CHAPTER 13FACILITY LOCATION DECISIONS 1 (a) The center-of-gravity method involves finding the X,Y coordinates according to the formulas: !V R X X= !V R i i i i i i i and !V R Y Y = !V R i i i i i i i These formulas can be solved in tabular form as follows. Point P1 P2 M1 M2 M3 X Y 3 8 8 2 2 5 6 4 8 8 Totals Vi 5,000 7,000 3,500 3,000 5,500 Ri .040 .040 .095 .095 .095 ViRi ViRiXi ViRiYi 200.0 600.0 1600.0 280.0 2240.0 560.0 332.5 665.0 1662.5 285.0 1710.0 1140.0 522.5 4180.0 4180.0 1620.0 9395.0 9142.5 Now, X= and 9,395.0 = 5.8 1,620.0 Y = 9,142.5 = 5.64 1,620.0 This solution has a total cost for transportation of $53,614.91. This problem may also be solved using the COG module in LOGWARE. (b) Solving for the exact center-of-gravity method requires numerous computations. We now use the COG module of LOGWARE to assist us. A table of partial results is shown below. 160 Iteration number 0 1 2 3 . . . 50 X coord 5.799383 5.901199 5.933341 5.941554 . . . 5.939314 Y coord 5.643518 5.518863 5.446919 5.402429 . . . 5.317043 Total cost 53,614.91 ⇐ COG 53,510.85 53,483.97 53,474.60 . . . 53,467.71 After 50 iterations, there is no further change in total cost. The revised coordinates are X = 5.94 and Y = 5.32 for a total cost of $53,467.71. (c) The center-of-gravity solution can be one that is close to optimum when there are many points in the problem and no one point has a dominant volume, that is, has a larger volume relative to the others. Otherwise, the best single location can be at a dominant location. The exact center-of-gravity approach has the capability to find the minimal cost location. Although the COG model only considers transportation costs that are constant per mile, the transportation cost can be the major consideration in single facility location. However, other costs such as labor, real estate, and taxes can also be important in selecting one location over another. These are not directly considered by the model. Although the COG model may seem of limited capability, it is a useful tool for locating facilities where transportation costs are dominant. Location of oil wells in the Gulf, truck terminals, and single warehouses are examples of application. It also can be quite useful to provide a starting solution to more complex location models. (d) Finding multiple locations by means of the center-of-gravity approach requires assigned supply and demand volumes to specific facilities and then solving for the center of gravity for each. In this problem, there are 3 market combinations that need to be considered. This creates three scenarios that need to be evaluated. They can be summarized as follows. Point volumes appear in the body of the table. Scenario I 2 1 II 2 1 III a Whse 1 P1 1458a 3542 2708 2292 3750 P2 2042 4958 3792 3208 5250 M1 3500 M2 3000 M3 5500 3500 3000 5500 3500 5500 2 1250 1750 3000 Allocated as a proportion of the volume to be served through the warehouse. That is, 5000×3500/(3500 + 3000 + 5500) = 1458. The volumes associated with other supply points are computed similarly. 161 The COG module in LOGWARE was used to find the exact centers of gravity for each warehouse in each scenario. The computational results are: Scenario I II III Warehouse 1 X Y 2.00 5.00 5.84 4.04 7.06 7.28 Warehouse 2 X Y 7.88 7.80 8.00 8.00 6.00 4.00 Total cost $39,050 35,699 ⇐ 46,568 Scenario II appears to be the best. 2 (a) The center-of-gravity formulas (Equations 13-5 and 13-6) can be solved using the COG module of LOGWARE, or they can be solved in tabular form as shown below. Point A B C D E F G H I J X Y 50 0 10 10 30 15 40 20 10 25 40 30 0 35 5 45 40 45 20 50 Totals Vi 9,000 1,600 3,000 700 2,000 400 500 8,000 1,500 4,000 Ri .75 .75 .75 .75 .75 .75 .75 .75 .75 .75 ViRi 6,750 1,200 2,250 525 1,500 300 375 6,000 1,125 3,000 23,025 ViRiXi 337,500 12,000 67,500 21,000 15,000 12,000 0 30,000 45,000 60,000 600,000 ViRiYi 0 12,000 33,750 10,500 37,500 9,000 13,125 270,000 50,625 150,000 586,500 Now, X= and 600,000 = 261 . 23,025 Y = 586,500 = 25.5 23,025 The total cost of this location is $609,765. The exact center-of-gravity coordinates are: X = 2351 . , Y = 26.98 with a total cost of $608,478. (b) The number of points, even in this small problem, requires us to apply some heuristics to find which patient clusters should be assigned to which warehouses. We will use a clustering technique whereby patient clusters are grouped by proximity until two clusters are found. The procedure works as follows: 162 • There are as many clusters as there are points, which are 10 in this case. • The closest points are found and replaced with a single point with the combined volume located at the center of gravity point. There is now one less cluster. • The next closest two points/clusters are found, and they are further combined and located at their center of gravity. • The process continues until only two clusters remain. The centers of gravity for these two clusters will be the desired clinic locations. Applying the clustering technique, we start by combining points D and F into cluster DF. X= and 40( 700)( 0.75) + 40( 400)( 0.75) = 40.00 700( 0.75) + 400( 0.75) Y = 20( 700)( 0.75) + 30( 400)( 0.75) = 23.64 700( 0.75) + 400( 0.75) Continuing this process, we would form two clusters containing A, C, D, and F, and B, E, G, H, I, and J. The centers of gravity would be: Cluster 1 - ACDF X = 50.00, X = 0.00 Cluster 2 - BEGHIJ X = 5.00, Y = 45.00 for a total cost of $241,828. These are the same results obtained from the MULTICOG module in LOGWARE. (c) The second clinic can save $608,278 − 241,828 = $366,458 in direct costs annually. This savings does not exceed the annual fixed costs of $500,000 required to maintain a second clinic. On economic grounds, it should not be built. 3 (a) The center-of-gravity location can be determined by forming the following table or by using the COG module in LOGWARE. The coordinates for each location must be approximated. 163 930 ViRiYi 6.4 1.080 7.000 40.966 (c) Additional costs can be included in the analysis.0 5.000/2)0.10 .400 ViRiXi 1.000 3.10 .000 10.2 + 2 + 4 = 9.5 6.000 30.5 1.500 3.400 with a total cost of $195.350 56.10 .1 1.75.000 90.260 1. The cost matrix is developed in the same manner as that in text Figure 13-11.10 .10 .367.0 4. although not necessarily in the model.200 13.000 5.5 4.8 Totals Vi 10.000 27.700 12.10 .Point A B C D E F G H I X Y 1.34 27. The COG model can evaluate the variable costs of location.7 7.650 49. the exact center of gravity was found to be: X = 4.000 10.000 19.000 Ri . Y = 4.930 = 4.000 4.600 6.7]/(200.966.966 − 195.367 × 100 = 0.700 1. The cell cost for W1-C1 would be: [100(200.120 11.000 500 7.9 8.500 2.000 4.000/2 )+ 2 + 4 + 0 = 3. (b) After 100 iterations in the COG module.800 112.3 4.10 .430 = 410 .8 1.000 7. using the exact center of gravity coordinates as compared with the approximate ones reduced costs by only: 195.400 Y = 112.600 3. 4 We begin by developing a 3-dimensional transportation problem.10 .6 4.000 12.1 1.000 1.8 7.430 The center-of-gravity coordinates are: X= and 118.62 with a total cost of $195.200 9.000 700 1.500 20.000 70.2 164 .3% 195.3 8. In this case. 27.5 5. The initial throughput of W1 and W2 is found by assuming that an equal amount of the customer demand flows through each warehouse.800 118.0 7. Other costs are compared with these.10 ViRi 1. 2 W2 799. Using the transportation method of linear programming (e.000 60.999 50. TABLE 13-1 Summary Information for Solution to Problem 4 Whse 1 Whse throughput Costs: Transportation Inbound Outbound Inventory Warehousing Fixed Production Total 0 Whse 2 200. Solving the problem by means of the transportation method shows the solution given in Figure 13-2.000 100. and the product is produced in plant P2 and stocked in warehouse W2.000 50.999 0 200. No further iterations are needed since only one warehouse is used and no further dropping of warehouses is possible. as shown in Figure 13-2. the TRANLP module in LOGWARE). 5 We begin by forming the cell cost matrix of a three-dimensional transportation problem.000 0 0 0 99 0 6.000.000 100. The total cost is $2.2 10. the cell costs are identical to those in text Figure 13-11.213.In fact.000 300.000 999. the cell cost and solution matrix for iteration 1 is shown in Figure 13-1.213.000 a 0 99 9.2 6. FIGURE 13-1 Cell Cost and Solution Matrix for Iteration 1 of Problem 4 Warehouses Customers W1 W2 C1 C2 C3 Capacity a 4 9 99 99 99 60.999 50.2 8. The solution shows that only W2 remains and the solution process can be terminated. Stop iterating.000 513.714 Iteration 2 A repeat of the iteration 1 solution.999 Capacity/ Req’mts 60.714 200.714.2 Whses W1 60.000 Plants P1 0 0 8 6 99 99 99 P2 999. 165 .2 5.000 999.000 50. A summary of the costs is shown in Table 13-1.g..000 0 800.000 $2. except that there is no fixed cost element.000 a High rate of $99/unit for an inadmissible cell.000 $0 0 0 0 0 0 $400. It is similar to the text Figure 13-1 except that the capacity for warehouse 1 is set at 75. 2 W1 W2 Revising the warehouse-customer cell costs and solving gives the same warehouse throughputs.000 Capacity/ Req’mts 999.000)0.00/unit $400.000 Plants P1 60.000 100.000 Given the solution from iteration 1.000 50. so cell costs will no longer change. A stopping point is reached.7 = $3. A summary of the costs is: 166 .00/unit Adding outbound transportation and warehouse handling to per-unit inventory and fixed costs gives the following cell costs.000 100.2 9.999 P2 40.000 units W2 Fixed W1 W2 $100.000 100.2 10.999 Whses W1 899.000 units $100(100.000 100 0 8.7 = $3.16 / unit 100.000 = $1.2 8.000/100. C1 10.000 50.FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 5 Warehouses Customers W1 W2 C1 C2 C3 Capacity 4 9 100 100 100 60.16 / unit 100.2 C3 11.2 W2 50.2 7.000 = $4. The solution is the same as that in Figure 13-2.000 0 100 9. Inventory W1 $100(100.7 10.999 100.999 100.000 50.000 8 6 100 100 100 999. the per-unit inventory and fixed costs are revised.2 C2 9.7 999.000)0.7 8.000/100.2 10. 000/200.000 Warehouse 2 100.714 = $478. $200.7 = 316.000 100.999c W2 Warehouse capacity & customer demand a b 60.000×2 = 50.228 $3.456 − 2.228 100(100. 100.000 100(100.000 cwt.2 + 2 + 4 + 0. warehousing.000 100.228 100.000 8 P2 Warehouses W1 0 60.000 100. c Used to represent unlimited capacity.7 60.000 100.000/200.000 99b 6 200.000 40.999 99 9. e 3.999c 50. d Inventory carrying.2 50.7 d W1 4a Plants P1 W2 9 C1 99b C2 99 99 8.000 50.000 $1.000 7.000)0.000×2 = 200.2 + 1 + 2 + 1 = 7.456 Compared with the costs from the text example.613.2.000×2 = 50.000 400. 167 .476.616.2 100.7 6.000 = $2/unit.000 Production plus inbound transport rates.2e 50. This is the penalty for restricting a warehouse with economic benefit to the network.000×2 = 200. 3.000 = $1/unit instead of $400.7 = 316.000×1 = 100. the cost difference is $3.000 C3 99 99 999. Used to represent an infinitely high cost. 6 Prepare a matrix for a 3-dimensional transportation problem like that in text Figure 1311. 60.000 cwt.000 100.000 60.000 100. that is.092.228 $1.742.000×4 = $400.000)0.000×3 = 300.000×4 = $240.999c 10. The matrix setup and first iteration solution are shown in Figure 13-3.000 999. That is.5 = 9. that is.000 100.000×4 = 160. FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 6 Warehouses Customers Plant & warehouse capacities 60. and fixed rates.000×0 = 0 40. except that the per-unit cell costs for warehouse 2 to customer are reduced by $1/unit to reflect the reduction in that warehouse’s fixed costs. outbound transportation.000 99 0 799.000 7.000×4 = 160. 4 + 0 = 4.000 999.092.7.Cost type Production Inbound transportation Outbound transportation Fixed Inventory carrying Handling Subtotal Total Warehouse 1 100. 000.714. then the fourth year alternatives are evaluated.The results show that one warehouse is to be used. The total network costs are the same as those in the text example minus the $200.524 = 1.324 The best action in the beginning of the fifth year. solve for the number of locations from one to nine.676 144.924 = 1. This is an entry in Table 13-2.341. The action will be to move (M) or to stay (S). as further warehouse consolidation is not possible.676 Net profit = $1.398. we evaluate each course of action.312.611. 9 (a) Using PMED software in LOGWARE and the PMED02.413.336. The best locations for each number of sites are given in the table below.000 1.336.924 = 1.526. if we are already in location A.000 Moving cost 0 144. given the discounted moving cost of 300.000/(1 + 0.000 reduction in fixed costs for a total coat of $2.200 1. We now include the profits for the subsequent years in our calculations.600 1.2)(3) = $173. This is initially to locate in location D and remain there throughout the subsequent years. The only change is that the cost of moving from one location to another is now $300.20)(4) = $144.676.486. After Table 13-2 is completed.000 instead of $100. 8 This problem requires us to rework the dynamic programming solution to the example problem given in the text.000/(1 + 0.676 144. we generate the following table for location A.381.600 1. assuming that we are in location alternative A at the end of year four.457. For example. We begin with the last year and determine the best action based on the highest net profits.DAT database. 168 .000 = 1. Alternative (x) A B P5(A) = max C D E Location profit $1. Once each of the five alternatives is evaluated for the fifth year.676 144. we search the first column for the highest cumulative profit. The moving cost is 300. is to move to location E. Further computations are not needed.253. From the location profits of text Table 13-6. 767 SA $3.355.268.379.556.007.457.557.100 SE a Strategy symbol refers to “staying” (S) in the designated location or “moving” (M) to a new location as indicated. b Arrows indicate maximum profit location plan when warehouse is initially located at D.600 D b 3.TABLE 13-2 Location-Relocation Strategies Over a Five-Year Planning Horizon with Cumulative Profits Shown from Year j to Year Five for Problem 8 Year from present date j Warehouse 5th 1st 4th 3rd 2nd location StraStraStraStraaltertegya P2(x) tegya P3(x) tegya P4(x) tegya P5(x) P1(x) natives (x) A $3.800 SE 1.007.459.268.800 SC 1.767 SA $3.319.381.900 SC 3.673.553.000 C 3.200 SC 2.156.289 MD 1.200 SC 3.667 SB 3.720.526.300 SD 3.500. Strategya ME SB SC SD SE 169 .374.486.534.000 SD 2.667 MD 2.600 3.324 B 3.363.900 SD 1.289 MD $1.071.167 SB 3.700 SE 3.000 E 3.398.667 MD $2.600 SD 3.300 SE 2.216. Cincinnati.740 = × 100 = 35. 170 . Chicago.360 30.836. (c) Increase the annual volume for Los Angeles and Seattle markets by a factor of 10 and re-solve the problem as in part a. Denver.035. 10 (a) We can apply Huff's model of retail gravitation to this problem. Selected results are as follows: Number of sites 6 7 8 9 Total cost 30.340 24.624.410 25.660 without any major investment.340 = $4. branch A can be expected to attract 11.Number of sites 1 2 3 4 5 6 7 8 9 Total cost $95. Phoenix.720 25.780. and Seattle.152.060 Seven sites is optimal.400 37.340 − 23.624. and branch B should attract the remaining 50.340 as found in part a.735 + 11. The savings better the optimal four sites and the optimal seven sites is $30. An additional site at Los Angeles is needed.163.250 27.600 = $6.000.382.468.5% Investment 18.311. The company operates the same sites as found in the optimal solution for four sites. For an additional three sites.230.250 27.018.060 The optimal number of sites is eight.040 24.9 percent of the customers.163.000.163. the cost savings comes from a between assignment of customers to the sites. Summarizing. the simple return on investment is: ROI = Savings 6.018. (b) The optimal cost for four sites is $30.740.220 49.000 − 30.760 23. The solution table (Table 13-3) can be developed. The savings of $35.735/(11. Therefore.1 percent. compared with the seven locations found in part a. and they are to be located at New York.892.163.739.600 25.00 This is a benefit certainly worth considering.382.780.765) = 49. 25 0.16 0.5 2.4 2.15 0.20 0.20 0.440 300 480 11.74 0.of customers) Operating expenses Profit $1.10 0.67 B 0.820 850 885 1. and the reputation of the bank may be just as important in estimating patronage.700 520 11.4 !S j S j / Tij2 j / Tij2 B 0.04 0.50 0. we create a table showing the counties that are adjacent to each county.52 Eij = Pij Ci A 900 1.28 0.TABLE 13-3 Estimate of the Number of Customers Attracted to Each Branch Bank Pij = Time to ja Customer i 1 2 3 4 5 6 7 8 9 a Tij2 A 0. 171 .173.500 300.0 5.560 1.000 = 116.26 0.08 0.01 0.25 0.64 0.57 0.8 3. How likely is this to happen? Are the customer numbers stable? Will a third or fourth bank be locating branches in the region? Can we expect that customers will drive such long distances to seek banking services? 11 This problem can be solved as an integer linear programming problem similar to the Ohio Trust Company example in the text.08 0.8 17.15 0.45 0.08 0.5 100.735 B 100 60 880 5.000 $ 873. The nature of the services offered.5 4.4 2.50 0.0 3.78 0.000/750.20 0.32 0. First.5 6.41 0.59 0. That is.1 2. so the branch should be constructed.940 3.0 4.85 0.29 S j / Tij2 A 12.765 A 0. (c) The size of a branch and its proximity to customers may be too simple to explain the market share of each.40 0.3 25.45 B 0.25 0.180 150 625 2. What will the countermoves be of the competing branch? Adding another branch and locating near branch A could substantially reduce its market share.45 0.16 0.2 B 1.120 1.8 8.85 0.50 0.97 0.03 0.28 0.22 0.28 0.36 0.04 0. the accessibility of the site.40 0.04 0.4% The ROI seems sufficiently high.0 12.54 A 0.000 Return on investment 873.90 0.20 0.72 0.10 0.5 17.48 Time = ( X i − X )2 + (Yi − Y )2 / 50 (b) The economic analysis of the sites would be: Revenue (100×No.20 0.52 0. 8 1. Putnam 11.5. and Auglaize counties.7 3.7.12 4. An X of one means that the branch is located in the county.19.15.20.15.DAT.10.14. Sandusky 9.3.8. Marion 20.10.19 13.6. Wood.7. Lucas 4.14. The problem formulation is shown in Table 13-4.Counties under consideration 1.8. according to the problem formulation given in the Ohio Trust Company example.10.11.18 11.6.14 6. Williams 2.6.16 18.11.11.4. Auglaize 19. Hancock 12.14.10.18 13. Mercer 18.7.3.15. Shelby 21.18. The Xs take on the values of zero or one.20 Next.11 3. They should be located in Henry.16.16 7.21 11.17. Wyandot 17.9. Allen 15.12.6 1.12. we can build the matrix as given in the prepared database called ILP03.13.15. Van Wert 14.16 9.7. Paulding 10.21 15. Henry 7.21 15.18 10.12 5. Wood 8. A coefficient of one is given to each county and its adjacent counties. 172 .6. Hardin 16. Fulton 3.11. Solving this problem using the integer-programming module in LOGWARE shows that a minimum of five principal places of business are needed.17. Ottawa 5.13.9. Putnam. Logan Adjacent counties by number 2.10.4.14.2. Defiance 6.7. The sum of all constraints must be one or greater.10 1. Hardin. This matrix can be solved by the integerprogramming module (MIPROG) in LOGWARE for solution.18.13 5.5. Note that all coefficients are ones or zeros. Seneca 13.6 2. TABLE 13-4 Coefficient Matrix Setup for Problem 11 Obj Fun Constraints Williams Fulton Lucas Ottawa Defiance Henry Wood Sandusky Paulding Putnam Hancock Seneca Van Wert Allen Hardin Wyandot Mercer Auglaize Marion Shelby Logan X1 1 1 1 X2 1 1 1 1 X3 1 X4 1 X5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X6 1 1 1 1 1 1 1 X7 1 X8 1 X9 1 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 1 1 1 1 1 1 1 1 1 1 1 1 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ RHS 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ≥1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 173 . 000 2.196 5.750 713 0 1.000 Miami 35. A database has been prepared for it called PMED04.067 2.751 2.775.041. an estimation of vendor to laboratories transportation cost.079.067 2.000 10.886.000.067 2.241 2.000 L.067 2.000 L.750 15.479.886.351.696. The fixed cost must be adjusted for the number of sites being evaluated.500 FOC.913. The vendor to site transportation cost is included externally.180.000 3.000x310x0.000 10.000 Total 680.852.241 2. Calculating the distances between vendor and the selected sites easily can be done by using the MILES module in LOGWARE with a scaling factor of one.000 Atlanta 145.535. of locations 1 2 Outbound cost.000 Total 680.000 4 New York 200.250 Inbound cost.067 11.500.122.000 2.000 6 New York 200.641.041.02=3.247. as shown in the previous table. and an enumerative search.041.750 7. The fixed operating cost must be calculated for each possible number of locations.041.000 L.446 Total cost.852. It should be recognized that the model handles only the outbound leg of the network (sites to serve laboratories). inbound transport costs are a product of site volume. the FOC must be recalculated and entered into the database.775.000 1.220.126.968.000 Chicago 170.515. When other numbers are to be evaluated.000 5 New York 200.193.241 2.534 The PMED program is used to find the best combination of sites for a particular number of sites to be found.170. Angeles 150.500 826.000 Total 680.250.000 3.636. lb.568 28.000 Chicago 170.000 10.12 This problem can be solved with the aid of the PMED program in LOGWARE.000 1.000.041.000 Chicago 280.193.000. The scaling factor is set at one for this problem.029.241 2. $ 33.249 713 585 1180 0 790 1.250 713 585 0 790 1.241 12. distance.000 0 948.000 Atlanta 100. $ 5.083.000 Total 680.000 Dallas 60. Then. 174 .100 0 5. $ 28.000 Dallas 60.236.068 2.236.071.000 L.660.000 0 948.350. Using PMED in LOGWARE gives the following results based on recalculated FOC values.236. mi.750 5.100 2.000 2.584 27.500.350.534 7.999 713 585 0 1.000 2.352 27.000 2.751 2.000 8.000 1 515.000 9. Angeles 165.000 5.500.500.0001 5.500 10.750 27.041. 0 310 1.000 31.775.236. Volume.500 0 5.852. The database shows the fixed costs for a single site.000 2 / 2 = 3.535.000 L.500 2. $ 0 3.000 Atlanta 65.000 Total 680.886.751 8.000 5.DAT.535.428. Angeles 165.000 10.000 Chicago 170.750 6.000 3 New York 235.241 2.250.828.236. and the inbound rate.000 Cleveland 515. Sites Chicago 680. Angeles 165.253 2. Angeles 150.000 Inbound distance.000 2 No.5342 3.313.335 2.000 760. 9. 18.479. and 12 4 and 5 6. 2. The total cost for 5 sites is $27. 3. and 16 15. 7. 175 .000 150.000 60. We select five sites as economically the best number. Their customer assignments are Location number 1 2 3 4 5 Assignments New York Atlanta Chicago Dallas Los Angeles Volume 200. the following solution is found.000 170.584. 13 After changing the fixed costs in the problem setup matrix. and 20 A map of the solution is as follows.Searching from one to N sites shows that outbound transportation costs decrease while inbound and fixed costs increase with increasing numbers of sites. 19. 8. 10. 17.000 100. Initially. 14.000 Customers 1. total cost declines until five sites are reached after which total cost increases. and 11 13. 14 (a) The demand of customer 1 for product 1 increases to 100.0000 . In the problem matrix of ILP02.0000 140000.0000 40000.0000 260000.0000 1.0000 7.0000 .0000 . yW1C1 Cap-W2.0000 11.0000 1.0000 1..0000 .0000 .0000 90000.0000 7.0000 110000.0000 .0000 20000.0000 .0000 130000.0000 400000.0000 Rate 8.0000 70000.0000 140000.0000 .0000 .0000 . yW2C1 Obj.0000 100000.0000 500000.0000 .0000 .0000 .0000 9.0000 .0000 .0000 .0000 1.0000 .0000 .0000 11..0000 .0000 10.0000 7.0000 . yW2C1 Obj.0000 8.0000 . However.0000 100000.0000 6.0000 .0000 10.0000 11. yW2C1 From -50000 -50000 70000 70000 140000 70000 To -100000 -100000 120000 120000 240000 120000 The result shows that warehouse 2 is still the only warehouse used.0000 .0000 180000. 176 .0000 7.0000 990000.0000 .0000 .0000 220000.0000 10000.0000 .0000 3700000.0000 8.000.0000 70000.0000 5.0000 11.0000 6.0000 13.0000 . and the products are sourced from plant 2.0000 7.0000 12.0000 . yW2C1 Cap-W1.0000 .0000 .0000 450000.0000 Variable label P1S1W1C1 P1S1W1C2 P1S1W1C3 P1S1W2C1 P1S1W2C2 P1S1W2C3 P1S2W1C1 P1S2W1C2 P1S2W1C3 P1S2W2C1 P1S2W2C2 P1S2W2C3 P2S1W1C1 P2S1W1C2 P2S1W1C3 P2S1W2C1 P2S1W2C2 P2S1W2C3 P2S2W1C1 P2S2W1C2 P2S2W1C3 P2S2W2C1 P2S2W2C2 P2S2W2C3 zW1 zW2 yW1C1 yW1C2 yW1C3 yW2C1 yW2C2 yW2C3 Objective function value = Note that warehouse 2 is no longer used in favor of all products flowing through warehouse 1.0000 .0000 30000.000 cwt.0000 .0000 9.500. Cell Dem P1W1C1.DAT the following cell values are changed. coef. the costs have increased to $3.0000 140000.0000 220000.0000 .0000 .0000 .00 Cost .0000 50000.0000 50000.0000 .0000 .OPTIMAL SOLUTION Variable X(1) = X(2) = X(3) = X(4) = X(5) = X(6) = X(7) = X(8) = X(9) = X(10) = X(11) = X(12) = X(13) = X(14) = X(15) = X(16) = X(17) = X(18) = X(19) = X(20) = X(21) = X(22) = X(23) = X(24) = X(25) = X(26) = X(27) = X(28) = X(29) = X(30) = X(31) = X(32) = Value .0000 260000.0000 150000. coef.0000 .0000 8.0000 600000.0000 10.0000 11.0000 20000. yW1C1 Dem P1W2C1. .000) × $4/cwt.000 110.000 90.DAT can adjust the capacities on plant 1. plant 1 supplies warehouse 1 and customer 1 with 20. The remainder flows from plant 2 through warehouse 2 and on to customers 2 and 3. P2S2W2C1 Obj. Cell Obj. The remaining 90. (c) Using the ILP02. (d) Making some slight revisions in file ILP02.DAT file in MIPROG of LOGWARE. P2S2W1C1 Obj.000 cwt. Coef. For product 2.000 cwt..999 To 150.e.000a 130. Coef. yW2C1 Obj. Coef.000. However. The cell changes to make are: Cell Cap-P1S1. (50. summarized as: The solution can be 177 ..000. and the products are sourced from plant 2..920. P2S2W2C2 Obj. Coef. P2S2W2C1 From 9 8 10 7 6 7 To 12 11 13 10 9 10 The result shows that warehouse 2 is still the only warehouse used. P2S2W1C2 Obj.000 520. yW2C2 Obj.000 999. The total cost is $3. the changes are made to the following cells. Coef.000 The sum of demand for the same customer for all products multiplied by the handling rate. Coef. flows from plant 2 through warehouse 2 to customers 2 and 3. Coef. the following changes are made to the following cells.DAT file in MIPROG of LOGWARE.380..000 cwt.000 + 20.. i.. Cell Obj. RHS From 60.000 To 280. The solution for product 1 shows that 50.. Coef..000 Both warehouses are now used for both products.000 440.000. P2S2W1C3 Obj. = 280. Coef.(b) Using the ILP02. flows from plant 1 through warehouse 1 and on to customer 1.. yW2C3 a From 70. RHS Cap-P1S2. the costs have increased to $3. 000 3 – 50.9.000 cwt.9.000 3 – 60.000 2 – 60. P2S2W2C3 to have a very high cost (999).10 1.2.4.3. these cells are locked out of consideration. The solution is the same as the text example except that both the customers products are served from warehouse 1.6.7.2.3.7. Coef.000 2 – 40.5.7.6.10 2. 2 – 60.2.5.3.9.6.000 2 – 30. coef.000 Warehouse 1 – 50. identify the zones that are within 30 minutes of any particular zone.9.000 Customer 1 – 50.000.10 1.2.9.000 2 – 30.000 1 – 20.4.4.4.4.000 2 – 50.3.Product 1 1 1 1 2 2 2 Plant 1 – 50.000 2 – 30.000 2 – 40.2.000 cwt.270.8 4.5.7.000 The total cost is $3. Zone no. 178 .000 1 – 20.5.000 2 – 60.10 1.3.000 1 – 20. That is.5.10 3.000 2 – 60.000 (e) Again revising the ILP02..7. First.3.8.. 1 2 3 4 5 6 7 8 9 10 Zones within 30 minutes 1. 1 – 60.8.2.340.7.8 1. The total cost is $3. P2S1W2C3 and Obj.8.8. 15 This problem follows the form of the Ohio Trust Company example in the text. the following matrix can be defined.5.9.000 2 – 40.9.6.4.7.6.10 1.8.DAT file by changing cells Obj.000 2 – 50.10 Using the MIPROG module in LOGWARE.10 1. 500. or P4.0000 X(5) = .0000 .0000 X(4) = 1.0000 1. A minimum of nine trucks are required to meet all constraints on the problem.0000 1. The total daily cost for this location is the route cost + vehicle costs + yard operating cost.0000 X(7) = .0000 1.The solution from MIPROG is: OPTIMAL SOLUTION Variable Value Rate X(1) = 1.87. A database file for this problem (RTR13.0000 1.0000 1.0000 1.DAT) has been prepared. Optimizing the routing from the current material yard gives the route design shown in Figure 13-3.0000 .87 + 9 x P200 + P350 = P6.0000 .0000 X(8) = .0000 X(2) = . Solving this problem requires balancing the cost of transporting the merchandise to the customers with the operating cost of the material yards at various locations.0000 X(3) = . 16 This is a location problem where the dominant location factor is transportation cost is and this cost is determined from optimizing the multi-stop routes originating at the material yard.0000 X(6) = .0000 1.0000 X(10) = .0000 .0000 1. The ROUTER module in LOGWARE can be used to generate these routes for each yard location.0000 .650.0000 1.0000 .0000 1.0000 Objective function value = Cost 1.0000 2.0000 .0000 .00 Variable label 1 2 3 4 5 6 7 8 9 10 The optimal solution is to place claims adjuster stations in zones 1 and 4.0000 1. FIGURE 13-3 Optimized Routing from Current Material Yard Location 179 .0000 X(9) = . 65 4500.8 46.5% 9999 0 0 8 1 1000 1000 0 100.9 2.0% .9 6.3 1.8 08:00AM 05:35PM 4 189 5.5 1.6 1.0% . FIGURE 13-4 Route Design for Yard Location A 180 .9 2. The total daily cost for this location is 3872.3 1.0% .0 .0% Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Substituting yard location A for the current yard location and solving for the route design in ROUTER yields Figure 13-4.0 5.2 1.0 2. so the minimum number of trucks remains at nine.0% 9999 0 0 6 1 1000 950 0 95.0% .0% .6 6.8 2.48 311.152.5% 9999 0 0 3 1 1000 900 0 90. time.0 5.6 5.3 08:00AM 05:22PM 4 185 9.092.0% .0% . time.2 08:00AM 05:34PM 2 214 9.9 1.6 1.0 2.0 6.05 599.3 08:00AM 05:37PM 3 204 9.96 625.0% 9999 0 0 5 1 1000 900 0 90. time.1 30 1476 Route no 1 2 3 4 5 6 7 8 9 Total Route cost.64 552.0 3.2 08:00AM 03:24PM 3 130 9. time.5 2.1 08:00AM 05:58PM 4 207 7.25 608.1% 89991 0 0 Cube util .5 08:00AM 01:56PM 3 89 74.7 8.0% 9999 0 0 7 1 1000 825 0 82. No routes can be put end-to-end so that one truck can be used instead of two.87 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 925 0 92.0 1.$ 602.4 2.0% 9999 0 0 4 1 1000 950 0 95.0 31.5 1.0% 9999 0 0 Total 9000 7925 0 88.0% .1 20.79 563.4 1.8 2.0 4.8 .0 08:00AM 05:56PM 4 205 3.0% 9999 0 0 9 1 1000 850 0 85.7 1.Mi 9.4 5.4 4.5% 9999 0 0 2 1 1000 625 0 62.59 416.7 1.*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION Route Run Stop Brk Stem time.6 6.0 6.0% .1 2.02 +9 x 200 + 480 = P6.4 2.45 220.7 08:00AM 11:25AM 3 52 10. Start Return No of Route hr hr hr hr hr time time stops dist. 15 321.42.04 410.4 5.*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION Route Run Stop Brk Stem time.0% .3 2.02 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 475 0 47.0 3.2 1.5% 9999 0 0 6 1 1000 1000 0 100.0% .2 4.4 1.0 2.0 25. Start Return No of Route hr hr hr hr hr time time stops dist.0% .1 2.0% 9999 0 0 3 1 1000 1000 0 100. time. nine trucks are required and the total daily cost for the route design in Figure 3 is 3370.5% 9999 0 0 2 1 1000 950 0 95.42 + 9 x 200 + 450= P5.620.0% 9999 0 0 Total 9000 7925 0 88.0% .0 3.9 5.6 4.0% 9999 0 0 4 1 1000 975 0 97.7 2.5% 9999 0 0 5 1 1000 875 0 87.7 08:00AM 04:51PM 3 183 5.4 5.57 470.82 3872.7 2.8 2.0% Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Continuing with location B. time.5 30 1225 Route no 1 2 3 4 5 6 7 8 9 Total Route cost.6 08:00AM 02:18PM 3 93 5.4 1.1 1.27 254.3 20.2 1.3 1.8 7. FIGURE 3 Design for Yard Location B 181 .68 227.6 08:00AM 01:03PM 3 55 68.9 1.8 08:00AM 04:26PM 3 163 8.3 2.7 9.0 .0 1.0% .1 1.0% 9999 0 0 7 1 1000 875 0 87.88 498.1% 89991 0 0 Cube util .4 08:00AM 05:47PM 4 224 8.0 38.0 2.0% .0 5.6 08:00AM 03:12PM 3 128 6.0 2.$ 649.40 491.0 2.2 08:00AM 04:37PM 5 152 7.5% 9999 0 0 8 1 1000 825 0 82.Mi 9.8 1.6 08:00AM 01:15PM 3 66 8.0% .5% 9999 0 0 9 1 1000 950 0 95.9 08:00AM 04:26PM 3 161 8.0% .0% .9 2. time.3 1. time.1 2.0 1.0 3.0 1.22 548. 46 557.9 1. Start Return No of Route hr hr hr hr hr time time stops dist.42 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 825 0 82.5 2.0% .8 30 1024 Route no 1 2 3 4 5 6 7 8 9 Total Route cost.84 300.0% .97 311.0 08:00AM 05:40PM 5 187 61. FIGURE 4 Design for Yard Location C 182 .9 2.9 2.0% 9999 0 0 5 1 1000 925 0 92.8 08:00AM 02:27PM 3 100 7.53 454.0% .*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION Route Run Stop Brk Stem time. time.3 08:00AM 03:06PM 4 120 6.0 08:00AM 03:56PM 4 146 5.0 2.0 20. time.6 2.3 1.8 2.5 3.6 2.3 1.5% 9999 0 0 2 1 1000 950 0 95.0% .699.1 3.3 08:00AM 02:41PM 3 114 5.7 3.1 3.99.4 1.0 2.1 1.8 1.0% . The total daily cost is 4479.31 3370.0% .7 32.0 2.2 1.Mi 7.4 08:00AM 01:56PM 3 89 5.5% 9999 0 0 9 1 1000 950 0 95.22 350.3 1.7 5.30 340.0 3.5% 9999 0 0 4 1 1000 850 0 85.1 2.5% 9999 0 0 7 1 1000 950 0 95.1 08:00AM 01:55PM 3 84 6.9 4.41 375.7 9.0 19.0% Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Finally.1 08:00AM 01:50PM 3 80 6.4 1.0 3. time.99 + 9 x 200 + 420 = P6.0% .0 1.0 1.6 2. time.0% 9999 0 0 8 1 1000 825 0 82.0 1.8 2.7 2. the route design from location C is shown in Figure 4.8 2.39 290. Although 10 routes are in the design.$ 389.1% 89991 0 0 Cube util .0% .3 1.8 08:00AM 02:07PM 2 104 9.5% 9999 0 0 6 1 1000 825 0 82.0% 9999 0 0 Total 9000 7925 0 88.0 2.0% 9999 0 0 3 1 1000 825 0 82. two of these can be dovetailed so that only nine trucks are needed.0% . 9 08:00AM 03:39PM 3 136 7.9 .6 7.5% 9999 0 0 3 1 1000 975 0 97.53 663.91 389. time.3 1.9 08:00AM 02:41PM 2 120 8.0 6.Mi 1.8 08:00AM 02:59PM 3 117 7.*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION Route Run Stop Brk Stem time.5 .3 1.5 4.7 4.0 30.4 1. it appears that yard location B is the best choice.3 1.0% 9999 0 0 8 1 1000 950 0 95.0 .40 679.4 1.0% Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 From an economic analysis.$ 140.7 2.6 08:00AM 09:30AM 1 20 6.3% 99990 0 0 Cube util .0 1.7 3.10 383.0% 9999 0 0 Total 10000 7925 0 79.7 08:00AM 03:29PM 3 134 7.5 08:00AM 03:18PM 3 127 9.0 2.0 1.0% .7 20.0% .6 08:00AM 05:35PM 2 229 9.7 9.3 08:00AM 05:47PM 3 236 74.0 4.0 3.0% 9999 0 0 10 1 1000 400 0 40.99 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 375 0 37.5 08:00AM 04:00PM 5 132 9.0% .1 2.0 2.8 7.2 2.0 5.0% .0 3.21 420.4 5.69 429.0% .0% .0 2. time.4 1.04 408.0% .4 08:00AM 05:25PM 5 180 7.5 44. time.0 1.8 1.0% .5% 9999 0 0 6 1 1000 1000 0 100.5% 9999 0 0 4 1 1000 925 0 92.9 1.5% 9999 0 0 5 1 1000 925 0 92.48 425.5% 9999 0 0 2 1 1000 875 0 87.2 1.3 4.6 2.0% 9999 0 0 9 1 1000 550 0 55. 183 .0% .0% .0 3.0% 9999 0 0 7 1 1000 950 0 95. time.6 .20 540. Start Return No of Route hr hr hr hr hr time time stops dist.42 4479.3 30 1432 Route no 1 2 3 4 5 6 7 8 9 10 Total Route cost.7 2.0 1.2 2.4 1. both inbound to the warehouse and outbound from it. ft.104 $ Using the COG module with inbound transport rates from Phoenix set at $16.40/1188 = $0.0235/cwt.082 The annual cost savings can be projected as $7.75/sq.404.SUPERIOR MEDICAL EQUIPMENT COMPANY Teaching Note Strategy The purpose of the Superior Medical Equipment Company case study is to encourage students to apply the center-of-gravity methodology to a problem involving a single warehouse location./mile and from Monterrey set at $9.532. Although the methodology is somewhat elementary. the primary location costs are transportation.754.61 and Y = 4. which varies with the location.022.404. The total relevant cost would be: Transportation Lease $3. Phoenix output will increase by 5 percent. and the warehouse lease. Total $ 6.535 4. what are the coordinates for a better location? What cost improvement can be expected from the new location? When a single warehouse is to be located.104 − 7./mile. or approximately Oklahoma City.014/cwt.73/1163 = $0.000 sq.162. and Denver markets to grow by 5 percent. × 200. ft. and the outbound transport rate from the unknown warehouse location set at $0. It allows students to evaluate the financial implications of alternative network designs. That is. but the remaining markets to decline by 10 percent.51. Transport costs are expected to be unchanged.000 7.532.082 = $128. Answers to Questions (1) Based on information for the current year. ft. and it is intended to be solved with the aid of the COG module provided in the LOGWARE software. how? 184 .819.082 650.569 550.754. Would your decision about the location of the warehouse change? If so. based on information given in the case. the coordinates for the best location are X = 7.082. management expects that Seattle.000 sq. Los Angeles. it can be useful in providing some first approximations to good warehouse locations./mile. ft.008/cwt.000 $ 7. Total transportation cost for this location would be $6. The current location serves as a benchmark against which the costs for other locations can be compared. (2) In five years. the total relevant cost for the Kansas City location is: Inbound transportation Outbound transportation Lease $2. × 200.25/sq. Total relevant cost $ 2. is Kansas City the best location for a warehouse? If not. and Monterrey's output will decrease by 10 percent. 718.114.000 $ 7.636 670. (3) If by year five increases are expected of 25 percent in warehouse outbound transport rates and 15 percent in warehouse inbound rates.40 33.150.277.206 650.150 .122 371.850 4 33.75/sq.52.170 10 7.32 25. would your decision change about the warehouse location? 185 . Total $ 6.600 5 13. $ $1.75 18.277)/2 = $141.69 30. the fifth year benchmark costs can be computed as follows: Volume.000 sq.000 Management must now judge whether 47.556 269. Point cwt.367 1.080.483 The total 5th-year benchmark relevant cost would be: Transportation Lease $2.022 + 154. The simple annual return on investment (moving cost) can be computed as: ROI = $141. × 100 = 471% $300. ft.448 337.000 sq.17 27.230 Rate.483 − 7. × 200.114. After adjusting the plant and market volumes according to the changes indicated.718.73 9.575 2 108.020.540 3 17.125 6 8.550 8 18.276 601.740 11 9.464.550 7 26. The average annual cost savings are (128.206 = $154.05 and Y = 4.000 $ 7. Total $ 6.340 1.447 $6.268.574 985.022.268. $/cwt.25/sq.206 The annualized cost savings would be $7.A new benchmark for the fifth year can be computed from the data given in Tables 1 and 2.52 26.630 Totals 346. ft.969 156.43 25. 1 64.66 26.900 9 37. × 200. The relevant costs for this location are: Transportation Lease $3. ft.1 percent annual return is worth the risk of changing warehouse locations.483 550.748 202.483 Optimizing the location with the fifth year data gives a location at X = 7.98 Transport cost.24 19. ft. $16. However. 186 .867 = Subtotal Total $8. A revised fifth year benchmark can be recomputed by applying the cost growth factors to the overall fifth year transport costs. (2) transportation rates on a per-mile basis are constant.617. Such costs as inventory carrying. central cities.497.15 × 2. That is. (3) volumes are known and constant for given demand and source points. what are its benefits and limitations for locating a warehouse? The center-of-gravity method locates a facility based on transportation costs alone.545 The annualized cost savings would be $8. This is reasonable when only one facility is being located and the general location for it is being sought.042 − 8.334 Running COG shows that the minimum transport cost location would be at coordinates X = 7. The shift in location is minimal.000 5.589.738.000 Total $ 8.616 = $2.62. The increase in possible cost savings further encourages relocation from Kansas City and toward a site near Oklahoma City. The cost for this location is: Transportation $ 7.939. and (5) it gives precise locations through a coordinate system. (2) it considers all possible locations (continuous).772. but they are not particularly relevant to the problem. or restricted lands. It can be concluded that: 1.708 Outbound 1.042 550. costs such as warehouse storage and handling. ft. and other costs that vary by the particular site are not included but may be relevant in a given situation. Transportation costs are assumed linear with distance. This may not be strictly true.25×4.545 Lease $3.415. (4) If the center-of-gravity method is used to analyze the data. production. 650.25/sq.000 sq.589.738. 2.20 and Y = 4. OK. and (4) locations may be suggested that are not feasible such as in lakes.It is assumed here that the fifth year demand level applies.545 = $148. and warehouse fixed are not included.042 Lease $8.100. Some potential limitations are (1) coordinates need to be linear. The obvious benefits of the method are (1) it is a fast solution methodology. × 200. (3) it is simple to use. (4) its data are readily available. Location is similar to the optimized fifth year location.188. which is near the previous location in question two. Inbound 1. although distance may be nonlinear. ft. 0140 0.90 61.600 11 14.70 7.0235 0.300 10 12.Concluding Comments The analysis in the case seems to suggest a move from Kansas City to a region around Oklahoma City would be advantageous.55 41.0235 0.10 17.5 Scale factor: 230 Point X-coordinate Y-coordinate Volume 1 3.25 10.00 120.000 4 1. In any case.0080 0.500 2 6.20 6.60 3.90 29.000 9 14.000 5 5.0235 0.80 3.30 3.10 12.95 4.90 9.30 8.0235 0. however management must now seek a particular site in the area whose choice may add or detract from this savings potential.60 6.0235 0.0235 0.95 21. A return on investment of 47 percent or higher is possible.500 6 7.600 3 0.500 8 11.700 187 . APPENDIX 1 LOGWARE COG Module Input Data for the Current Year Rate 0.20 32.00 6.0235 Title: SUPERIOR MEDICAL EQUIPMENT COMPANY Power factor: .90 1.80 8.500 7 10.0235 0.60 9.0235 0. the COG method has assisted in the selection of good potential locations and testing their sensitivity to changes in costs and volumes. It also provides an application for the multiple center of gravity location methodology. the benchmark remains the base for comparison. Therefore. what improvements can be made on the existing eight locations? Besides moving the locations of the bureaus. staff salaries. which cannot be known until the problem is solved. for 5 bureaus. the bureau locations. establish a benchmark against which changes to the network can be compared. The cost for residents traveling to the bureaus is not known because the bureau territories are not known.706. Since MULTICOG attempts to optimize bureau location. It will be assumed that all bureaus are of the same size. a possible approach to the analysis is outlined as follows.340. Hence. Since the costs for a particular network design depend on the size of each bureau. how should the network be configured? The nature of the costs and the number of possible alternative network designs make it impractical to seek an optimal solution. First. 188 . A reduced number of bureaus. an initial assumption must be made. The case may be assigned as a homework problem. an estimate can be made of travel costs by solving the problem in MULTICOG for eight bureaus. a short case study project. it is now necessary to estimate the approximate number of bureaus that are needed to serve the area. there are no obvious improvements to be made. the average number of residents in each bureau’s territory would be the total number of residents divided by the number of bureaus. this travel cost is probably understated. Table 2 is developed to show the bureau size and the number of staff for one to 10 bureaus. The module allows students to quickly evaluate alternatives as to the number of bureaus to use. and utility expenses can be derived from this estimate. The location costs for the current operation are estimated to be $1. Therefore. Much of the data for this is given in the case write up. in the range of two. The MULTICOG module in the LOGWARE software can effectively be applied.700/5 = 138. which results in resizing the facilities and adjusting the staff numbers. The costs can simply be applied to the size of each bureau and its associated staff.355. and the size of the territory that each bureau should serve. and what concerns he should have about changing the existing network design. or as a case for class discussion.OHIO AUTO & DRIVER'S LICENSE BUREAUS Teaching Note Strategy The purpose of this case study is to introduce students to a logistics problem in a service area. Second. Answers to Questions (1) Do you think there is any benefit to changing the network of license bureaus in the Cleveland area? If so. what data he needs and where to obtain it. The later would be appropriate especially if adequate attention is given to the issues of how Dan should go about solving a problem such as this. Rent. However. or 691. This would encourage students to think beyond the computational aspects of the problem. The benchmark costs are summarized in Table 1. Table 3 extends the average costs from these estimates. is about right. Third. Staff sq.800 1.000 5 6 115.800 1.600 84.000 8. no. bureau bureaus bureau sq.800 2.170 1. ft.200 756.000 7.400 220.500 4 189 .000 105.855 1.700/(1) Avg.000 84.814 1.000 6.500 4 8 86.200 4 26. Utilities.800 2.400 84.500 4 9 78.000 7 3 230.TABLE 1 Bureau 1 2 3 4 5 6 7 8 Totals Benchmark Costs for the Current Network of License Bureaus Customer Size.500 10 2 345. TABLE 2 Average Size and Staff for Various Numbers of Bureaus (1) 691.000 2.000 6.283 2.700 4 37.925 2.000 1.106* *Approximated by running MULTICOG at 8 bureaus.500 6 4 172.400 105.500 5 33.000 10.463 1.700 5 59. Rent.000 105. Travel.000 58. ft. of bureau Staff per No.200 5 48. of residents per size.700 4.500 4 10 69.000 8.000 6. $ $ $ $ 1.600 36 321. 1 691.000 5 44.340 2.850 3.000 5 5 138.000 14.200 1.500 4 33.800 4 39.400 105. Avg. Salaries.400 84.567 2.000 4.000 5 7 98. 000 48.239 926.313.000 630.922 = $474.181 1.000 Resident travel.500 691.000 + 24. Annual variable cost savings compared with the benchmark would be $1.600 Annual total cost.106 1.000 264.000 294.000 132.000 420.204.000 + 20. Compared with the benchmark.000 168.600 The cost estimates can now be refined around two bureaus. $ 662.000 264.636 1.098.000 378.922 Bureau 1 2 Totals Residents 290. 36 − 14 = 22 staff members will be separated for a cost of 22 × $8.000 Staff salaries.784.191.000 756. A simple return on investment would be: ROI = $474.000 165.000 14.000 Utilities.000 132. $ 166.000 176.000 Similar calculations are carried out for various numbers of bureaus. $ 18. sq.635 220.000 525.106 206.000 Resident travel cost.287 237.922.000 77.000 297.000 = $196.000 24.000 672.239 298.000 + 430.332 264.000 294. Total movement costs would be $176. $ 99.000 60.TABLE 3 Average Costs by Number of Bureaus No.706 − 880.000 278.000 220.000 24.700 Size. $ 210.496 1. Equipment movement costs to two bureaus would be 2 × 10. 190 .590 430.000 10.000 231.000 40.922⇐ 927.000 42.287 1.000.355.965 1.181 249.000 Staff 6 8 14 Rent.200 401.000.319 880. $ 989.000 588. of bureaus 1 2 3 4 5 6 7 8 9 10 Rent.496 198.784 × 100 = 242% $196.000 54.922 = $880.319 430. A sample analysis for two bureaus.000 32. Staff salaries. $ 55.000 840.000 = $176.428.000 30.063. There are one-time costs due to staff separation and equipment moves.500 6. $ $ 126.000.000 + 294. 2.922 354. ft.000 The total annual variable cost is $132.000 = $20.000 330.500 3. is shown below. These results are tabulated in Table 4. A comparison with the benchmark costs and a return of the initial investment (costs related to changing the network design) are sought. Utilities. based on a design provided by MULTICOG.000 48. 00 3.48.66 195.635 110.181 146.000 34 1.82.500 29 1.000 83. although they are not always easy to fulfill.36 2.75.43.157.000 42.000 31 1.10.56.000 428.200.5.55.37.467 243 4 8.70. This would likely be a problem with any other solution approach as well. The results of this methodology can only be used as a first approximation at best.49. The details for a design with four bureaus are given in Table 5.00 109.25.525 223 6 11.63. locations across a continuous space are desired. The design is shown pictorially in Figure 1 of this note.36.60.355.706 --------- The maximum annual savings occurs with a network containing two bureaus.28.741 246⇐ 5 9.59.600 36 1.40.31.800 57. investNo.000 394.784⇐ 3 7.8.50.965 160.69. Moving $ ment.13.44. sq.84.029.23. Several criticisms of this particular approach can be offered as follows: 191 .27.12.071 141 8 13.33 2 7.54. However. 45.00 2.000 366. and transportation cost (except for some fixed costs) is the primary location variable. The center-of-gravity approach is appropriate in this problem since there are no capacity limitations on the facilities. staff $ cost.319 218.61.106 96.74 5.239 176.000 14 880.62.15.22.26.500 18 927.39.387 168 2 6. 83.9.500 21 960.58.78.000 326.41.46.17.600 87 Benchmark 14.32. 4 10.66.2. The data requirements are relatively straightforward.81.500 24 1.51.287 106.72.68.52.4.6.272.19. 79.TABLE 4 Cost Savings and Return of Investment for Alternate Network Designs as Produced by MULTICOG Return on Total Annual Savings. the maximal return on investment occurs with four bureaus.73.7. % bureaus ft. 24.29.18.00 168.53 3 9. 67.200 1.80 (2) Do you think Dan Roger's study approach is sound? Overall Dan can be praised for the simplicity of the methodology that he has chosen.000 198. ROI is selected as the appropriate measure on which to base this economic decision.922 196.65.16.000 242 474.700 34.35.47.71.38.64.76.77. of size.11.500 10 989. $ 1 4.20.74 218. TABLE 5 Design Details for a Network with Four Bureaus Column grid Row grid coordicoordiBureau nate nate Residents Grid box number assignment 1 3.419 187 7 12. 14.74.30. Total variable cost.21.000 155.3. Since staff is a large expense in the operation. • Good facilities may not be available at the indicated location coordinates. and reasonable neighborhood reactions to this type of operation. He may find that the design does not change a great deal over a wide range of assumed values. If not. it provides a starting point for further analysis. • Travel to the bureaus is assumed straight line. there are site selection factors to be taken into account such as the availability of adequate space near the location coordinates. although this is not likely to be a strong issue if relocation were to occur in the same area. location in the area is likely to be influenced by a road network. They may not strictly do this. he should seek to find a more precise value. • The fixed costs associated with location are not handled directly by the center-ofgravity approach. This assumes their willingness to be relocated. Dan should conduct a sensitivity analysis around this cost. Reducing the number of locations will result in a releasing some of the staff. In addition. (3) What concerns besides economic ones should Dan have before suggesting that any changes be made to the network? A quantitative approach to location will rarely give the precise locations to be implemented. There are a number of other factors to be considered before the revised network design can be implemented. Dan may experience some political resistance to this. transferring staff to other governmental operations may be a way of dealing with this issue. there are political concerns. If this is the case. • Residents are assumed to travel to the locations within their assigned territories. proximity to good highway linkages. Of course. retention of a larger number of bureaus may be required. However. the cost of resident travel is subject to much interpretation.• The effect of bureau location on the resident's perception of service is not as well known as portrayed in the case. Rather. currently about 2/3 of the costs. service may need to be represented by more than location. there may be difficulty in demonstrating the economics of network redesign. Third. 192 . and utilities. Although others may appreciate the costs of rent. salaries. First. he can feel comfortable that his recommendation is fundamentally sound. Those favoring many bureaus may argue the high cost while those wanting to reduce the number of bureaus may perceive it as not very significant. While an exact cost is not likely to be known. Time may be more important than distance to residents. Second. The analysis is particularly weak around the estimate of the resident travel cost. 12 .12 .12 .12 .5 POINT X-COORDINATE Y-COORDINATE 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 1 7 8 2 1 9 2 2 10 2 3 11 2 4 12 2 5 13 2 6 14 2 7 15 3 1 16 3 2 17 3 3 18 3 4 19 3 5 20 3 6 21 3 7 VOLUME 4100 6200 7200 10300 200 0 0 7800 8700 9400 11800 100 0 0 8100 10500 15600 10500 200 0 0 RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 193 .12 .12 .12 .FIGURE 1 Four Bureau Locations and Their Territories Grid row number Current bureau locations Revised bureau locations Grid column number SUPPLEMENT Sample Input Data File for MULTICOG in LOGWARE for the Ohio Auto & Driver's License Bureau Case Study Title: LICENSE BUREAU Number of sources: 4 Number of demand points: 84 Scaling factor: 2.12 .12 .12 .12 .12 . 12 .12 .12 .12 .12 .12 .12 .12 .SUPPLEMENT (Continued) POINT 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 X-COORDINATE 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 9 10 10 10 10 10 10 10 11 11 11 11 Y-COORDINATE 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 VOLUME 10700 12800 13800 15600 400 0 0 11500 13900 14500 13700 600 0 0 9300 14900 13700 10200 1200 0 0 10100 12600 16700 15800 12400 2600 0 8800 13700 15200 14100 10800 17200 500 5300 16700 13800 11900 13500 18600 12000 5100 17400 10300 9800 10300 15500 11700 7700 9200 7500 8500 RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 194 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 . 12 .12 .12 .12 195 .12 .12 .12 .12 .12 .12 .POINT 75 76 77 78 79 80 81 82 83 84 X-COORDINATE 11 11 11 12 12 12 12 12 12 12 Y-COORDINATE 5 6 7 1 2 3 4 5 6 7 VOLUME 7800 9900 8700 4300 6700 5800 6800 5400 7100 6400 RATE . Rough estimates of its profitability can be made.000 = 0. The overhead and sales expense is 27 percent.000. what plan for distribution would you suggest? If growth is uniform over the next five years. they should be encouraged to apply the transportation method of linear programming to assist in the analysis of alternatives using the TRANLP module in LOGWARE.000 − 500. Southern can expect that demand for its products will exceed the currently available plant capacity.60 = $224.015. they should consider factors other than those in the analysis that might alter the course of their recommendation and be sensitive to the limitations and benefits of linear programming as a solution methodology.128. This is an average cost per barrel of $60.000 = 95. Therefore. Let's assume that she has this approval.128. That is.20 × [(0 + 95. Next. would you agree with the proposal to build the new brewery? If you do.5 years.000 barrels of beer would not be sold annually in the fifth year if additional capacity is not constructed.000 ÷ 403. she may wish to explore the opportunities available by improving upon the existing distribution system without the presence of the new plant. Total costs per barrel are $148.000 barrels will be used up in (500.92.92 + 75. We know that 595.SOUTHERN BREWERY Teaching Note Strategy The purpose of this case study is to provide students with the opportunity to design a distribution network where plant location is at issue.000. The 20 percent profit margin seems valid.015.60 per barrel. which is about 80 percent of the sales dollar. the current annual capacity of 500. Finally. Second. A major concern is whether it would be profitable to construct the new plant.5] = $2. the benefit of serving the potentially lost demand with a new brewery can be estimated using on a simple return on investment: ROI = $2.) From the benchmark1 costs for the current system.000 If management feels that this is an adequate return for such a project. as shown in Table 1. demand is increasing at the rate of (0 + (595.21. Carolyn should proceed with her analysis.5 years assumes that the new brewery can be brought on stream at approximately the time when capacity will be used up in the existing plants. (The figure of 2.000 barrels at a total cost of $60. 196 .000. or $280 × 0.000 − 403. Thus.000) ÷ 2.000 − 403.52.27 = $75. projecting profits with and without the new plant.000)) ÷ 5 = 38.400 barrels per year.000 = $148. Answers to Questions (1) If you were Carolyn Carter. This is an improved 1 A benchmark refers to the costs of producing and distributing demand as currently allocated throughout the network. This represents a potential average lost revenue of $280/barrel × 0.400 = 2. Southern is currently producing and distributing 403. or 21% $10.000) ÷ 38. They first should identify the major costs and alternatives that are important to such a design problem. 54 151.000 − 59.80 94 SAVANNAH 154. The results given in the TRANLP module of LOGWARE can be summarized in Table 3.562 355 3.015 $60.000 while transportation costs will be increased by $189.840 $475 $8.65 144.38 157.64 154.98 152. production and distribution costs can be reduced by $60.80 144.015 TABLE 2 TRANLP Problem Setup RICHFrom\To RICHMD COLMBA MONTGM JACKVL Demand MOND 148.25 31 KNOXVILLE 156.000 savings can be realized without any capital investment.69 148.373 8 Tallahassee Montgomery 26 3. TABLE 1 Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Producport Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7.72 26 JACKSONVL 158.206 577 5.000 = $211.54 13 MONTGOMRY 159.96 149.20 148.315 2 Raleigh Richmond 31 4. production costs will be reduced by $400.54 156.85 147. Note that this $211.18 150.98 157.18 142.960 7 Montgomery Montgomery 79 10.84 157.benchmark2 and it can be found by solving a transportation-type linear programming problem of the type shown in Table 2. respecting plant capacity restrictions. This shows that with some slight reallocation of demand among the plants.80 150.823 550 11.81 150.27 152.49 157.70 154.672 3 Knoxville Columbia 22 3.35 149. That is.48 22 COLUMBIA 152.000 $4.48 156.000 barrels of Columbia's market should be shifted from the Columbia plant to the Montgomery plant.015.16 44 ATLANTA 155.80 79 TALL AHAS 164.380 306 6.93 146.837 6 Savannah Montgomery 13 1.78 153.686 5 Atlanta Montgomery 94 12. 197 .783 9 Jacksonville Montgomery 38 Total 403 $56.000 annually.340 332 4.917 5.54 150.472 4 Columbia Columbia 44 6.30 160.68 38 Supply 100 100 300 0 2 An improved benchmark refers to a reallocation of current demand in an optimal way.000.878 959 13. A comparison of the benchmark results in Table 1 and the improved benchmark results in Table 3 shows that Knoxville's demand should be shifted from Columbia to Montgomery and 28.93 143.13 56 RALEIGH 150. The Columbia plant is a high-cost producer and cost savings are achieved by trading production costs for transportation costs.781 179 1.190 282 3.804. TABLE 3 Improved Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Producport Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7. note that the Columbia plant is needed once again although its capacity is not required until the last one-half year of the fifth year planning horizon.340 332 4.837 6 Savannah Montgomery 13 1.783 9 Jacksonville Montgomery 38 Total 403 $55.000 − 59.000. Therefore.204 $59. Costs will rise only slightly.600 $4. We do not know the savings for the fifth year demand level since not all demand can be served without the presence of the new plant.198 5 Atlanta Montgomery 94 12. a future-year benchmark cannot be determined. Also. A careful inspection of this plan shows that only one barrel of demand in the Knoxville region is assigned to Richmond.318 4 Columbia Columbia 16 2.840 $475 $8. could be made as efficient as the new brewery.781 179 1.000 barrels per year provides enough capacity to satisfy demand out to the fifth year. this option might be attractive.373 8 Tallahassee Montgomery 26 3. The new plant is not likely to be brought on stream immediately nor is it needed for two and a half years. If the new plant were constructed and producing immediately.917 5. such as subcontracting beer production to a non-competing company.000 per year (compare the total costs in Tables 3 and 4. through modernization.000.000.320 111 2. this is $1.014 304 3.000 = $714.000 in cost savings. we would need to resolve the linear programming problem with Columbia's per barrel costs at $135 plus transportation costs. If the modernization where to cost no more than $5.836 362 4. This suggests that the Columbia plant should not be sold. Of course.878 959 13.315 2 Raleigh Richmond 31 4. we do know how the new plant should be utilized within the system (see Table 5) and how demand allocation should be adjusted to accommodate it.431 4 Columbia Montgomery 28 3.090.000-barrel capacity.960 7 Montgomery Montgomery 79 10. total costs could be reduced from the improved benchmark by $59.804.804 Adding a plant at Jacksonville with a capacity of 100. so Carolyn might suggest a distribution plan similar to that in Table 5. An interesting question is whether the Columbia plant. and what the implications for distribution might be.562 355 3. We know that this could potentially save $145 − 135 = $10 per barrel in production costs.672 3 Knoxville Montgomery 22 3.206 577 5. However. At a 100. Splitting demand to this extent is probably not practical and can be assigned to Columbia where there is excess capacity.) The Columbia plant would not be needed if the lower-cost Jacksonville plant were on line. This would tell us how and to what extent demand would be allocated to Columbia and give a more accurate basis for determining 198 .823 550 11. but perhaps some alternate use could be made of the facility in the interim. 340 332 4.373 8 Tallahassee Jacksonville 26 3. portions of the demand in Tallahassee and Jacksonville should not be served.510 253 3.105 257 3. As can be seen in Table 6.318 4 Columbia Montgomery 21 2.315 2 Raleigh Richmond 31 4.840 $475 $8.823 550 11.422 9 Jacksonville Jacksonville 38 Total 403 $55.877 272 3. what distribution plan would you propose to top management? Table 6 shows a linear programming solution where the new plant is not brought on stream and the demand in the markets is set at the five-year level.755 124 1. 199 .149 4 Columbia Jacksonville 23 3.672 3 Knoxville Montgomery 22 3. Similarly. The most costly demand region to serve is not assigned to any plant.837 6 Savannah Jacksonville 13 1.362 5 Atlanta Montgomery 94 12.878 959 13.763 5. it would be interesting to explore what it means to expand the capacity of an existing brewery at a lower investment cost per barrel than the construction of a new facility.the cost savings.090 (2) If the new brewery is not to be constructed.130 292 5.818 $59. TABLE 4 Production and Transportation Costs ($000s) for Current Demand with the Jacksonville Plant Demand Transin 000s Producport Brewery of Total costs origin barrels tion costs costs Market area 1 Richmond Richmond 56 $7.014 304 3. Top management may wish to adjust this plan for reasons other than economic ones. An interesting solution occurs when demand exceeds capacity. and essentially the entire Knoxville market should not be served at all.272 $3.879 7 Montgomery Montgomery 79 10. 260 584 10.755 6 Savannah Columbia 20 2.975 383 8.900 375 5. • Production is assumed limited to exactly the values given without the possibility for expansion through overtime. For example.503 2 Raleigh Richmond 35 35 5.091 7 Montgomery Montgomery 119 119 17.240 233 3.844 9 Jacksonville Jacksonville 76 Total 595 $82.473 10.900 256 3.810 4 Columbia Columbia 55 7.931 9 Jacksonville Columbia 76 Total 595 500 $74.358 5 Atlanta Montgomery 141 19.091 7 Montgomery Montgomery 119 16.156 3 Knoxville Montgomery 12 1.503 2 Raleigh Richmond 35 4.275 3 Knoxville Richmond 1 140 16 156 3 Knoxville Columbia 20 2.770 TABLE 6 Production and Transportation Costs ($000s) for Projected Fifth Year Demand Without the Jacksonville Plant Expected Served demand in demand in Brewery of 000s 000s Market area origin barrels barrels Total costs 1 Richmond Richmond 64 64 $9.960 $543 $9. 200 . or subcontracting.131 8 Tallahassee Montgomery 52 40* 6.226 *Indicates market demand is not fully served due to inadequate plant capacity.438 20.358 5 Atlanta Montgomery 141 141 20.644 166 1.836 382 4.755 6 Savannah Columbia 20 20 3.375 $5. additional shifts.131 8 Tallahassee Montgomery 28 3.218 8 Tallahassee Jacksonville 24 3.275 3 Knoxville Richmond 33 1* 156 4 Columbia Columbia 55 55 8.395 $87.026 25* 3. • Demand has been assumed to grow at a constant rate in the markets. (3) What additional considerations should be taken into account before reaching a final decision? A number of assumptions have been implied in the analysis shown above.900 191 3.317 1.TABLE 5 Production and Transportation Costs ($000s) for Projected Fifth Year Demand With the Jacksonville Plant TransDemand port Producin 000s Brewery of Total costs costs barrels tion costs origin Market area 1 Richmond Richmond 64 $8.303 828 17. vested interests. is a facilitating vehicle for analysis. 201 . It does not provide the final answer.• Per-unit production and transportation costs are assumed to remain unchanged with the reallocation of demand throughout the network. etc. Mentioning that linear programming does not consider such factors as fixed costs. at best. • There is no change in per-unit costs throughout the five-year planning horizon. or the many subjective factors (top management's intuition about location. • Customer service effects are not considered in reallocation of demand. return on investment. This case might end with a discussion of the appropriateness of using linear programming as a vehicle for analysis in a problem such as this.) that are typically a part of such problems means that linear programming.