Supply Chain Network OptimizationProf. Steve De Lurgio

BA 544

The concept of optimization has been mentioned several times this semester. The methods that can be used to optimize a system vary depending on the mathematical, logical, and probabilistic complexity of the system being modeled. I have decided to develop an example of a Supply Chain Network Optimization under conditions of relative certainty. Generalization of this network to a complex nonlinear, logical, and probabilistic model is not difficult to do when using EXCEL. To solve a more realistic problem, you can use the underlying structure of management science models that we presented in the context of problem 2 of chapter 10, Fish Forwarders, and the general transportation model. Thus, in an actual application you should be able to incorporate the necessary model structure from the above applications plus the following network model.

Consider a supply chain network consisting of five levels (what is commonly called a multi-echelon supply chain), four suppliers (S), who can supply four manufacturers (M), who can supply four regional distribution center (R) who can supply four warehouses or if your prefer, four wholesalers (W), who can supply four large retail outlets (O). The word “can” is used because the model that is developed below determines the number and location of the four suppliers, four manufacturers, four regional distribution centers, four warehouses, and the quantities shipped to the each of the those locations as well as fulfillment of all demands at each of the four outlet malls.

You will find an EXCEL spreadsheet model on the web at: http://bloch.umkc.edu/forecast/ba544/xls/Here is the underlying model:

Table 1. Names of the Levels in the NetworkSUPPLIERS (S) MANUFACTURER

S (M)REGIONAL DC’S (R)

WAREHOUSE (W) OUTLETS(O)

SA ME RI WM OQSB MF RJ WN ORSC MG RK WO OSSD MH RL WP OT

A graphical display of the supply chain network is shown in Figure 1. While it may not be obvious from Figure 1, there are thousands of different paths through this Supply Chain Network. There are 64 different paths from Suppliers to Manufacturers, 64 different paths from Manufacturers to Regional Distributions, 64 different paths from RDCs to Warehouses, 64 different paths from Warehouses to Outlets. These combinations of paths result in an extremely difficult network design problems. In fact, as we will see, MicroSoft Excel’s solver is unable to consistently find an optimal solution. We will discuss this further in a moment.

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Figure 1. Supply Chain: Suppliers > Manufacturers > Regional DCs > Warehouses > Outlets.

The fixed sources of supply in this network are the capacities of each link in the supply chain. In addition, the demands at the outlets (outlets Q, R, S, T) are determined in a forecasting system. In other applications, there may be more constraints.

We will work three problems using this spreadsheet with three different levels of demands:Low: 600, 300, 200, 300Medium: 700, 400, 300, 500High: 700, 400, 600, 500 at the Q, R, S, and T outlets respectively.

The following mathematical statement is presented for the use and information of users, but it is not necessary for you to understand this complete formulation for this class. However, you should understand the underlying structure of the model.

A Mathematical Statement of the problem is.

Minimize: Total Cost = FCA*A + CA,E*SA,E + CA,F*SA,F + CA,G*SA,G + CA,H*SA,H + FCB*B + CB,E*SB,E + … + FCD*D + CD,H*SD,H + FCE*E + CE,I*SE,I + … + FCH*H + CH,L*SH,L + FCI*I + CI,M*SI,M + … + FCL*L + CL,P*SL,P + FCM*M + CM,Q*SM,Q +… + FCP*P + CP,T*SP,T

Where Ci,j is the cost of shipping from location i to location j.FCi is the fixed cost of opening location i.Si,j is the number of units shipped from Location i to Location j.A, B, C, … , R, S, T represent whether a site is open (=1) or closed (=0).

The above objective function is minimized subject to the following constraints:

Note: In this formulation, it is assumed that each of the sources of supply has a limited (finite) capacity.

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SA

SB

SC

SD

ME

MF

MG

MH

RI

RJ

RK

RL

WM

WN

WO

WP

OQ

OR

OS

OT

Supply Constraints – The following constraints assure that the number of units shipped from a facility does not exceed it capacity.

Supply from the Suppliers (A-D) is less than or equal to Suppliers Capacity. SA,E + SA,F + SA,G + SA,H - Capacity of Supplier A <= 0 SB,E + SB,F + SB,G + SB,H - Capacity of Supplier B <= 0 SC,E + SC,F + SC,G + SC,H - Capacity of Supplier C <= 0 SD,E + SD,F + SD,G + SD,H - Capacity of Supplier D <= 0

In order to assure that a facility is actually open and fixed costs are incurred, the above equations are modified as shown below:

SA,E + SA,F + SA,G + SA,H – A*CA <= 0SB,E + SB,F + SB,G + SB,H – B*CB <= 0SC,E + SC,F + SC,G + SC,H – C*CC <= 0 SD,E + SD,F + SD,G + SD,H – D*CD <= 0

Where CA is the capacity of supplier A, and A is a binary variable denoting that supplier A is open (A=1) or closed (A=0). Similar notation is used for each facility, CB is the capacity of supplier B, … , CD is the capacity of supplier D, … , etc. for all facilities as shown below.

Units shipped from Manufacturers (E-H) do not exceed the capacity of the manufacturers.SE,I + SE,J + SE,K + SE,L – E*CE <= 0 SF,I + SF,J + SF,K + SF,L – F*CF <= 0SG,I + SG,J + SG,K + MG,L – G*CG <= 0SH,I + SH,J + SH,K + MH,L – H*CH <= 0

Units shipped from Regional Distribution Centers (I-L) do not exceed the capacities at each RDC.SI,M + SI,N + SI,O + SI,P – I*CI <= 0SJ,M + SJ,N + SJ,O + SJ,P – J*CJ <= 0 SK,M + SK,N + SK,O + SK,P – K*CK <= 0 SL,M + SL,N + SL,O + SL,P – L*CL <= 0

Units shipped from the Warehouses (M-P) do not exceed their capacties.SM,Q + SM,R + SM,S + SM,T – M*CM <= 0SN,Q + SN,R + SN,S + SN,T – N*CN <= 0 SM,Q + SN,R + SO,S + SO,T – O*CO <= 0 SM,Q + SN,R + SO,S + SP,T – P*CP <= 0

1000 = CAPACITYSA = CA = Supplier A’s Capacity600 = CAPACITYSB …600 = CAPACITYSC900 = CAPACITYSD400 = CAPACITYME = CE = Manufacturer E’s Capacity600 = CAPACITYMF …600 = CAPACITYMG900 = CAPACITYMH500 = CAPACITYRI = CI = Regional Distribution Center I’s Capacity

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700 = CAPACITYRJ700 = CAPACITYRK900 = CAPACITYRL400 = CAPACITYWM = CM = Warehouse M’s Capacity600 = CAPACITYWN . . .600 = CAPACITYWO900 = CAPACITYWP = CP = Warehouse P’s Capacity

Demand Constraints – All the previously forecasted demands at the retail Outlets are to be fulfilled. Thus, demand constraints will work backwards from Demand at the retail Outlet backward to Demands imposed at the Supplier level. However, we will work forward from Suppliers to Outlets in presenting these equations.

Units shipped from Suppliers (A-D) to Manufacturers (E-H) equal the units shipped from Manufacturers (E-H) to Regional Distribution Centers (I-L).SA,E + SB,E + SC,E + SD,E = SE,I + SE,J + SE,K + ME,L = Demand at Manu. E. SA,F + SB,F + SC,F + SD,F = SF,I + SF,J + SF,K + MF,L = Demand at Manu. F.SA,G + SB,G + SC,G + SD,G = SG,I + SG,J + SG,K + MG,L = Demand at Manu. G.SA,H + SB,H + SC,H + SD,H = SH,I + SH,J + SH,K + MH,L = Demand at Manu. H.

However, to solve these equations, it is necessary to include variables only on the left hand side of the equation. Thus, the following equations are the above equations where all variables are shown on the left hand side:

SA,E + SB,E + SC,E + SD,E - SE,I - SE,J - SE,K - ME,L = 0.0 SA,F + SB,F + SC,F + SD,F - SF,I - SF,J - SF,K - MF,L = 0.0SA,G + SB,G + SC,G + SD,G - SG,I - SG,J - SG,K - MG,L = 0.0SA,H + SB,H + SC,H + SD,H - SH,I - SH,J - SH,K - MH,L = 0.0

Units shipped from Manufacturers (E-H) to Regional Distribution Centers (I-L) equal units shipped from Regional Distribution Centers (I-L) to W/H’s (M-P).SE,I + SF,I + SG,I + SH,I - SI,M - SI,N - SI,O - SI,P = 0.0SE,J + SF,J + SG,J + SH,J - SJ,M - SJ,N - SJ,O - SJ,P = 0.0 SE,K + SF,K + SG,K + SH,K - SK,M - SK,N - SK,O - SK,P = 0.0 SE,L + SF,L + SG,L + SH,L - SL,M - SL,N - SL,O - SL,P = 0.0

Units shipped from RDCs (I-L) to Warehouses (M-P) equal units shipped from Warehouses (M-P) to outlets (Q-T).SI,M + SJ,M + SK,M + SL,M - SM,Q - SM,R - SM,S - SM,T = 0.0SI,N + SJ,N + SK,N + SL,N - SN,Q - SN,R - SN,S - SN,T = 0.0 SI,O + SJ,O + SK,O + SL,O - SM,Q - SN,R - SO,S - SP,T = 0.0 SI,P + SJ,P + SK,P + SL,P - SM,Q - SN,R - SO,S - SP,T = 0.0

Units shipped from W/Hs (M-P) are supplied to Retail Outlets (Q-T) to fulfill fixed demands at (Q-T).SM,Q + SN,Q + SO,Q + SP,Q = Demand at Outlet Q which is fixedSM,R + SN,R + SO,R + SP,R = Demand at Outlet R which is fixedSM,S + SN,S + SO,S + SP,S = Demand at Outlet S which is fixedSM,T + SN,T + SO,T + SP,T = Demand at Outlet T which is fixed

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Again, to make these equations the equivalent of the previous equations, we move the Demand variable to the left side of the equation:

SM,Q + SN,Q + SO,Q + SP,Q – (Demand at Q)> 0.0SM,R + SN,R + SO,R + SP,R – (Demand at R)> 0.0SM,S + SN,S + SO,S + SP,S – (Demand at S)> 0.0SM,T + SN,T + SO,T + SP,T – (Demand at T)> 0.0

Integer Variables.All the above variables can be constrained or specified to be integer variables; however the computational times may become excessive depending on your computer speed.

Fixed Charge Constraints for Each Level.A = Binary = {0,1} If Supplier A is closed or open.B = Binary = {0,1} If Supplier B is closed or open.C = Binary = {0,1} If Supplier C is closed or open. D = Binary = {0,1} If Supplier D is closed or open.E = Binary = {0,1} If Manu. E is closed or open.F = Binary = {0,1} If Manu. F is closed or open.G = Binary = {0,1} If Manu. G is closed or open.H = Binary = {0,1} If Manu. H is closed or open.I = Binary = {0,1} If RDC I is closed or open.J = Binary = {0,1} If RDC J is closed or open.K = Binary = {0,1} If RDC K is closed or open.L = Binary = {0,1} If RDC L is closed or open.M = Binary = {0,1} If W/H M is closed or open.N = Binary = {0,1} If W/H N is closed or open.O = Binary = {0,1} If W/H O is closed or open.P = Binary = {0,1} If W/H P is closed or open.

A Note about Integer SolutionsSource: http://www.solver.com/tutorial7.htm#Integer%20Variables

“Integer Variables

Integer variables (i.e. decision variables that are constrained to have integer values at the solution) in a model make that model far more difficult to solve.  Memory and solution time may rise exponentially as you add more integer variables.  Even with highly sophisticated algorithms and modern supercomputers, there are models of just a few hundred integer variables that have never been solved to optimality.

This is because many combinations of specific integer values for the variables must be tested, and each test requires the solution of a "normal" linear or nonlinear optimization problem.  The number of combinations can rise exponentially with the size of the problem.  "Branch and bound" and "branch and cut" strategies help to cut down on this exponential growth, but even with these strategies, solutions for even moderately large mixed-integer programming (MIP) problems can require a great deal of time.

Different formulations of the same model with integer variables can take radically different amounts of time to solve.  With such models, expert consulting help with the formulation may make the difference between an easily solvable and a practically unsolvable model.  With a well-designed formulation, it is often possible to solve linear programming problems with hundreds, or even thousands of integer variables in a modest amount of time.

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Implementing Large-Scale Optimization Models in Excel Using VBALARRY J. LEBLANC

Vanderbilt University - Owen Graduate School of ManagementMICHAEL R. GALBRETH

University of South Carolina - Moore School of Business

July 15, 2006http://papers.ssrn.com/sol3/papers.cfm?abstract_id=917805#PaperDownload

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