 Applied Mathematics, 2011, 2, 551-555 doi:10.4236/am.2011.25072 Published Online May 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM Multi-Item EOQ Model with Both Demand-Dependent Unit Cost and Varying Leading Time via Geometric Programming Kotb A. M. Kotb1, Hala A. Fergany2 1Department of Mathematics and Statistics, Faculty of Science, Taif University, Taif, Saudi Arabia 2Department of Mathematical Statistics, Faculty of Science, Tanta University, Tanta, Egypt E-mail: kamkotp2@yahoo.com Received March 10, 2011; revised March 17, 2011; accepted March 21, 2011 Abstract The objective of this paper is to derive the analytical solution of the EOQ model of multiple items with both demand-dependent unit cost and leading time using geometric programming approach. The varying purchase and leading time crashing costs are considered to be continuous functions of demand rate and leading time, respectively. The researchers deduce th e optimal order quantity, the demand rate and the leading time as de-cision variables then the optimal total cost is obtained. Keywords: Inventory, Geometric Programming, Leading Time, Demand-Dependent, Economic Order Quantity 1. Introduction The problem of the EOQ model with demand-dependent unit cost had been treated by some researchers. Cheng  studied an EOQ model with demand-dependent unit cost of single-item. The problem of inventory models involv-ing lead time as a decision variable have been succinctly described by Ben-Daya and Abdul Raouf . Abou- El-Ata and Kotb  developed a crisp inventory model under two restrictions. Also, Teng and Yang  exam-ined deterministic inventory lot-size models with time-varying demand and cost under generalized holding costs. Other related studies were written by Jung and Klein , Das et al.  and Mandal et al. . Recently, Kotb and Fergany  discussed multi-item EOQ model with varying holding cost: a geometric programming approach. The aim of this paper is to derive the optimal solution of EOQ inventory model and minimize the total cost function based on the values of demand rate, order quan-tity and leading ti me using geometric programmin g tech-nique. In the final a numerical example is solved to illus-trate the model. 2. Notations and Assumptions To construct the model of this problem, we define the following variables: rD = Annual demand rat e (decision variabl e). prCC = Unit pu rchase (pr oduction ) cost. hr = Unit holding (inventory carrying) cost per item per unit time. orC = Ordering cost. SS = rKL= Safety stock. n = Number of different items carried in inventory. rLQ = Leading rate time (decision variable). r = Production (order) quantity batch (decision vari-able). TC,,rrrDQL = Average annual total cost. For the rth item. The following basic assumptions about the model are made: 1) Demand rate is uniform over time. r2) Time horizon is finite. D3) No shortages are allowed. 4) Unit production cost ,bpr rpr rCD CD1,2 3,,,1r, nbb is inversely related to the de-mand rate . W here is called the price elasticity. 5) Lead time crashing cost is related to the lead time by a function of the form ,1,2,3,,,rrRLL rn0,00 5.. where , are real constants selected to provide the best fit of the estimated cost func- K. A. M. KOTB ET AL. 552 tion. 6) Our objective is to minimize th e annual relevant to-tal cost. 3. Mathematical Formulation The annual relevant total cost (sum of production, order, inventory carrying and lead time crashing costs) which, according to the basic assumptions of the EOQ model, is:  12nrrrrrpr rorrrrrrhrrrDTC D,Q,LDCDCQQDKLC RLQ   (1) Substituting prrCD and rRL into (1) yields: 112nbrrr rrprorrrrrrhrrrDTC D,Q,LDCCQQDKLC LQ   (2) To solve this primal objective function which is a convex programming problem, we can write it in the form: 11min2nbrpr rorrrrrrhrrrDTCC DCQQDKLC LQ  (3) Applying Duffin et al.  results of geometric pro-gramming technique to (3), the enlarged predual function could be written in the form: 13245213541112345112 345122rrrrrrrrrrWWWbnrpr ror rhrrrrr rWWhr rrrrrrWWWnpr or hrrrr rWWhrrrbWrDC DC QCGW WQWWKC LDLWQWCCCWW WKCWWD  125 2354512rrr rrrrrWW WWWrWWrQL  (4) Since the dual variable vector jr, is arbitrary and can be chosen according to convenience subject to: W,1,2,3,4, 5,1,2,3,,jrn0123451rrrrr jrWWWWW ,W (5) We choose jr such that the exponents of rr are zero, thus making the right hand side of (4) independent of the decision variables. To do this we re-quire: WD,Qand rL12 523 545100102rrrrr rrrbW WWWW WWW  (6) These are called the orthogonality conditions which together with (5) are sufficient to determine the v alues of ,12,3,4,5,jrWj, 1,2, 3,,rn. Solving (5) and (6) for jrW, we get: 1525354511221211121 21112 and212rrrrrrrrWWbbbWWbbbWWbWW   (7) Substituting ,1,2,3,4, 1,2,3,,jrWj rn5rgW in (4), we get the dual function . To find 5rW which maximize 5rgW , the logarithm of both side of 5rgW5rW, and the partial derivatives were taken relative to . Setting it to equal zero and simplifying, we get: 2325515 51512 1212 0bbrrr rbrfWW AbWWAb W   (8) where:  12 13112,,21 211212 and12bbbbbbbbb 1212121221 121bhrorbbbbor hrprKCACbCC Cb  It is clear that 00f and which means that there exists a root 10f5r. The trial and error method can be used to find this root. However, we shall 0,1WCopyright © 2011 SciRes. AM K. A. M. KOTB ET AL.553 first verify the root 5 calculated from (8) to maxi-mize . This is confirmed by the second deriva-tive to with respect to , which is always negative. *rW*r,r*jrWj*rW5rgWlngW1,2,Wj5rWgW3,4,5rW,n2,3,4**DQThus, the root calculated from (8) maximize the dual function . Hence, the optimal solution is , where 5 is the solution of (8) and are evaluated by substituting value of in (7). 55r5, 1,2,3,*jr ,1,*rW*5To find the optimal values rrr, we apply Duffin et al.  of geometric programming as indicated below: ,,L1525*r**rWg1*br*r*r*rrWg,DCWW,QprorCD 45**rrgWgW35and2*rr***r rCW,L WhrhrQCK By solving these relations, the optimal demand rate is given by: 112b21223,2*or hrr**prr rCC WD.CWW*r (9) The optimal order quantity is: 212112223,2b**b*rQorprrorhrr***rprr rCW CCWCWC WW (10) The optimal lead time is: 2221214 1223 232*rL2b*** bor hrrr r** **pr rrpr rrCCCWW WWWCWWorKC (11) By substituting the values of in (3), we deduce the minimum total cost as: and** *rr rD,Q L121123222 111 22122112223min 2122b*n*or hrrr pr**rpr rr** *rr rhr** *rr rbrr**or hrrr***rprrrCC WLC CWWWKCW WCCWWWCWW2312**rrro**pr*or hr ror hrprTC D ,Q ,WWWCCWWCC WCCC 12 431b*br*rWW (12) As a special case, we assume 0, 0rL and 0bRL , and 14**rrWW50,*rW2312**rrWW. This is the classical EOQ inventory model. 4. An Illustrative Example We shall compute the decision variables (optimal order quantity , optimal demand rate and optimal lead time ) whose values are to be determined to minimize the annual relev ant total cost for three items (n = 3). The parameters of the model are shown in Table 1. *rQ*rL*rDAssume that the standard deviation 6 unit/year and K = 2. For some different values of  and b, we use equa-tion (8) to determine 5, whose value is to be deter-mined to obtain *rW*jrW, j = 1, 2, 3, 4, r = 1, 2, 3 from (6). It follows that the optimal values of the production batch quantity Q demand rate *rDad time *rL and minimum annual total cost are given in Tables 2-7. *r,, leTable 1. The parameters of the model. r orC prC hrC r 1 \$ 200 \$ 10 \$ 0.8 \$ 1 2 \$ 140 \$ 08 \$ 0.5 \$ 2 3 \$ 100 \$ 05 \$ 0.3 \$ 3 Table 2. The optimal solution of and *rQ*rD as a func-tion of b (forall). b 1*Q 2*Q 3*Q 1*D 2*D 3*D 2 28.21 31.17 33.10 1.450 1.5501.540 5 26.41 28.55 31.00 1.395 1.4551.440 8 25.65 27.49 29.90 1.316 1.3501.340 1025.25 26.99 29.40 1.280 1.3001.295 2024.20 25.72 28.03 1.170 1.1801.178 Table 3. The optimal solution of and as a function of b (). *rLminTC=0.1b 1*L 2*L 3*L min TC 2 8.410–9 8.910–8 1.310–6 59.6932 5 4.210–17 2.410–14 4.910–11 66.0704 8 2.110–23 2.410–17 1.010–12 71.6195 10 1.210–26 8.110–20 2.710–13 83.8121 20 3.410– 396.910–26 1.510–14 422.838 Table 4. The optimal solution of and as a function of b (*rLminTC=0.2). b 1*L 2*L 3*L min TC 2 2.710–9 5.110–8 6.510–7 77.539200 5 2.510–17 2.210–14 2.310–11 243.99000 8 2.210–24 4.310–19 3.710–14 2906.5400 10 7.110–28 1.410–21 1.710–15 15175.300 20 6.810–43 7.710–31 9.210–20 1.3203107Copyright © 2011 SciRes. AM K. A. M. KOTB ET AL. 554 Table 5. The optimal solution of and as a function of b (). *rLminTC=0.3 b 1*L 2*L 3*L min TC 2 2.6 10–9 2.510–8 5.510–7 112.62400 5 2.4 10–17 2.110–14 1.210–11 6729.1300 8 1.7 10–24 8.710–20 4.1 10–15 75148.000 10 6.4 10–29 5.410–23 2.310–17 1.49510720 1.2 10–46 2.810–35 1.9 10–25 2.8581012 Table 6. The optimal solution of and as a function of b (*rLminTC=0.4). b 1*L 2*L 3*L min TC 2 5.9 10–10 1.210–8 2.010–7 309.50100 5 3.9 10–17 1.610–14 5.4 10–12 99640.100 8 3.4 10–25 3.910–20 4.6 10–16 1.317108 10 7.9 10–29 2.910–24 8.7 10–19 2.050109 20 6.8 10–51 5.010–41 1.5 10–32 5.6291018 Table 7. The optimal solution of and as a function of b (*rLminTC=0.5). b 1*L 2*L 3*L min TC 2 1.2 10–10 3.610–9 7.410–8 1013.00000 5 6.1 10–18 7.210–15 8.410–12 2.5485106 8 5.2 10–26 6.010–21 7.110–17 2.2880101010 4.9 10–30 5.610–25 6.710–19 2.31429101220 4.3 10–52 4.810–44 5.710–38 2.38881024 Figure 1. The optimal order quantity against b (for all ). Figure 2. The optimal demand rate against b (for all ). Figure 3. The minimum total cost against b (for all ). Figure 4. The minimum total cost against  (for all b). Solution of the problemmay be determined more re adily by plotting (min**rrQ,D ,TC) against b and min TC against , for each values of . 5. Conclusions his paper is devoted to study multi-item inventory tal cost is found at Tmodel that consider the order quantity, the demand rate and the leading time as three decision variables. These decision variables ,and,1,2,3,,** *rr rQD Lrn are evaluated and the is deduced. The classical system is derived as special case and a numerical example is solved. The smallest value of the minimum tominimum annual total cost min TC the smallest values of b and . 6. References ] T. C. E. 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