Lab Exercise 7

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Industrial Engineering

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Dec 6, 2023

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Lab Exercise 7 Location Selection Problems ‘When selecting a location for a new facility, planners need some metric for evaluating how good any particular location is. One of the most widely-used metrics is distance, which most commonly measures the distance from the new location to its customers, its suppliers, or both. Another option is using gravity models that, essentially, measure the total "attractive force" of a potential location to the entire set of customers. In this exercise, we will apply three different models to location selection problems. These models all assume a known set of demand data, which are contained in the "DemandData.xlsx" file. 7.1 Distance Models We'll start with distance models that measure the total distance between a business location and the distribution of expected demand. These models assume that distance is a strong factor in customer choice; travel costs would be incurred when customers travel to the facility, or service providers travel to customers, or a physical product is delivered to customers. In all cases, the goal is to minimize these total travel costs, which vary depending on whether travel is unrestricted or limited to a rectangular road grid. Distance models may be implemented with weighted or unweighted customer demand. In the unweighted models, customers are considered to be undifferentiated and, accordingly, the location with the smallest sum of distances is considered to be the best. In the weighted models, customers are differentiated. For example, some might buy more often, in greater quantities, or more profitable items. The weighted approach is also suitable for aggregate demand, like the total customers in a zip code. The two distance models we consider are: o Euclidean: This model uses the total straight-line distances between all customers and the business location. For a customer at (x. y.) and a potential operation location (xo, ), the distance between them is \/(x, X0)? + ¥0)?. The Euclidean model is appropriate when customers can travel reasonably directly to the business. e Metropolitan model: This model uses the distances between the x- and y-coordinates of each customer-location pair. For a customer at (xc,yc) and a potential operation location (xo, ¥o), the distance between them is simply | x. xo | + | yc ¥o |. The metropolitan model is appropriate when all travel is confined to east-west or north-south roads, which is reasonably accurate for many cities in the United States. Additionally, more sophisticated models (not considered here) might also consider travel time variations, alternate travel modes (like walking or using public transportation), or route selection (as the most direct route might not be the fastest). Still, the general principle of choosing the location with the lowest total travel costs remains the same. 7.1.1 Euclidean Models Open the DemandData file to the "Euclidean" worksheet. There are 1200 customer demand data points (which might be derived from past purchase activity, demographic predictors of 139
shopping behavior, or other sources). Each data point represents a household with location coordinates and income. We’ll use Euclidean models to evaluate a potential facility location whose coordinates are given in the range Q3:R3. The various data points will be grouped into zones and spending will be estimated as a function of distance (from the zone centers to the new store location) and household income. 1. In cell Q5 (under the label "Distance") enter the formula "=SQRT((O5-$Q8$3)"2+(P5- $R$3)"2)". 2. Select RS and use autofill to copy the formula first over the range Q6:Q13. This fills out the distances from the store location to the zone centers. 3. In the next few steps, we’ll use VLOOKUP functions to obtain household zone codes. In cell D3, enter the formula "=VLOOKUP(B3,$K $4:5L$6,2)". A VLOOKUP function will compare a cell’s value against a table and return a value from a specified row and column in that table. The column is specified in the function (in the previous function, 2 means the second column). The row is found by comparing the cell’s value against the left column of the table. If the cell’s value is less than the first row, it will generate an error. Otherwise, VLOOKUP checks the next table value. If that value exceeds the table value, VLOOKUP will return the value from the first row (and specified column). Otherwise, the process continues until a table value is greater (and VLOOKUP returns the value from the previous row and specified column) or the cell’s value exceeds the last table value (and the value from the last row and specified column is returned). 4. Incell E3, enter the formula "=VLOOKUP(C3,$K$9:8L.$11,2)", then copy the formula to cell H4. 5. Incell F3, enter the formula "=D3+3*E3". This combines the X and Y codes into a single value. Then select D3:F3 and use autofill to copy their formulas down to D1202:F1202. 6. Now we’ll use another VLOOKUP to convert income brackets to estimated household income values. In cell G3, enter the formula "=VLOOKUP(G3,$K$15:$1.$19,2)". Use autofill to copy this formula down to G1202. Our worksheet already has a formula in place for the relationship between distance and sales. Naturally, any given scenario could have a different formula depending on factors like weather, the prevalence of car ownership, and the distribution of competitors. 7. Now we’ll combine household income, distance, and propensity to spend on groceries (the third column in the income code table) into estimated sales. In cell I3, enter the formula "=H3*VLOOKUP(G3,8K$15:5M$19,3)*VLOOKUP(F3,SN$5:3R$13,5)", then copy it down to 11202 using autofill. 8. Incell R15, enter the formula "=SUM(I3:11202)". This gives the estimated total sales. Discuss: Suppose your firm was considering four new store locations: (40, 40), (40, 60), (60, 60), and (60, 40). Using this model, which would be predicted to have the highest sales? 140
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