CHAPTER 08 LOCATION PLANNING AND ANALYSIS

CHAPTER 08 LOCATION PLANNING AN ANALYSIS Solutions 1. Given: We have the following information shown below for two plant location alternatives: Omaha Kansas City R $185 $185 v $36 $47 Annual FC $1,00,000 $1,400,000 Expected annual demand (units) (Q) 8,000 1,000 etermine the expected profits per year for each alternative: Profit = Q(R v) FC Omaha: 8,000($185 $36) $1,00,000 = -$8,000 Kansas City: 1,000($185 $47) $1,400,000 = $56,000 Conclusion: Kansas City would produce the greater profit. 8-1 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

. Given: We have the following information shown below for three potential locations for a new outlet: A B C R $.65 $.65 $.65 v $1.76 $1.76 $1.76 Monthly FC $5,000 $5,500 $5,800 a. etermine the monthly volume necessary at each location to realize a monthly profit of $10,000 (round to 1 decimal). Location A Volume: Location B Volume: Location C Volume: b. etermine the expected profits at each facility given the expected monthly volumes: A = 1,000 per month, B =,000 per month, & C = 3,000 per month. Profit = Q(R v) FC Location A: 1,000($.65 $1.76) $5,000 = $13,690 per month Location B:,000($.65 $1.76) $5,500 = $14,080 per month Location C: 3,000($.65 $1.76) $5,800 = $14,670 per month Conclusion: Location C would yield the greatest profits. 8- Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

3. Given: There are two alternatives for which costs and revenue are listed below: A B R $17,000 $17,000 v $14,000 $13,000 Annual FC $800,000 $90,000 a. Find the volume at which the two locations have the same total cost (TC). TC = FC + VC TC = FC + (Q x v) TC (Location A) = $800,000 + $14,000Q TC (Location B) = $90,000 + $13,000Q Set the two cost equations equal and solve for Q: $800,000 + $14,000Q = $90,000 + $13,000Q $14,000Q $13,000Q = $90,000 $800,000 $1,000Q = $10,000 Q = $10,000 / $1,000 Q = 10 units b. Range over which A and B would be superior: Location A has the lowest fixed costs; therefore, it is preferred at lower volumes. Conclusion: Location A Preferred: 0 < 10 units Location B Preferred: > 10 units 4. Given: There are three alternatives for which costs are given below: A (new) B (sub) C (expand) v $500 $,500 $1,000 Annual FC $50,000 --- $50,000 a. Step 1: etermine the total cost equation for each alternative. TC = FC + VC TC = FC + (Q x v) A: TC = $50,000 + $500Q B: TC = $,500Q C: TC = $50,000 + $1,000Q 8-3 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Step : Graph the alternatives. 500 TC ($000) [50] 400 300 00 B (subcontract) C (expansion) A (new location) 100 [50] 0 33.3 100 00 300 400 B C A No. of Boats/yr. Step 3: etermine over what range each alternative is preferred. Looking at the graph, we can tell that Alternative B is preferred over the lowest range, Alternative C is preferred over the middle range, and Alternative A is preferred over the highest range. First, we find the indifference (break-even) point between Alternatives B & C by setting their total cost equations equal to each other and solving for Q. B: TC = $,500Q C: TC = $50,000 + $1,000Q $,500Q = $50,000 + $1,000Q $,500Q $1,000Q = $50,000 $1,500Q = $50,000 Q = $50,000 / $1,500 Q = 33.33 units Second, we find the indifference (break-even) point between Alternatives C & A by setting their total cost equations equal to each other and solving for Q. C: TC = $50,000 + $1,000Q A: TC = $50,000 + $500Q $50,000 + $1,000Q = $50,000 + $500Q $1,000Q $500Q = $50,000 $50,000 $500Q = $00,000 Q = $00,000 / $500 Q = 400 units 8-4 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Conclusion: Alternative B preferred: < 33.33 units Alternative C preferred: > 33.33 units to < 400 units Alternative A preferred: > 400 units b. Expected volume of 150 boats: Based on the graph, Alternative C would yield the lowest total cost (TC) at a volume of 150 boats. c. Other factors that might be considered when deciding between the expansion and subcontracting alternatives include subcontracting costs will be known with greater certainty, subcontracting provides a secondary (backup) source of supply, and expansion offers more control over operations. 5. Rework Problem 4b using this additional information: Alternative A (New Location) will have an additional $4,000 in fixed costs per year. Alternative B (Subcontracting) will have $5,000 in fixed costs per year. Alternative C (Expansion) will have an additional $70,000 in fixed costs per year. Step 1: Change the costs in the table. A (new) B (sub) C (expand) v $500 $,500 $1,000 Annual FC $54,000 $5,000 $10,000 Step : etermine the total cost equation for each alternative. TC = FC + VC TC = FC + (Q x v) A: TC = $54,000 + $500Q B: TC = $5,000 + $,500Q C: TC = $10,000 + $1,000Q Step 3: Find TC for 150 units. A: TC = $54,000 + $500(150) = $39,000 B: TC = $5,000 + $,500(150) = $400,000 C: TC = $10,000 + $1,000(150) = $70,000 Conclusion: Alternative C (Expand) would yield the lowest total cost for an expected volume of 150 boats. 8-5 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

6. Given: Expected annual volume (Q) = 10,000 units. There are three lease alternatives for which costs are given below: Memphis Biloxi Birmingham Lease building & equipment $40,000 $60,000 $100,000 Transportation $50,000 $60,000 $5,000 v $8 $4 $5 Step 1: etermine fixed cost (FC) for each alternative & add FC to table. FC = Lease cost + transportation cost. Memphis Biloxi Birmingham Lease building & equipment $40,000 $60,000 $100,000 Transportation $50,000 $60,000 $5,000 Annual FC $90,000 $10,000 $15,000 v $8 $4 $5 Step : etermine the total cost equation for each alternative. TC = FC + VC TC = FC + (Q x v) Memphis: $90,000 + $8Q Biloxi: $10,000 + $4Q Birmingham: $15,000 + $5Q Step 3: Find TC for 10,000 units. Memphis: $90,000 + $8(10,000) = $170,000 Biloxi: $10,000 + $4(10,000) = $160,000 Birmingham: $15,000 + $5(10,000) = $175,000 Conclusion: The Biloxi alternative yields the lowest total cost for an expected annual volume of 10,000 units. 8-6 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

7. Given: There are two alternative shop locations for which costs are shown below: a. (1) Monthly profit for Q = 00 cars: City Outside R $90 $90 v $30 $40 Monthly FC $7,000 $4,700 Step 1: etermine total profit equation for each alternative. Total profit = Q(R v) FC City: Q($90 $30) $7,000 Outside: Q($90 $40) $4,700 Step : etermine total profit for each alternative at the expected monthly volume. City: 00($90 $30) $7,000 = $5,000 Outside: 00($90 $40) $4,700 = $5,300 Conclusion: Outside location yields the greatest profit if monthly demand is 00 cars. () Monthly profit for Q = 300 cars: City: 300($90 $30) $7,000 = $11,000 Outside: 300($90 $40) $4,700 = $10,300 Conclusion: City location yields the greatest profit if monthly demand is 300 cars. b. etermine the indifference (break-even point) between the two locations. Set their total profit equations equal to each other and solve for Q: Q($90 $30) $7,000 = Q($90 $40) $4,700 $60Q $7,000 = $50Q $4,700 $60Q $50Q = -$4,700 (-$7,000) $10Q = $,300 Q = $,300 / $10 Q = 30 cars 8-7 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

8. Given: We are provided the location factors below for four different types of organizations. Factor Local bank Steel mill Food warehouse Public school Convenience for customers Attractiveness of building Nearness to raw materials Large amounts of power Pollution controls Labor cost and availability Transportation costs Construction costs Student answers will vary regarding how they rate the importance of each factor in terms of making location decisions using L = low importance, M = moderate importance, and H = high importance. One possible set of answers is given below. Factor Local bank Steel mill Food warehouse Public school Convenience for customers H L M H M H Attractiveness of building H L M M H Nearness to raw materials L H L M Large amounts of power L H L L Pollution controls L H L L Labor cost and availability L M L L Transportation costs L M H M H M Construction costs M H M M H 8-8 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

9. Given: We are given factors, weights, and factor rating scores for three locations. Scores range from 1 100 (best). Location Score Factor Wt. A B C Convenience.15 80 70 60 Parking.0 7 76 9 isplay area.18 88 90 90 Shopper traffic.7 94 86 80 Operating costs.10 98 90 8 Neighborhood.10 96 85 75 1.00 Multiply the factor weight by the score for each factor and sum the results for each location alternative. Weight x Score Factor Wt. A B C Convenience.15.15(80) = 1.00.15(70) = 10.50.15(60) = 9.00 Parking.0.0(7) = 14.40.0(76) = 15.0.0(9) = 18.40 isplay area.18.18(88) = 15.84.18(90) = 16.0.18(90) = 16.0 Shopper traffic.7.7(94) = 5.38.7(86) = 3..7(80) = 1.60 Operating costs.10.10(98) = 9.80.10(90) = 9.00.10(8) = 8.0 Neighborhood.10.10(96) = 9.60.10(85) = 8.50.10(75) = 7.50 1.00 87.0 8.6 80.90 Conclusion: Based on composite score, Location A seems to be the best. 8-9 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

10. Given: We are given factors, weights, and factor rating scores for three locations. Scores range from 1 100 (best). Location Score Factor Wt. East #1 East # West Initial Cost 8 100 150 140 Traffic 10 40 40 30 Maintenance 6 0 5 18 ock space 6 5 10 1 Neighborhood 4 1 8 15 Multiply the factor weight by the score for each factor and sum the results for each location alternative. Weight x Score Factor Wt. East #1 East # West Initial Cost 8 8(100) = 800 8(150) = 100 8(140) = 110 Traffic 10 10(40) = 400 10(40) = 400 10(30) = 300 Maintenance 6 6(0) = 10 6(5) = 150 6(18) = 108 ock space 6 6(5) = 150 6(10) = 60 6(1) = 7 Neighborhood 4 4(1) = 48 4(8) = 3 4(15) = 60 1518 184 1660 Conclusion: Based on composite score, Location East # seems to be the best. 8-10 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

11. Given: We are given factors and factor rating scores for three locations. Scores range from 1 10 (best). Location Score Factor A B C Business services 9 5 5 Community services 7 6 7 Real estate cost 3 8 7 Construction costs 5 6 5 Cost of living 4 7 8 Taxes 5 5 4 Transportation 6 7 8 a. Assume that the manager weights each factor equally. Because there are seven factors, each factor will have a weight of 1/7. Therefore, we can sum the scores and divide by 7 to determine the weighted score for each alternative. Factor A B C Business services 9 5 5 Community services 7 6 7 Real estate cost 3 8 7 Construction costs 5 6 5 Cost of living 4 7 8 Taxes 5 5 4 Transportation 6 7 8 Total 39 44 44 Total / 7 5.57 6.9 6.9 Conclusion: Location B or C is best, followed by Location A. 8-11 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Chapter 08 - Location Planning and Analysis b. Two of the factors (business services and construction costs) are given weights that are double the weights of the other factors. We will give these two factors weights of /9 and the other five factors weights of 1/9. Then, we will multiply each factor s rating by that factor s weight. Factor Location Score Weight Weight x Score Business Services 9 5 5 /9 18/9 10/9 10/9 Community Services 7 6 7 1/9 7/9 6/9 7/9 Real Estate Cost 3 8 7 1/9 3/9 8/9 7/9 Construction Costs 5 6 5 /9 10/9 1/9 10/9 Cost of Living 4 7 8 1/9 4/9 7/9 8/9 Taxes 5 5 4 1/9 5/9 5/9 4/9 Transportation 6 7 8 1/9 6/9 7/9 8/9 1.0 53/9 55/9 54/9 Conclusion: Location B is best, followed by Location C, and then Location A. 1. Given: A toy manufacturer produces toys in five locations and will ship raw materials from a new, centralized warehouse. The monthly quantities to be shipped to each location are identical. The coordinates for all five locations are shown below. Location X Y A 3 7 B 8 C 4 6 4 1 E 6 4 We know that the quantities to be shipped to each location are identical so we can eliminate quantities from consideration. The correct formulas for the center of gravity are shown below: 8-1 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Chapter 08 - Location Planning and Analysis Sum the values in each coordinate s column. n = 5 locations. Location X Y A 3 7 B 8 C 4 6 4 1 E 6 4 Sum 5 0 (round to 1 decimal) (round to 1 decimal) Conclusion: The new warehouse should be located at 5.0, 4.0. 13. Given: A clothing manufacturer produces clothes at four locations. The manufacturer must determine the location of a central shipping point. The coordinates and weekly shipping quantities to the four locations are shown below. Location X Y Weekly Quantity (Q) A 5 7 15 B 6 9 0 C 3 9 5 9 4 30 The correct formulas for the center of gravity are shown below: Sum the values in the quantity column. Location X Y Weekly Quantity (Q) A 5 7 15 B 6 9 0 C 3 9 5 9 4 30 Sum 90 8-13 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Chapter 08 - Location Planning and Analysis ( ) ( ) ( ) ( ) (round to 1 decimal) ( ) ( ) ( ) ( ) (round to 1 decimal) Conclusion: The central shipping point should be located at 6.0, 7.0. 14. Given: A company handling hazardous waste wants to minimize shipping cost for shipments to a disposal center from five stations that it operates. The coordinates for each of the five stations and the volumes shipped to the new disposal center are shown below. Location X Y Volume in Tons per ay (Q) 1 10 5 6 4 1 9 3 4 7 5 4 6 30 5 8 7 40 The correct formulas for the center of gravity are shown below: Sum the values in the quantity column. Location X Y Volume Tons per ay (Q) 1 10 5 6 4 1 9 3 4 7 5 4 6 30 5 8 7 40 Sum 130 ( ) ( ) ( ) ( ) ( ) (round to 1 decimal) ( ) ( ) ( ) ( ) ( ) (round to 1 decimal) Conclusion: The disposal center should be located at 6.0, 6.0. 8-14 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Chapter 08 - Location Planning and Analysis 15. Given: A company is considering two locations for a distribution center: L1 & L. The five shipment destinations and the monthly shipments (Q) to all five destinations are shown below. estination Q 1 900 300 3 700 4 600 5 100 Step 1: Use the map to determine the coordinates of each destination (use a ruler if necessary). Add those coordinates to the table. Then, sum the values in the quantity column. estination X Y Q 1 1 900 4 300 3 3 1 700 4 4 600 5 5 3 100 Sum 3700 Step : etermine the center of gravity for the optimal location for the distribution center. The correct formulas for the center of gravity are shown below: ( ) ( ) ( ) ( ) ( ) (round to 1 decimal) ( ) ( ) ( ) ( ) ( ) (round to 1 decimal) Step 3: Use the map to determine the coordinates for L1 & L. Use a ruler if necessary. L1 coordinates.6,.4 L coordinates 3.5,.5 Step 4: etermine the distance between each proposed location and the center of gravity. istance between two points = ifference in X coordinates + ifference in Y coordinates istance between L1 & Center of Gravity.6 3. +.4.3 = 0.6 + 0.1 = 0.7 istance between L & Center of Gravity 3.5 3. +.5.3 = 0.3 + 0. = 0.5 Conclusion: L is closer to the center of gravity and is the better site. 8-15 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Case: Hello Walmart? [This is a good case for in-class discussion. Three groups can be formed: one to take the position of residents, one to take the position of small businesses, and one to take the position of Walmart.] 1. Owners of small businesses: Pro: Restaurants and other businesses that do not compete directly with Walmart generally would welcome the additional traffic that Walmart would attract. Con: Businesses that compete directly would not fare well unless they provided a product or service value that would offset Walmart s lower price.. Residents: Pro: Another shopping option, lower prices, and other, non-competing businesses that would be attracted by the increased traffic that Walmart would generate. Con: Increased traffic and noise, construction inconveniences, loss of small-town atmosphere, and loss of local businesses and jobs. Walmart responses: The company would be a good neighbor, supporting the community and providing jobs for low-skilled and handicapped workers. Construction would create construction jobs and generate taxes and revenues for the community. Shoppers would benefit from Walmart s low prices. In addition, there would be an increase in the tax base. Enrichment Module A. istance Measurement B. Center of Gravity Method with Predetermined Sites C. Factor Scoring Model. Emergency Facility Location A. istance Measurement The companies measure distance when making two important decisions: 1. Facility Layout ecision: istances are estimated/measured in determining the best layout of equipment or departments within a manufacturing facility such as a plant, a distribution facility such as a warehouse, or a service facility such as a department store. istance is an important input in determining the best possible layout that minimizes the total distance traveled between departments or workstations.. Facility Location ecision: istance measurement also is a very important input measure in determining the best location for a new service or a manufacturing facility, relocation of an existing facility, or elimination of an existing facility. In most instances, distance measurements are used to estimate the distances between existing warehouses or plants and the newly proposed potential location sites. The estimated distance measures then are used to estimate the transportation costs. Transportation cost is considered a critical factor in the facility location decision. 8-16 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

istance Measurement Methods: There are four methods of measuring distance: 1. Empirical Method. Rectilinear Method 3. Euclidean Method 4. Weighted Average Method 1. Empirical Method: The empirical method is the most accurate distance measure. Using the empirical method, distance can be measured in one of two ways. The first type of empirical measurement is the actual recorded travel distance, where a driver records the distance based on a vehicle odometer reading. The second type of empirical measurement of distance is estimation from a map. When there are no actual travel data available, then the map estimation can be very useful. However, the actual travel data obviously provides a more accurate measure of distance than the map estimation. The main advantage of using either empirical method is that they generally provide the most accurate distance measures. On the other hand, the disadvantage is that it can be very time consuming to gather the data and use it as a part of a computerized layout or a location technique, especially if there are many existing and/or proposed locations.. Rectilinear Method This mathematical method is very easy to compute and lends itself to easy implementation of computerized layout or location techniques. Rectilinear method requires the use of a two dimensional space with a horizontal axis, X, and a vertical axis, Y. Rectilinear distance often is called Manhattan distance because it requires going around the block when no straightline route is available. If A and B are the locations in question, the Rectilinear distance between A and B is given by the following formula: r = X A X B Y A Y B Where: r = Rectilinear distance measure between location A and location B; X A = X coordinate of location A; X B = X coordinate of location B; Y A = Y coordinate of location A; Y B = Y coordinate of location B. 8-17 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Example Problem Calculation of the Rectilinear istance: When answering this question, refer to textbook, Chapter 8, Problem 13. a. etermine the Rectilinear distance between location A and B. b. etermine the Rectilinear distance between location A and. c. etermine the Rectilinear distance between location B and. Solution to Example Problem Measurement of Rectilinear istance: a. The Rectilinear distance between location A and B is: r = X A X B Y A Y r = 5 6 7 9 3 b. The Rectilinear distance between location A and is: r = X A X Y A B Y r = 5 9 7 4 7 c. The Rectilinear distance between location B and is: r = X B X Y B Y r = 6 9 9 4 8 The main advantage of this method is the ease of calculations, while the disadvantage is that it generally overestimates the distance. Because of its propensity to overestimate distance, Rectilinear distance is sometimes referred to as the pessimistic distance measurement method. Rectilinear measure of distance is more commonly used in solving facility layout problems. 3. Euclidean Method This mathematical method is based on the Pythagorean Theorem. Euclidean distance measures distance as the crow flies and is most applicable when a straight-line route is possible. The following formula for the hypotenuse of a right triangle provides us with the Euclidean distance: C = A + B or C = A B C is the side of a right triangle opposite the right angle, and A and B are the right degree sides of a right degree triangle. The Euclidean distance method also requires the use of a two dimensional space with a horizontal axis, X, and a vertical axis, Y. If F and G are the locations in question, then the Euclidean distance between F and G is given by the following formula: 8-18 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

e X X Y Y F G F G Where: e = Euclidean distance measure between location F and location G; X F = X coordinate of location F; X G = X coordinate of location G; Y F = Y coordinate of location F; Y G = Y coordinate of location G. Example Problem Calculation of the Euclidean istance: Refer to textbook, Chapter 8, Problem 13. a. etermine the Euclidean distance between location A and B. b. etermine the Euclidean distance between location A and. c. etermine the Euclidean distance between location B and. Solution to Example Problem Calculation of the Euclidean istance: a. X X Y Y e A B A B b. c. e e e e e e e e e e 5 6 7 9 5.36 X X Y Y A 5 9 7 4 16 9 5 5 X X Y Y B A 6 9 9 4 9 5 34 5.831 B The Euclidean distance is the direct or the shortest distance between two given sites. Therefore, for any given pair of sites, the Euclidean distance provides the most direct or the straight-line connection. In most cases, however, there is no straight route between a pair of locations. Therefore, the Euclidean measure provides an optimistic measure of distance and in most cases underestimates the true distance. 8-19 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

4. Weighted Average Method As we discussed in the previous sections, the Rectilinear method is a conservative, pessimistic measure of distance, and the Euclidean method is an optimistic measure of distance. Both Euclidean and Rectilinear measures of distance are frequently the only standard distance options used as a part of many layout and location software programs. However, some managers prefer the weighted average method that provides a compromise distance between the optimistic Euclidean and the pessimistic Rectilinear methods. If A and B are the locations in question, then the weighted average distance between A and B is given by the following formula: wa w ) ( ) ( w ) ( ) ( e e r r Where: r = Rectilinear distance measure between location A and location B; e = Euclidean distance measure between location A and location B; w r = weight associated with the Rectilinear distance; w e = weight associated with the Euclidean distance. w i 1 i i w 0 Example Problem Calculation of the Weighted Average istance: Refer to textbook, Chapter 8, Problem 13 and the two examples given earlier in this section. Assume that the pessimistic (Rectilinear) weight is 0.4 and the optimistic (Euclidean) weight is 0.6. a. etermine the weighted average distance between location A and B. b. etermine the weighted average distance between location A and. c. etermine the weighted average distance between location B and. Solution to Example Problem Weighted Average a. The weighted average distance between location A and B is: WA = (0.4)(3) + (0.6)(.36) =.5416 b. The weighted average distance between location A and is: WA = (0.4)(7) + (0.6)(5) = 5.8 c. The weighted average distance between location B and is: WA = (0.4)(8) + (0.6)(5.831) = 6.7 8-0 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Exercise Problems 1. Refer to textbook Problem 15 in Chapter 8, and measure the Rectilinear distance between destination 1 and destination 5.. Refer to textbook Problem 15 in Chapter 8, and measure the Euclidean distance between destination 1 and destination 5. 3. Refer to textbook Problem 15 in Chapter 8, and measure the Weighted Average distance between destination 1 and destination 5 (w r = 0.3 and w e = 0.7). 4. Refer to textbook Problem 15 in Chapter 8 and measure the Rectilinear distance between destination 1 and destination 3. 5. Refer to textbook Problem 15 in Chapter 8, and measure the Euclidean distance between destination 1 and destination 3. 6. Refer to textbook Problem 15 in Chapter 8, and measure the Weighted Average distance between destination 1 and destination 3 (w r = 0.5 and w e = 0.5). Solutions to Exercise Problems 1.. 3. 4. 5. 6. r e r e 1 5 3 5 WA WA 17 4.131 (0.3)(5) (0.7)(4.131) 4.386 1 3 1 3 5.361 (0.5)(3) (0.5)(.361).6181 B. Center of Gravity Method with Predetermined Sites: (Weighted istance Technique) The center of gravity method in its original form identifies a central location relative to its distance to other existing locations and the amount of goods shipped to and from other existing locations. However, the center of gravity method in its original form is unable to evaluate specific location alternatives. If we have narrowed the choice of the new location to a few known specific sites, then the weighted distance technique can evaluate these known locations. For each candidate site, a weighted distance value is computed using the following formula: n W w i d where: i i 8-1 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

W = weighted distance value for a given site w i = weight associated with existing location i d i = the distance between existing site i and the proposed location For a typical facility location problem, w i (weight) represents the amount of goods or units shipped between the proposed location and the existing location i. For a facility layout problem, w i (weight) represents the number of trips between the existing location i and the proposed location. istance is calculated using either Rectilinear, Euclidean or Weighted Average methods. The site with the lowest weighted-distance would be selected, because the lowest weighted-distance usually results in the lowest transportation costs. Problem 1 Based on the destination locations and quantities given in textbook problem 15, the company is considering two locations for a new plant. The coordinates of the first plant location are: (x =, y = 3) and the coordinates of the second location are: (x = 4, y = 3). a. etermine the Euclidean distance from the first proposed plant location to all of the destination locations. b. etermine the Rectilinear distance from the first proposed plant location to all of the destination locations. c. etermine the Euclidean distance from the second proposed plant location to all of the destination locations. d. etermine the Rectilinear distance from the second proposed plant location to all of the destination locations. e. etermine the Weighted istance value for the first and second proposed plant locations based on Euclidean distance and decide where the new plant should be located. f. etermine the Weighted istance value for the first and second proposed plant locations based on Rectilinear distance and decide where the new plant should be located. Solution to Problem 1 estination X coordinate Y coordinate Q 1 1 900 4 300 3 3 1 700 4 4 600 5 5 3 100 8- Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

a. Plant Site 1 Euclidean istance From Plant Site 1 (P 1 ) to estination j ( j ) Euclidean istance ( P ) Euclidean istance ( P Euclidean istance ( P ) Euclidean istance ( P ) ) Euclidean istance ( P ) 1 1 1 1 1 1 3 4 5 ( 1) ( ) ( 3) ( 4) ( 5) (3 ) (3 4) (3 1) (3 ) (3 3) 1.414 1 1 5.36 5.36 9 3 b. Plant Site 1 Rectilinear istance From Plant Site 1 (P 1 ) to estination j ( j ) Rectilinear istance( P ) 1 3 1 1 Rectilinear istance( P 1 ) 3 4 1 Rectilinear istance( P ) 1 3 3 3 1 3 Rectilinear istance( P 1 4 ) 4 3 3 Rectilinear istance( P ) 1 5 5 3 3 3 c. Plant Site Euclidean istance From Plant Site (P ) to estination j ( j ) Euclidean istance( P ) 1 (4 1) (3 ) 10 3.16 Euclidean istance( P ) (4 ) (3 4) 5.36 Euclidean istance( P ) 3 (4 3) (3 1) 5.36 Euclidean istance( P 4 ) (4 4) (3 ) 1 1 Euclidean istance( P ) 5 (4 5) (3 3) 1 1 8-3 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

d. Plant Site Rectilinear istance From Plant Site (P ) to estination j ( j ) Rectilinear istance( P ) 4 1 3 1 4 Rectilinear istance( P ) 4 3 4 3 Rectilinear istance( P ) 3 4 3 3 1 3 Rectilinear istance( P 4 ) 4 4 3 1 Rectilinear istance( P ) 5 4 5 3 3 1 e. Weighted istance Values Based on Euclidean istances P 1 = Plant 1 P = Plant k = # of existing locations k i1 w i d i W P1 = 900(1.414) + 300(1) + 700(.36) + 600(.36) + 1,00(3) = 8,079.4 W P = 900(3.16) + 300(.36) + 700(.36) + 600(1) + 1,00(1) = 6,881.8 Because 6,881.8 < 8,079.4, choose proposed Plant Location. f. Weighted distance values based on Rectilinear distances W P1 = 900() + 300(1) + 700(3) + 600(3) + 1,00(3) = 9,600 W P = 900(4) + 300(3) + 700(3) + 600(1) + 1,00(1) = 8,400 Because 8,400 < 9,600, choose proposed Plant Location. C. Factor Scoring Model This simplistic selection procedure has many areas of application. The two most common areas of application are: 1) Facility Location; ) Product Selection. Factor Scoring is a very flexible method that considers both tangible and intangible factors. It has the capability to consider multiple decision criteria simultaneously. Steps of the Factor Scoring Model: 1. evelop a list of (factors) criteria to be considered. The decision maker should consider these factors important in evaluating each decision alternative.. Assign a weight to each factor that describes the factor s relative importance. 8-4 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Let w i = the weight for factor i, where i = 1,,.., F F = the number of factors considered. The higher the weight, the more important the criterion. We will use a five-point scale to establish the relative importance of the factors considered. The following table provides an interpretation of the weight scale: Factor Weight Interpretation Table Importance Weight Very important 5 Somewhat important 4 Average importance 3 Somewhat unimportant Very unimportant 1 For example, if a factor has a weight of 4, it is somewhat more important than the average factor. If a factor has a weight of 1, relative to the other factors being considered, the factor in question is very unimportant. 3. etermine a list of decision alternatives. Let d j = decision alternative j and assign a rating for each factor/decision alternative combination. Let r ij = the rating for factor i and decision alternative j. Where j = 1,,, and = number of decision alternatives considered. 8-5 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

For the purposes of rating each decision alternative/factor combination, we use the following 9-point scale. Level of Satisfaction Rating Extremely High 9 Very high 8 High 7 Slightly High 6 Average 5 Slightly Low 4 Low 3 Very Low Extremely Low 1 For example, a score of 7 for a given decision alternative would indicate that the decision maker (manager) rates this decision alternative (location) high with respect to a given factor. 4. Compute the factor score for each decision alternative. Let S j = factor score S j F i w r i ij 5. Sequence the decision alternatives from the highest score to the lowest score. The decision alternative with the highest factor score is the recommended decision alternative. The decision alternative with the second highest factor score is the second choice decision alternative and so on. Three popular areas of application for the Factor Scoring model are product selection, facility location, and job selection. Product selection and facility location are company-related problems while the job selection is an individual problem. An example for each of the three applications is provided in the following pages. PROUCT SELECTION EXAMPLE An appliance manufacturing company is considering expanding its product line. It has sufficient capital to introduce only one of the three following products: 1. Microwave Ovens. Refrigerators 3. Stoves Management thinks that the following decision criteria should be used in selecting the product: 1. Manufacturing capability/cost 8-6 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

. Market demand 3. Unit profit margin 4. Long term profitability/growth 5. Transportation costs 6. Useful life The company has determined the following weights for the decision criteria (factors): FACTOR WEIGHT Manufacturing capability/cost 4 Market demand 5 Unit profit margin 3 Long term profitability/growth 5 Transportation costs Useful life 1 The decision factor ratings for each criteria are given in the following table: ECISION FACTOR RATINGS FACTOR MICROWAVE REFRIGERATOR STOVE 1 4 3 8 8 4 3 6 9 5 4 3 6 7 5 9 4 6 1 5 6 Based on the information provided, determine the factor scores for all three products. What is the best choice for the appliance manufacturing company? What is the second best choice? 8-7 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

FACILITY LOCATION EXAMPLE ue to a significant sustained increase in demand, the ITM fastener manufacturing company has decided to build a new plant. After an initial study, it has narrowed its choices to four locations: 1. Albany, New York. Indianapolis, Indiana 3. Akron, Ohio 4. Mason, Georgia ITM also identified the following as being important factors in terms of the facility location decision: 1. Transportation costs. Construction/land costs 3. Labor climate 4. Availability of qualified labor 5. Production costs On a five-point scale, the company has determined the following weights for these factors: FACTOR WEIGHTS FACTOR WEIGHTS Transportation costs 5 Construction/land costs 3 Labor climate 1 Availability of qualified labor Production costs 4 The decision factor ratings for each criterion are given in the following table: ECISION FACTOR RATINGS FACTOR ALBANY INIANAPOLIS AKRON MASON 1 3 8 6 1 5 4 7 8 3 3 5 9 4 8 7 6 5 4 5 6 7 etermine the factor scores for all four locations. What is the recommended site? If the first site is not available, what is the second choice? 8-8 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

JOB SELECTION EXAMPLE Assume that you are a senior student majoring in Marketing. You have been involved in the job search process for the last few months. At this time you have three job offers from three different firms. Your potential job titles and locations of the jobs are presented below: 1. Pharmaceutical Sales Chicago. Marketing Research St. Louis 3. Advertising/Promotions San iego The salaries of the jobs are as follows: 1. Chicago: $37,000. St. Louis: $30,000 3. San iego: $40,000 The following table provides additional information about each job and company: COMPANY/JOB CHARACTERISTICS CHARACTERISTIC CHICAGO ST. LOUIS SAN IEGO Culture Formal Mix Informal Job Security High Medium Low Future Earnings & Advancement Potential Limited Medium High Job Expectations Reasonable Medium Very High a. etermine the weights for all the listed factors (5 being very important and 1 being very unimportant) and rate the importance of the factors from one to five. FACTOR # FACTORS WEIGHT 1 Type of Job Salary 3 Culture 4 Job Security 5 Future Earnings & Advancement Potential 6 Job Expectations 7 Location Proximity to Family & Friends 8-9 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

b. Complete the following table indicating your decision factor ratings for each criteria. FACTOR # CHICAGO ST. LOUIS SAN IEGO 1 3 4 5 6 7 c. etermine the factor score for each location. d. Based on your answer to part c, indicate the best choice and the second best choice. Solutions to Example Problems SOLUTION TO PROUCT SELECTION PROBLEM S micro = 4(4) + 5(8) + 3(6) + 5(3) + (9) + 1(1) S micro = 16 + 40 + 18 + 15 + 18 + 1 = 108 S refrigerator = 4(3) + 5(4) + 3(9) + 5(6) + () + 1(5) S refrigerator = 1 + 0 + 7 + 30 + 4 + 5 = 98 S stove = 4(8) + 5() + 3(5) + 5(7) + (4) + 1(6) S stove = 3 + 10 + 15 + 35 + 8 + 6 = 106 Therefore, the first choice is the microwave oven and the second choice is the stove. SOLUTION TO FACILITY LOCATION PROBLEM S Albany = 5(3) + 3(5) + 1(3) + (8) + 4(4) = 65 S Indianapolis = 5(8) + 3(4) + 1(5) + (7) + 4(5) = 91 S Akron = 5(6) + 3(7) + 1() + (6) + 4(6) = 89 S Mason = 5(1) + 3(8) + 1(9) + () + 4(7) = 70 Indianapolis is the first choice and Akron is the second choice.. Emergency Facility Location The factors considered in deciding where to locate a facility differ drastically among different types of facilities depending on the type of facility considered and the location of the existing facilities. For example, in deciding where to locate a distribution warehouse, a firm would consider a number of factors including the location of the existing warehouses, the location of 8-30 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

the manufacturing facilities that would supply the warehouse, and the location of existing stores that the warehouse would supply. On the other hand, if we were deciding the location of a new restaurant franchise, we would consider the location of existing restaurants, both competitive eateries as well as other branches of the franchise. We also would consider the population of the surrounding communities and, more importantly, the forecasted demand from each community. The factors considered in locating service facilities tend to be different than the factors considered in locating manufacturing facilities. In addition, there is further variation among the location factors depending on the type of service provided. In this section, we will solve emergency facility problems, where the most important factor (objective) is saving human lives. One way to minimize loss of life in the context of emergency facility location is through minimization of emergency response time. The response time is defined as the amount of time between receiving the emergency call and an emergency vehicle arriving at the site of emergency. The emergency facility objectives (methods) can be categorized as follows: 1. Minimization of maximum response time. Minimization of average response time 3. Minimization of weighted average response time In minimizing the maximum response time, we are concerned about the worst-case scenario response. In minimizing the average response time, the chosen location will be the site with the lowest average response time, even though the worst-case response may not be desirable. These first two objectives may be acceptable if the need or the demand for emergency vehicles do not differ among the different communities served. However, in many cases the demand for vehicles will differ among different locations. A retirement community or a community with a high crime rate experiences higher demand for emergency vehicles than a safer and younger community. The third category, minimizing weighted average response time, specifically considers variable demand for emergency vehicles. To utilize any of the above methods, we must first determine the feasible sites that require emergency service, and then we must decide on the proposed sites that can be used to locate the emergency facility. Finally, the response time between each site needs to be represented as a network with nodes and arcs where each node represents a community or location while each arc represents the estimated response time between two locations. A two-stage procedure is used in determining the location of an emergency facility using any of the three methods introduced above. The first stage is the same for all three methods and consists of determining the shortest route (response time) between each of the nodes in the emergency network. There are formal algorithms for determining the shortest route, however we will assume that the networks are simple enough to obtain all of the shortest routes using simple intuition. The second stage of the procedure depends on which method is utilized. The following section summarizes the second stage for each method. Minimization of maximum response time: After determining the maximum response time from each proposed location, we simply choose the location with the smallest of the maximum response times. Minimization of average response time: After determining the average response time from each proposed location, we simply choose the location with the smallest average response time. 8-31 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Minimization of weighted average response time: First, we determine the weighted-average response time using the following formula: Let : w j weight of location j (# of trips to location j) t ij shortestresponsetime from location i to location j WAT i weighted averageresponsetime from location i k number of l number of locations served proposedlocations WAT i k j1 k j1 w t j w j ij for all i 1,,...l The best way to develop a further intuitive understanding for emergency vehicle location is through solving problems. Therefore, in the next section we provide three simple example problems. Problem 1 The local hospital in northeast Ohio has just received approval from its executive board to build a new emergency facility. The facility will serve six communities in northeast Ohio. The response times in minutes are given in the network figure below. Network iagram 1 1 8 10 7 4 6 5 5 11 9 4 3 3 7 6 a. Based on the emergency response network given above, determine the location of the emergency facility that minimizes the maximum response time. b. Based on the emergency response network given above, determine the location of the emergency facility that minimizes the average response time. 8-3 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

c. Assume that community 4 is a retirement community and community 6 is a high crime area. Therefore, these communities experience higher demand for emergency vehicles than the other communities do. Based on historical data from last year, the average number of weekly emergency trips to the six communities is given in the following table. Average Number of Weekly Emergency Trips Community 1 Community Community 3 Community 4 Community 5 Community 6 8 trips 5 trips 10 trips 0 trips 1 trips 4 trips Use the weighted average response time criteria and the # of trips as the respective weights and determine the best location for the emergency center. Problem The city of Hampton is in the process of making a decision on where to locate a fire department that will serve the entire city. The city is divided into seven communities. The network of cities with fire truck response times is given in the following figure. 1 4 7 3 8 Network iagram 9 5 6 4 10 6 a. etermine the best location for the fire station based on the objective of minimizing the maximum response time. b. etermine the best location for the fire station based on the objective of minimizing the average response time. 5 3 7 6 7 8-33 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Problem 3 The mayor of a small town is planning to select one of three sites as the location for its ambulance facility. The town consultant has determined the following response times from each of the proposed locations to the four main areas of the town that must be served (North, South, East, and West). Area Served Proposed Site North South East West A 4 8 10 6 B 9 5 7 3 C 11 6 7 a. If the mayor selects the site for the ambulance facility based on minimizing the maximum response time, which site should be selected? b. If the mayor selects the site for the ambulance facility based on minimizing the average response time, which site should be selected? c. The number of ambulance trips to each of the four areas in the past year has been: North = 10; South = 80; East = 80; and West = 340. Use the weighted average response time and determine the best site location for the mayor. Solution to Problem 1 Shortest Response Time Matrix From 1 3 4 5 6 To Longest Response Time Average Response Time 1 0 8 9 15 10 13 15 9.17 8 0 10 7 6 9 10 6.67 3 9 10 0 9 4 7 10 6.50 4 15 7 9 0 5 8 15 7.33 5 10 6 4 5 0 3 10 4.67 6 13 9 7 8 3 0 13 6.67 a. Using the longest response time criteria, we can choose community, 3, or 5 (minimum maximum response time of 10 minutes). b. The calculation of average response time to community 5 is: 10 6 4 5 0 3 4.67 min. 6 Using the average response time method, we select location 5 (4.67 minutes is the smallest average response time). 8-34 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

c. 8(0) 5(8) 10(9) 0(15) 1(10) 4(13) WAT1 10.91 8 5 10 01 4 8(8) 5(0) 10(10) 0(7) 1(6) 4(9) WAT 7.49 8 5 10 01 4 8(9) 5(10) 10(0) 0(9) 1(4) 4(7) WAT3 6.56 8 5 10 01 4 8(15) 5(7) 10(9) 0(0) 1(5) 4(8) WAT4 6.9 8 5 10 01 4 8(10) 5(6) 10(4) 0(5) 1(0) 4(3) WAT5 4.08 8 5 10 01 4 8(13) 5(9) 10(7) 0(8) 1(3) 4(0) WAT6 5.5 8 5 10 01 4 Select location 5 because it has the lowest weighted average response time (4.08 minutes). Solution to Problem Shortest Response Time Matrix To From 1 3 4 5 6 7 Longest Response Time Average Response Time 1 0 4 7 9 13 15 17 17 9.9 4 0 8 5 14 11 13 14 7.86 3 7 8 0 9 6 9 11 11 7.14 4 9 5 9 0 9 6 8 9 6.57 5 13 14 6 9 0 3 5 14 7.14 6 15 11 9 6 3 0 15 6.57 7 17 13 11 8 5 0 17 8.00 a. Using the longest response time criteria, we choose location 4 because the minimum longest response time is 9 minutes. b. The calculation of average response time to location 6 is: 15 11 9 6 3 0 6.57 min. 7 Using the average response time method, we select location 4 or 6 (6.57 minutes is the smallest average response time). 8-35 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill

Solution to Problem 3 a. Maximum response times from proposed locations A, B, and C are 10, 9, and 11, respectively. Choose site B because it has the lowest maximum response time. b. Average response times from proposed locations A, B, and C are 7, 6, and 6.5, respectively. Therefore, because 6 < 6.5 < 7, choose site B. c. 10(4) 80(8) 80(10) 340(6) WATA 6.78min. 10 80 80 340 10(9) 80(5) 80(7) 340(3) WATB 4.95min. 10 80 80 340 10(11) 80(6) 80() 340(7) WATC 6.76 min. 10 80 80 340 Choose site B because it has the lowest weighted average response time (4.95 minutes). 8-36 Copyright 015 McGraw-Hill All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill