# A Note on two inventory models

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1 A Note on two inventory models J.Colpaert E.Omey Introduction In inventory models the major objective consists of minimizing the total inventory cost and to balance the economics of large orders or large production runs against the cost of holding inventory and the cost of going short. In the present paper we analyse the fluctuations in the stock and starting from some basic assumptions we obtain bounds between which the stock varies. The main purpose and use of our results is that we are able to determine the exact upper and lower stockbounds. In the paper we formulate a deterministic and a stochastic version of our model. 1 The deterministic model Consider an inventory process involving one item and suppose that the initial stock is equal to a, a positive real number. During regular time-intervals the stock decreases because of demand. We assume that the demand isin units of size b. When inventory is below the level b, the policy consists of ordering p = a + b new units. This type of inventory process has also been analyzed by Andres and Emmons (1975) and by Zoller (1977). Formally the model is the following. Let 0 <b<adenote arbitrary real numbers. We define a sequence {x n } of stock-levels as follows : x 1 = a; x n+1 = x n b if x n b and x n+1 = x n b + p = x n + a if x n <b. Research Fellows, K.U.Leuven Received by the editors July Communicated by M. Hallin. Bull. Belg. Math. Soc. 4 (1997),

2 404 J.Colpaert E.Omey We analyse the sequence {x n } and want to find sharp lower and upper bounds between which the sequence takes its values. Obviously we have to find an upper bound by examining those x-values greater than a. We shall denote by z n the x- value for which x i a for the n-th time. We can find an upperbound by studying {z n,n 1}; we can find a lower bound by studying {z n a, n 2}. Now suppose that x m a and that x m = z n,then-thtimethatwehaveavalue a. We determine x m+1,x m+2,..., x m+k and k such that x m+k = x m kb b and x m+k+1 = x m (k +1)b<b. Obviously k +1=[x m /b] with[x m /b] the integer part of x m /b. Using this value of k we find a new x-value a : z n+1 = x m+k+2 = x m (k +1)b + a = x m [x m /b]b + a. It follows that z 1 = a and that z n+1 = zn [z n /b]b + a, n 1. Now divide by b and let u n := z n /b so that u 1 = d = a/b and u n+1 = d + u n [u n ],n 1. A further simplification we get by replacing u n by y n = u n [d] : y 1 = c = d [d] andy n+1 = c + y n [y n ],n 1. In the next lemma we find a simple expression for y n : Lemma 1.1 For n 1:y n = nc [(n 1)c] =:f(n). Proof. By induction on n. We study the sequence {y n } and distinguish two cases. In case 1 we assume that c is rational. The second case is devoted to irrational c. 1.1 Case 1 : c is rational We suppose that c = p/q where p, q N, p < q and GCD(p, q) =1. Itiseasyto see that f(q + 1) = f(1) and consequently that f(n + q) = f(n). Hence {f(n)} is periodic with period not greater than q. Now suppose that {f(n)} is periodic with period r, i.e. f(r +1)=f(1). From this it follows that rc =[rc] is an integer. It follows that the period of {f(n)} equals q because q is the first integer r for which rc is a natural number. Summarizing our findings we have Lemma 1.2 a) If c = p/q with GCD(p, q) =1,then{f(n)} is periodic with period q. b) {f(n)} is periodic if and only if c is a rational number.

3 A Note on two inventory models 405 Under the conditions of Lemma 1.2 we can find max {f(n)} by looking at {f(1),f(2),..., f(q)}. Obviously all numbers f(i) (i = 1,..., q) are different - since otherwise the period of {f(n)} would be less than q. Moreover for each i we have c f(i) <c+1 and f(i) isoftheformf(i) =c + n i /q where 0 n i q 1. Hence {f(i),i = 1, 2,..., q 1} = {c + j/q,j =0,..., q 1} and hence max {f(i)} = c +(q 1)/q = (p + q 1)/q. For min{x n } we study min{z n a, n 2}. Introducing y n and f(n) as before, we find that min{f(i),i 2} = c + 1/q. Summarizing, we have proved the following result. Theorem 1.3 If c = p/q Q, withgcd(p, q) =1,then max{y n,n 1} =(p + q 1)/q =1+c 1/q. min{y n,n 2} =1/q. Returning to the starting point of our analysis, we obtain Corallary 1.4 Max{x n } = a + b GCD(a, b) and min{x n } = GCD(a, b). Proof. For the first part, use Lemma 1.3 and z n = b(y n +[d]), where d = a/b. For the second part, recall that lower bound can be found by considering the minimum of the values z n a. Remarks. 1. Consider the following extension of the previous model. Consider an inventory process involving m items. For each item j the inventory policy consists of the pair (b j,a j ) as in the previous model. If x n,j denotes the inventory level of item j at stage n, then the total inventory level equals I n = x n, x n,m. Corollary 1.4 states that m m GCD(a j,b j ) I n (a j + b j + GCD(a j,b j )). j=1 j=1 Whereas the bounds in Corollary 1.4 are sharp, the question rises whether the bounds for I n are sharp. 2. If we assume that the demand is not in units of b but in fractions of b (i.e. the demand is tb per time unit, where 1/t is an integer and t<1), then the sequence of inventory levels (with c = tb + a) is defined as follows : x 1 = a; x n+1 = x n tb if x n b and x n+1 = x n + a if x n <b. Among similar lines the analogue of Corollary 1.4 follows. Using u n = z n /tb and then y n = u n [a/tb] it follows that y 1 = c = a/tb [a/tb]andy n+1 = y n [y n ]+c+1 1/t. We obtain Corallary 1.5 For the sequence {x n } we have max{x n } = a +(2t 1)b GCD(a, tb) andmin{x n } =(t 1)b + GCD(a, tb). 3. The results of remark 2. can be generalized as in remark 1.

4 406 J.Colpaert E.Omey 1.2 Case2:cirrational In the case where c is a irrational number we prove the following result, cf. lemma 1.3. Lemma 1.6 For irrational c, 0 <c<1, there holds sup{nc [nc],n 1} =1 and inf{nc [nc],n 1} =0. Proof. If c is irrational we represent c as c = q q q or as c =(q 1,q 2,q 3,...), where q 1 =[c],q 2 =[1/(c q 1 )] > 0, etc. Using this notation, for each integer n we consider A n /B n =(q 1,q 2,..., q n ). Obviously A n and B n are integers with the following properties : (i) A 1 = q 1,B 1 =1,A 2 = q 1 q 2 +1,B 2 = q 2 ; (ii) for n>1:a n+1 = A n q n+1 + A n 1 and B n+1 = B n q n+1 + B n 1 ; (iii) using (ii) :A n B n+1 A n+1 B n =( 1) n+1 ; (iv) {B n } is strictly increasing with limit +. Now let a =(q n+1,q n+2,...) sothata>0andc =(q 1,..., q n,a). From (ii) it follows that (v) c =(A n a + A n 1 )/(B n a + B n 1 ). A combination of (iii) and (v) yields (vi) A 2n /B 2n <c<a 2n+1 /B 2n+1. Using (v) and (vi) we obtain 0 <B 2n c A 2n < 1/B 2n+1. By using (iv), for each β>0 we can find N sufficiently large so that 0 <B 2n c A 2n <β, for all n N. Since A 2n is an integer, it follows that 0 B 2n c [B 2n c] <β. Since B 2n is an integer also, it follows that inf{nc [nc],n 1} =0. In a similar way from (v) and (vi) we obtain that 0 <A 2n+1 B 2n+1 c<1/b 2n. For arbitrary β>0it follows that 0 1 (B 2n+1 c [B 2n+1 c]) <β, for all n N. Hence the other statement of the lemma follows. For the original sequence {x n } we have the following analogue of Corollary 1.4. Theorem 1.7 If a/b is irrational, then sup{x n } = a + b and inf {x n } =0.

5 A Note on two inventory models The Stochastic model Suppose a, b and c(a >b) are positive real numbers and suppose that X, X 1,X 2,... are i.i.d. positive random variables (r.v.) with distribution function (d.f.) F (x). A new sequence Y 0,Y 1,.., Y n of r.v. is constructed as follows. Let Y 0 = a and for n 0, let Y n+1 = Y n X n+1 if Y n b and Y n+1 = Y n X n+1 + c if Y n <b. The interpretation of Y n is clear : we start with an initial inventory (a) and demand is stochastic. When the inventory at time n is below level b the policy consists of ordering a fixed number c of new units during the next time interval. Example 1. A waterreservoir is initially at level a; from time to time a random amount X i of water is demanded. Once the level is below b, anamountofwatera is added to the reservoir. Example 2. A firm has an amount of cash a and at certain times bills have to be paid and the cash decreases with an amount X i. Once below a level b, thecashis increased by an amount c. For this sequence relevant questions are : what is the maximal value? what is the smallest value? What is the mean re-order time? Let S 0 = 0 and for each n 1, let S n = X X n. Clearly we have Y 0 = a. For n =0, 1, 2,... we have Y n = a S n as long as S n 1 a b. The first time S n 1 >a b we have Y n = a S n + c. The corresponding value of n is related to the the renewal counting process for the process S n. As usual we define N(x), the renewal process, as follows : N(x) =n if and only if S n 1 x<s n. It follows that S n 1 a b<s n iff N(a b) =n. At the first renewal the inventory equals m 1 = Y N(a b) = a S N(a b) and at the next time point the inventory is M 1 = Y N(a b)+1 = a + c S N(a b)+1. From this time on, the process starts again and a new renewal occurs at step N(a + c b). Then we have m 2 = Y N(a+c b) = a+c S N(a+c b) and M 2 = a+2c S N(a+c b)+1. Continuing this way we find that at step k we have mk = a + kc S N(a+kc b) and M k = a +(k +1)c S N(a+kc b)+1.

6 408 J.Colpaert E.Omey In the following discussion we will concentrate on m k. We can do this because the event {N(x) =n} does not depend on X n+1, and this implies that the distributins of m k and M k are related by the following obvious relation : (1) Mk = d = mk + c X. Remark. If the r.v. X is degenerate and takes only the value b, thens n = nb and N(x) =[x/b]+1. Inthiscasewehavem k = a + kc N(a + kc b)b. Ifc = a + b, this leads to m k = a(k +1) [(k +1)a/b]b, a relation which was treated in section 1. In order to find the mean values of m k (and M k ) we use the renewal function. The renewal function H(x) is defined as the mean of N(x) : H(x) = E(N(x)). It is well-known (see e.g. Feller (1971) or Ross (1970)) that H(x) = n=0 F n (x) where * denotes convolution. Now Wald s identity states that E(S N(x) ) = E(X)H(x) =µh(x) whenever E(X) =µ is finite. Using Wald s identity and (1), we find Lemma 2.1 If E(X) =µ is finite, then E(m k )=a + kc µh(a + kc b) and E(M k )=a +(k +1)c µ(h(a + kc b)+1). The asymptotic behaviour of H(x) is well-known and can be used to determine the asymptotic behavior of E(m k )ande(m k )ask a tends to infinity. Lemma 2.2 (a) Suppose that F (x) is not lattice and that Var(X) <. Thenask tends to infinity, lim E(m k )=b E(X 2 )/2µ and lim E(M k )=b+c µ E(X 2 )/2µ. (b) If F is lattice with GCD{n P {X = n} > 0} = 1 and if Var(X) <, thenask tends to infinity, E(m k )=b E(X 2 )/2µ +(a + kc b [a + kc b] 1/2) + o(1). Proof. (a) Since F is not lattice and X has finite variance, we have (Feller, 1971, IX.3) H(x) =x/µ + E(X 2 )/2µ 2 + o(1). The result (a) now follows from Lemma 2.1. (b) For lattice distributions with GCD {n P {X = n} > 0} = 1 and with finite variance, it follows from Feller (1970, XIII,3 Theorem 3 and eq.(12.8),p.341) that H(x) =[x]/µ +(EX 2 + µ)/2µ 2 + o(1). Using Lemma 2.1 it follows that E(m k )=a + kc µ{(e(x 2 )+µ)/2µ 2 +[a + kc b]/µ + o(1) where the o(1)-term tends to 0. Hence result (b) follows. In the next result we determine the d.f. of m k and M k. To this end let Y (t) =S N(t) t denote the excess or residual life at time t. Clearly m k = b Y (a + kc b). For x b it follows that (2) P {m k <x} = P {Y (a + kc b) >b x} =1 P {Y (a + kc b) b x}.

7 A Note on two inventory models 409 Now the distribution function of Y (t) is known (see e.g. Ross (1970), p.44): (3) P {Y (t) x} =1 t 0 [1F (t xy)]dh(y), x,t 0. Furthermore, if F is not lattice and X has a finite mean µ, then the limit distribution of Y (t) isknown: x (4) lim P {Y (t) x} = 1 [1 F (y)]dy =: m(x). t µ 0 In the next result we obtain the limit distributions of m k and M k. Recall that f g denotes the (Stieltjes-)convolution of the two measures f and g. Lemma 2.3 If F (x) is not lattice and X has a finite mean µ, then: (a) lim P {m k <x} =1 m(b x), k x b and (b) lim P {M k <x} =1 m F (b x + c), k x b + c. Proof. Result (a) follows from (2) and (4). Result (b) follows from (a) and relation (1). Remarks. 1. It follows from the result that lim P {m k < 0} = 1 m(b) and lim P {M k <b} =1 m F (c). When b and c are sufficiently large, then these probabilities can be made as small as desired. 2. If X is concentrated in the interval [0,b ], then 1 m(b) =0foreachb>b. 3. Suppose X has the exponential distribution F (x) = 1 exp( x/µ) for x > 0. In this case H(x) =F (0) + x/µ and the d.f. of Y (t) equalsf (x). For this special case we have P (m k x) =exp( (b x)/µ) (x b) ande(m k )=b µ. 4. If E(X) is not finite, then the previous results do not apply. In this case instead of limit laws for Y (t) one can consider limits for Y (t)/t, see e.g. Bingham et.al. (1987, p ). Now only limit results for m k /k are available. Finally we consider the rate of convergence in Lemma 2.3. To this end for u, v 0 we define R(u, v) = H(u + v) H(u) v/µ. Using (3) and the definition of m(x) and R(u, v) it is straightforward to prove the following identity : (5) P {Y (t) x} m(x) = x z=0 R(t + x z,z)df (x)+r(t, x)(1 F (x)). For non-lattice d.f. F (x) and for fixed x, R(t, x) tends to zero as t tends to infinity. The rate at which this occurs determines the rate of convergence in Lemma 2.3. In order to formulate the next result, recall that a measurable positive function is regularly varying at infinity and with index α (notation RV α )ifff(tx)/f(t) tends to x α for each x>0. From Frenk (1983) we recall the following result. Lemma 2.4 (Frenk (1983) Theorem 4.1.9) Suppose that X has an absolute continuous d.f. F (x) and that 1 F (x) is RV α with α>1. Then as t,r(t, x) (1 m(t))/µ locally uniformly (l.u.) in x.

8 410 J.Colpaert E.Omey Under the conditions of Lemma 2.4, 1 m(t) isrv 1 α and 1 m(t + x) 1 m(t) l.u. in x. It follows from (5), Lemma 2.4 and Lebesgue s theorem that as t, (6) (P {Y (t) x} m(x))/(1 m(t)) F (x)/µ +(1 F (x))/µ =1/µ, l.u. in x. Specializing to m k and combining (2) and (6) we obtain the following result. similar result holds for M k. A Lemma 2.5 Under the conditions of lemma 2.4, for x b as k we have P {m k <x} =1 m(b x) (1/µ + o(1))c 1 α (1 m(k)). Acknowledgement. We thank the referee for his comments on an earlier version of the paper. References [1] Bingham, N.H., Goldie, C.M. and Teugels,J.L., Regular Variation, Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge [2] Andres, F. and Emmons, H., A multiproduct Inventory System with Interactive Setup Costs, Management Science, Vol.21, Nr. 9, pp , [3] Feller, W., An Introduction to Probability Theory and Its Applications, Vol.I, Wiley, New York [4] Feller, W., An Introduction to Probability Theory and Its Applications, Vol.II, Wiley, New York [5] Frenk, J.B.G., On Renewal Theory, Banach Algebras and Functions of Bounded Increase, Ph.D.Thesis, Math. Centre Amsterdam, [6] Ross, S.M., Applied Probability Models with Optimization Applications, Holden- Day, San Francisco, [7] Zoller, K., Deterministic Multi-Item Inventory Systems with Limited Capacity, Management Science, Vol. 24, Nr. 4, pp , EHSAL, Stormstraat 2, 1000 Brussels-Belgium

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