Appl. Math.-JCU 10B (1995), 325-336

A S E C R E T A R Y P R O B L E M O N F U Z Z Y S E T S

LI XIAOJIE AND JIN ZHIMING

A b s t r a c t . This paper deals with a secretary problem on fuzzy sets, which allows both the recall of applicants and the uncertainty of a current applicant receiving an offer of employment. A new decision criterion is given to select a satisfactory applicant. This result extends the works of M.C.K. Yang and M.H. Smith.

1. In troduct ion

The secretary problem is typical of optimal stopping theory. The problem with no backward selection was excellently discussed in [2,6,8], which gave optimal se- lection rule for the choice of the best secretary (decision criterion I) and minimized the expected rank of the chosen applicants (decision criterion II). The problem with backward selection was first studied by Yang[9]. He assumed that the current appli- cant was sure to receive an offer after being interviewed and gave the optimal rule of decision criterion I. On this basis, Petruccelli[7] and Choe and Bai[1] combined and generalized the results of both Smith and Yang. However, the problem with backward selection in decision criterion II remains to be solved.

By using a new decision criterion III to maximize the probability of selecting a satisfactory applicant and recalling only the relatively best applicants, the present paper generalizes the previous works and partly solves the recall problem of decision criterion II.

Satisfactory applicant sets axe Fuzzy sets

A = { f ( 1 ) f(2) f (N)} 1 ' 2 ' ' ' " N

where f(i) is the membership grade (or the satisfactory degree), which is non- increasing. Our decision criterion III is to find an optimal stopping rule t* such that

N N

P(bt. �9 A) = ~ f(k)P(bt. = k) = sup ~ f(k)P(bt = k) k = l tET k = l

where (51, b2,.. . , b~,) is a permutation of {1, 2 , . . . , N}.

Received March 30, 1994. 1991 MR Subject Classification: 60G40. Keywords: Optimal stopping rule, secretary problem, satisfactory applicant.

325

5 LI X I A O J I E A N D J I N Z H I M I N G

Section 2 gives assumptions and model. Section 3 discusses the optimal s topping rules for f ( k ) > O, 1 ~ k ~ N . Section 4 considers the results for f ( k ) = - k , 1 ~_ k _< N. An example of opt imal procedure is given in Section 5, which extends Petruccelli 's results.

2. Assumpt ions and Model

Let

= the set of all permuta t ions (51 ,b2 , . . . , b N) of { 1 , 2 , . . . , N } ,

Piw} = ~.,,

F = the set of all subsets of fL

wn = the posit ion function of the relatively best applicant among bl, b2 , . . . , bn. For example, w,~ = n - i implies that bi is the relatively best applicant.

=

z,~ = 1 if the relatively l: es,: applicant among bl, b2,. �9 �9 b,~, receives an offer of employment, and z~ = 0, otherwise. We assume that

(1) P(z,~ = 1 I w,~ = n - i) = q(n - i), where q(r) is non-increasing with respect to r .

(2) Once having rejected an offer, the first n applicants are no longer available for selection. For all j > n,

P(z j = 1 ]wj = oan + j - n,z,~ = O) = O.

L e m m a 1. {wl ,w2, . . . ,wn} is a Markov chain, and

1 P(wn+l = 0 [ w n = n - i ) - - - 0 < i < n < N - 1 , (2.1)

n + l ' n

P(wn+l = n + l - i I w n = n - i ) - m , 0 < i < n < N - 1 , (2.2) n + l 1

P(wn = n - i ) = - , O < i < n < N - 1 . (2.3) n

The proof is obvious.

Now we consider the recall secretary problem. This recall technique consists of s topping at stage n and recalling the relatively best applicant, but if this person refuses the offer, then investigation of the remaining applicants is resumed. An offer is made by using some opt imal rule. Let the reward function be

xn = P(stopping at stage n and obtaining the satisfactory applicant by using the recall technique)

= P(at stage n, recalling the relatively best applicant IFn) + P(recall fails and selection continues IF,~).

A SECRETARY PROBLEM ON FUZZY SETS 327

Hence

P(at stage n, recalling the relatively best applicant [F,~) n

= E E(f(bi)I(z' ,=l)[Wn = n - i) j - - 1

n

= E E(f(bi)lwn = n - i ) q ( n - i)I(.~=,~_i) j = l

n N

= E E f ( k ) P ( w N = N - k[wn = n - i )q(n - i ) I(~.=n_ i i = 1 k = l

N N--1 ,

= ~ f ( k ) - - ~ q t n - i ) I ( ~ = , ~ _ 0 i=l k=l

n

= E h(n, 1)q(n - i)I(~on=n--i) i----1

where

With {hi = 1} = {w N

N N - k k-1 ( n - i ) ( i - - 1 ) h(n , i ) = f ( k ) (2.r,)

k=i (n N)

= N - i} and by induction on i, we see that

"/,~ ---- sup E(xt[F~) = sup E(xt[wn) tECn teen

w h e r e C n = { t E F : t >_ n, E x t < e c } .

-'~ is the reward if the interviewer decides to Let -~,~ = sup E(~?~IFn) in which x t tECn+l

select further to stage k > n and not recall the relatively best one prior to n. For N finite, the opt imal rule 5 E C,~+1 must exist, so E(2,FIF~ ) -- #n. Then

n

xn = E ( h ( n , 1 ) q ( n - i) + ( 1 - q ( n - i )~n , i ) I (~ .=~_i ) (2.6) i = 1

where ~n,i = "Yn[(~,=,~-i). So xn has the Markov property. Hence 7~ is a a(wn)- measurable function. Obviously ~N,i = 0.

0 Let Xn,i = Xn](w.=,~-i), Xn = Xn,n, 7n,i = 7,~[(.,~=n--i), and 7 ~ = ~'n,n. T h e n we

have

L e m m a 2 . For any i < n < N - 1 ,

E ( ' T n + l [ ( . ~ n = n - i ) ) -= 1 o n n + 13',~+1 + rT-~i-~n+l, i , (2.7)

328 LI X I A O J I E A N D J IN Z H I M I N G

1 0 It ~ zY"'i = n + 1%+1 + 7 " ~ ")%+1'i'

N ~ It 0

5 ' ~ - % , i = Z j ( j _ l ) Y J " j = n + l

(2.s)

(2.9)

Proof. We have

E(%+IIw~ = n - i) 0 = " f n + l e ( w n + l = 0 [ w . = It - i ) + " T n + l , i e ( w n + i -~ n + 1 - i[OJn = n - i )

1 o n = n + 1 ") 'n+l q- - ~ " Y n + l , i .

By the Markov property, for A �9 Borel sets) and t �9 C n + l ,

P(&~ E Alw. = n - i)

1 - e ( w , ) = n - i ) [ P ( ~ e A, O3n+1 = 0, 5d n = It - - i)

+ P(5:~ c A , w.+l = n + l - i , ~,~ = n - i ) ]

n 1 P(&r e AIw~+l 0) + P(:~t. n+l �9 AIw.+ 1 n + 1 i).

n + l n + l

Therefore 1 n

;Yn,i - - It + 1 + 7 . + 1 Or- - ' ~ Z / n + l , n "

By recursion, we obta in (2.9). From (2.9) we know that ~,~,i is not dependent on i, so "~,,i = "~. []

J

Throughout the paper, we shall assume that if j < n, then ~ ak = 0 and k =n

Y

H a k = l . k..~n

i = h ( N , 1)q(N i ) , r n n and L e m m a 3 . L e t ~ N - - = ~rn,

(h(n, 1) ~ / . , i ) q ( n - i ) V n ~rn = - n + 17rn+l ' (2.1o)

where a V b means max(a , b}. Then, for any n <_ N - 1, we have

1 ~i I t i E("/n+lIFn)lw~=n-i + - - "~- ~/n n ~- 17rn-k 1, (2.11)

A S E C R E T A R Y P R O B L E M O N F U Z Z Y S E T S 329

xn# = ~ , i + (h(n , 1) -

N

k = n + l

~ i ~n,i = "/'n,i q- "fin,

N 7r k

#o= E T. k = n + l

(2.12)

(2.13)

(2.14)

Proof. We only need to prove (2.11), (2.12), and (2.14). We shall do so by using backward induct ion on n. Since

N

~ = XN = E h ( N ' I ) q ( N - i ) I ( ~ = N - 0 , i= l

hence

E(7~IF~_,)I~,N_ =N_I_ i = N h ( N , 1 ) q ( 0 ) + ~ - ~ l h(N, 1)q(N - i),

~/N-1 = N h( N, 1)q(0),

XN--l,i = h(N - 1, 1 )q (N - 1 - - i ) + (1 - q(N - 1 - i));y~_,,

= l h ( N , 1)q(O) + ( h ( i - 1 , 1 ) - l h ( N , 1)q(O))q(N- 1 - i), 1u Iv

7N-1,, = h(N, 1)q(0) + ~ - 1 "

Therefore (2.11), (2.12) and (2.14) are valid for n = N - 1. Assume tha t they are valid for N " 1, N - 2 , . . . n + I. We shall show tha t they are true for n. In fact

1 o n

= n + 1 n§ + 1 o n _ 1 o

= n + 1%'+1 + - ~ - f " ) ' n + l , i -ff ~ - - ~ ? n + l q-

~ 7 2 i

•Tn , i "~- 7rn+ 1, n + l

71. i n + 1 7rn+l

1 0 n ~ "~n = n + 1 "Yn+l + ~ - ~ - ~ ' ) ' n + l - -

1 1 ~'n+l + "~n+l ~-~

n + l n + l

1 1 n

N N

7r..+1 + ~ = k k:=n+2 k = n + l

330 LI XIAOJIE AND JIN ZHIMING

From "~ = ")n+l + ~ and Lemma 2, we know tha t n4-1

X,~,i = (h(n, 1) - ;~=)q(n - i) + ;y,~ = ~n + (h(n, 1) - N

71" k - - ~ - ) q ( n - i),

k=n+l

3',~,i = x,~,i v E(~ , ,+IIF ,~ ) ]~ , ,=n ' i N

( k ) n i =~, ,~+ h(n, 1 ) - Z q ( n - i ) V n + l % ~ + l k----n+l

i : "~n "4- 7rn. []

By Lemma 3, we see tha t the opt imal s topping rule is

N { 7rnw 1 ~ ~7rk } n i (h(n, 1) )q(n - i) t * = inf n : > -

n + l g

k : n + l

3. T h e M a i n R e s u l t for 0 < f(k) < 1, 1 < k < N

In this section, we consider the recall secretary problem for 0 < f ( k ) <_ 1, k < N .

1 <

N-1 T h e o r e m 1. For the general problem with recall probabilities {q(r) }~=o, the opti- mal rule is not to select any applicant until all applicants have been interviewed if and only if

q(r + 1) > (N - 1)f(1) -4- f (2) - q(0)f(1) (3.1) q(r) ( N - 1)f(1)

Proof. The necessity is obvious. We prove the sufficiency.

By (3.1) we have for i < N - 1

E(~rN [F~_I)I~N_ = g _ l _ i > xN_,.,.

Assume tha t for k = N - 1 , . . . , n + 1

E ( ~ k + l I~ ,=k-~) > Xk,~.

We shall show tha t E(~,~+I I~n = n - i) > z , , i . (3.2)

By Lemma 3, we know tha t (3.2) is equivalent to

g - 1 7~+1 h(n, 1) h(N, 1) q(N - i) (k + i ) k > - - . ( 3 . 3 / n N q(n - i)

k = n

A SE CRE T ARY PROBLEM ON FUZZY SETS 331

By (3.2), we know that 7rn+i i ---- nNtl h(N, 1)q(N - i). It follows that

o

")'k+l = '~k+l + 7['k+1

N

= --:-- + 7rk+l

j=k+l #

T k + l

k + l

= k h ( N , 1 ) q ( N - k - 1 ) + l h ( N , 1) N

E q(N - j ) j = k + l

>_ k h(N, 1 ) q ( N - k - 1 ) + ~ - ~ h ( N , 1 ) q ( N - k - 1)

= h(N, 1)q(N - k - 1).

Then

( k + 1)k > h(N, 1)q(N - k - 1) - (k + 1)k

q(0) - 1)f(1) + f(2) - q(O)f(1)] N-k-1 >- k(k + 1 ) [ ( N (N- - 1--~1) -

q(0) [1 - ( Y - k - 1)(q(0)f(1) - f (2))] >-k(k + 1) ( N - 1)f(1) > q(0)f(1)- f(2) - ( k + 1 ) ( N - 1)f(1)

1 [1 q ( k + l - i ) l ->k + 1 q--~- ~ Tj

> q ( k - i ) , i)[1 q(k + l - i ) - (k + l) q t n - q ~ - - ~ ]

1 - ( k + 1 ) q ( n - i ) [ q ( k - i ) - q ( k + 1 - i ) ]

- q ( n - i) q(k - i) h(k + 1, 1) k + 1 q(k + 1 - i)]. (3.4)

From (3.2)-(3.4), we see that the result holds for k = n. []

N - - 1 "

T h e o r e m 2. For the general recall probabilities {q(r)}~= o , q(0) < 1, there exists a positive integer s such that the optimal ru/e is to reject the/irst s - 1 applicants. For

q(r + 1) < (N - 1)f(1) + f(2) - q f(1) (3.5) q(r) - (N - 1)f(1) '

let ~ - = i n f { n > l : q ( r + l ) < h(n, 1 ) - ~ n } (3.6)

- q(r) - n~_l(h(-n-~l, 1-)---~n+l) "

332 LI X I A O J I E A N D J IN Z H I M I N G

Then the optimal rule is to select the new relatively best applicant without stopping before stage ~'. I f the offer is refused, the employer should continue to make an offer to each present relatively best applicant. Here

~ n = q ( O ) ~ l h ( j + i , 1 ) I I k + l - q ( O ) j=n ( J + l ) 2 k=n k

(3.7)

Proof. Let N

{ -k } s= in f n = l : h(n, 1 ) - E -ff>O .

We see t h a t for n < s,

k = n + l

N n ( _ ~ )

E(" fn+ l [Fn) lw ,=n- i -Xn , i - n - + l " f i n + l , i - - h(n , 1 ) - E _ > 0 . k=n-4-1

Therefore no select ion should be m a d e pr ior to s tage s. To prove the second resul t , define r as in (3.6). By backward induc t ion , if for all T < k = n + 1 , . . . , N - 1,

7k,i < Xk,i, i < k, (3.8)

t hen

7r~k = (h(k, 1) - z~k)q(k - i),

+ zy,~+l _ q(O) - h(n + 1, 1) + 7r,~+1 "~n -- n + l n + l

n + 1 - q(0)~ " ) ' n + 1 �9

n + l

Hence

h(n, 1) - -~,~ - n +-~ ~wq(O) [h(n-+- l, 1) - ~n+l] 1

n + 1 [ h ( n + 1, 1) - h (n + 1,2)].

Therefore

h(n, 1) - 7 n

~ l ( h ( n + 1, 1) - "7n+1)

1 =1 - n + l [q(O)h(n + 1, 1) - h(n + 1, 2)] + 1-q(0)~ n + l / n + l

~-~1 (h(n + 1, 1) - 7=+1) (3.9)

) ) ~*/,=-/" On the o ther hand , it is easy to prove t h a t for all n _> r, {~+~ ( h ( n + l , 1)--~n+l N h(n+l,1) ~f l--q(O) ~ 1g are de- h(,~+l,2)}N=r are increasing, and { n+l }nN--r and t ,~+1 . n + l ~ n = r and { n+l

h(n,1)-~,~ "iN creasing in n, so t h a t { .--~(h(n+l,1)-#.+l) Jn=r is increas ing in n for all n > T. []

A S E C R E T A R Y P R O B L E M ON FUZZY SETS

T h e o r e m 3. Let q(O) = q < 1 and for every r >_ 1, q(r) = p < q.

(1) For every i < n < N - 1

= n - i ) >

333

Then we have

(3.10)

This implies that the optimal rule is to stop at the current relatively best applicant.

(2) There exists a positive integer s = s(N, p, q) such that the optimal rule is to reject the first s - 1 applicants. In fact

N - - 1 N - - 1 u - 1 n

s = i n f { n : I l a k f ( i ) q + E b u H a k > - ~ f ( i ) p }. (3.11) k ~ n 'tt~n k----n

(3) The optimal stopping rule is

t * = i n f { k _ > s : w k = 0 } . (3.12)

Proof. (1) We prove (3.10) by backward induct ion. For n = N - 1

N - l h (N , 1)p - (h(N - 1, 1) - l h ( N , 1)q)p - N

= N h ( N , 1)q(O)- l h ( N , 2)p >_ O.

Assume tha t (3.10) is t rue for k = N , N - 1 , . . . , n + 1 . T h e n we have r n + l = ~+Nlh(N, 1)p. Hence

n ) > _ _ - n + l

n

n + l

n r i - (h(n, 1) - ~ n ) p n + l n+l

(h(n+ 1,1) - ~ n + l ) p - (h(n, 1) - % ) p

1 (h(n + 1, 1) - 3'n+l)p - [ ~ ( h ( n + 1, 1) + ---:-:h(n + 1, 2) - "~n]p

n - t - I n t l

7rnw1 = n + l l ~ n + l p n + l l h ( n + l , 2 ) P + n + l

_ 1 1 l h ( n + l , 2 ) p + N h ( N , 1)p - n + l z/n+lp n +

> N h ( N , 1)p - N h ( N , 2)p > O.

(2) Let s = inf{n : (h(n, 1) - ;Yn)q > ~h(N , 1)p}. We assume tha t for k <_ s

(h(k, 1) --~k)q ~ k h ( N , 1)P. (3.13)

334 LI X I A O J I E A N D J IN Z H I M I N G

t h e n we see that

Then

(h(k - 1,1) --~k-1)q = [ h(k, 1) + h(k,2) k

=[~J-h(k,a)+�88 2 ) - l h ( N , X ) p - ~ k ] q

1 = [ ~ - - ~ ( h ( k , 1 ) - - ~ k ) + �88 2) - Nh(N, 1)p - ~k]q

k - 1 k - 1 <----~(h(k, 1 ) -~k)q~ ~ h ( N , 1)p.

On the other hand, for k > s, we assume that

(h(k, 1) -~k)q> kh(N, 1)p.

- - - - 5 k ] q

k + 1 k + l . (h(k + 1, 1) - ZYk+l)q -- > k (h(k, 1) - ~k)q > ----~-h(N, 1)p.

From the above, we note that there exists a unique s such that

k + l (h(k + 1, 1) - ~/k+l)q > -----~h(N, 1)p

if and only if k > s. Hence for n > s

N 7rk

7r,~=(h(n, 1 ) - E - ~ - - ) q = ( T r , ~ + l - - - 7rn+l q) - (h(n + 1,1) - h(n, 1))q n + l

k = n + l

= ( 1 q )Trn+l-(h(n+l ,1)-h(n, 1))q=anTrn+bn. n + l

Solving this difference equation, we have

N - 1 N - 1 u - 1

7rn= ( H ak)f(1)q+ E bu Hak" k = n U = n + l k = n

This yields (3.11).

(3) From the general theory of optimal stopping, we have

F = inf{n _> 1

(3.14)

(3.15)

o 1 0 } . : X . > % ~ } = i n f { n _ > l : X,~>"/~}=inf{n>s : an=

A SECRETARY PROBLEM ON FUZZY SETS 335

T h e o r e m 4. For q(r) = qpr, O < p < q < l, and O < r < N - l, let

{ h(n, 1) - 5n } T = i n f n > 1 : n > p . (3.16)

n'+l (h(n + 1, 1) - "~n) -

Then the optimal rule is to pass over the (T -- 1) applicants and to select "the relatively best applicant at stage v. I f the at tempt at stage T is unsuccessful, then the process is to be continued by selecting the relatively best applicant at the subsequent stage.

Proof. From the definit ion of 7 and ~ -- p, we have for n < T q(~)

h(n, 1) - ~,~ P > n~l (h(n + 1, 1) - ~n)" (3.17)

We only have to prove t h a t x,~,i < %~,i for n < T. In fact

k i - (h(k, 1) - ~k)q(k - i) E ( ' T k + l [ F k ) [ ~ k = k - i -- X k , i = k + 1 7rn+l

k >- k + 1 (h(k + 1, 1) - ;yk)q(k + 1 - i) - (h(k, 1) - ;yk)q(k - i) > O.

This implies (3.18). []

4. T h e O p t i m a l S t o p p i n g R e s u l t for f(k) -- - k

In this section, we consider the problem for f ( k ) = - k . We see t h a t

N + 1 h(n , j ) = - j - -

n + l (4.1)

From section 2, we have the opt imal s topping rule '

n i N + I t * - - i n f n : n + l 7'l'n+l "> ( - - n T i -

N 7rk

E - ~ ) q ( n - i ) } . k=nq-1

Obviously,

E('~NIFN-1)I~N_~=N-I-i-- XN_x,,

N : N - 1 - i) - l q ( N - i ) - [ N + N J N + 1 q(O) lq (N

>_NN + l q ( i - l - i) - N - 1 N - q(O) q(N _ l _ i) > O.

This implies tha t the sa t is fac tory applicant is sure to be employed before the N - t h applicant .

336 LI XIAOJIE AND JIN ZHIMING

5. E x a m p l e

Let f(1) = 1, f ( k ) = 0, and 2 _< k < N. Then

n x n = ~ - ~ [ - ~ q ( n - i) + (1 - q ( n - i ) ~ , i ) ] I ( ~ =,~_i).

i-~1

This is the secretary problem discussed in [9]. For more details on this model, see [4].

R e f e r e n c e s

[1] Choe, K.L. and Bai, D.S., A Secretary problem with backward solution and uncertain employ- ment, J. Appl. Probab., 20 (1983), 891-896.

[2] Chow, Y.S., Moriguti, S., Robbins, H., and Samuals, S.M., Optimal selection based on relative rank (the secretary problem), Israel J. Math., 2 (1964), 81-90.

[3] Chow, Y.S., Robbins, H., and Siegmund, D., Great Expectation: The Thoery of Optimal Stopping, Houghton Mifftin Company, Boston, 1971.

[4] Jin Zhiming and Li Xiaojie, A secretary problem with backward solicition, Journal of National University of Defence Technology, 14:1 (1992), 76-85.

[5] Li Xiaojie, The best choice problem on fuzzy sets, Preprint, 1988.

[6] Mucci, 'A.G., Differential equations and optimal choice problems, Ann. Statist., 1 (1973), 104- 113.

[7] Petruccelli, J.D., Best-choice problems involving uncertainly of selection and recall of observa- tions, J. Appl. Probab., 18 (1981), 415-425.

[8] Smith, M.H., A secretary problem with uncertain employment, J. Appl. Probab., 12 (1975), 620-624.

[9] Yang, M.C.K., Recognizing the maximum of a sequence based on relative rank with backward solicition, J. Appl. Probab., 11 (1974), 504-512.

China Securities Company Ltd., Beijing 100036.

National University of Defence Technology, Changsha 410073.