that K(x, y) have continuous second derivatives with respect to both its arguments.

N. I Grachev and Yu. G. Evtudtenko

104

The functions K(a, y) and K(b, u) are evaluated for all y E Y and the following functions Zr (Y) , qi( y) is constructed: (3.1)

CP~(Y)=~(~(~),Y)=

max K(s,y)=maxEKta,y),K(b,y)l. =-z Denote by Ni the set of points y, such that

(3.2)

0%(if.) = miqi(y). UEY

At least one point y1 E Ni will be found. We find the set of values XI, 52, . . . , Zj, for which (3.3)

i = $2,.

@i = Inn; K(2, Yi) = Kbi, Y*)9

. . , j.

Obviously, ED,> qpi( yi) . If Q* =v(l/i)t

(3.4)

the minimax problem will be solved: the set Nr is found, and the points xi are given by condition (3.3). For, (3.5)

yw~Z’P~(~).

Ii = y;y” ““” K(s, Y>> qpi(Yi) 0

On the other hand, from (3.4) t 3.6)

max R(X, yi) = cpi(yi) 2 min max K(z, y) = 1%. XGX yEY SCX

Comparing (3.5) and (3.6), we get Yi = “pi(yr). If (3.4) is not satisfied, then the system (2.9) is solved j times with the initial conditions s(t0) =

xi,

We find functions G(y) the relations~ps (3.7)

Y (Ll)

=

31,

i=

1. ,*..,I*

and VPZ(Y)along the solutions

cpzb> = w%(!/I

9Y>= maxf.cp&),

*

G(Y) of system (2.9) from

y=jWd~hdl.

We then find mincpz(y) =

qh(Y2).

llEY

A new set x1, . . . , xl is found, from the conditions % = mix K(z, y2) = K(2, ye),

i=i,...,t.

XEX

If @. = cp2(&) , the problem is solved, Otherwise, the process is continued until for some p, we obtain

105

Numerical method for solving minimax problems

nfl

0

08

I y

FIG. 2

FIG. 1

It is not possible to state here that

for any y E Y.Nevertheless, Z,(Y) =5(y) can easily be seen that any function x(j~) could be

= 5 (yp) . It Y, E &r and UP taken as the initial approximation ZI (y) .However,

of the straight lines r(y) mation

~6

=

a or x(y)

=

since the solution

can lie on either

b, it is advisable to take an initial approxi-

of pieces of these lines.

Xr (y) consisting

The true mlue of II can be found by means of the computational

scheme described.

Instead of N, a set Np is found, of which N is a part. The set N may be obtained excluding from Np all the values of y for which max “E=, uUp The method

K(z,

y) > 1,.

thus enables a wider class of solution

than that required by the definition Then the method approximately.

of solution

is realized numerically,

For instance,

by

of problem

in Section problems

(1 .l) to be obtained

1. (3.1) and (3.2) are solved

when solving (3.2), the set Nr is made to include all the

ye for which

cpi(y.) -

min rF1(y) Q e. I/EY

Here, E is the accuracy

of the solutions.

global sampling over a uniform methods

The simplest way of solving (3.1) and (3.2) is

mesh; to improve the solution

of (3.1), various local

may be used (e.g., the method of steepest descent, etc.).

A computer

was programmed

in ALGOL 60 to realize the algorithm

described,

and numerous computations have been performed at the Computational Centre of the Academy of Sciences. The solutions of problems (1.1) and (1.4) will be described

106

N. Z. Grachev and Yu. G. Evtushenko

for the following

example:

1)‘exp [-g(z - 0.3)“].

- 0.7)‘] +

+ yz exp [-g(z

.x = [O,11,

y= [O,11,

g=

K(s,

y) =

(y -

The results are illustrated

g=

50,

5.

in Figs. 1 and 2, where Fig. 1 refers to the case g = 50,

and Fig. 2 to g = 5. The thin lines represent

the coordinate

axes and the boundary

of

the graphs of Zi;wp’(y),p = 2, and

the region Q = X x‘ Y. The heavy lines represent

the crosses mark the points of the solution of problem (1 .l). The graph of VP (z) , p = 2, for problem (1.4), is represented by points. The large point is the solution of problem (1.4). The curve made up of circles is the solution of the equation K, = 0 for problem (1.1). The data obtained

are

I, = 0.25,

I, = 0.07, M = {l/z} for g = 50, N = {Vz}, Ii = I?,= 0.41, M = (‘12) for g = 5. N= {‘L},

In the latter case, therefore,

(M, 31) is a saddle-point.

4. The multidimensional

case

The method of the last section can be extended

without

problem

serious modification

(1.1) in the case when n > 1 and m = 1, and to problem

m > 1. The algorithm

for solving problem

assumed that X is the ndimensional

to

(1.4) when n = 1 and

(1.1) will be briefly described.

It will be

parallelepiped n

x =

xc9

x xw x . . . x XC”) =

IIj=i X(f).

Here, XC’, =

[&‘, b”l],

Denote by I’i the boundary

which has 2” elements.

r,=

i i=i

Y=

[c, d].

of Xc’), i.e.,ri =

The boundary

pi x

{a(‘), b(‘)}and introduce

of X can be written

fixq, j=i

where the prime denotes the absence of the factor In the general case, it can prove extremely function

f, (y) from the condition

the set

as

X(9.

laborious

to obtain the initial vector

Numerical method for solving minimax problems

107

K(Z,(y), y) = max K(s. y). xer We therefore performed CD, >

put fi (y) =

as described

a, where a is a point of r. The further

in Section 3. However, improvements

computations

are

are required when

rpr (yi) ~Let xi be one of the set of points x1, . . _ , xi, determined

from condition

(3.3). Three cases are possible. Case 1. xi Ej$ I’S*The Cauchy problem is then solved for the system

(4.1)

D(F*,...,&) 3P--i),y,&+i),. ** 23"') ' D(d'),..., D(P*,..*,C)

qf+

-

4J -=

D (d”, . . .

dt

)

s=f,2,...,?2;

xq ’

here, D(F 1,. . . , Fn)

@iihq

L)(d’(‘), * *. , xw) = with the initial conditions

9. f * dF,ld;zc”~

:. : ,>j,;&Q

*I . i+&Pi,

x(‘)(t,)

= x8’), y(&)

I

= Yi, 8 = I,&.

. . n. In (4.1), F’,

denotes &s) (5,~). Gzse 2. zi E I’. We take xi as xib),

i.e., xib)

3 xi.

Case 3. xi E I‘, \ I’. For clarity, suppose that x$) E I’, s = 1,, . . , r < E. We then put the first r components of the vector function xi(~) equal to s =

x&+)(y) EE z(“,

1,. . . , r, while the remaining

solving the Cauchy problem

=

-

D(,P+V,

with the initial conditions X2(y) is obtained

. . . , &-9,

Check computations

y, xP+‘), . , . , d”))

y (&,) = ,zft, A”)(&)

from a condition

K(~~(Y),y)=max[rpi(y),

present time.

are found by

. . . , Fn)

D(F,.,,,

ax(“) -dt

n -. r components

for the system

in accordance

=

’

ry’, s = 1”+ 1, . . . , n. The function

similar to f3.7):

~~-jK(ri(Y)7P)l* with the scheme described are in progress at the

108

N. J. Gmchevand

Yu. G. Evtushenko

REFERENCES 1.

HELLER, J. E., CRUZ, J. B. and MEDANIC, J. An algorithm for minimax design, Proc. Internat. Con5 Theory and appl. Differ. Games, Massachusetts, 1969.

2.

DEMYANOV, V. F. Algorithms for some minimax problems, J. Comput. System. Sci., 2, 342-380, 1968.

3.

GRACHEV, N. I. and EVTUSHENKO, Yu. G. Some properties of min-max problems, in: Opemtional research (Issl. opemtsii),

2, VTs Akud. Nuuk SSSR,

Moscow, 1911.