1.1 Generalized Matrices

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1 Asian ]3ulletln II (1980) AN ILLUSTRATION OF THE DUALITY TECHNIQUE IN SEMI-CONTINUOUS LINEAR PROGRAMMING* F. DE VYLDER We give a complete parametric solution of the following problem Find a claml size distnbuhon F on the finite interval [o, ~], maxitmzmg the stop-loss premmm corresponding to a given excess e, under the constraints that the first inoment of F be at most cqual to p. and the second at most equal to v The method used is the duality technique m semi-continuous hnear prograinmmg described in DE VYLDER (197 8) This technique is summarized, without proofs, m the first part of the paper 1. THE DUALITY TECHNIQUE IN SEMI-CONTINUOUS LINEAR PROGRAMMING 1.1 Generalized Matrices Let A~ be defined for, e I = {1,2..., m} and x e K c_ R. In most situations K is a finite interval, but in many considerahons it might be a rather general set in R We consider A as a generahzed matrzx, with the set I of row-indices and the set K of cohunn-mdices. The augmented matrix A. = (A, 1) is formed by the matrix A followed by the umt matrix I of dinaension m We suppose that the columns of tlns unit matrix are indexed by any indmes at, c~2... am (not necessarily numbers), of course none m K. These indices are called slack indices. Let we denote by S the set of slack indices. If M = (x~, x2... Xr) is any finite sequence of elements in K + S, we denote by A M the usual matrix with columns A. z', A. z:,... A~.,, to be defined m a moment. If x~= x e K, then A x is the column (A~, A~,... A,,,) z,. If x~ is a slack index, say xj =,g (1 ~< k ~< m), then the column A k is the colulnn with m elements (0, 0..., o, 1, o,..., o, 0), where the 1 is at the k-th place. Similarly, if b* is defined for x ~ K, we consider b as a generalized row with the set K as column-indices The augmenled row b. = (b, o) is formed by b followed by m zero's. The usual row b M is the row where bzj = 0 if xj is a slack index. b M = (b ~', b*',... b~,), In the sequel, the sequence M will have exactly m elements. Thus A M will be a square matrix. The order of the elements in 33I is irrelevant, but usually we shall suppose that the lndmes m K are written hrst, m the natural order, followed by the slack mdmes in the order induced by their subscripts. * Presented at the 14th ASTIN Colloquium, Taormina, October

2 18 F. DE VYLDER J.2. Problems in Semi-Continuous Linear Programming In the sequel K is always fixed. F is an unknown distribution function on R (not necessarily a probabihty distribution function, the total mass induced by F on R may be different from 1, but it must be finite, for simplicity). In fact, the mass induced by F on R-K is completely irrelevant, since all integrals considered in the sequel will be taken over K (then this set is generally not indicated) or over a subset of K, relatively to the distribution function F. Let A, b be defined as in 1.1. and let a be the column a = (al, a2... am)'. Then the problem noted (A, a, b, F, MAX) is to find a distribution function F maxmfizmg f b z dfz under the constraints S Af dffz ~< a, (i I). The dual problem is to find a row y = (yl, y~... ym) minnnizing y a = ~ y* a, under the constraintsy* ~ o (i I), X y* A~ ~ b z (x K) A Fundamental Theorem If F satisfies the constraints of the problem (A, a, b, F, MAX), if y satisfies the constraints of the dual problem and if (1) fbzdf~ = ~ yea,, then F, y are solutions of the problem and its dual respectively The Duality Techmque Various tests have proved that the following technique for solwng the problem (A, a, b, F, MAX) works in rather general circumstances, provided step 1 can be executed. It would be highly interesting to know the most general conditions on K, A, a, b for the technique to succeed. STEP 1. Find a solution y of the dual problem. STEP 2. Find M = (xl, x2,..., xm) satisfying y A M = b ~. The indmes xj (slack or not) can be searched for the one after the other. Each x satisfying y A z = b z may be an element of the sequence M. In view of the following step try to find as many different radices as possible. Ideal, but not always realizable (nor necessary) is that A M be inverslble. STEP 3- Find a column z= (z~, z2... zm)' satisfying A M z = a, zt ~ o (i I). Then define the discrete distribution F in the following way. For each xj in M, xj K (thus xj not a slack index), place the mass z~ at the point xj. If e.g. x~ = x2, then the mass zt + z2 must be placed at xt = x2 and similarly if more indices are equal. STEP 4. Verify (1). Then, by the fundamental theorem, it follows that F and y are solutmns (for y this is in fact a confirmation) since the relations A M z = a, z, ~ o (i I) imply that F satisfies the constraints of the given problem.

3 DUALITY TECHNIQUE Original Problem 2. THE MAIN PROBLEM CONSIDERED IN THIS PAPER Find a claim ~ize probability distribution F on K = [o, col maximizing the stop-loss premmm co (2) I (x- e)df~ corresponding to the excess e, under the constraints (3) < Ix df < 2.2. Equivalent Problem Find a distribution F on K maximizing (2) under the constraints (4) I dfz ~< 1, I x df. ~< Ix, J'x2 df, ~< v. e That this problem is equivalent results from the fact that the first members in the relations (3) do not depend on the probability mass at the origin of K. If we have a distribution F, solution of the second problem, we can make a probabdity distribution of it by placing, if necessary, the missing probability at the origin. (It will turn out that this modification will not even be necessary.) Thus, our main problem is the problem (5) (A,a,b,F, MAX), where I = {I,2,3},K = [o, co] (o) ia i = (x K), a =, = \4a/ x= x efor e ~< x ~< co 2.3. Dual Problem The dual problem is to find y = the constraints (u, v, w) minimizing u + v~. + wv under (7) u i> o,v i> o,w i> o, u + vx + wx2 >1 b~(o.< x.< oo). Of course, these constraints are equivalent to (8) u i> o,v t> o,w >1 o, u + vx + wx 2 >1 x - e (e ~< x < o~) The Parameters of the Problem The problem is one with four parameters co, e, p, v and we shall give its solution whatever be the relations among these parameters. For the sake of completeness, unrealistic cases have been included. Evidently, we always assume (9) ~ > o, v > o, o~< e < co

4 20 F. DE VYLDER and also, until section 5, the relations (lo) ~ < V~,,, < ~oo implying (I I) ~ < ~o~, v- < co. Suppose for a momnent that we have not the first relation (lo), i.e. that we have ~/~ < tx. Then it follows from the last relation (4) that (12) Ix df~ ~< [I x2 dfx~ ½ ~< 1~ ~< ~z and the second constraint (4) becomes superfluous. Thus, the case ~ ~< tz leads to a simpler problem that shall be constdered separately. Note that the first relation (12), well-known when F is a probability distribution, is also vahd if.f is defective. Similarly, suppose for a moment that we have not the last relation (to), i.e. that we have V.o~ ~< ~. Then it follows from the second relation (4) that and then the last constraint (4) is superfluous. Again we are faced w~th a smlpler problem, in fact a problem equivalent to the GACLIARDI-STRAUB (1974) problem. 3. STEP 1. SOLUTION OF THE DUAL PROBLEM For each x(e ~< x ~< oo), considered as parameter, we imagine that we represent the part of the plane u + vx + wx 2 = x-e situated in the positive octant of the (u, v, w) space (fig. I). Let Pz = (x- e, o, o), Qz = ( o,, o ), R~ = o,o, X be the intersection with the u, v, w-axis respectively. When x increases from e to ~o, Pz and Qz remove monotonously from the origin. For Rx, two cases must be considered: O.)--e Case C1" 2e >/o~. Then Rz goes up from the height 0 to the height ~o Case C2: 2e < co. Then R~ goes up from the height 0 to the height I/4e (.O--g (for x = 2e), then comes down to the height o~ 2

5 DUALITY TECHNIQUE 21 w R~. ~ V U Fig. l w ~s =(o,o,~ 2Q e U=(o, 1--~,~) s u ~ Fig. 2

6 22 F. DE VYLDER 3.1. Study of Case C~ It is clear, in the case C1, that the mimmum of u + vbt + wv, subject to the given constraints, is obtained for (u, v, w) in the triangle P,,Q,~R,o. For reasons of linearity it must be obtained at one of the corners. From (10), (11) it is imreed_late that the minimum is obtained at R,o. Thus we have the solution and corresponding minimum : 3.2. Study of Case C2 U -~- 0, 7) = O, W -- (O--e ~2, MIN = (co-e) o2. Since Rz goes up and down when x increases from e to oo, the triangles PzQzR~ have a superior envelope. We can find its equation by eliminating x between the relation u + vx + wx 2 = x-e and its derivative in x. The equation of the envelope is ('4) w - 1 (1 -- V) 2 4 (u+e) " Of course, only the part in the positive octant must be used, and even not completely, because we have the restriction x ~< o). It is calculated that the triangle P,oQoR~,, corresponding to the maximum value of x, is tangent to the envelope along the straight segment TU (fig. 2), where T = - e,o,,u = o, I (t) Then the minimum of u + vbt + wv, under the given constraints, is obtalned for (u, v, w) in the part STU of the envelope, or for (u, v, w) in the plane portion PooQ,,UT. In order to study the variation of u + va + wv on STU, let we take u and v as independent variables and let we use (14). Then the quantity to be minimized equals, on STU: (l-v) 0- (15) f(u,v) = u (u + e) It is easily calculated that there can be no u, v annulating the partial derivatives f,~, f. We conclude that there can be no minimum at the interior of STU. Therefore, and also for reasons of linearity, the minimum of u + vvt + w,j is only possible for (u, v, w) on ST, on SU, at Po, at Qo Study of the Variation of J(u, v) along ST We have, for ST: = = ~ '/~..< e, g(u) f(u, O) U + 4 U + e' 2

7 DUALITY TECHNIQUE 23 g'(**) = 1 4 (u+e),' g"(u) 1> o. g'(u) = 0 foru = ½ ~/~- e. Therefore, if½ l/~ - e ~< o, then: Mmlmum on ST = Value ats - "9 4 e" if o < ~ ~/~ - e, then: Minimum on ST = Value at /~ - e,o, Study of the Vamalion off(u, v) along SU = 1/~ e, because½ ~ - e < - - eby(ll). 2 We have, for SU: h(v) = f(o, v) = vv. + ~ (~- v)~, ( o <. v <. h'(v) = V. -- 2e --(l--v), h"(v) t> O, e~ h'(v) = oforv =1-2--., Therefore, ea "9 if "9 ~< o, then: Minimum on SU = Value at,5 = e ifo < then : Minimum on SU = Value at o, 1-2 'q ~ "9 J "92 ] 1 becausel by () We note that the value of u + va + w'9 at Po = (~o - e. o. o) is (~o - ~) and at Q~ -- o, co- -, o it is (co- e) ~. O) ~ 3.3- Subclassifica~ion of Case Ca The subcases of Ca indmated in table 1, with, in the right column the possible mimmum, result from the discussion in and Note that, since a/co < l, the value (co- e) at Poo can already be abandoned against the value (co - e) ~/oo at Q~.

8 24 F DE VYLDER TABLE 1 Subcases of C~ (2e < co) Cat: vn 4 e2, v< 2etx C,~' ~< 4eL v> 2ev. C~" ~> 4 e -, ~ 2eve C~4: ~> 4 ea, ~> 2e~ Possible minimum 2_ 60 - e) -~ 4e ' 4e' ~ ~ - (o-e) tz o3 Irrelevant (see &scussmn) l~-e, ix(, - -~), (~o-e) ~o The first relation (lo) and the relation v~< 2e~. appearing twice in table 1, imply ~ = 2e 1/7~, ~ ~< 4 ~2. This means that case C2~ is impossible and that in case C2t, the condition v 44 4 e~ is superfluous. In case C21, we can abandon the ~z potential minimum (oo - e) =, because, by adding the relations 4e ~< 2' -- < O) -- 0.) 2' we have4 -- e < ((,~-e)-. The relation ~x 1 - < (~o-e) - a is seen to be 0J O) eqmvalent to,~ < p.o. Therefore, by (m), (co-e) - V- can also be abandoned O0 in C22 and C24 From o 44 (2elz- v) 2 results that -- can be abandoned in C22. 4e e e~ ~. Finally, m case C24, let we add the relations -~ <.~, -- < We obtain e - (~/~ + a) < 1. After multlphcatmn by ~/~- ~ > o, we have a relation equiva- lent to~x I - ~ < 1/~ - e, showing that 1/~ - e can be abandoned Then C2.o and C24 can be amalgamated in one case, say C'm, defined by ~ > 2e~., 2e < co Final Table The preceding discussion is resumed in table 2. TABLE 2 Point (u, v, W) giving the Cases minimum MIN co--e C~ : 2ea ~ o, o, -~---) (co-e) -- cos C-~: 2e<o, v<2e~ o, o, 4e

9 DUALITY TECHNIQUE FURTHER STEPS 4.1. Case C~ STEP 2. \Ve look for x ~ K satisfying ( ~ - e ) (~)= )o ifx < e o,o, ~ x-eifx >i e The solutions arex = oandx = ~.Smce O, O, ~2 ] = O, the slack index ~r2 can also be used. Then M = (o, o~, era). STEP 3- (lo) (z)ti) 0 oo I Z2 =. 0 co 2 0 \Ve look for zl, z2, ~ (the last is a slack var, able corresponding to ~2) satisfying The solutton is zt = 1 ~ z, z2 -- ~ = bt. f,)2 ' (D Thus, ff has two atoms, one at o, the other at oo, with respective masses ~//Zo = "t1~ ' 0)2 STEP 4. f (x- e)df. ~J = (co-e) 00 2 Value ~[IN in table Case C2t STEP 2. We look for x a K satisfying O, O, ~ X2 ~0 lfx < e x-elfx >1 e

10 26 F. DE VYLDER Solutions: x = o and x = 2e. The slack index,2 can also be used. Thus M = (o, 2e, *2). STEP 3. We look for zt, z2, ~ satisfying (o) (/(it o 2e 1 z2 =. o 4 e2 o Solution:z1 = z2- = ~ e2 ' 4 e2 ' 2e Here F has the atoms o and 2e with masses STEP 4. ~ V mo = m2e -- 4e2 ' 4e2 o (x-e)df~ =e - -- = Value MIN in table 2. 4 e2 4e 4-3- Case C'22 STEP 2. We look for x 8 K satisfying v ' v~/ x-elfx >~ e. V Solutions: x = o, x -- - Only the slack index -1 can also be used, but then we obtain the column (I, o, o)' already obtained for x = o. Since the use of the first slack index gives a defective distribution, X ~ M= o,~, (11 1)0 ()22 STEP 3. We search for zl, z2, z8 satisfying we prefer to try o [x 2 tx 2 Zo

11 DUALITY TECHNIQUE 2 7 ~2 [j2 The system is equivalent to" z2 + zs - --, z~ = l ---. There is an indetermination in the choice of zo., zs, but the resulting distribution F has anyway the two atoms o and v/a with corresponding masses ~2 ~2 ~0 = l -- --, ~l~vl ~ -- STEP 4. = Value MIN in table THE CASES v >/ ~ AND ~2 1> v These cases in which one constraint &sappears are left as exercmes for the reader. The answers can be found m table 3. TABLE 3 Maximum Conditions on Maximal stop-loss stop-loss the parameters dmtnbution premmm 2e~ 6) mo = t ~, m e = ~ 6) (o, -- e) ~2< v< btco 2e<~ 2evt<~ mo = I - --, m~/~= ~ ~ 1 - M v v v 4e2' 4ea 4 e bt Vt t*<c... m0 = 1 --,m~ =- (o~-e)-a Ca) it) 6) ~ 0... "g~ = I (CO -- B) 2e> co 2e< co 'q V V ~ oj2 m e = 1 (o)--e) v ~ xl N 4 e2 mo = 1 -- ~, m~e -- 4e 2 4e I - 4 e2 < v < oo"- m4~ = t V ~-e 6) 2

12 28 F. DE VYLDER 6. SUMMARY A complete solution of the original problem in 2.1 is given in table 3. It might seem surprtsmg that we solved the modffxed problem in 2.2 and nevertheless always found a probabdity distribution This is due to the fact, already observed m step 2 of the case C~2, that we never used the first slack index. It is clear that several cases could be regrouped in table 3. For clearness, we leave it as it is Indeed, given numerical values of the parameters, the table permits immediatly to situate the case to be used. Moreover, remind that the conditmn,j >/ p.~ amounts to the absence of the constraint on the second moment of F and that the condition p.~ >/,J amounts to the absence of the constraint on the first moment of F. See the discussion in 2 4. REFERENCES ~fvsv = Mlttellungcn dor Verelmgung schwelzenscher Versmherungsmathematlker. BUHLMANN, H (1974) Em anderc Bowels fur dle Stop-Loss-Unglcmhung In der Arbett Gaghardl/Straub MVSV, 74. DE VYLDER, F. (1978) Scml-contmuous hnear programming MVSV, 78 GAGLIARDI, B. and E STRAUB. (t974) Eine obere Grcnze fur Stop-Loss-Prammn 3~rVSV, 74