Constructing the EOQ Models Network

Acknowledgments

6.2 Deterministic Models

6.2.3 Economic Order Quantity Model for a Multi-Product Assembly System

6.2.3.1 Constructing the EOQ Models Network

To transform the problem into EOQ models, each node j in Ni considers the part j designated to the production of product i still in the system, either as part j or assembled inside other parts already, as its inventory. As an illustration, the echelon inventory of part 4 for the production of part 2 includes the quantity of part 4 that is designated for the production of part 2, the quantity of part 4 inside part 5 that is designated for the production of part 2, and the quantity of part 4 inside part 2 that is still in the system.

Since the holding cost of each part l in Pⁱj required for the production of part j is already accounted for by its predecessors in Ti, the echelon holding cost for node j in Ni is

hⁱ_j Hⁱ_{j l p}i _lⁱ

j H

= −

∑

_{∈ .}

Thus, each node j in Ni corresponds to an EOQ model with a setup cost of kj, a holding cost of hⁱj per unit product per unit time, and a demand rate of 2 units per unit time for the product.

However, a part j might be required by different products for production. To avoid multiple counts of setup cost for an assembly, each part j that is required for the production of multiple products is identi-fied with an EOQ model with a setup cost of kj, no holding cost, and a demand rate of 2 units per unit time for the product. At the same time, for each Ni that includes j, the setup cost is removed from the EOQ model corres ponding to node j in Ni. That is, node j in Ni corresponds to an EOQ model with zero setup cost, a holding cost of hⁱj per unit product per unit time, and a demand rate of 2 units per unit time for the product.

These EOQ models are presented in the EOQ models network for the system. The EOQ models network for the system is a directed network G^E = (N^E, A^E) with

N^E = Ui{ij: j ∈Ni}U{j: j ∈Ni for at least 2 different i} and A^E = Ui{(ij, il): (j, l)∈Ai}U{(ij, j): j ∈Ni and j ∈N^E}.

Associated with each node x in N^E is an ordered pair (k^E(x), h^E(x)) that represents the EOQ model, with a setup cost of k^E(x), a holding cost of h^E(x) per unit product per unit time and a demand rate of 2 units per unit time for the product, associated with node x. In particular,

k^E(x) = kj if x = j ∈Ni for some i, or x = ij for some j ∈Ni and j ∉N^E; otherwise,

k^E(x) = 0

and h^E(x) = hⁱj if x = ij ∈ Ni for some i;

otherwise,

h^E(x) = 0.

To illustrate this with an example, the EOQ models network for the two products with the assembly networks in Figure 6.2 is shown in Figure 6.3. In Figure 6.3, x:k^E(x),h^E(x) is shown inside each node x.

Redistributing the Echelon Holding Costs

The optimal reorder interval for an EOQ model with a setup cost of k^E(x), a holding cost of h^E(x) per unit product per unit time and a demand rate of 2 units per unit time for the product is [k^E(x)/h^E(x)]^1/2 . [k^E(x)/

h^E(x)]^1/2 is infinite if h^E(x) = 0. For any (x, y) in A^E, the optimal assembly policies corresponding to nodes x and y cannot be synchronized if [k^E(x)/h^E(x)]^1/2 < [k^E(y)/h^E(y)]^1/2. As discussed earlier for the two-facilities in-series system, the problem that h^E(x) is too large can be rectified by redistributing some of the echelon holding cost h^E(x) from node x to node y. Note that while redistributing some of the echelon hold-ing cost h^E(ij) from node ij to node il for some product i and (j,l) ∈ Ai is equivalent to reducing the holding cost of part j, that is, designated for the production of product i, redistributing some of the echelon hold-ing cost h^E(ij) from node ij to node j for some product i and j ∈ Ni is equivalent to not changing that part of the holding cost of part j that is designated for the production of product i.

For any subset N of N^E and the corresponding subnetwork G = (N, A) of G^E with A = {(x,y): x,y ∈ N and (x,y) ∈ A^E}.

It is optimal to assemble the parts corresponding to the nodes in N simultaneously, if the echelon hold-ing costs can be redistributed from predecessors to successors in G until the resulthold-ing echelon holdhold-ing costs h(x) satisfies [k^E(x)/h(x)]^1/2 is a constant for all the nodes in G with h(x) = 0 in the case k^E(x) = 0. That is, for each x ∈ N with k^E(x) > 0,

h x k x( )/ ( )^E = ∑_{x N∈} h x( )/∑_{x N∈} k x^E( )= ∑_{x N∈} h x^E( )/∑_x∈NNk x^E( ).

Such an even redistribution of the echelon holding costs is possible if and only if we can flow the amount of excess echelon holding cost, h^E(x) − h(x), from the source nodes x with h^E(x) > h(x) to cover the lack of echelon holding cost, h(y) − h^E(y), at the sink nodes y with h^E(y) > h(y) through the network G. In particular, the maximum flow network is G^F = (N^F, A^F) with

N^F = N  {s,t},

A^F = A{(s,x): h(x) > h^E(x)}  {(y,t): h(y) < h^E(y)}.

1₁:k₁,h¹₁ 2₂:k₂,h₂²

1₄:0,h¹₄

1₃:0,h¹₃ 2₅:k₅,h²₅

2₃:0,h²₃

2₆:k₆,h²₆ 2₄:0,h²₄

3:k3,0

4:k4,0

FIGuRE 6�3 The EOQ models network for products 1 and 2.

In addition, the capacity c(x, y) associated with each arc in A^F is infinite if (x, y) ∈ A. Each (s, x) ∈ A^F has a capacity c(s, x) = h^E(x) − h(x), while each (y, t) ∈ A^F has a capacity c(y, t) = h(x) − h^E(x). The objec-tive is to maximize the flow from node s to t through the network G^F, where there is a capacity c(x, y) on the flow to send along arc (x, y). The maximum s-t flow problem is a typical application of linear program-ming (LP). In solving the LP for the maximum s-t flow problem, either the optimal objective flow value = Σ(s,x)∈_A^F c(s,x), then an even redistribution of the echelon holding cost is possible. Otherwise, the dual minimum s-t cut (X, X′) with s ∈X and t ∈X′ partitions N into two sets N¹ = X\{s} and N² = X′\{t}. Since excess echelon holding costs, that cannot flow to cover the lack of echelon holding cost at the nodes in N², are still available at the nodes in N¹; nodes in N² are predecessors of the nodes in N¹. That is, predecessors do not have enough while successors have too many echelon holding costs. In other words, there is no problem of a predecessor having a smaller optimal inter-setup interval than its successor between the nodes in N² and N¹, and redistribution of the echelon holding cost can be considered separately for the two sets of nodes.

Start with N = N^E. Solve the maximum flow problem for the network subnetwork G, and if an even redistribution of the echelon holding cost is possible for G, then set h^F(x) = h(x) for each x ∈N. Otherwise, partition N into two sets N¹ and N² according to the optimal dual minimum cut and repeat the process for N = N¹ and N = N² until h^F(x) is determined for each x ∈N^E. As indicated by the earlier discussion, this redistribution of echelon holding costs results in h^F(x), x ∈ N^E that satisfy h^F(x) = 0 in the case k^E(x) = 0, and [k^E(x)/h^F(x)]^1/2 > [k^E(y)/h^F(y)]^1/2 for each (x, y) ∈ A^E, with k^E(x) > 0 and k^E(y) > 0.

Adjusting the Inter-setup Intervals of the Assemblies

Since [k^E(x)/h^F(x)]^1/2 > [k^E(y)/h^F(y)]^1/2 for each (x, y) ∈ A^E with k^E(x) > 0 and k^E(y) > 0, min{[k^E(x)/

h^F(x)]^1/2: k^E(x) > 0 and x ∈ N^E} = [k^E(zz)/h^F(zz)]^1/2 for some product z = 1, 2, . . . , n. Assemble product z every Tz = T^*z = [k^E(zz)/h^F(zz)]^1/2units of time.

For any x = ij ∈ N^E with k^E(ij) > 0 for some part j and product i, or x = j ∈ N^E for some part j, let T^*j = [k^E(x)/h^F(x)]^1/2 and2^m(j)T^*z < T^*< 2^m(j)+1T^*z for some positive integer m(j). If T^*j /(2 ^m(j)T^*z) < 2 ^m(j)+1T^*z/ T^*j, then part j is assembled every T j = 2^m(j)T^*z units of time. Otherwise, the part j is assembled every T j = 2^m(j)+1T^*z units of time. For any ij ∈ N^E with k^E(ij) > 0 for some part j and product i, let Tⁱj = Tj.

For any x = ij ∈ N^E with k^E(ij) = 0 for some part j and product i, assembly inter-setup time is set back-ward for successors first and then predecessors up the assembly network for product i. Let l be the unique immediate successor of j in the assembly network for product i; then part j designated for the production of product i is assembled every Tⁱj = max{Tj, Tⁱl} units of time.

The assemblies are synchronized by assembling 2Tⁱj units of part j designated for the production of product i simultaneously. Then, 2Tⁱj units of part j designated for the production of product i are assem-bled every Tⁱj units of time. Since [k^E(x)/h^F(x)]^1/2 > [k^E(y)/h^F(y)]^1/2 for each (x, y) ∈ A^E with k^E(x) > 0 and k^E(y) > 0 implies that Tⁱj > Tⁱl for any product i and (j, l) ∈ Ai, inventory is down to zero at every assembly of part j designated for the production of product i.

Since echelon holding cost is redistribution from a node x to a node y in N^E only when they have the same corresponding assembly inter-setup time, accounting for the redistributed part of the holding cost at the assembly corresponding to node x or that at node y makes no difference to the average cost of the assembly policy. Hence, the average setup and holding cost of the power-of-two policy is Σ{k^E(x)/Tj + h^E(x) Tj: x = ij ∈ N^E with k^E(ij) > 0 for some part j and product i, or x = j ∈ N^E for some part j}.

For any x = ij ∈ N^E with k^E(ij) > 0 for some part j and product i, or x = j ∈ N^E for some part j, since 2^-1/2 < Tj/T^*j < 2^1/2 by the choice of Tj. Σ{k^E(x)/Tj + h^E(x)Tj: x = ij ∈ N^E with k^E(ij) > 0 for some part j and product i, or x = j ∈ N^E for some part j} < [(2^-1/2 + 2^1/2)/2]Σ{2[k^E(x)h^E(x)]^1/2: x = ij ∈ N^E with k^E(ij) > 0 for some part j and product i, or x = j ∈ N^E for some part j}.

In other words, it is a 94% optimal policy. A 98% optimal power-of-two policy can be obtained using a more complicated adjustment of the assembly inter-setup intervals.