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Carnegie-Mellon UniversityPITTSSURGH, MNNMSYLVANIA 15213

GRADUATE SCHOOL OF INDUSTRIAL ADMINISTRATIONWILLAM LAMIMER MELLON, FONOER

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.AppNo Aplicable

A LINEAR PROGRAMMING FOR,1ULATION OF A SPECIAL QUADRATIC ASSIGNMENT PROBLEM

L . " ., rt ... •t . t e .. ,. . . .,'

Management Sciences Research Report November 1971

V. J. BowmanD. A. PierceR. Ramsey

F P t.:- "All " I. ' AL '.C OF ;'AS'?$ Ie Or f.' '

November 1971 20 64'""'ýa .C. ftte.CN A TOWS H? FI t ",I tf HIS)

N00014-67-A-0314-0007 Management Sciences Respirch Report No. 277

NR 047-048it W I F SC 4)PC) P "O' ,try ,rf I el , riumhrr ea t thl.V n ) OSSNlln( ld

WP - 39-71-2

This document has been approved for public release and sale;its distribution is unlimited.

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Lcogistics and Mathematical Statistics Br.Office of Naval Research

__Washington, D. C. 20360

A special quadratic aqsignme.at problem is shown to be equivalent to a

linear progranmming problem with n 3 constraints and n 2 variables where n is

the number of elements to be assignre. A labeling algorithm similar to that

for the linear transportation problem is presented for solving the problem.

An example is presented that deals with "triangularizing" input-output matrices.

FO 1 11 A4 I

DD,'.,1473 Uncassfie*t ~~~' )?IUclassCC-itC*~fie

L• 11 1. A: 1S .- ... IN4~~~~ ~ * 1.'9 L.I A44 I RC'Lr F, 0 L r

Quadratic Assigrnent

Linear Programuing

Labeling Algorithm

Permutation Polyhedra

Matrix Generation

Dual Simplex Methods

Input-Output Matrix

tI

DD ,PO0V 61473 (BACK) Unclaasified(F'Ati, 2) Security Classification~

"WP 39-71-2

Manigement Sciences Research Report No. 277

A LINEAR PROGRAMMING FORMULATION

OF A

SPECIAL QUADRATIC ASSIGNMENT PROBLEM

by

V. J. Bowman*

D. A. Pierce**

F. Ramsey**

November 1971

*Graduate School of Industrial Administration,Carnegie-Mellon University and

Department of Industrial Engineering, MAY 30 t972University of Pittsburgh

** Department of Statistics L-L-:*3U UOregon State Univeraity

This report was prepared as part of the activities of the Management SciencesResearch Group, Carnegie-Mellon University, under Contract N00014-67-A-0314-0007NR 047-048 with the U. S. Office of Naval Research. Reproduction in whole.orin part is permitted for any purpose of the U. S. Government.

Management Sciences Research GroupGraduate School of Industrial Administration

Carnegie-Mellon UniversityPittsburgh, Pennsylvania 15213I~~~ ~~ 9ThUI~ ATT~~ A

r

A LINEAR PgOCRAM4ING FORMU4ATION OF A SPECIAL QUADRATICI

A SSIGNMENT PROBLD(

Abstract

A special quedratic assignaent problem is ahown to be equivalent3 2

to a linear programming problem with n constraints and n variables where

n is the number of elements to be assigned. A labeling algorithm similar

to that for the linear transportation problem is presented for solving

the problem. An example is presented that deals with "triangularizLng"

input-output matrices.

.-1-

I ntrodc t ion

This raper shows that a special quadratic assignment problem can

be reformulated as a linear programming problem.

The general quadratic assignment problem can be written as

max T" y'AY

st Yij 1

r. = 0,1i

where the matrix dot product is defined as T.C t y iici i.e.,(i,j)

is sum of products of all paired indices, see Lawler 13 1. in Section 1,

we show that for a special form of T, (1) can be reforimlated as a

linear programming problem, i.e., all extreme points of a linear set

of constraints are solutions to (1) and vice-versa. In Section 2,

develop a labeling algorithm :or solving the linear programming formulation.

In Section 3 we discuss an application of this formulation to the triaugu-

iarization of Input-output matrices.

Section 1: Linear Programing 2Quivalent

Consider the quadratic assignment problem (1) where T is an tipn'er

triangular matrix oi ones, i.e.,

.0 l.........1'0 0 . . . . .

T - . . .. . .. .I

0 .... 0 1

0 .... 0.

b,.

This quadratic assignment problem can be viewed a3 finding a permuta-

Lion P I (rl,...r) of the rows and columns of A such that the sum of

the above diagonal elements is maximized. That is, if .A(P) is the

valu• of (1) for permutation P then

0-I n

A I- J ri+I rLr

The relationship of the permutation P to the variables y in (1) is

that Y is the permutation matri:- of P, L:at is

10 1 irj

1 irI

The structure of the problem yields the following easily verified

properties:

Property 1: For any permutation either aij or a £, i J j

must appear above the diagonal so that either anj or a,,

is in the function fA(P).

Property 2: If A is a symmetric matrix, Lhen f A(P)-constant

for all P.

Property 3: If A*=A+B then fA*(?) - fA(P) + fB(P) for all P.

-3.-

We can assume by Properties 2 and 3 that a .j 0 for all i and J.

Let us consider the graph G - (X,U) where X is a set of n vertices and

U is the set of directed arcs U - {(i,j) 1 4 j1. We will associate

with ea.h arc (i,J) of G a cost a j. Let r be the set of partial

graphs formed with n arcs of G such that (ij)sG'r = (ji) A G'.2

Let h(G') - rV a for G'-f, G' - (X',U). If we consider the

following problem

min h(G')

st C'wc (2)

and G' has no circuits,

then (1) and (2) are related by the following theorem.

Theorem 1. if z is the optimal solution to (t) aud w the optimal

solution to (2), then z - w

Proof: In order to prove this theorem we will show that every feasible

graph, G', in (2) corresponds to a permutation P' in (1) and that

there exists no permutation in (1) for which this correspondence does

not hold.

Lemma 1: If G'er then (i,j) % G' a (J,i) c C' for 1 4 J.

Proof: Consider the sets of arcs S; - ((ij),(J,i) I i > JI; There

are n(n-l)/2 such sets and by hypothesis only onu .rc .

from each set in constructing G'. If there exists a set Sij such that

S jnU' - 0 then IU'I< n(n-l)/2 contradicting the face that G' C.

-4-

Lemma 2: If there exists a circuit in G'c" of length g > 3 then there

exists a circuit of length 3.

Proof: Let (i 1 ,i 2 ,...,£ q-1I ) denote the circuit G'er. Now suppose

(0l0iq.-2)f G' since otherwise we are done. By Lemma 1, (iq.- 2 *il)*G';

thus there exists a circuit of length g-1. This indLrective step can

be repeated to prove the lemma.

Let us define the out degree N(i) of a vertex i in G' as N(i)=(ic'

Lemma 3: If N(i) - N(j) i 0 J then G' has a circuit.

Proof: Let N(i) NO(J1 ) - K and let (ijI),(i'J2),..(iJK)cc,

Now suppose (Jljt)eG' and t 1 Jil i - 2,...,K, then (t,i)eJ' by Lemma 1.

Therefore t - Jij i - 2,...,K, but there are only K-i such arcs and

since N(JI)=K there is one arc (Jlpt) for which t 4 ji, i 3 29...,K

and there is a circuit.

Proof of Theorem 1.

Let C' be feasible in (2). Then the nodes of G' can be ordered such

that i < j if m(i) > N(j). Let this ordering be P' (rl1r2 1...,r) i.e.,

r k if N(k) - n - j. Since 0 <N(i) < n-1.N(i ) -N(i 1 ) + 1, and thus,

P' is a permutation of the n integers (0,l,...,no-l), which is isomorphic to

the permutation of the integers (l, 2 ,...,n). Consequently P' is a permuta-

tion for (1). Notice that any permutation in (1), introduces the ordering

on G' by the out degree values. To show that z a wp we note that for ther. n

permutation P', z - P y a while for the associatad graph G'¢:,i-l J~i÷1 rrJ

n-1 nw-h(G')- r a - 7 ! ar r

(ij)•G i-I Ju-liQ.E.D.

S. . . .. .. . i l ... .. | 1 le .. .. . . . . . . . - - - ...

-5-

Corollary: The optimal solutions to (1) asd (2) are related as follows:

(j I ii N(i) - n-j

Yij 0 if otherwzv6

Proof: Obvious from the proof of Theorem 1.

-From Lemma 2 we can reformulate (2) as follows:

min h(G')

st G'er

G' has tio circuits of length 3.

It we let xij = I if (i,J) e C'

- 0 otherwise,

then the following equations do not allow circuits of length three:

Xii + xjk + X k< 2 i k

x ji + i'-k + xk. < 2 ;

in addition, Property 1 stipulates x.j + xji 1 i j. Thus the

quadratic assignmenc problem has been reduced to the following integer

linear programming problem

n nmax E ;7 a j'jxji-I J-1a/xt

st xij+ xjk+ Xki_2 i k €j 1 (4)X ikk + X kkJ + Xj _

X Ij Xjk J

xi 0 for all i

Loin= I_ _

6.

We now use the following theorem of Bownma 1 to reduce (4)

and correspondingly the quadratic assignment problem (1.) to a Iznear

program.

Theorem 2: the extreme points of the polyhedron formc~d by the coU-tA.rijilts

XiK + +x 2 i k j

Xik + xkj + xji 2

x ij + x = i j

i.j "" k

x i

x.. = 0 for all i1-1

are all integer.

The result of this theorem is that we can drop the integer con-

straints to (4) which reduces the problem to a linear programming

problem. However, therc are n(a-l) variables and n(n-l)/2 + n(n-l)(n-2)/3

constraints which causes problems with solving by conventional L.P.

methods as n grows large. In the next section, we develop a labeling

technique to solve (4).

Section 2: AIgorithu'

In this section we describu a labelling type algorithm for solving

the linear programming problem (4). This algorithm corresponds to a

dual simple:, routine, but the anmount of computer storage required is

smaller than cha standard linear progI'anming i'equiremenLs.

We first note that the constraint xi-- + . can bL replaced

by upper bcund consLraints on one of the variables. Therefore, let

us reformulate (4) as

n-I nmax E a.i=l J=i+l i ixi

x]j +'jk - 'ik.- 0 i j.k

"ij - k + <j

Ii <

We have thus reduced the problem to one involvLng • variables,2

but there are sLi].! n(n-i)(n-2)/3 constraints excluding the upper

bou:,ds. This, of course, implies solving the dual of (5). Let (5)

have the matrix representatiii

Max cx

s.t, Bx K e

and -Bx " 0

0 f., x 1 I

then the dual to (i) is

fiji •w 4 •v

,.t. B'w - B'u + Iv > c

w,u,v, _> 0

'-8-

or

min ew + ev

s.t. -B'w + B'u - Iv + Is - -c (6)

s,W,UV, > 0

Now since the columns of B' correspond to the constraints of (5) every

column can be denoted by a triple (i,j,k) where i < j < k. Of course,

identifying a column of B' is equivalent to identifying variables

in (6) and we therefore, use the following notation:

(i,j,k) denotes variable uijk £< J < k

and (7a)

(i,j,k) denotes variable wijk i < j < k.

In a similar manner we let

(O,i,j) denote variable s j i < j

and (7b)

(0-ij) denote variable v j i < J.

Observe that the variable designation also allows one to generate the

appropriate column representation in (6). Because of this fact we

will develop an algorithm that keeps track of the basis elements of

(6), but not their matrix inverse. Then when we decide to enter a

new variable, we will find its representation by a labelling algorithm.

This is quite similar to the approach used in transportation algorithms.

In order to determine incoming variables (or optimality) we will

use the fact that if we know the dual solution associated with a

i

-9-

basis of (6), i.e., values for x. then the reduced costs for w and

u are the slacks of the respective constraints for xij + X jk - Xjk.

and -xij - Xjk + Xik - 0 and the reduced costs for v and s are respec-

tively l-xjj and x ij. When these reduced costs are all positive we

are optimal, otherwise we introduce the appropriate variSable into the

basis. In order to understand the following algorithm we again point

out the variable designations (7a) and (7b) allow for the generation

of the appropriate column of (6) and thus, there is no need to ever

have the complete representation of (6).

We denote as follows various ntorage units used in the algorithm.

X: Upper triangular matrix whose values x.. correspond to the

solution of (5) associated with a basis of (6) i.e., they

are the reduced costs associated with variable s.

C: Right-hand side values associated with current basis of (6).

B: Upper triangular matrix bij is the identification of the

basic variable associated with row (i,j) of (6).

S: n(n-l)/2 lists denoted Sij. List Sij contains the basic

variables that have non-zero entries in row (i,J) of (6).

The other elements used are appropriate temporary storage units.

Step : lnitialization

Ct : aij "si ajil'

If (aij - ji <0, x)ij O 0, b ij . (0ij), Sij = t(O.iJ)

Otherwise, xij 1 1, bij - (0-.ij), Sj.j = ((O77j))

Remark: It is easy to confirm that this produces a feasible basic

solution to (6) made up of the variables v and s.

Smlndm m mu --, mmu m . ... ... ..... .. I

-10-

Step 1: Optimality Check

If 0 K xij I 1 for all i < j and

0 Y + x - X I or a]l i K j k, then

stop, this solution is optimal.

Remark: Let v = ((r,st) or (r-sc)3 be the entering variable corresponding

to violation of one of the above restrictions. Note r - 0 is

permissable. Go to Step 2.

Remark: When r = 0, addition on index 0 implies no operation and is

used only for notational convenience.

Step 2:

2A: Let zij = IS.j! (the number of elements in list Sij).Let z = z + i, z =z +I,z = +1I.

rs rs rt zrt zst Zst

2b: Reduction Step

2B1: D = 0 (D is a list)

2B2: If zij 1 for any zi., let P -S si, n D)

{(a,b,c) or (a,b,c)] and go to Step 2B3.

Otherxwise go to Step 3.

283: D = DU P and Zab Zab 1, Z Z - 1 a Zb Zbc

a30 a a b

(;o to 2P2.

Remark: Step 2B eliminates basic variables that cannot be used to

represent (r.s,t) or (r,s,t) because of having only one non-

zero clement in a row of the basis ,iatrix that has a zero in the

entering column.

.Step 3:

3A: Let E -- 0, i < J and zrs - z - 1 rt = zrt - ,

Ast = Ast - 1.

-1. Go L

Ste' 3C.-

Otherw iS. f se (rst) 0 Fr 1s r

C~o to 3C.

3C: Let (an ord~t-ed lisit)

3D:

31)1'. If ' i id Fý j Ofor some i-j, let P=SO Sý ~D.

to 31-,.

3D2~ Ei 0 for all i,j; go to Step 4.

303: f ~.. C for solili: E.. 4 0. Go to Step 5

j~ z. 2. for all E~ . 0, lef- P =somte cPieto

- ý 1. f orE 40 Le t b -P. L 1(b )q 0.

jj ij '.3 pq 1p

Go to 3E.

.3E1: If< (,a,b,c) then.i either (a~b) (i,j) and s'-t

PIZ

L 2(b pq) sign (E )or (a,c) =(i,j) and set

L,} () sg (E.) or (b,c) =(i,j) and set

-pq Uj

11b u, q sig (h E )

E2 p+

ac ac - (hpq) 1

Lbc Eb + L chw3

3E' (thrlL5- if h - (a~b,c)thefl cither (a,b) (i)

rInd .e L, b t) sign (F.) or (a,c) (.j)and set

(b ) = sign (-.)or (b,c) =(i,j) and set

(1, ) s Lig (E1 .)

E b 1 )(4

ac c 2p'

3F: Za Zab - ie a Z Zac - , ZbU P

Q = Q U (bpq, LI(bpq), L2 (bpq)) Go to 3D.

Remark: Step 3 finds the representation of the incoming column v

in terms of the present basis. L1 is a label that is one

if there is only one non-zero entry in a row which has a

non-zero value in v or its subsequent representation.

Step 31A always tries to force a permanent labelling since

this is a forced relationship.

L1 is a label that gives the sign of the basis element

b in its representation of v. Since the set is unimoiular,pq

we know the coefficient of a basia vector is 0 or + 1,

Finally, we note that 302 stops our labelling procedure

whei we - have found a linear combination of the basis vectors

that yields v.

Step 4: Change of Basis.

4A: Let Q+ be the set of Q such that J 2(b.j) = + for b i E Q and (-

be the set of Q such that L 2(bij) = - for ijE Q. Let mn C.

c and W = b.pq pq

c ij c . 4 c pq for all b • Q" and

c -i.. cij - cpq for all bij C Q (i,j) i (p,q).

b = v and S = S Uv, Srt = SrtUVSst _ SstLUvpq r s rs rt r t t

and i.f W (a,b,c) or (a,b,c) we have 9 ab - Sab - W,

S -= - W and S S - W.Sac ac Sbc Sbc

4B: For (O,i,j) E B, set x ij 0, for (O,ij) E B set xj i .

Co to 4C.

-13-

4C: For (i,j,k) E B•, i ý Q set, x + xjk - Xik ' 0 and for

(ij,k) G B set xtj + xjk - xik 1. Solve these constraints

with the conditions of 4U to obtain new X matrix. Go to Step 1.

SLep 5: Back-tracking.

5A: Back-track in list Q to the last element labelled 0 (temporary)

say b . All elements encountered in the back-tracking priormn

to bran say bpq must have L (b ) = +1 and are freed i.e., for

all p,q. Perform 5AI, 5A2, and 5A3.

5Al: if b = (ab,c), thenpq

E = E + L(b IEab ab 2 pqE*c Eac + L2(bpq) 1

Ebc = Ebc - L 2(b b) 1 . Go to Step 5A3.

5A2: Otherwise, if b = (a,b,cl, thenPq

Eab =Eab + L2 (bpq) I

Eac = Eac - L2 (bpq) i

Ebc = Ebc + L2 (bpb) 1 1. Go to Step 5A3.

5A3: D = D - bpq, Q=Q (b- pqLz( pq), SL(bpq))

1ab z ab + 1a z a , a + 1.

5B: [, 1(b) = +I. Go to 3D.

Remark: Note that the only change for bran is that it is permanently

labelled rather then temporary. This makes sure that it will be

freed when we back-track to some point prior to its assignmenL.

While this algorithm may seem quite involved compared to simple

simplex operations, one should note that its advantage comes in its

small core requirement compared to a linear programming approach.

If we assume that the original matrix is generated rather than stored,

-14-

our core requirement for simplex solution of (6) is essentially that ofn(n-1) 2 4

the basis iverse. This is ( 2 . or on the order of n . For the

above algorithm, oene ineeds to store n elements for X,C,B,Z,E, and2

2 bels and a maximum of n(- for S, that is approximately Sn(n-l).2

For large n this savings is considerable. For n - 20 the L.P. may re-

quire over 36,000 storage spaces while the proposed algorithm requires

r aughly 1900.

lia the ne:t section, we discuss the application of this algorithm

to a quadratic assignment problem involving the triangularization of

Input-Output matrices.

-15-

Section 3: Example - Triangularization In2.t-Output Matrices

The permutation of roys and columns of an input-output matrix,

denoted A, that maximize the sum of the above diagonal elements provides

a hierarchical listing of the sectors of the economy as "users" and

"suppliers ".

For the economic interpretations and a heuristic method for triang-

ularization, see Simpson and Tsukui [6). For an implicit enumeration

algorithm, see 15).

it is easily verified that the problem of permuting the rows and

columns of A uakes it a quadratic programming problem and since we are

concerned with the sum of the above diagonal elements we have the T

matrix as in Section 1.

The following example is a 5x5 subset of an original 37x 3 7 input-

output matrix presented by Leontief [4 1.

Agric. and Fish 2453 3896 2195 15 317

Food and Kindred Prod. 538 1427 61 8 0

Textile Mill 14 0 1321 2913 0

Appaifel 9 50 0 1471 0

Lumber and Food 34 25 20 0 1817

Step 0. (3358 2181 6 283\ /1 1

61 42 25 x 1 0 0

2913 20 y 0

0) 0

-16-

(0,1,2) (0,1,3) (ot,14) (o,1,ý)

(0,2,3) (0,2,4) (0,2,5)

(0.3,4) (0,3,5)

(0,4,5) /

st..p_1 x 2 3 + x3 4 - X24 a 2 an4 v , (2,3,4)

Step 2:

2A/1 1 1

2 2 12-

2 1

2DI: D,1

282: z 1 ,2 P 1, - (0,1,2)

2B3: D1- •t(O,2-1)2

2 2 1Z 0

2 1

2B2: zl,3 is, P - (0.1,3)

2B3: D ((-0,1,2), (,0l-,3))

iA 0 1 1

Z 2 2 1

.2 1

2B2: %1,4 1. Pto (0-1,4)

213: L) = 1(0,1,2), (0,1,3), (0,1.4)

/0 o 0 1a2 2 1

Z2

2

-17-

2B2 and 2B3 repeat for z1 , 5 ,z),5 z3t,5 and z4,5 ending with

2B3: D = •(O,!,2--- -),(0,1,3),(0,1,4),(0,1,5),(0,2,5),(O,3,5),(O,4,5)j

0 0 0 0

2 2 0

2 0

0

a 0 0 0 0 0 0 0

0 0 0 1 1 03A: E -

, 0 1 0

0 0

:o 0 0 0

1 -1 03B: E 0

1 0

0

3C: Q= 0

3Dl: z2&3 0 PdEr E2,22,3

3E: L2 (b 2 3 ) = + Q ((b 2 3 , 1, +)]

0 0 0 0 0 0 0 0

0 -1 0 0 1 0

1 0 1 0

0 0

3Dl: z = I and E24 • 0 p (0,2,4)

3E: L2 (h 2 +, Q (b(b 2 3 ,1,), (b 2 4 ,L,).

10 0 0 0 0 0 0 0

E 0 0 0 0 0 0

1 0 1 0

0 0

-18-

3DI: &3,4 I i3,4 940 P. (0,)

3E: L (b + Q - ''(b 2 ,3 ,,+),(b 2 ,4.1 +)(b3,.t,+)1

0 0 ) o 0 0 0E* 0 0 0 a oi o

0~ ~ 0)',,

0 0

3D2: En 0

4A: Q Q -

Sq- min(51,42,2.1 3 ) - 2,4 W (0.2,4)

/3358 2181 6 283

19 4; 25

2871 20

0

D02,- (293,4)

• , 2,3),(2,3,4)., 82," * .(2-",4)i, (34 .(O,3,4),(2,3,4))

/1 1 1 ! \•

, l x 0 :I 1 x f

0

4C:. x2 , 3 + x3 , 4 x 2 4 g2,3 139X2,4

1 1 01

0

1: All coditi~on. atisfied. Solacion optImal.

-19- i

The optimal solution has

0o 1 1 0 2 i 2

X* 0 0 x I 0 j, -

0 0 0 x o0 1 i o

1 1 3

so that the optimal arrangement Is (1,5,2,3,4) i.e.,

Agric. & Fish /2453 317 3896 2195 15

Lumber & Food 34 1817 25 20 0

Food & Kindred Prod. 538 0 1427 61 8

Textile Mills 14 0 0 1321 2913

Apparel , 9 0 50 0 1471

or in reduced form

x 283 3358 2181 6

0 X 25 20 0

0 0 X 61 0

0 0 0 x 2913

0 0 42 0 x

While the algorithm may seem quite cumbersome for such a small

problem, efficient use of bit processaog routines and proper programming

techniques greatly simplify the updating procedures.

-. i.

-20-

Conc lusion

We have shown that a special forai of the quadratic assignment

problem can be reformulated a, a linear programmning problem, and that

because of the structure of the ILnear program an algorithm can be

developed that parallels the simplex m4thod but at a greatly reduced

storage requirement. This might indicate that other quadratic assign-

ment problems may have equivalent linear progranwiing formulations for

specific T rmatrices in the formulation (I). These of course, will not

necessarily have to correspond to inteser solutions as in the case oi

the xij in the above problem, but rather, may fall in the category of

searching fur appropriaL.e permutation polyhedra which are discussed by

Bowman [1]. One should further note that the algorithm of Section 2

is really depandent on the constraint set of (5) and not its objective

function and thus, is applicable to any problems where this constraint

set arises whether oc not it comes from the Quadratic Assignment problem

(1). For a discussion of this constraint set in the context of transi-

tivity, see Bowman and Colantoni (2].

Reference.

1. Bowman, V.J., "Permutation Polyhedra" forthcoming in SIA__M ooornalon Applied Mathermatics.

2. Bowman, V.J. and C.S. Co'Tantoni, "Majority Rule Under TransitivityConstraints", Working Paper '89-70-1, C, SIA, Carne~ie-MellolUniver-sity.

3. Lawler, E., "The Quadratic Assignment Problem, Management Science 9(1963), pp. 586-599.

4. Leontief, W., "Input-Output Economics", Scientific Amnerican 185,(1951), pp. 15-21.

5. Ramsey, F.L., D.A. Pierce and V.J. Bowman, "Triangularivotion o0Inpit-Outpdt Matrices", Technical Report 016, Department of Sta-tistics, Oregon State University, 1969.

6. Simpson, D. and J. Tsukui, "The Fundamental Structure of Input-OutputTables, An Internatiional Comparison", Review of Economics and Sta-tistics 47, (1965), pp. 434-446.