A Closed-Form Approximation Solution for an Inventory Model with Supply
                     Disruptions and Non-ZIO Reord...
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receiver. In Tomlin [2005], two supply sources exist, one
cheap and unreliable and one expensive and reliable. The above
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if r** ≥ 0. If r** < 0 then the minimum occurs at (q*(0),0),                                                        D2
i.e...
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production and inventory textbooks. These parameter sets are
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A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and a Non-ZIO Reorder Policy
A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and a Non-ZIO Reorder Policy
A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and a Non-ZIO Reorder Policy
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A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and a Non-ZIO Reorder Policy

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A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and a Non-ZIO Reorder Policy

  1. 1. A Closed-Form Approximation Solution for an Inventory Model with Supply Disruptions and Non-ZIO Reorder Policy David I. HEIMANN MSIS Department, University of Massachusetts Boston Boston, MA, 02125, USA and Frenck WAAGE MSIS Department, University of Massachusetts Boston Boston, MA, 02125, USA ABSTRACT When the purchaser’s inventory reaches the re-order quantity r, then the purchaser will order q units from the supplier. The In supply chains, domestic and global, a producer must decide demand is D units per time period, the ordering cost is K per on an optimal quantity of items to order from suppliers and at order, and the inventory holding cost is h per unit per time what inventory level to place this order (the EOQ problem). period. If the supplier is in the “ON” state, the q units are We discuss how to modify the EOQ in the face of failures and instantly shipped by the supplier and received by the purchaser, recoveries by the supplier. This is the EOQ with disruption with the inventory rising to q + r units. problem (EOQD). The supplier makes transitions between being capable and not being capable of filling an order in a If the supplier is in the “OFF” state when the purchaser’s Markov failure and recovery process. The producer adjusts the inventory reaches the re-order quantity r, no orders can be reorder point and the inventories to provide a margin of safety. placed and the purchaser has to wait till the supplier returns to the “ON” state. When the supplier returns to the “ON” state, Numerical solutions to the EOQD problem have been the purchaser’s inventory level will have been reduced to s′ ≤ r developed. In addition, a closed-form approximate solution has through sales. If the purchaser’s inventory level is reduced to 0 been developed for the zero inventory option (ZIO), where the and further demands occur, a stock-out penalty of π dollars per inventory level on reordering is set to be zero. This paper unit of unmet demand is incurred. When the supplier returns to develops a closed-form approximate solution for the EOQD the “ON” state, the purchaser orders a quantity adequate to problem when the reorder point can be non-zero, obtaining for bring inventories from s’ up to S′ = q + r units. An (s′, S′) that situation an optimal reorder quantity and optimal reorder policy is therefore used when the supplier returns to the “ON” point that represents an improvement on the optimal ZIO state following a stay in the “OFF” state. No back orders are solution. The paper also supplies numerical examples allowed, so s′ and S′ are both non-negative. demonstrating the cost savings against the ZIO situation, as well as the accuracy of the approximation technique. All of this introduces a new trade-off not considered in the traditional EOQ model. If the purchaser decides on a small q Keywords: Supply Chain Management, Reliability, and a small r, they risk stock-outs and lost sales. If the Systemics, Operations Management, Optimization purchaser attempts to avoid these stock-out costs by increasing the re-order point r or by ordering a larger q, they will increase their inventories and thus their inventory carrying costs. The purchaser must trade off between both q and r to identify a cost 1. INTRODUCTION minimizing policy (q*, r*). The fundamental EOQ model assumes that the supplier is A sample pattern of the rise and decline in inventory levels perfectly reliable so that supply disruptions will never occur. over time is shown in Figure 1. Parlar and Perry [1996] presented in their paper an extension of this model to take supply disruptions into account. EOQ with disruptions has come to be labeled EOQD in the literature. 2. BUILDING BLOCKS Their model assumes that the supplier will, at any moment in time, be either in an “ON” state from which they will deliver an The Unitary-Demand Numerical Cost Model: order instantaneously the moment the order is received, or in an Parlar et al [1996] assumed that the demand D ≡ 1. With this “OFF” state from which they incapable of delivering anything assumption, they show that the expected cost of placing orders, at all. The “ON” time is exponentially distributed at a rate λ. carrying inventories and incurring outage losses is given by The “OFF” time is exponentially distributed with a rate µ. A transition into the “OFF” state means a disruption of the supply ( K + hq 2 / 2 + hqr ) + β 0C10 ( r ) , (1) g 0 (q , r ) = chain. β q + µ0 SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4 1
  2. 2. ( K + hq 2 / 2 D ) + β 0 Dπ / µ (2) g 0 (q) = q D +β µ β 0 = β (1 − e − ( λ + µ )( q / D ) ) ( K + hq 2 / 2 D ) + βDπ / µ (3) g (q) = D q +βµ Our Approach: In this paper, we start with the cost equation (2) and expand it to address the general case where r ≥ 0, i.e., the non-ZIO reorder policy, obtaining a two-variable function g0(q, r). We then obtain the corresponding version of equation (3), denoting the function by g(q, r), and from it Where β0 is the probability that a supplier is in the “OFF” state develop cost-minimizing formulas in closed form. Specifically, when the purchaser’s inventories reach the re-order point r, i.e., we develop the following: In Section 4: Theorem 1 develops the closed form formula β 0 = β 0 (q) = β (1 − e − ( λ + µ ) q ) which calculates the cost-minimizing order quantity q* for a β is the fraction of time that the supplier is in the “OFF” state, given r ≥ 0. i.e., In Section 5: Theorem 2 develops the closed form formula which calculates the cost-minimizing re-order point r* for a λ β= given q ≥ 0. λ+µ In Section 6: Theorem 3 develops the closed form formula that C(r) is the expected cost incurred from the time when the calculates the unique global cost-optimal point (q**,r**) inventory reaches r and the state is “OFF” to the beginning of In Section 7: Theorem 4 proves that the approximation g(q, r) the next cycle when the supplier has transitioned into the “ON” is an upper bound for the exact value g0(q, r), up to a critical state. They demonstrated that ˆ value r = r , above which the reverse is true. In Section 8: Theorem 5 proves that the goodness of the C (r ) = 1 µ2 (h( µr − 1) + e − µr (πµ + h)) . relative approximation of g to g0 is bounded above by the They then found the cost-minimizing pair (q, r), solving their relative approximation between β and β0 model by numerical methods. In Section 9: Theorem 6 proves that if q0 is the value of q that minimizes the exact cost function g0(q,r) for a given value of r, The basic EOQ model (without supply disruptions) provides and q* is the value of q which minimizes the approximating solutions in closed form, explicitly giving formulas for the ˆ function g(q,r), then q* > q0 for r up to r , the critical value of optimal order quantity q and the re-order quantity r, thereby r identified in Section 7, above which the reverse is true. showing the interrelations among q and r, the average cost, and In Section 10: We conclude with a set of application example the model parameters. However, for the EOQD model (with drawn from the literature and used in Snyder [2005], as well as supply disruptions), closed form solutions that calculate the a replication of the numerical example in Parlar et al [1996]. minimum of the exact cost function (1) have not been developed in the literature, including Parlar et al [1996], with numerical solutions being provided instead. However, closed- 3. LITERATURE REVIEW form solutions for the cost minimum as a function of the variables for q and r and the model parameters would have Early papers to address the effect of disruption include Meyer, important and widely appreciated benefits, including the Rothkopf, and Smith [1979]. Chao [1987] discusses dynamic following: programming to address this problem for electrical utility Explicitly showing how values of the model’s parameters and companies facing possible market disruptions. Groenevelt, the independent variables q and r combine to generate the Pintelon, and Seidmann [1992] discuss a deterministic dependent variable of optimal cost, thereby providing a strong economic lot-sizing problem in view of machine breakdowns contextual explanation for the EOQD dynamics. and corrective maintenance. Facilitating parametric analysis and sensitivity analysis for the optimal solution. Parlar and Berkin [1991], corrected by Berk and Arreola-Risa Exploring the model’s asymptotic behavior for extreme values [1994], begin a number of articles that incorporate disruptions of the various parameters. into classical inventory models. Parlar et al [1995], mentioned Providing building blocks for richer and more complex models previously, extends these results. Parlar and Perry [1996] extend the results in Parlar et al [1995] to the situation with two The ZIO Closed-Form Cost Model: Snyder [2005] suppliers and provide direction for analysis with more than two allowed the demand D to take on other values than 1, but suppliers. Gürler and Parlar [1997] allow more general failure restricted the reorder point r to 0 in (4), obtaining the exact cost and repair processes in the two-supplier model, presenting function (2). Since r = 0 this is known as the zero-inventory asymptotic results for large order quantities. ordering (ZIO) policy. Equation (2) was then approximated by replacing β0 with β, obtaining equation (3), and closed-form Arreola-Risa and DeCroix [1998], as well as Moinzadeh and solutions of (3) were then obtained. Aggarwal [1997] consider (s,S) models with supplier disruptions. Song and Zipkin [1996] consider a situation where the availability of the supplier is partially known to the 2 SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4
  3. 3. receiver. In Tomlin [2005], two supply sources exist, one cheap and unreliable and one expensive and reliable. The above β 0 = β (1 − e − ( λ + µ )( q / D ) ) (8) papers (except for Tomlin [2005]) discuss solving the models using a numerical approach rather than in closed form, due to As was done in Snyder [2005] we approximate the function the complexity of the model equations. g0(q,r) by the function g(q,r), where β0 is replaced by β as follows As mentioned previously, Snyder [2005] developed an approximate closed-form model for the ZIO situation with a ( K + hq 2 /( 2 D ) + hqr ) + βDC (r ) (9) single supplier, where the demand is allowed to take on values g ( q, r ) = q β other than unity. The paper develops bounds on the + approximation error and addresses a power-of-two ordering D µ policy. Note that if r = 0, i.e., the ZIO policy, then C(r) = π/µ and equations (6) and (9) reduce to the equations for g0(q) and g(q) in Snyder [2005]. Also, if D = 1, equation (6) reduces to the 4. OPTIMAL REORDER QUANTITY Q* FOR GIVEN equation for g0(q,r) in Parlar et al [1995]. In addition, if λ = 0, REORDER POINT R i.e., no disruptions occur, then β = β0 = 0 and both equations reduce to the standard EOQ formulation. The average cost objective function is derived from Parlar et al [1995] as We shall impose three reasonable assumptions on the problem parameters. First, we assume that all costs and other problem parameters including q and r are non-negative. Second, we ( K + HQ 2 / 2 + HQR ) + β 0C ( R ) , (4) G0 (Q, R ) = β assume that λ < µ, i.e., the supplier “ON” states last longer than Q + Μ0 the “OFF” states. where Third, we assume that 2 KhD + hrD < πD . If there β 0 = β (1 − e − ( Λ +Μ )Q ) were no disruption, the model would reduce to the classical EOQ model with a reserve inventory of r, whose optimal Λ β= annual cost is 2 KhD + hrD . This optimal annual cost is Λ+Μ therefore a lower bound on the optimal cost of the system with and disruptions. A feasible solution for the system with disruptions C ( R) = 1 Μ2 ( H (Μ R − 1) + e −ΜR (πΜ + H )) . is never to place an order and maintain the reserve of r, instead stocking out on every demand; the annual cost of this strategy is πD. Since a lower bound for the optimal cost is clearly a In Parlar et al [1995], the demand is set as D ≡ 1, so the values lower bound for one of the feasible costs, this establishes the H, Λ, and Μ are set in terms of demand time, e.g., M = 0.2 assumption. means that an average of 0.2 recoveries occur in the period of With these parameters, formulations, and assumptions, we time it takes for one unit of demand to occur, which, if the address the question of, for a given reorder point r ≥ 0, what demand is D = 100 per year, is equivalent to 20 recoveries per quantity q*(r) the purchaser should order so as to minimize the year. Let h, λ, and µ be the equivalent rates for H, Λ, and Μ set average cost. The closed-form result is given in Theorem 1 in terms of calendar time, so that h = HD, λ = ΛD, and µ = MD. below: Then equation (4) becomes ( K + D Q 2 / 2 + D QR ) + β 0C ( R ) h h (5) Theorem 1: The optimal reorder value q*(r) for a given value G0 (Q, R ) = β0 Q+ µ/D of r ≥ 0, i.e., the value of q where g(q,r) is a minimum, is given by where, µ − R − β Dh + ( β Dh) 2 + 2h( KD + β D 2 (−h + e − µr (πµ + h))) C ( R) = µ 1 2 D 2 ( D ( D R − 1) + e h µ D (π µ D + D )) h q * (r ) = µ µ µ2 (10) h for all Let q = Q and r = R/D. Then (5) becomes   2 Kµ h ( K + q 2 / 2 + hqr ) + β 0C ( rD ) ~ = − 1 ln h − β r<r  (11) G0 ( q, rD ) = D D µ  h + πµ  q β0   + D µ If the argument of the “ln” function in (11) is negative, then Let g0(q,r) = G0(Q,rD)/D. Then (10) is true for all non-negative values of r. Remark: If λ = 0 then β = 0 and (10) reduces to the standard 2 (6) EOQ result. If r = 0 then (10) reduces to the result shown in ( K + hq /( 2 D ) + hqr ) + β 0 DC ( r )) Snyder [2005]. g 0 (q , r ) = q β + 0 D µ Proof: To find the optimal reorder value q*(r) for a value of r where we differentiate (9) with respect to q and equate to zero. Differentiating: C (r ) = C (rD) / D = 1 µ2 (h( µr − 1) + e − µr (πµ + h)) (7) and SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4 3
  4. 4. q β 1 1 1 2 5. OPTIMAL REORDER POINT R* FOR A GIVEN ∂g ( D + µ )(D hq+ hr) − ( D )(K + 2 hq / D + hqr+ DβC(r)) (12) = ORDER QUANTITY Q ∂q ( D + β )2 q µ We next address the question of, for a given order quantity q ≥ 0, at what inventory point r* the purchaser should reorder so as to minimize the average cost. The closed-form result is given 1 2 hq2 / D2 + ( µ 1 1 D βh)q − ( K + µβ (−h + e−µr (πµ + h))) D 2 in Theorem 2 below: = q β 2 ( + ) D µ Theorem 2: For a given reorder quantity q ≥ 0, the reorder inventory point r*(q) that minimizes the cost function g(q,r) is Equating to zero and using the quadratic formula, the value for given by q* becomes  h( D ( µ ) + 1)  q β q * (r ) = − β µ Dh ± ( β Dh ) 2 + 2 h( KD + µ β µ2 D 2 ( − h + e − µr (πµ + h ))) . (13) r * (q ) = − µ ln 1  (17) h  πµ + h    If q > πβD/h, then r*(q) = 0. Since q* must be nonnegative, the “±” in (10) is actually “+”. In order for q*(r) to be real and positive, we must have that Proof: We differentiate g with respect to r and equate to zero, so that KD + β µ2 D 2 (− h + e − µr (πµ + h)) > 0 , (14) hq + βD (h − e − µr (πµ + h)) ∂g µ = . (18) Solving the above for the quantity e − µr , condition (14) ∂r q + β D µ becomes 2 Equating the numerator to zero and solving for r we obtain h − Kµβ e − µr > D , (15)  h( D ( β ) + 1)  q µ h + πµ r * (q) = − µ ln  1  Kµ 2  πµ + h    If h− Dβ ≤ 0, then (15) holds for all r, so that q*(r) is Kµ 2 Differentiating (18) with respect to r we obtain the second positive for all r. If h− Dβ > 0, derivative as then (15) holds for ∂ 2 g βDe − µr (πµ + h) h− Kµ 2  = , (19) r < ~ = − ln   ∂r 2 q+ β 1 Dβ r µ µ  h + πµ    which is positive. Therefore r* minimizes the cost function The numerator of the right-hand side of (15) is clearly less than g(q,r). ~ the denominator. Therefore, r is positive and so q*(r) is ∂g If q > πβD/h, then > 0 for all nonnegative r, so that ~ positive for all values of r from zero to r . ∂r r*(q) = 0. To establish that the value of q*(r) defined in (10) is in fact a minimum, we need to establish that the second derivative of QED g(q,r) with respect to q is positive. Differentiating (12) with respect to q, we have that β 2 2β ∂2g (µ ) h + (2 K / D 2 + Dµ 2 (− h + e − µr (πµ + h))) 6. GLOBAL MINIMUM COST POINT (Q**,R**) = D . (16) ∂q 2 q ( + ) D β 3 µ The previous sections addressed optimal values of q for specified values of r, and vice versa. The manager for the From the discussion above, the numerator of (16) is positive for purchaser may instead wish to fully optimize the average cost all r where Theorem 1 holds so that q*(r) is indeed a minimum. by determining and using the pair of values (q**,r**) that QED provides the global optimum. We now address that question, i.e., the optimal solution for the overall EOQD problem. The Remark The value of having a closed-form formula for q*(r) closed-form result is given in Theorem 3 below: can be seen by noting the effect on q*(r) of small values of h. From inspection of (13) we have that Theorem 3: The global minimum for the EOQD function g(q,r) occurs at q * ( r ) = −( β D / µ ) + [( β D / µ ) + O (1 / h)] , D2 q ** = D µ (1 − β ) + 2 KD h + µ2 (1 − β ) 2 showing not just that q*(r) increases as h → 0 but also showing and the asymptotic properties of this increase. 1   2  r ** = − ln  h 1 + 2 Kµ + (1 − β ) 2  µ  β (πµ + h)   Dh    4 SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4
  5. 5. if r** ≥ 0. If r** < 0 then the minimum occurs at (q*(0),0), D2 i.e., the ZIO policy (r = 0) is optimal. q ** = D µ (1 − β ) + 2 KD h + µ2 (1 − β ) 2 (23) Proof: To determine the global minimum for g(q,r) we need to Substituting the expressions for Y and Z into (21) we have ∂g ∂g h set the two partial derivatives and , given by µ q ∂q ∂r s= (1 + β D ). πµ + h equations (12) and (18) respectively, to zero and solve simultaneously for q and r. From (23), therefore, The numerators for (12) and (18) are quadratic in q and linear in e-µr. Let s = e-µr. Then equation (10), which sets (12) to  2  zero, can be expressed as s ** = h 1 + 2 Kµ + (1 − β ) 2  , q = V + W + Xs , (20) β (πµ + h)   Dh   where and β V =− D, r * * = − µ ln s * * . 1 (24) µ 2 β  2 hβ Since s** is positive, r** is real. W =  D  + ( KD − 2 D 2 ) , µ  ∂2g ∂2g   h µ From equations (16) and (19) we have that and 2 β 2 ∂q 2 ∂r 2 and X = D (πµ + h) , are nonnegative. In addition, we have that h µ2 2 2 − µr ∂ g ∂ g ( Dβ / µ )e (πµ + h) = = , which is Furthermore equation (17), which sets (18) to zero, can be ∂q∂r ∂r∂q ((q / D) + ( β / µ )) 2 expressed as clearly nonnegative, so that equations (23) and (24) do yield a s = Y + Zq , (21) minimum for g(q,r). where If s** > 1, implying a negative value for r**, then consider the h function where r*(q) is defined in (17). Y= , and πµ + h Then 1  h  Z=   ~ Dβ / µ  πµ + h    dg ∂g ∂g dr * = + (25) dq ∂q ∂r dq Solving (20) for s in terms of q we obtain From (17) (q − V ) 2 − W s= dr * X = −1 /( µD( D + β )) q µ (26) dq Substituting into (21) we obtain Substituting (12), (18), and (26) into (25) q 2 − (2V + XZ )q − (W + XY − V 2 ) = 0 ~ 1 2 2 dg 2 hq / D − µD (1 − β )q − ( K / D) h Using the quadratic formula, we have that = (27) dq ( D + β )2 q µ q ** = 1 2 {(2V + XZ ) ± (2V + XZ ) 2 + 4(W + XY − V 2 ) } From equation (23), q** is a root of the numerator of (27), so ~ dg that =0 at q = q**. Let q^ be defined such that Note that V + XY − V 2 = 2K h D > 0. Therefore q** is dq real and positive and its value is governed by the “+” part of dr * the ± sign. After some algebra, r*(q^) = 0, i.e., q^ = πβD /h > 0. Then since <0 from q * * = V + ( 1 XZ ) + 2V ( 1 XZ ) + ( 1 XZ ) 2 + W + XY (22) dq 2 2 2 equation (26) and r*(q**) = r** < 0, we have that q^ < q**. 2 2 From inspection of (27) we have that the two roots of (27) are Taking ½ XZ = D/µ and XY = 2βD / µ and substituting into ~ (22) dg q** and a negative value, with <0 for q in between dq SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4 5
  6. 6. ~ dg  h + µ 2 Kh / D  ) those values, and therefore <0 for all 0 < q < q^ (note for r < r = − 1 µ ln   and  dq  πµ + h  that this is the interval in which q > 0 and r*(q) > 0), so that the ~ minimum of g in this interval occurs at q = q^, or µC (r ) − hr < 2 Kh / D (31) equivalently at r = r*(q^) = 0. ) QED r>r, for with both (30) and (31) being equations when ) r=r. Remark: Note that q** is independent of the stockout cost π. The purchaser’s optimal order quantity therefore does not ) ~ ~ Remark: Note that r < r , where r is given in Theorem 1, depend on the stockout cost (though the reorder point does ) so that Theorem 1 applies for all r < r . Note also that since depend on it) by assumption λ < µ so that β < ½, it also follows that ) r * * < r , where r** is given Theorem 3. Therefore it is the 7. ABSOLUTE DEGREE OF APPROXIMATION first parts of Lemma 3 and the other lemmas below including Theorem 4 (as well as the second part of Theorem 5) that apply Having established the minimizing properties of the at the global minimum r = r** (and q = q**). approximate cost function g(q,r), we wish to examine its relationship to the exact cost function g0(q,r). In this section Proof: From equation (7), and using some algebra, we have we establish that g is an upper bound for g0 for all values of r that below a specified critical value that is greater than the global µC (r ) − hr = − µ + e − µr (π + µ ) h h (32) optimum r**. Therefore, using q and r based on g(q,r) as against g0(q,r), and especially using the globally-optimizing values (q**,r**) for g(q,r), will not result in cost “surprises” The right-hand side of (32) is when r = 0 , equal to π, which by due to cost underestimates. assumption is greater than 2 Kh / D , a decreasing We prove three lemmas, followed by the overall theorem (for function of r, when r = ∞, equal to − µ h , which is less brevity in each lemma and the theorem, the value q*(r) is denoted by q*): than 2 Kh / D . ) Lemma 1: There is therefore a unique value r > 0 where (30) and (31) q *2 = 2D ( Kµ + µ ( Dβ (e − µr (πµ + h) − h)) − βhq*) (28) 1 ) hµ are equalities when r = r , where (30) holds for all Proof: Equate the numerator of (12) to zero and solve for ½hq* ) ) r < r ,and where (31) holds for all r > r . Equating (32) to and then q*. ) 2 Kh / D and solving for r we obtain r as QED )  h + µ 2 Kh / D  r = − µ ln 1    Lemma 2:  πµ + h  g(q*,r) = h(q*+Dr) (29) QED Proof: Substituting (28) into (9) Lemma 4: K+ 1 2D h( 2 D ( Kµ + Kµ 1 µ ( D β (e − µr (πµ + h) − h)) − βhq*)) + hq * r + βDC 10 ( r ) g ( q*, r ) = q* β 1 2 − µr D + µ 2 KhD + hrD < g (q*, r ) < DµC (r ) (33) 2 ( Kµ D + µ β D ( e (πµ + h) − h) − Dβhq*) + Dβhq * + hrD( µq * + βD ) = ) q * µ + βD for r < r , and hµq *2 + Dβhq * + hrD( µq * + βD) 2 KhD + hrD > g (q*, r ) > DµC (r ) (34) = (from Lemma 1) q * µ + βD ) ) for r > r . Both (33) and (34) are equations when r = r hq * (q * µ + Dβ ) + hrD(q * µ + βD) ) = Proof: For r < r , 2 Kh / D + hr < µC ( r ) by Lemma q * µ + βD 3. Substituting the definition of C(r}and carrying out some algebra, we have = h(q * + Dr ) µ 2 Kh / D < − h + e − µr (πµ + h) QED Lemma 3: Multiplying both sides by 2 βh, adding 2Khµ2/D+ (βh)2 to both sides, completing the square on the left-hand side and µC (r ) − hr > 2 Kh / D , (30) rearranging terms on the right-hand side, 6 SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4
  7. 7. ) Ifr < r , we need to establish that (µ ) 2 2 Kh / D + βh < ( βh) 2 + 2hµ[ Kµ / D + µ (−h + e − µr (πµ + h))] . β q * DµC (r ) − ( KD + 1 hq *2 + hq * rD ) > 0 2 Further algebra leads to This is equivalent to ( 2KhD + hrD < − µ Dh + ( µ Dh) 2 + 2h[ KD + β β β µ2 ) (−h + e − µr (πµ + h))] + hrD h 2 q *2 −2q * hY + X < 0 By equation (13), therefore, where X = 2KDh and Y = DµC ( r ) − hrD . Completing the square, this is equivalent to 2 KhD + hrD < hq * (r ) + hrD , hq * + Y 2 − X > Y and by Lemma 2, and 2 KhD + hrD < g (q*, r ) hq * − Y 2 − X < Y establishing the left-hand inequality of Lemma 4. so that For the right-hand inequality of Lemma 4, again by Lemma 3, 2hK / D < µC (r ) − hr . Multiplying both sides by µ, Y − Y2 − X Y + Y2 − X < q* < squaring, and completing the square on the right-hand side, we h h have Therefore, q* must lie between the two values 2 2 2 ( βh) + 2hKµ / D + 2hβ ( µ C (r ) − hµr ) < ( µ 2 C (r ) − hµr ) 2 + 2hβ ( µ 2 C (r ) − hµr ) + ( βh) 2 ( DµC (r ) − hrD ) ± ( DµC (r ) − hrD) 2 − 2 KhD (38) h or after some algebra g (q*, r ) > g 0 (q*, r ) is µ 1 (− βh + 2 ( βh) + 2hµ[ Kµ / D + β ( µC (r ) − hr )] ) From (29) and (38) the condition satisfied if and only if < µC (r ) − hr ) ( DµC ( r ) − hrD) − ( DµC (r ) − hrD) 2 − 2 KhD < g ( q*, r ) − hrD From equation (13) and the definition of C(r), this implies that < ( DµC ( r ) − hrD) + ( DµC (r ) − hrD) 2 − 2 KhD hq*/D + hr < µ C(r). By Lemma 2, hq* + hrD = g(q*,r), so the right-hand inequality of Lemma 4 is established. The right-hand inequality is satisfied by the right-hand inequality of (33). To prove the left-hand inequality, we note ) that for any a, b, and c such that a < b and c ≤ a2, If r > r , a similar argument establishes the reverse b − b 2 − c < a − a 2 − c by the concavity of the square-root inequalities. function. ) If r = r , a similar argument establishes the equations (note Setting that since ˆ ˆ ˆ 2 KhD + hrD = g (q*, r ) = hq * + hrD , a = 2 KhD , b = DµC (r ) − hrD, c = 2 KhD , we it follows that ˆ q * (r ) = 2 KD / h , i.e., the EOQ value have that a < b by Lemma 4 and noting the two inequalities in for q). (33) establishes the left-hand inequality and therefore the theorem. QED ) If r > r , then we need to establish Theorem 4: ) g(q*,r) - g0(q*,r)> 0 for r < r , and (35) (h 2 q *2 −2q * hY + X ) > 0 (39) ) g(q*,r) - g0(q*,r)< 0 for r > r . (36) ) If r = r , then g(q*,r) - g0(q*,r)= 0. From Lemma 3 we have that Y2 – X < 0, establishing (39) and proving the theorem. Proof: From equations (6) and (9), the approximation ) difference is given by r = r then g (q*, r ) = DµC (r ) by Lemma 4, so that If ( β − β 0 )( q * D 2 µ 2 C ( r ) − µD 2 ( K + 21D hq *2 + hq * r )) (37) g − g0 = (q * µ + βD )( q * µ + β 0 D) q* = 1 ( DµC (r ) − hDr ) by Lemma 3. Therefore h SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4 7
  8. 8. hq*=Y. Since Y2 = X by Lemma 4, we have that ( β − β 0 ) Dµ (qDµC (r ) − ( KD + 1 hq 2 + hqDr)) 2 2 2 g ( q, r ) − g 0 ( q, r ) = h q * −2q * hY + X = 0 , and therefore (qµ + βD)(qµ + β 0 D) Therefore, q * DµC (r ) − ( KD + 1 hq *2 + hq * rD ) = 0 , 2 g ( q, r ) − g 0 (q , r ) ( β − β 0 )( DµC ( r ) − ( KD / q + 1 hq + hDr )) 2 KD / q + 1 hq + hDr 2 = establishing that g(q*,r) = g0(q*,r) and proving the theorem. g 0 ( q, r ) ( qµ / D + β )( KD / q + 1 hq + hDr ) 2 KD / q + 1 hq + hDr + D 2 β 0 C (r ) / q 2 QED ( β − β 0 )( DµC (r ) − g E (q, r )) g E (q, r ) = 8. RELATIVE DEGREE OF APPROXIMATION ( qµ / D + β ) g E ( q , r ) g E (q, r ) + D 2 β 0 C (r ) / q (β − β0 )  DµC (r )  =  − 1 We wish to investigate the relative degree of approximation of (qµ / D + β )(1 + D 2 β 0C (r ) / qg E (q, r ))  g E ( q, r )    g to g0, i.e., (β − β0 )  DµC (r )  <  − 1 g (q*, r ) − g 0 (q*, r ) (qµ / D )( D 2 β 0C (r ) / qg E (q, r ))  g E ( q, r )    ( β − β 0 )  g E (q, r )  ( β − β 0 ) exp(−(λ + µ )(q / D )) g 0 (q*, r ) = 1 − β 0  DµC (r )   <  β0 = 1 − exp(−(λ + µ )(q / D)) in particular the degree to which the quantity (β- β0)/ β0 is an upper bound to this degree of approximation. Since, for many values of λ, µ, and q*, the values β and β0 are very close to b) From the quadratic formula, (38) is equivalent to each other, such an upper bound will be a powerful statement on the value of the approximation. KD hq * + + hrD = g E (q*, r ) < DµC (r ) Theorem 5: Let gE(q,r) = KD + hq + hr be the classical q* 2 q 2 EOQ cost function with a nonnegative reorder point. Then: The result therefore follows from part (a). a) For all q>0 such that gE(q,r) < DµC(r), c) From part (a) we have that g ( q, r ) − g 0 ( q, r ) β − β 0  g ( q, r )  < 1 − E  DµC (r )   g 0 ( q, r ) − g ( q, r ) g 0 ( q, r ) β0   = g 0 ( q, r ) β − β0 e − ( λ + µ )( q / D ) (β − β0 )  DµC (r )  < = 1 −  β0 1 − e − ( λ + µ )( q / D ) (qµ / D + β )(1 + D 2 β 0C (r ) / qg E (q, r ))   g E ( q, r )   ) b) If r < r then gE(q*,r) < DµC(r), so that Therefore, g 0 (q, r ) − g (q, r ) ( β − β 0 )  DµC (r )  g (q*, r ) − g 0 (q*, r ) β − β 0  g (q*, r )  < 1 −  g ( q, r )   < 1 − E   g 0 ( q, r ) β   g 0 (q*, r ) β0  DµC (r )  E e − ( λ + µ )( q * / D ) (β − β 0 ) < β − β0 = < = e −( λ + µ )( q*/ D ) β0 1 − e − ( λ + µ )( q * / D ) β ) d) This part follows directly from Theorem 4 c) If r > r then QED g 0 (q*, r ) − g (q*, r ) β − β 0  DµC (r )  < 1 −  g 0 (q*, r ) β  g E (q*, r )    9. UPPER (OR LOWER) BOUND ON THE EXACT β − β0 REORDER LEVEL < = e − ( λ + µ )( q*/ D ) β In addition to the approximating cost function g being an upper ) bound for the exact cost function g0, we can also establish for If r = r then (g(q,r) – g0(q,r))/ g0(q,r) = 0. what values of the reorder point r the optimal order quantity q* for the approximating cost function is an upper (or lower) Proof: bound for the optimal order quantity q0 for the exact cost function. a) From (37) Theorem 6: If ˆ r < r , then q* > q0, where q0 minimizes g0(q,r). If ˆ ˆ r > r , then q* < q0., and if r = r then q* = q0. 8 SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4
  9. 9. production and inventory textbooks. These parameter sets are dβ 0 λ − ( λ + µ )( q / D ) displayed in Table 1. For each pair of failure rate λ and Proof: Since from equation (8) = e , the recovery rate µ, we therefore have 10 sets of parameters. We dq D selected 16 failure and recovery rate pairs, using 4 values of λ first derivative of g0 with respect to q is given by (i.e., 0.5, 1, 2, and 5) and 4 values of µ (i.e., 2λ, 4λ, 10λ, and  q β0 h  20λ) for each value of λ. This results in 160 total application  ( + )( q + hr + λ exp(−(λ + µ )(q / D))C (r ))  examples. ∂g 0 1  D µ D  = + µ  − ( K + h q 2 + hrq + β DC (r ))( 1 + λ exp(−(λ + µ )(q / D)))  q β0 ∂q D  2D 0 D µD  We also tested our formulas on the numerical example given in   Parlar et al [1995], shown in the last rows of each of the tables The numerator can be rewritten as that follow. h q2 + β hq + β hr − K − βC (r ) Table 1 – Problem parameters 2 D2 Dµ µ D ( β + e −(λ + µ )( q / D ) − 2 µD 2 q 2 + D C (r )q − Dµ hq − λhr q − λh λ Dµ λK Dµ β − µ hr + βC (r ) ) Set h K π D 1 0.8 30 12.96 540 and further rewritten as 2 15.0 10 40.00 14 h q + 2 β hq + β hr − K − βC ( r ) 3 6.5 175 12.50 2000 2D2 Dµ µ D (40a,b,c) +e − ( λ + µ )( q / D ) (− h 2 q − β β hq − µ hr + K + βC ( r ) ) 4 2.0 50 25.00 200 2 D2 Dµ D +e − ( λ + µ )( q / D ) ( h (1 − λ ) q + 2 λ C ( r )q − λhr q− K (1 + λ ) ) 5 45.0 4500 440.49 2319 2D2 µ D Dµ D µ ∂g 6 5.0 300 50.00 3000 The expression in (40a) is the numerator for ∂q and the 7 0.0132 20 0.34 1000 parenthetical term in (40b) is its negative, so that when q = q* 8 5.0 28 80.00 520 both of these terms are zero. Using (29), the second factor of 9 0.005 12 0.12 3120 (40c) can be rewritten as 10 3.6 12000 65.73 8000 ( 1 µD (1 − λ )( Kµ + ( µC (r ) − hr ) β − βhq) + D C (r )q − λhr q − K (1 + λ ) µ λ Dµ D µ ) Parlar & 5 10 260 1 Perry = ( β µD (1 − µ )( µC (r ) − hr − hq) + λ (−2 K + ( D C (r ) − hr ) q) (41) λ µ D µ D ) For each application example, we computed (1 − λ ) > 0 We assume that λ < µ, so that ˆ . If r < r , then µ • The optimal ZIO (i.e., r = 0) order quantity q*(0), using by Lemma 3, hDr + hq* = g(q,r*), and by Lemma 1, g(q,r*) < the closed-form approximating function g, and the value DµC(r), so that DµC(r) - hDr - hq* > 0. Similarly, DµC(r) - of g and the exact function g0 for q = q*(0) and r = 0 hDr > hq*, so that q*(DµC(r) – hDr) > hq*2. Also by Lemma • The optimal reorder inventory point r*(q) for the order 2 and Lemma 3, 2 KhD + hDr < hq + hDr , so that quantity q = q*(0), and the value of g and g0 for hq 2 > 2 KD . Hence the expression in (41) and therefore • q = q*(0) and r = r*(q*(0)) the second term of (40c) are both positive. Therefore • The globally optimal point (q**,r**) using the closed- ∂g 0 / ∂q is positive when form approximating function g, and the value of g and g0 at (q**,r**) q = q*. Since g0 is unimodal (by a similar argument as • The globally optimal point (q0,r0) using the exact function Proposition 2b in Berk et al [1994], it must attain its minimum g0 (obtained via numerical methods) to the left of q*, therefore q* > q0. If ˆ r > r, then a similar argument shows that ∂g 0 / ∂q is A Sample Cost Analysis: Figure 2 shows a sample plot of negative when q = q*, so that g0 must attain its minimum to the the approximate and exact cost functions g and g0 for one of the right of q*, so that q* > q0. sample sets (λ = 2, µ = 20, and the sixth set of parameters in Table 1). The upper set of curves shows the approximate and If ˆ r = r, then a similar argument shows that ∂g 0 / ∂q is exact cost functions g and g0 using the ZIO policy (i.e., R = 0 for each value of Q), while the lower shows the two functions zero when q = q*, so that q* = q0. using the optimal reorder policy (i.e., R = R*(Q) for each value of Q). QED • Text box 0 shows the EOQ policy in the absence of supplier failures. 10. COMPUTATIONAL RESULTS • Text box 1 shows the EOQ policy in the presence of supplier failures We tested our formulas using the 10 sets of parameters h, K, π, • Text box 2 shows the ZIO policy (Q = Q* and R is zero) and D adopted in Snyder [2005], which are in turn derived from sample problems for the “(Q,R)” model found in SYSTEMICS, CYBERNETICS AND INFORMATICS VOLUME 5 - NUMBER 4 9

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