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**Chapter 5 Network Design in the Supply Chain**

Ok… I hope every one is ready for the first chapter. I’m guessing that you thought we would start on Chapter 1… well the meat of the book is in the middle, so we are going to start there. With Chapter 5. 5-1

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**Outline The Role of Network Design in the Supply Chain**

Factors Influencing Network Design Decisions Framework for Network Design Decisions Models for Facility Location and Capacity Allocation The Role of IT in Network Design Making Network Design Decisions in Practice Notes: What can you expect to learn in the next few hours? We are going to look at the role of network design, factors influencing the design, the framework, models, we will briefly discuss “it” in network design and lastly network design in practice. We will spend the majority of the day on the last sections working on problems. 5-2

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**Network Design Decisions**

Facility role Facility location Capacity allocation Market and supply allocation When we talk about the facility role, what do we mean. We want to know what processes are performed at each facility. For example, Toyota built in the flexibility to change car designs into each manufacturing plant. The thought was the plant could serve markets other than the local one. The plants would produce both SUVs and cars… this flexibility helped a lot in the 2008s when gas prices shot up. The next aspect is facility location. Why is facility location a big issue in network design? ANSWER? You want to have your factory near your consumers. If you are making fishing equipment in the US (with labor cost, land and taxes being equal) you want to minimize your transportation costs. Once a decision to build a facility is decided on…. It will be the location for a very long time. Lets talk about capacity allocation. What do you think is better? Two factories that are both running at 50% or one factory running at 100%? ANSWER If the answer is based on a steady demand stream than 100% is likely the best, because you are maximizing efficiencies. If lost sales have a significant negative result and demand is plagued by peaks and valleys, then the 2 factories solution is probably the best. I talked about the market allocation when I discussed facility location, but you also need to be concerned with your suppliers locations. You can easily get spare parts for a machine supplied from the same city… no the case if it is supplied from the other side of the globe. It is best to continueally review your network locations. Making a decision today on a factory location may have been the best decision… but 20 years from now your suppliers and market might have moved. You probably need to re look at the decision. P&G announced that it would reevaluate its network which was designed when oil was $10 a barrel. The current price (Dec 2010) is about 88 dollars a barrel. 5-3

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**If you are interested… you can see the price of oil in 2008 dollars**

If you are interested… you can see the price of oil in 2008 dollars. The P&G cost of $10 was an uninflated price probably in the flat section from the 40s to the 70s.

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs We are going to individually examine some factors effecting the network design… We will start with strategic factors. 5-5

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**Strategic Factors Firm’s competitive strategy Cost Responsiveness**

What is the competitive strategy of walmart? Low Cost…. What type of location are they looking for… Low cost. If you are selling Prada… you don’t mind a the high rent district on Rodeo Drive in California. How about responsiveness. If you are looking for a bag of chips or a soda… how far are you willing to drive. The convience store strategy is that you will not drive far, so they have locations all over the place. If you want a better deal… then Costco will give you a bulk price. But for the discount, you may have to drive much farther. There are probably dozens of conveniece stores in the service area of a Sam’s club or Costco.

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs The next factor we will look at is technological factors. How much is the production facility to design and build. If you are making an expensive high tech facility… you are not going to make a lot of them. In contrast, if youhave very low fixed cost, then it may be very feasible to create a lot of production facilities. For example, Coke-cola sets up many bottleing plants because the nature of their production. Low costs, but high transportation costs. 5-7

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs When we talk about macroeconomics we are including taxes, tariffs, exchange rates, and other external factors that are not internal to the organization. I will very briefly talk about each. Tariffs are the duties paid to pass international or state boundaries. If the tariffs are high, then you may want to build your facility inside… if low (such as with NAFTA or the European Union… then you do not need to include that factor with your location. What are tax incentives? ANSWER. Low taxes to encourage the location in a particular area. General Motors built its Saturn facility in Tennessee primarily based on tax incentives. Why does exchange rate matter regarding facility location? When the goods are sold can’t you just use the current exchange rate? ANSWER Well the reality is that time passes between agreements and payments. You may create a contract for 100 million dollars when the yen is at Well the payment is due over the next 2 years… in which the yen value rises to 132 (which it did between 2002 and 2004). This increase in yen, devalued the price they were receiving (in their currency). It is reported that every 1 cent rise in the Euro cost BMW and Mercedes roughly $75 million each year. You will lean about the tricks to counteract the exchange rate issue in your international logistics class. 5-8

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs Political stability is a big issue in the location of a facility. If your country has well established laws then it is less risky than one that is often curropt of changes leadership regularly. Building a facility in a risky area could mean that the facility it lost during a coup or change in power. 5-9

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs If you are going to build a facility in a location you are going to need good infrastructure… what is infastructure? We are talking about roads, rail, electricity, waterways, ports, airports, water, etc. Without infrastructure requirements, it is very costly to set up a facility. 5-10

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs Do you want to place your factory near your competitors… this is a question you need to answer. Far away, you have your market and they have theirs. Close you compete against each other, but maybe this will be advantageous to suppliers locating is the same regional area or shopper traveling to the location to look for the best deals (such as the mall concept). Many times competitors set up shop in the same location… look at silicon valley. The experience built up in one company helps the others as people transfer jobs and discuss similar topics in social situations while off work. This benefits all companies and employees. Lets math matically determine which is better regarding location from your competitor. 5-11

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**Competitor location A 1-B 1**

1 You are looking at two store locations… A and 1-B.

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**Competitor location A 1-B 1**

1 A gets all the sales from 0 to halfway between A and 1-B. 1-B gets the rest of the sales to 1. If yo ulook the 1-B has a little more area. So which direction do you think that A wants to move… towards 1-B.

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**Competitor location A 1-B 1 D1 = A + (1 – B – A)/2 D2 = (1+ B – A)/2**

1 Mathematically you can see the Demand for A is … the distant to A plus half way to 1-B (D1). For (1-B) the formula is a little more difficult to see. We know the distance between A and 1-B is (1-B-A)/2. [1 – (A+ (1-B-A)/2)]2 = 2-(2A+1-B-A) = 2 – (1-B+A) = (1+B-A)/2 The optimal location is when A=B or ½… as close together as possible. This is all based on market share. If they compete on price and there is transportation cost to the customer… it is best to be on opposite sides. D1 = A + (1 – B – A)/2 D2 = (1+ B – A)/2

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**Factors Influencing Network Design Decisions**

Strategic Technological Macroeconomic Political Infrastructure Competitive Logistics and facility costs The last factor is logistics and facility cost. As the number of facilities increase… inventory and facility costs increase. (inventory increases because the lack of parts pooling that exists with fewer facilities.) Transportation costs decrease as the number for facilities increase (closer to the customer). If the costs for processing are high (like in steel production)… then you want to be closer to the source of supply. This reduces the distance the large quantities of ore must travel. We will look at a few graphs next… 5-15

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**Service and Number of Facilities**

Response Time Notes: The response time decreases as you increase the number of facilities. If you have one factory in Seattle… it takes a long time to service the east coast customers. One in New york and one in Seattle decreases the response time. Number of Facilities 5-16

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**Costs and Number of Facilities**

Inventory Facility costs Costs Transportation Notes: Total cost is the sum of three costs: inventory, facility costs, and transporation costs. You can see inventory increases with the increase of facilities… and constantly increases costs. The facility costs increase with the number of facilities and keeps increasing costs. The transportation cost go down with the number of facilities and the start to increase. You lose your economy of scal when you stop using full truckloads because you have too many facilities. Number of facilities 5-17

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**Cost Buildup as a Function of Facilities**

Total Costs Cost of Operations Facilities Inventory Notes: You can also add the cost of labor which increases with the number of facilities. Add everything together and you get the total costs. You are looking for the low point… I’m going to guess in the middle somewhere. Transportation Labor Number of Facilities 5-18

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**Framework for Network Design Decisions**

Phase I – Supply Chain Strategy Phase II – Regional Facility Configuration Phase III – Desirable Sites Phase IV – Location Choices Our next topic is the framework for the network design. The concept is to maximize a firm’s profits while satisfying your customers needs (both demand levels and responsiveness). There are four phases to the framework, we will go through each one. The phases are the strategy, facility configuration , desirable sites, and location choices. 5-19

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**A Framework for Network Design Decisions**

Competitive STRATEGY GLOBAL COMPETITION PHASE I Supply Chain Strategy INTERNAL CONSTRAINTS Capital, growth strategy, existing network TARIFFS AND TAX INCENTIVES PRODUCTION TECHNOLOGIES Cost, Scale/Scope impact, support required, flexibility REGIONAL DEMAND Size, growth, homogeneity, local specifications PHASE II Regional Facility Configuration COMPETITIVE ENVIRONMENT POLITICAL, EXCHANGE RATE AND DEMAND RISK The first phase is the supply chain strategy. A company needs a clear definition of how the company plans to compete in the competitive environment. Will they be the low cost leader or the innovator (where you pay top dollar for new products). The strategy takes in a number of factors including the competition, Competitive strategy, and the internal company constraints. The internal constraints are how much money the company has to invest in the supply chain, current facility locations and the growth strategy. Phase 2 is the regional facility configuration… this is the plan on where you plan the region, capacity and role for each of the potential facility locations. The first factor you need to examine is demand. Who will be purchasing the product and how much do they want. The manager needs to examine if economies of scale can help out with increasing the bottom line. IF so, you may want to combine multiple locations together. If the economies of scale are not significant then it is best to serve each facility with a separate facility. You then need to look at tariffs, taxes, exchange rates and the political environment for each of the locations. These risk must be accounted for. The last part to examine is the competitors facilities locations. We are on to phase 3. WE are looking into potential sites in each region. We need to examine hard infrastructure requirements such as availability of suppliers, transportation services, utilities and communication infrastructure. Next we examine the soft infrastructure requirements: such as skilled workforces, community receptivity to the new industry. For example, I live near a naval air station and they want to put an outlying field in out community. When you drive around you see NO OLF in everyone's yard because they do not want the jet noise to decrease their quality of life and house value. The last phase is the selection. Here you are looking for the precise location and capacity for each facility. We are going to closely look at the process of completing Phase 2 – 4 tonight. PHASE III Desirable Sites AVAILABLE INFRASTRUCTURE PRODUCTION METHODS Skill needs, response time FACTOR COSTS Labor, materials, site specific PHASE IV Location Choices LOGISTICS COSTS Transport, inventory, coordination 5-20

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**Linear programming (LP)**

Optimization of linear objective functions Normally formulated: Maximize X Subject to y < z Please give me a green check if you are familiar with linear programming. If not, I will get you through it. Linear programming is optimization based on linear functions (things that increase of decrease in a linear manner). You can linearly increase the amount of fertilizer spread on your lawn. There is also something called integer programming which looks only at integers (or whole numbers)… often linear programming is used for this function and then just rounded to the next whole number. I’m not going to make you an expert in linear programming, but I’m going to teach you enough to get through the class. The LP is normally stated Maximize X subject to y < z. This could be maximize revenue subject to fertilizer spread cannot be greater than 500 pounds. We are going to learn by walking through a basic example.

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**LP example (Joe’s chop shop)**

Joe takes plain vans and converts them into custom vans and can produce either fine or fancy vans. Both types require a $25,000 plain van. Fancy vans sell for $37,000 and Joe uses $10,000 in parts to customize them yielding a profit margin of $2,000. Fine vans use $6,000 in parts and sell for $32,700 yielding profits of $1,700. Joe figures the shop can work on no more than 12 vans in a week. Joe hires 7 people including himself and operates 8 hours per day, 5 days a week and thus has at most 280 hours of labor available in a week. Joe also estimates that a fancy van will take 25 hours of labor, while a fine van will take 20 hours. To maximize profit, how many of each van should Joe produce? Read… So if we wanted to figure out how to maximize profit we are subject to some rules. We find the rules in the write up. The first thing we want to figure is our maximize formula. If we want to maximize profit in terms of car sales what figures would we use? The profit margin for fancy is $2000 and fine is $1700. We write that like this.

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**LP Example (Joe’s chop shop)**

Maximize Z = 2000 Xfancy Xfine WE want to maximize Z where 200 times the number of fancy car made times the number of fine car made. Now we have to constrain this formula with other information.

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**LP example (Joe’s chop shop)**

Joe takes plain vans and converts them into custom vans and can produce either fine or fancy vans. Both types require a $25,000 plain van. Fancy vans sell for $37,000 and Joe uses $10,000 in parts to customize them yielding a profit margin of $2,000. Fine vans use $6,000 in parts and sell for $32,700 yielding profits of $1,700. Joe figures the shop can work on no more than 12 vans in a week. Joe hires 7 people including himself and operates 8 hours per day, 5 days a week and thus has at most 280 hours of labor available in a week. Joe also estimates that a fancy van will take 25 hours of labor, while a fine van will take 20 hours. To maximize profit, how many of each van should Joe produce? The first constraint is a logical rule. Number for fine vans must be greater or equal to 0… we can’t make negative vans.

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**LP Example (Joe’s chop shop)**

Maximize Z = 2000 Xfancy Xfine Subject to: Xfancy ≥ 0 Xfine ≥ 0

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**LP example (Joe’s chop shop)**

Joe takes plain vans and converts them into custom vans and can produce either fine or fancy vans. Both types require a $25,000 plain van. Fancy vans sell for $37,000 and Joe uses $10,000 in parts to customize them yielding a profit margin of $2,000. Fine vans use $6,000 in parts and sell for $32,700 yielding profits of $1,700. Joe figures the shop can work on no more than 12 vans in a week. Joe hires 7 people including himself and operates 8 hours per day, 5 days a week and thus has at most 280 hours of labor available in a week. Joe also estimates that a fancy van will take 25 hours of labor, while a fine van will take 20 hours. To maximize profit, how many of each van should Joe produce? The next rule is the maximum number of cars he can make a week. It is 12.

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**LP Example (Joe’s chop shop)**

Maximize Z = 2000 Xfancy Xfine Subject to: Xfancy ≥ 0 Xfine ≥ 0 Xfancy + Xfine ≤ 12 25 Xfancy + 20 Xfine ≤ 280 This is it … it is all of our equations. Well what to we do with this mess. Solve it with Excel, but first I want to show you a graphical representation. You will not need to graph anything out ever… but it helps some people understand the concept.

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**Xfancy + Xfine ≤ 12 Xfancy To graph just zero out each individyally.**

10 8 Xfancy 6 To graph just zero out each individyally. Xfancy 12, Fine 0 Then Xfancy 0, Fine 12… connect dots. 4 2 2 4 6 8 10 12 Xfine

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**25 Xfancy + 20 Xfine ≤ 280 Xfancy Same 0 process… Xfancy 11.2, fine 0**

12 10 8 Xfancy 6 Same 0 process… Xfancy 11.2, fine 0 Fancy 0, fine 14 4 2 2 4 6 8 10 12 Xfine

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**The usable area is below each… the lines represent the max amount.**

12 10 8 Xfancy 6 The usable area is below each… the lines represent the max amount. 4 2 2 4 6 8 10 12 Xfine

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2000 Xfancy Xfine 12 10 8 Xfancy 6 Now we need the max line… find the equaliblium between Xfancy and fine. 8.5 fancy cars is the equivalent to 10 fine cars… pricewise. This helps determine where the max is. You slide the red line to the right. 4 2 2 4 6 8 10 12 Xfine

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**Now lets let excel do the work.**

12 10 8 Xfancy 6 The last point it hits in the highlighted area is 8 fancy and 4 fine. If the max formula was different you could have hit 12 fancy 0 fine or the reverse. Now lets let excel do the work. 4 2 2 4 6 8 10 12 Xfine

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**LP Example- Excel Solver**

Steps to add the Solver add-in in Excel 2007 1. Click the Microsoft Office Button , and then click Excel Options. 2. Click Add-Ins, and then in the Manage box, select Excel Add-ins. 3. Click Go. 4. In the Add-Ins available box, select the Solver Add-in check box, and then click OK. Tip: If Solver Add-in is not listed in the Add-Ins available box, click Browse to locate the add-in. If you get prompted that the Solver Add-in is not currently installed on your computer, click Yes to install it. After you load the Solver Add-in, the Solver command is available in the Analysis group on the Data tab If you have never installed solver… this is what you need to do.

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**LP Example Maximize Z = 2000 Xfancy + 1700 Xfine # of Fancy produced**

# of Fancy produced # of Fine produced Subject to Xfancy + Xfine <= 12 25 Xfancy + 20Xfine <= 280

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**LP Example Maximize Z = 2000 Xfancy + 1700 Xfine 22800**

# of Fancy produced 8 # of Fine produced 4 Subject to Xfancy + Xfine <= 12 12 25 Xfancy + 20Xfine <= 280 280

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**LP example #2 (calculators)**

A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day. If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits? Here is your chance to try it… the problem is similar.

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**LP Example #2 (Calculators)**

Maximize Z = -2 Xscientific + 5 Xgraphing Subject to: Xscientific ≥ 100 Xgraphing ≥ 80 Xscientific + Xgraphing ≤ 200 Xscientific ≤ 200 Xgraphing ≤ 170 This is it … it is all of our equations. Well what to we do with this mess. Solve it with Excel.

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**LP Example #2 (Calculators)**

Maximize Z = -2 Xsci + 5 Xgraph 650 # of Sci produced 100 # of Graph produced 170 Subject to Xsci + Xgraph <= 200 270

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**Network Optimization Models**

Allocating demand to production facilities Locating facilities and allocating capacity Key Costs: Fixed facility cost Transportation cost Production cost Inventory cost Coordination cost Now we are going to look at facility location and capacity allocation. We are going to use solver and the process we just learned to help us with each problem. Which plants to establish? How to configure the network? 5-39

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**Plant Location with Multiple Sourcing**

Notes: We are going to look at plant location with multiple suppliers. If you look at page 117 the valables will be discribed. n = the number of potential plants m = the number of markets (or demand points) D = annual demand for each market K = capacity of each plant f = fixed cost for each plant c = the cost to ship from each plant to each market (for each combination) This includes production, inventory transportation and tariffs. y = plant open (if open 1… if not 0). This is important use of anything times 0 is 0. x = quantity shipped from plant to market. So lets look at the formulas… We want to minimize ( fixed cost for open plants plus the cost to ship times the quantity). We are finding the best combination that will minimize the production and shipping costs. This is subject to some constraints…. We need to ship an amount equal to the demand. (if we didn’t do this we would ship 0 to minimize costs) We also need to have enough plants open to meet capacity requirements to make the needed shipping amount The last constraint is just saying that each plant is either open of closed… it must have a 1 or 0 value. Lets look at the first problem from the book… 5-40

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**Demand Region Production and Transportation Cost per 1,000,000 Units**

Figure 5-3 Inputs - Costs, Capacities, Demands Demand Region Production and Transportation Cost per 1,000,000 Units Fixed Low High Supply Region N. America S. America Europe Asia Africa Cost ($) Capacity 81 92 101 130 115 6,000 10 9,000 20 117 77 108 98 100 4,500 6,750 102 105 95 119 111 6,500 9,750 125 90 59 74 4,100 6,150 142 103 71 4,000 Demand 12 8 14 16 7 If you will look at the purple… this is our facility list. The question is question is what facilities do we need to have open and the capacity of the open facilities. If it is not open it costs us nothing in this model. The low capacity produces less, but also costs less. High produces more, but costs more. The low is highlighted in pink, high in orange and the demand for each region is highlighted in blue. The numbers in the middle are the costs for producing and moving product from one area to another. Producing in N. America for N. America costs 81… producing in N. America for Africa costs There is a lot more costs with shipping and tariffs in this example. We need to set up some decision variables for Excel to solve for us.

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**Demand Region - Production Allocation (1000 Units)**

Figure 5-4 Decision Variables Demand Region - Production Allocation (1000 Units) Plants Supply Region N. America S. America Europe Asia Africa (1=open) The decision variables starts with where do you want to produce the items. It is currently zeroed out… because solver is going to find the values. In the homework assignments you are going to need to create these tables. It is going to be challenging. I’m going to set up an discussion board for questions. I will review it at least every other day…. I would like for you guys to help each other out as a primary and I will be the secondary. The learning will happen as to dig into the process. Off my soap box and back into the problem. Solver will need the selected values to equal demand for each region and not exceed capacity. Additionally the plants will be determined if they are open or closed. Note there are multiple plants at each location. The reality is that one is a low capacity plant the other is a high capacity plant. You can’t have two plants open (one the other or neither) If you are producing something you need to be open, otherwise you will be closed.

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**Figure 5-5 Constraints Supply Region Excess Capacity N. America**

N. America S. America Europe Asia Africa Unmet Demand 12 8 14 16 7 Objective Function Cost = $ - Now we are looking at the constraints. We are going to want to make sure that all demand is met (right now we have unmet demand in each of the areas). The formula for this cell is region demand minus the sum of all the production from the previous slide. The excess capacity is calculated by multiplying an open low capacity plant times its capacity + open high capacity with high capacity levels and then subtract the amount produced for the region. The last part is the objective function. Here we calculate the cost… the sum of the open plant fixed costs, the production and transportation costs, and the transportation and production costs times the region supplying.

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**Figure 5-6 Constraints: All decision variables are ≥ 0**

All excess capacity is ≥ 0 All unmet demand = 0 Plants are open or closed (0,1) The first hard part is complete… the second hard part is setting up the constraints in solver. All decision variables are greater or equal to 0… b14 to h18 >= 0 All excess capacity is ≥ 0… b22 to b26 >=0 All unmet demand is 0 ; b28 to f28 = 0 Plants are open or closed (g14 to h18 = binary) easy way to say = 0 or 1.

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**Figure 5-7 Here is the answer that solver came up with.**

Inputs - Costs, Capacities, Demands Demand Region Production and Transportation Cost per 1,000,000 Units Fixed Low High Supply Region N. America S. America Europe Asia Africa Cost ($) Capacity 81 92 101 130 115 6,000 10 9,000 20 117 77 108 98 100 4,500 6,750 102 105 95 119 111 6,500 9,750 125 90 59 74 4,100 6,150 142 103 71 4,000 Demand 12 8 14 16 7 Decision Variables Demand Region - Production Allocation (1000 Units) Plants (1=open) 1 4 Constraints Excess Capacity 3 Unmet Demand Objective Function Cost = $ 23,751 Here is the answer that solver came up with. North America and Europe are closed. All open factories are at high capacity… We use S. America for N. America Asia produces for Asia and some Europe Africa produces for some Europe and Africa. Africa has a little excess capacity (3) Ready to try it yourself??? Here is your chance.

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Gravity Method C C C So the regions have been determined in Phase 2… we need to look at potential locations in each region. This is slightly different from the book… I’m having you do this by hand. The complexty does not require solver for this problem…. But you could use it. Let look at the Gravity method for warehouse location selection. You have four customers located in Idaho, Arizona, Texas and Ohio. Where would you put the warehouse… C

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x = (60x5000) + (15x1000) + (80x3000) y = (15x5000) + (70x1000) + (110x3000) x = /9000 ≈ 62 y = /9000 ≈ 53 Here is the work… Ready for more? OK lets go.

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Gravity Method 100 90 80 C 70 C 60 50 40 C Here is some of the additional information (x,y coordinates)… The real information you need is the demand for each customer. We will say that the demand is 500 tons for Idaho, Arizona demands 2000, Texas 1200 and ohio 600. 30 C 20 10 10 20 30 40 50 60 70 80 90 100 110 120 130 140

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Gravity Method 100 90 80 C 70 C 60 50 40 Customer x y demand Idaho 75 40 500 Arizona 35 2000 Texas 25 80 1200 Ohio 65 105 600 C Here is a visual of the coordinates and the demand for each customer. Where do you thing the center of gravity is? 30 C 20 10 10 20 30 40 50 60 70 80 90 100 110 120 130 140

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**Gravity Method Ton Mile-Center Solution**

x : X coordinate center of gravity y : Y coordinate center of gravity dnx : X coordinate of the nth location dny : Y coordinate of the nth location Vn : Annual tonnage to delivery location n Notes: Here is the math we need to do. I need a show of hand on the individuals who see the information in this format and are completely overwhelmed. Honesty, will help direct me as I create future lectures. Give me an X is the screen is overwhelming? Well lets walk through the example… we are looking for an x and y coordinate. The formula for the x coordinate is: The sum of all the x coordinates times the volume divided by the volume. The same formula is used for Y… lets look at the math. 5-50

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x = (75x500) + (35x2000) + (25x1200) + (65x600) y = (40x500) + (40x2000) + (80x1200) + (105x600) x = /4300 = 41 y = /4300 = 60 OK lets look at the x coordinate times the demand (v for volume of tonnage) for each customer. Divide that by the total demand. Do the same for y. Now you get the chance to try.

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Gravity Method 100 90 80 C 70 60 C 50 40 You have three customers in California Texas and New York… where should you put the warehouse. 30 20 C 10 10 20 30 40 50 60 70 80 90 100 110 120 130 140

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**Customer x y demand California 60 15 5000 Texas 70 1000 New York 80**

110 3000 Here is the information that you need. I will give everyone a few minutes to figure it out. Round up the answer.

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**QUESTION>>>**

What is the location 41/70 Oklahoma (62/53 Colorado)XX 55/71 Kansas 73/82 Iowa

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Gravity Method C C C Let look at the Gravity method for warehouse location selection. You have four customers located in Idaho, Arizona, Texas and Ohio. Where would you put the warehouse… This is the demand part of the second phase. C

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x = (60x5000) + (15x1000) + (80x3000) y = (15x5000) + (70x1000) + (110x3000) x = /9000 ≈ 62 y = /9000 ≈ 53 You will use solver to solve a similar problem in the homework. Here is the work… Ready for more? OK lets go.

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Exercise #2 DryIce, Inc., is a manufacturer of air conditioners that ha seen its demand grow significantly. The company anticipates nationwide demand for the year 2010 to be 180,000 in the South, 120,000 units in the Midwest, 110,000 in the East, and 100,000 units in the West. Managers at DryIce are designing the manufacturing network and have selected four potential sites- New York, Atlanta, Chicago, and San Diego. Plants could have a capacity of either 200,000 or 400,000 units. The annual fixed costs are the four locations are shown in the table below, along with the cost of producing and shipping an air conditioner to each of the four markets. Where should DryIce build its factories and how large should they be? This problem is similar to the example we did earlier in the day. I’m going to have you start by determining the Objective function and constraints… if you want you can keep working on the rest of the problem also. New York Atlanta Chicago San Diego Fixed Costs 200k 6,000,000 5,500,000 5,600,000 6,100,000 400k 10,000,000 9,200,000 9,300,000 10,200,000 Variable Costs East 211 232 238 299 South 212 230 280 Midwest 240 215 270 West 300 225

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**Exercise #2 Objective function: Subject to :**

Minimize Z = fixed costs + Variable costs Subject to : All shipments are positive integers (≥0) All shipments are ≤ All shipment add up to the 2010 requirements. READ the Objective function and Constraints. Now you need to set up the excel document in an effort to use solver.

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Question Question Answer is: **265740 250150 231000 276090

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**Exercise #2 200k 400k East 211 110,000 232 - 238 299 South 212 180,000**

New York Shipped from NY Atlanta Shipped from Atlanta Chicago Shipped from Chicago San Diego Shipped from San Diego Requirements Supply Fixed Costs 200k 6,000,000 5,500,000 5,600,000 6,100,000 400k 10,000,000 9,200,000 9,300,000 10,200,000 Variable Costs East 211 110,000 232 - 238 299 South 212 180,000 230 280 Midwest 240 215 120,000 270 West 300 225 100,000 23,210,000 38,160,000 25,800,000 22,500,000 Fixed 29,210,000 43,660,000 31,400,000 28,600,000 132,870,000 TOTAL SYSTEM COST 265,740,000 Cells that are changed All non negative integers The fixed price depends on the production level. That is an if/then statement. If the sum of the shipped from NY column is less than then the value is Each city total is the fixed plus the variable shipping costs… * 211 = = The sum of all the cities is the system total This is the value we are minimizing while changing the values in grey.

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**Exercise #2 200k 400k East 211 110,000 232 - 238 299 South 212 180,000**

New York Shipped from NY Atlanta Shipped from Atlanta Chicago Shipped from Chicago San Diego Shipped from San Diego Requirements Supply Fixed Costs 200k 6,000,000 5,500,000 5,600,000 6,100,000 400k 10,000,000 9,200,000 9,300,000 10,200,000 Variable Costs East 211 110,000 232 - 238 299 South 212 180,000 230 280 Midwest 240 215 120,000 270 West 300 225 100,000 23,210,000 38,160,000 25,800,000 22,500,000 Fixed 29,210,000 43,660,000 31,400,000 28,600,000 132,870,000 TOTAL SYSTEM COST 265,740,000 Cells that are changed All non negative integers ADD SOLVER Picture….

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**The Role of IT in Network Design**

IT systems help with network design by: Making the modeling of the network design problems easier Containing high-performance optimization technologies Allowing for “what-if” scenarios Interfacing with planning and operational software As you can guess computers have significantly help in the world of optimization. In the past it was possible to do a few scenarios to determine if option A was better than B….. Now all scenarios can be examined. Of course, there are still some items 5-62

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**Making Network Design Decisions In Practice**

Do not underestimate the life span of facilities Do not gloss over the cultural implications Do not ignore quality of life issues Focus on tariffs and tax incentives when locating facilities We have looked at some of the process of optimizing the network design… Here are some major factors to consider. It is likely that the facility location you select will be there for your grand children to visit. The cultural issues can have a major impact on your workforce. This is the same case for quality of life issues. My current location ranks very low in the quality of life category. But it could be much worse. When you focus on placing a facility outside the country you are shipping to, pay special attention to the tariffs and potential tax incentives‘. 5-63

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**Summary of Learning Objectives**

What is the role of network design decisions in the supply chain? What are the factors influencing supply chain network design decisions? Describe a strategic framework for facility location. How are the following optimization methods used for facility location and capacity allocation decisions? Gravity methods for location Network optimization models Notes:OK to sum it up we looked at the role of network design… the factors influencing the design. The strategic framework for location. And lastly the gravity and optimization models. 5-64

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