CHAPTER 13
FACILITY LOCATION DECISIONS
1
(a) The center-of-gravity method involves finding the X,Y coordinates according to the formulas: X V R X V R i i i i i i i =
∑
∑
and Y V R Y V R i i i i i i i =∑
∑
These formulas can be solved in tabular form as follows.
Point X Y Vi Ri ViRi ViRiXi ViRiYi P1 3 8 5,000 .040 200.0 600.0 1600.0 P2 8 2 7,000 .040 280.0 2240.0 560.0 M1 2 5 3,500 .095 332.5 665.0 1662.5 M2 6 4 3,000 .095 285.0 1710.0 1140.0 M3 8 8 5,500 .095 522.5 4180.0 4180.0 Totals 1620.0 9395.0 9142.5 Now, X = 9 395 0 = 1 620 0 5 8 , . , . . and Y = 9 142 5= 1 620 0 5 64 , . , . .
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.
Iteration Total
number X coord Y coord cost
0 5.799383 5.643518 53,614.91 ⇐ COG 1 5.901199 5.518863 53,510.85 2 5.933341 5.446919 53,483.97 3 5.941554 5.402429 53,474.60 . . . . . . . . . . . . 50 5.939314 5.317043 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 Whse P1 P2 M1 M2 M3 1 1458a 2042 3500 I 2 3542 4958 3000 5500 1 2708 3792 3500 3000 II 2 2292 3208 5500 1 3750 5250 3500 5500 III 2 1250 1750 3000
The COG module in LOGWARE was used to find the exact centers of gravity for each warehouse in each scenario. The computational results are:
Warehouse 1 Warehouse 2 Total
Scenario X Y X Y cost
I 2.00 5.00 7.88 7.80 $39,050
II 5.84 4.04 8.00 8.00 35,699 ⇐
III 7.06 7.28 6.00 4.00 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 X Y Vi Ri ViRi ViRiXi ViRiYi A 50 0 9,000 .75 6,750 337,500 0 B 10 10 1,600 .75 1,200 12,000 12,000 C 30 15 3,000 .75 2,250 67,500 33,750 D 40 20 700 .75 525 21,000 10,500 E 10 25 2,000 .75 1,500 15,000 37,500 F 40 30 400 .75 300 12,000 9,000 G 0 35 500 .75 375 0 13,125 H 5 45 8,000 .75 6,000 30,000 270,000 I 40 45 1,500 .75 1,125 45,000 50,625 J 20 50 4,000 .75 3,000 60,000 150,000 Totals 23,025 600,000 586,500 Now, X = 600 000 = 23 025 261 , , . and Y = 586 500 = 23 025 25 5 , , .
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:
• 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 = + + = 40 700 0 75 40 400 0 75 700 0 75 400 0 75 40 00 ( )( . ) ( )( . ) ( . ) ( . ) . and Y = + + = 20 700 0 75 30 400 0 75 700 0 75 400 0 75 23 64 ( )( . ) ( )( . ) ( . ) ( . ) .
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.
Point X Y Vi Ri ViRi ViRiXi ViRiYi A 1.5 6.6 10,000 .10 1,000 1,500 6,600 B 4.7 7.3 5,000 .10 500 2,350 3,650 C 8.0 7.1 70,000 .10 7,000 56,000 49,700 D 1.5 4.0 30,000 .10 3,000 4,500 12,000 E 5.0 4.9 40,000 .10 4,000 20,000 19,600 F 8.5 5.1 12,000 .10 1,200 10,200 6,120 G 1.5 1.3 90,000 .10 9,000 13,500 11,700 H 4.4 1.8 7,000 .10 700 3,080 1,260 I 7.8 1.8 10,000 .10 1,000 7,800 1,800 Totals 27,400 118,930 112,430
The center-of-gravity coordinates are:
X =118 930 = 27 400 4 34 , , . and Y =112 430 = 27 400 410 , , .
with a total cost of $195,966.
(b) After 100 iterations in the COG module, the exact center of gravity was found to be:
X =4 75. ,Y =4 62.
with a total cost of $195,367. In this case, using the exact center of gravity coordinates as compared with the approximate ones reduced costs by only:
195 966 195 367
195 966 100 0 3%
, ,
, .
− × =
(c) Additional costs can be included in the analysis, although not necessarily in the model. The COG model can evaluate the variable costs of location. Other costs are compared with these.
4
We begin by developing a 3-dimensional transportation problem. The cost matrix is developed in the same manner as that in text Figure 13-11. The initial throughput of W1
and W2 is found by assuming that an equal amount of the customer demand flows through
each warehouse. The cell cost for W1-C1 would be:
In fact, the cell costs are identical to those in text Figure 13-11, except that there is no fixed cost element.
Using the transportation method of linear programming (e.g., the TRANLP module in LOGWARE), the cell cost and solution matrix for iteration 1 is shown in Figure 13-1. The solution shows that only W2 remains and the solution process can be terminated.
A summary of the costs is shown in Table 13-1. The total cost is $2,213,714, and the product is produced in plant P2 and stocked in warehouse W2. No further iterations are
needed since only one warehouse is used and no further dropping of warehouses is possible.
TABLE 13-1 Summary Information for Solution to Problem 4 Whse 1 Whse 2 Whse throughput 0 200,000 Costs: Transportation Inbound $0 $400,000 Outbound 0 300,000 Inventory 0 513,714 Warehousing 0 200,000 Fixed 0 0 Production 0 800,000 Total $2,213,714 Iteration 2
A repeat of the iteration 1 solution. Stop iterating.
FIGURE 13-1 Cell Cost and Solution Matrix for Iteration 1 of Problem 4
Warehouses Customers W1 W2 C1 C2 C3 Capacity Plants P1 4 0 9 0 99a 99 99 60,000 P2 8 0 6 200,000 99 99 99 999,999 Whses W1 0 60,000 99a 9.2 0 8.2 0 10.2 0 60,000 W2 99 0 799,999 6.2 50,000 5.2 100,000 6.2 50,000 999,999 Capacity/ Req’mts 60,000 999,999 50,000 100,000 50,000 a
High rate of $99/unit for an inadmissible cell. 5
FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 5 Warehouses Customers W1 W2 C1 C2 C3 Capacity Plants P1 4 60,000 9 100 100 100 60,000 P2 8 40,000 6 100,000 100 100 100 999,999 Whses W1 0 899,999 100 9.7 8.7 100,000 10.7 999,999 W2 100 0 8.2 50,000 7.2 8.2 50,000 100,000 Capacity/ Req’mts 999,999 100,000 50,000 100,000 50,000
Given the solution from iteration 1, the per-unit inventory and fixed costs are revised. Inventory W1 $100( , ) $3. / . 100 000 100,000 un 16 0 7 its = unit W2 $100( , ) , $3. / . 100 000 100 000 16 0 7 units = unit Fixed W1 $100,000/100,000 = $1.00/unit W2 $400,000/100,000 = $4.00/unit
Adding outbound transportation and warehouse handling to per-unit inventory and fixed costs gives the following cell costs.
C1 C2 C3
W1 10.2 9.2 11.2
W2 10.2 9.2 10.2
Revising the warehouse-customer cell costs and solving gives the same warehouse throughputs, so cell costs will no longer change. A stopping point is reached. The solution is the same as that in Figure 13-2.
Compared with the costs from the text example, the cost difference is $3,092,456 − 2,613,714 = $478,742. This is the penalty for restricting a warehouse with economic benefit to the network.
6
Prepare a matrix for a 3-dimensional transportation problem like that in text Figure 13-11, 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. That is, $200,000/200,000 = $1/unit instead of $400,000/200,000 = $2/unit. The matrix setup and first iteration solution are shown in Figure 13-3.
FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 6
Cost type Warehouse 1 100,000 cwt. Warehouse 2 100,000 cwt. Production 60,000×4 = $240,000 40,000×4 = 160,000 100,000×4 = $400,000 Inbound transportation 60,000×0 = 0 40,000×4 = 160,000 100,000×2 = 200,000 Outbound transportation 100,000×3 = 300,000 50,000×2 = 100,000 50,000×2 = 100,000 Fixed 100,000 400,000 Inventory carrying 100(100,000)0.7 = 316,228 100(100,000)0.7 = 316,228 Handling 100,000×2 = 200,000 100,000×1 = 100,000 Subtotal $1,476,228 $1,616,228 Total $3,092,456 Warehouses Customers W1 W2 C1 C2 C3 Plant & warehouse capacities P1 4a 9 99b 99 99 60,000 P2 8 6 200,000 99 99 99 999,999c W1 0 60,000 99 9.7d 8.7 10.7 60,000 W2 99b 0 799,999 7.2e 50,000 6.2 100,000 7.2 50,000 999,999c Warehouse capacity & customer demand 60,000 999,999c 50,000 100,000 50,000 a
Production plus inbound transport rates, that is, 4 + 0 = 4.
b
Used to represent an infinitely high cost.
c
Used to represent unlimited capacity.
d
Inventory carrying, warehousing, outbound transportation, and fixed rates, that is,
Plants
The results show that one warehouse is to be used. Further computations are not needed, as further warehouse consolidation is not possible. The total network costs are the same as those in the text example minus the $200,000 reduction in fixed costs for a total coat of $2,413,714.
8
This problem requires us to rework the dynamic programming solution to the example problem given in the text. The only change is that the cost of moving from one location to another is now $300,000 instead of $100,000. We begin with the last year and determine the best action based on the highest net profits. The action will be to move (M) or to stay (S). For example, given the discounted moving cost of 300,000/(1 + 0.20)(4) = $144,676, we evaluate each course of action, assuming that we are in location alternative A at the end of year four. From the location profits of text Table 13-6, we generate the following table for location A.
Alter- Location Moving Net
native (x) profit cost profit
A $1,336,000 - 0 = $1,336,000 B 1,398,200 - 144,676 = 1,253,524 P5(A) = max C 1,457,600 - 144,676 = 1,312,924
D 1,486,600 - 144,676 = 1,341,924 E 1,526,000 - 144,676 = 1,381,324
The best action in the beginning of the fifth year, if we are already in location A, is to move to location E. This is an entry in Table 13-2.
Once each of the five alternatives is evaluated for the fifth year, then the fourth year alternatives are evaluated. The moving cost is 300,000/(1 + 0.2)(3) = $173,611. We now include the profits for the subsequent years in our calculations.
After Table 13-2 is completed, we search the first column for the highest cumulative profit. This is initially to locate in location D and remain there throughout the subsequent years.
9
(a) Using PMED software in LOGWARE and the PMED02.DAT database, solve for the number of locations from one to nine. The best locations for each number of sites are given in the table below.
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
Warehouse location alter-natives (x) P1(x) Stra-tegya P 2(x) Stra-tegya P 3(x) Stra-tegya P 4(x) Stra-tegya P 5(x) Stra-tegya A $3,557,767 SA $3,363,767 SA $3,007,667 MD $2,268,289 MD $1,381,324 ME B 3,556,167 SB 3,379,667 SB 3,007,667 MD 2,268,289 MD 1,398,000 SB C 3,673,200 SC 3,500,900 SC 3,156,200 SC 2,319,800 SC 1,457,600 SC D 3,720,300 SD 3,553,600 SD 3,216,000 SD 2,459,900 SD 1,486,600 SD E 3,534,100 SE 3,374,700 SE 3,071,300 SE 2,355,800 SE 1,526,000 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.
1st 2nd 3rd 4th 5th
Year from present date j
Number
of sites Total cost
1 $95,035,220 2 49,311,400 3 37,152,360 4 30,163,340 5 24,739,040 6 24,230,760 7 23,780,600 8 25,018,250 9 27,624,060
Seven sites is optimal, and they are to be located at New York, Cincinnati, Chicago, Phoenix, Denver, and Seattle.
(b) The optimal cost for four sites is $30,163,340 as found in part a. The company operates the same sites as found in the optimal solution for four sites. Therefore, the cost savings comes from a between assignment of customers to the sites. The savings of $35,000,000 − 30,163,340 = $4,836,660 without any major investment.
The savings better the optimal four sites and the optimal seven sites is $30,163,340 − 23,780,600 = $6,382,740. For an additional three sites, the simple return on investment is:
ROI = Savings = × = Investment 6 382 740 18 000 00 100 35 5% , , , , .
This is a benefit certainly worth considering.
(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. Selected results are as follows:
Number
of sites Total cost
6 30,892,720 7 25,468,410 8 25,018,250
9 27,624,060
The optimal number of sites is eight. An additional site at Los Angeles is needed, compared with the seven locations found in part a.
10
(a) We can apply Huff's model of retail gravitation to this problem. The solution table (Table 13-3) can be developed. Summarizing, branch A can be expected to attract 11,735/(11,735 + 11,765) = 49.9 percent of the customers, and branch B should attract the remaining 50.1 percent.
TABLE 13-3 Estimate of the Number of Customers Attracted to Each Branch Bank Time to ja T ij 2 Sj/Tij2 P S T S T ij j ij j ij j =
∑
/ / 2 2 E P C ij = ij i Customer i A B A B A B A B A B 1 0.28 0.72 0.08 0.52 12.5 1.4 0.90 0.10 900 100 2 0.10 0.50 0.01 0.25 100.0 2.8 0.97 0.03 1,940 60 3 0.28 0.45 0.08 0.20 12.5 3.5 0.78 0.22 3,120 880 4 0.40 0.20 0.16 0.04 6.3 17.5 0.26 0.74 1,820 5,180 5 0.20 0.40 0.04 0.16 25.0 4.4 0.85 0.15 850 150 6 0.50 0.50 0.25 0.25 4.0 2.8 0.59 0.41 885 625 7 0.45 0.28 0.20 0.08 5.0 8.8 0.36 0.64 1,440 2,560 8 0.57 0.20 0.32 0.04 3.1 17.5 0.15 0.85 300 1,700 9 0.67 0.54 0.45 0.29 2.2 2.4 0.48 0.52 480 520 11,735 11,765 a Time = (Xi−X)2+(Yi−Y) /2 50(b) The economic analysis of the sites would be: Revenue (100×No.of customers) $1,173,500
Operating expenses - 300,000
Profit $ 873,000
Return on investment 873,000/750,000 = 116.4%
The ROI seems sufficiently high, so the branch should be constructed.
(c) The size of a branch and its proximity to customers may be too simple to explain the market share of each. The nature of the services offered, the accessibility of the site, and the reputation of the bank may be just as important in estimating patronage.
What will the countermoves be of the competing branch? Adding another branch and locating near branch A could substantially reduce its market share. 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. First, we create a table showing the counties that are adjacent to each county. That is,
Counties under Adjacent counties consideration by number 1. Williams 2,5,6 2. Fulton 1,3,6 3. Lucas 2,4,6,7 4. Ottawa 3,7,8 5. Defiance 1,6,9,10 6. Henry 1,2,3,5,7,10,11 7. Wood 3,4,6,8,10,11,12 8. Sandusky 4,7,12 9. Paulding 5,10,13 10. Putnam 5,6,7,9,11,13,14 11. Hancock 6,7,10,12,14,15,16 12. Seneca 7,8,11,16 13. Van Wert 9,10,14,17,18 14. Allen 10,11,13,15,18 15. Hardin 11,14,16,18,19,21 16. Wyandot 11,12,15,19 17. Mercer 13,18 18. Auglaize 13,14,15,17,20,21 19. Marion 15,16 20. Shelby 18,21 21. Logan 15,18,20
Next, according to the problem formulation given in the Ohio Trust Company example, we can build the matrix as given in the prepared database called ILP03.DAT. The problem formulation is shown in Table 13-4. This matrix can be solved by the integer-programming module (MIPROG) in LOGWARE for solution. Note that all coefficients are ones or zeros. A coefficient of one is given to each county and its adjacent counties. The sum of all constraints must be one or greater. The Xs take on the values of zero or one. An X of one means that the branch is located in the county.
Solving this problem using the integer-programming module in LOGWARE shows that a minimum of five principal places of business are needed. They should be located in Henry, Wood, Putnam, Hardin, and Auglaize counties.
TABLE 13-4 Coefficient Matrix Setup for Problem 11 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 RHS Obj Fun 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Constraints Williams 1 1 1 1 ≥ 1 Fulton 1 1 1 1 ≥ 1 Lucas 1 1 1 1 1 ≥ 1 Ottawa 1 1 1 1 ≥ 1 Defiance 1 1 1 1 1 ≥ 1 Henry 1 1 1 1 1 1 1 1 ≥ 1 Wood 1 1 1 1 1 1 1 1 ≥ 1 Sandusky 1 1 1 1 ≥ 1 Paulding 1 1 1 1 ≥ 1 Putnam 1 1 1 1 1 1 1 1 ≥ 1 Hancock 1 1 1 1 1 1 1 1 ≥ 1 Seneca 1 1 1 1 1 ≥ 1 Van Wert 1 1 1 1 1 1 ≥ 1 Allen 1 1 1 1 1 1 ≥ 1 Hardin 1 1 1 1 1 1 1 ≥ 1 Wyandot 1 1 1 1 1 ≥ 1 Mercer 1 1 1 ≥ 1 Auglaize 1 1 1 1 1 1 1 ≥ 1 Marion 1 1 1 ≥ 1 Shelby 1 1 1 ≥ 1 Logan 1 1 1 1 ≥ 1
12
This problem can be solved with the aid of the PMED program in LOGWARE. A database has been prepared for it called PMED04.DAT. The database shows the fixed costs for a single site. When other numbers are to be evaluated, the FOC must be recalculated and entered into the database. The scaling factor is set at one for this problem. The fixed operating cost must be calculated for each possible number of locations. Using PMED in LOGWARE gives the following results based on recalculated
FOC values, an estimation of vendor to laboratories transportation cost, and an enumerative search. No. of loca-tions Sites Volume, lb. Outbound cost, $ Inbound distance, mi. Inbound cost, $ FOC, $ Total cost, $ 1 Chicago 680,000 28,350,000 0 0 5,000,000 33,350,000 2 Cleveland 515,000 310 3,193,0001 3,535,5342 L. Angeles 165,000 1,750 5,775,000 3,535,534 Total 680,000 15,083,500 8,968,000 7,071,068 31,122,568 3 New York 235,000 713 3,351,100 2,886,751 Chicago 280,000 0 0 2,886,751 L. Angeles 165,000 1,750 5,775,000 2,886,751 Total 680,000 10,641,999 9,126,100 8,660,253 28,428,352 4 New York 200,000 713 2,852,000 2,500,000 Atlanta 145,000 585 1,696,500 2,500,000 Chicago 170,000 0 0 2,500,000 L. Angeles 165,000 1,750 5,775,000 2,500,000 Total 680,000 7,515,250 10,313,500 10,000,000 27,828,750 5 New York 200,000 713 2,852,000 2,236,067 Atlanta 100,000 585 1,170,000 2,236,067 Chicago 170,000 0 0 2,236,067 Dallas 60,000 790 948,000 2,236,067 L. Angeles 150,000 1,750 5,250,000 2,236,067 Total 680,000 6,079,249 10,220,000 11,180,335 27,479,584 6 New York 200,000 713 2,852,000 2,041,241 Atlanta 65,000 585 760,500 2,041,241 Miami 35,000 1180 826,000 2,041,241 Chicago 170,000 0 0 2,041,241 Dallas 60,000 790 948,000 2,041,241 L. Angeles 150,000 1,750 5,250,000 2,041,241 Total 680,000 5,029,250 10,636,500 12,247,446 27,913,196 1515,000x310x0.02=3,193,000 2 5,000,000 2/2=3,535,534
The PMED program is used to find the best combination of sites for a particular number of sites to be found. The fixed cost must be adjusted for the number of sites being evaluated. It should be recognized that the model handles only the outbound leg of the network (sites to serve laboratories). The vendor to site transportation cost is included externally, as shown in the previous table. 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. Then, inbound transport costs are a product of site volume, distance, and the inbound rate.
Searching from one to N sites shows that outbound transportation costs decrease while inbound and fixed costs increase with increasing numbers of sites. Initially, total cost declines until five sites are reached after which total cost increases. We select five sites as economically the best number. Their customer assignments are
Location
number Assignments Volume Customers
1 New York 200,000 1, 2, 3, and 12
2 Atlanta 100,000 4 and 5
3 Chicago 170,000 6, 7, 8, 9, 10, and 11
4 Dallas 60,000 13, 14, and 16
5 Los Angeles 150,000 15, 17, 18, 19, and 20
A map of the solution is as follows.
The total cost for 5 sites is $27,479,584.
13
After changing the fixed costs in the problem setup matrix, the following solution is found.
OPTIMAL SOLUTION
Variable Value Rate Cost Variable label
X(1) = .0000 8.0000 .0000 P1S1W1C1 X(2) = 10000.0000 7.0000 70000.0000 P1S1W1C2 X(3) = 50000.0000 9.0000 450000.0000 P1S1W1C3 X(4) = .0000 11.0000 .0000 P1S1W2C1 X(5) = .0000 10.0000 .0000 P1S1W2C2 X(6) = .0000 11.0000 .0000 P1S1W2C3 X(7) = 50000.0000 12.0000 600000.0000 P1S2W1C1 X(8) = 90000.0000 11.0000 990000.0000 P1S2W1C2 X(9) = .0000 13.0000 .0000 P1S2W1C3 X(10) = .0000 8.0000 .0000 P1S2W2C1 X(11) = .0000 7.0000 .0000 P1S2W2C2 X(12) = .0000 8.0000 .0000 P1S2W2C3 X(13) = .0000 6.0000 .0000 P2S1W1C1 X(14) = 30000.0000 5.0000 150000.0000 P2S1W1C2 X(15) = 20000.0000 7.0000 140000.0000 P2S1W1C3 X(16) = .0000 11.0000 .0000 P2S1W2C1 X(17) = .0000 10.0000 .0000 P2S1W2C2 X(18) = .0000 11.0000 .0000 P2S1W2C3 X(19) = 20000.0000 9.0000 180000.0000 P2S2W1C1 X(20) = .0000 8.0000 .0000 P2S2W1C2 X(21) = 40000.0000 10.0000 400000.0000 P2S2W1C3 X(22) = .0000 7.0000 .0000 P2S2W2C1 X(23) = .0000 6.0000 .0000 P2S2W2C2 X(24) = .0000 7.0000 .0000 P2S2W2C3 X(25) = 1.0000 100000.0000 100000.0000 zW1 X(26) = .0000 500000.0000 .0000 zW2 X(27) = 1.0000 140000.0000 140000.0000 yW1C1 X(28) = 1.0000 260000.0000 260000.0000 yW1C2 X(29) = 1.0000 220000.0000 220000.0000 yW1C3 X(30) = .0000 70000.0000 .0000 yW2C1 X(31) = .0000 130000.0000 .0000 yW2C2 X(32) = .0000 110000.0000 .0000 yW2C3
Objective function value = 3700000.00
Note that warehouse 2 is no longer used in favor of all products flowing through warehouse 1.
14
(a) The demand of customer 1 for product 1 increases to 100,000 cwt. In the problem matrix of ILP02.DAT the following cell values are changed.
Cell From To
Dem P1W1C1, yW1C1 -50000 -100000
Dem P1W2C1, yW2C1 -50000 -100000
Cap-W1, yW1C1 70000 120000
Cap-W2, yW2C1 70000 120000
Obj. coef., yW2C1 140000 240000
Obj. coef., yW2C1 70000 120000
The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,500,000.
(b) Using the ILP02.DAT file in MIPROG of LOGWARE, the following changes are made to the following cells.
Obj. Coef., P2S2W1C1 9 12 Obj. Coef., P2S2W1C2 8 11 Obj. Coef., P2S2W1C3 10 13 Obj. Coef., P2S2W2C1 7 10 Obj. Coef., P2S2W2C2 6 9 Obj. Coef., P2S2W2C1 7 10
The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,380,000.
(c) Using the ILP02.DAT file in MIPROG of LOGWARE, the changes are made to the following cells.
Obj. Coef., yW2C1 70,000a 280,000 Obj. Coef., yW2C2 130,000 520,000 Obj. Coef., yW2C3 110,000 440,000 a
The sum of demand for the same customer for all products multiplied by the handling rate, i.e.,
(50,000 + 20,000) × $4/cwt. = 280,000.
The solution for product 1 shows that 50,000 cwt. flows from plant 1 through warehouse 1 and on to customer 1. The remainder flows from plant 2 through warehouse 2 and on to customers 2 and 3.
For product 2, plant 1 supplies warehouse 1 and customer 1 with 20,000 cwt. The remaining 90,000 cwt. flows from plant 2 through warehouse 2 to customers 2 and 3. The total cost is $3,920,000.
(d) Making some slight revisions in file ILP02.DAT can adjust the capacities on plant 1. The cell changes to make are:
Cell From To
Cap-P1S1, RHS 60,000 150,000
Cap-P1S2, RHS 999,999 90,000
Both warehouses are now used for both products. The solution can be summarized as:
Cell From To
Product Plant Warehouse Customer 1 1 – 50,000 cwt. 1 – 50,000 1 – 50,000 cwt. 1 1 – 60,000 2 – 60,000 2 – 60,000 1 2 – 40,000 2 – 40,000 2 – 40,000 1 2 – 50,000 2 – 50,000 3 – 50,000 2 1 – 20,000 1 – 20,000 1 – 20,000 2 2 – 30,000 2 – 30,000 2 – 30,000 2 2 – 60,000 2 – 60,000 3 – 60,000
The total cost is $3,270,000
(e) Again revising the ILP02.DAT file by changing cells Obj. coef., P2S1W2C3 and Obj. Coef., P2S2W2C3 to have a very high cost (999), 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. The total cost is $3,340,000.
15
This problem follows the form of the Ohio Trust Company example in the text. First, identify the zones that are within 30 minutes of any particular zone. That is,
Zone
no. Zones within 30 minutes
1 1,2,4,7,8,9,10 2 1,2,3,4,7,9,10 3 2,3,4,5,8,9,10 4 1,2,3,4,5,6,10 5 3,4,5,6,7,8 6 4,5,6,7,8 7 1,2,5,6,7,9,10 8 1,3,5,6,8,9,10 9 1,2,3,7,8,9,10 10 1,2,3,4,7,8,9,10
The solution from MIPROG is:
OPTIMAL SOLUTION
Variable Value Rate Cost Variable label
X(1) = 1.0000 1.0000 1.0000 1 X(2) = .0000 1.0000 .0000 2 X(3) = .0000 1.0000 .0000 3 X(4) = 1.0000 1.0000 1.0000 4 X(5) = .0000 1.0000 .0000 5 X(6) = .0000 1.0000 .0000 6 X(7) = .0000 1.0000 .0000 7 X(8) = .0000 1.0000 .0000 8 X(9) = .0000 1.0000 .0000 9 X(10) = .0000 1.0000 .0000 10
Objective function value = 2.00
The optimal solution is to place claims adjuster stations in zones 1 and 4.
16
This is a location problem where the dominant location factor is transportation cost is and this cost is determined from optimizing the multi-stop routes originating at the material yard. The ROUTER module in LOGWARE can be used to generate these routes for each yard location. A database file for this problem (RTR13.DAT) has been prepared.
Solving this problem requires balancing the cost of transporting the merchandise to the customers with the operating cost of the material yards at various locations. Optimizing the routing from the current material yard gives the route design shown in Figure 13-3. A minimum of nine trucks are required to meet all constraints on the problem. The total daily cost for this location is the route cost + vehicle costs + yard operating cost, or P4,500.87 + 9 x P200 + P350 = P6,650.87.
*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION
Route Run Stop Brk Stem
Route time, time, time, time, time, Start Return No of Route Route
no hr hr hr hr hr time time stops dist,Mi cost,$
1 9.9 6.4 2.5 1.0 5.0 08:00AM 05:56PM 4 205 602.45 2 3.4 1.6 1.8 .0 .7 08:00AM 11:25AM 3 52 220.25 3 10.0 6.5 2.5 1.0 3.1 08:00AM 05:58PM 4 207 608.59 4 7.4 4.1 2.3 1.0 2.2 08:00AM 03:24PM 3 130 416.05 5 9.6 6.4 2.3 1.0 5.3 08:00AM 05:37PM 3 204 599.64 6 9.4 5.8 2.6 1.0 4.3 08:00AM 05:22PM 4 185 552.96 7 9.6 6.7 1.9 1.0 6.2 08:00AM 05:34PM 2 214 625.79 8 9.6 5.9 2.7 1.0 2.8 08:00AM 05:35PM 4 189 563.48 9 5.9 2.8 2.2 1.0 1.5 08:00AM 01:56PM 3 89 311.65 Total 74.8 46.1 20.7 8.0 31.1 30 1476 4500.87 VEHICLE INFORMATION
Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle
no typ capcty weight weight util capcty cube cube util description
1 1 1000 925 0 92.5% 9999 0 0 .0% Truck #1 2 1 1000 625 0 62.5% 9999 0 0 .0% Truck #1 3 1 1000 900 0 90.0% 9999 0 0 .0% Truck #1 4 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 5 1 1000 900 0 90.0% 9999 0 0 .0% Truck #1 6 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 7 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 8 1 1000 1000 0 100.0% 9999 0 0 .0% Truck #1 9 1 1000 850 0 85.0% 9999 0 0 .0% Truck #1 Total 9000 7925 0 88.1% 89991 0 0 .0%
Substituting yard location A for the current yard location and solving for the route design in ROUTER yields Figure 13-4. No routes can be put end-to-end so that one truck can be used instead of two, so the minimum number of trucks remains at nine. The total daily cost for this location is 3872.02 +9 x 200 + 480 = P6,152.092.
*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION
Route Run Stop Brk Stem
Route time, time, time, time, time, Start Return No of Route Route
no hr hr hr hr hr time time stops dist,Mi cost,$
1 9.8 7.0 1.8 1.0 5.4 08:00AM 05:47PM 4 224 649.88 2 8.4 5.1 2.3 1.0 3.8 08:00AM 04:26PM 3 163 498.40 3 8.4 5.0 2.4 1.0 3.9 08:00AM 04:26PM 3 161 491.57 4 8.6 4.8 2.9 1.0 2.2 08:00AM 04:37PM 5 152 470.04 5 7.2 4.0 2.2 1.0 2.6 08:00AM 03:12PM 3 128 410.15 6 6.3 2.9 2.4 1.0 1.6 08:00AM 02:18PM 3 93 321.27 7 5.3 2.1 2.2 1.0 1.6 08:00AM 01:15PM 3 66 254.22 8 8.9 5.7 2.1 1.0 3.7 08:00AM 04:51PM 3 183 548.68 9 5.1 1.7 2.3 1.0 .6 08:00AM 01:03PM 3 55 227.82 Total 68.0 38.3 20.7 9.0 25.5 30 1225 3872.02 VEHICLE INFORMATION
Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle
no typ capcty weight weight util capcty cube cube util description
1 1 1000 475 0 47.5% 9999 0 0 .0% Truck #1 2 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 3 1 1000 1000 0 100.0% 9999 0 0 .0% Truck #1 4 1 1000 975 0 97.5% 9999 0 0 .0% Truck #1 5 1 1000 875 0 87.5% 9999 0 0 .0% Truck #1 6 1 1000 1000 0 100.0% 9999 0 0 .0% Truck #1 7 1 1000 875 0 87.5% 9999 0 0 .0% Truck #1 8 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 9 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 Total 9000 7925 0 88.1% 89991 0 0 .0%
Continuing with location B, nine trucks are required and the total daily cost for the route design in Figure 3 is 3370.42 + 9 x 200 + 450= P5,620.42.
*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION
Route Run Stop Brk Stem
Route time, time, time, time, time, Start Return No of Route Route
no hr hr hr hr hr time time stops dist,Mi cost,$
1 7.1 3.7 2.4 1.0 2.3 08:00AM 03:06PM 4 120 389.30 2 6.5 3.1 2.3 1.0 1.8 08:00AM 02:27PM 3 100 340.53 3 7.9 4.6 2.4 1.0 3.0 08:00AM 03:56PM 4 146 454.97 4 5.9 2.8 2.2 1.0 1.4 08:00AM 01:56PM 3 89 311.84 5 5.9 2.6 2.3 1.0 1.1 08:00AM 01:55PM 3 84 300.41 6 6.7 3.6 2.1 1.0 2.3 08:00AM 02:41PM 3 114 375.39 7 5.8 2.5 2.3 1.0 2.1 08:00AM 01:50PM 3 80 290.22 8 6.1 3.3 1.9 1.0 2.8 08:00AM 02:07PM 2 104 350.46 9 9.7 5.8 2.8 1.0 3.0 08:00AM 05:40PM 5 187 557.31 Total 61.7 32.0 20.7 9.0 19.8 30 1024 3370.42 VEHICLE INFORMATION
Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle
no typ capcty weight weight util capcty cube cube util description
1 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 2 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 3 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 4 1 1000 850 0 85.0% 9999 0 0 .0% Truck #1 5 1 1000 925 0 92.5% 9999 0 0 .0% Truck #1 6 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 7 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 8 1 1000 825 0 82.5% 9999 0 0 .0% Truck #1 9 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 Total 9000 7925 0 88.1% 89991 0 0 .0%
Finally, the route design from location C is shown in Figure 4. Although 10 routes are in the design, two of these can be dovetailed so that only nine trucks are needed. The total daily cost is 4479.99 + 9 x 200 + 420 = P6,699.99.
*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION
Route Run Stop Brk Stem
Route time, time, time, time, time, Start Return No of Route Route
no hr hr hr hr hr time time stops dist,Mi cost,$
1 1.5 .6 .9 .0 .6 08:00AM 09:30AM 1 20 140.91 2 6.7 3.7 2.0 1.0 1.9 08:00AM 02:41PM 2 120 389.21 3 8.0 4.1 2.9 1.0 1.5 08:00AM 04:00PM 5 132 420.20 4 9.4 5.6 2.8 1.0 3.4 08:00AM 05:25PM 5 180 540.48 5 7.5 4.2 2.3 1.0 2.7 08:00AM 03:29PM 3 134 425.69 6 7.7 4.2 2.4 1.0 1.9 08:00AM 03:39PM 3 136 429.10 7 7.0 3.7 2.3 1.0 2.8 08:00AM 02:59PM 3 117 383.04 8 7.3 4.0 2.3 1.0 3.5 08:00AM 03:18PM 3 127 408.53 9 9.6 7.2 1.4 1.0 5.6 08:00AM 05:35PM 2 229 663.40 10 9.8 7.4 1.4 1.0 6.3 08:00AM 05:47PM 3 236 679.42 Total 74.5 44.7 20.7 9.0 30.3 30 1432 4479.99 VEHICLE INFORMATION
Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle
no typ capcty weight weight util capcty cube cube util description
1 1 1000 375 0 37.5% 9999 0 0 .0% Truck #1 2 1 1000 875 0 87.5% 9999 0 0 .0% Truck #1 3 1 1000 975 0 97.5% 9999 0 0 .0% Truck #1 4 1 1000 925 0 92.5% 9999 0 0 .0% Truck #1 5 1 1000 925 0 92.5% 9999 0 0 .0% Truck #1 6 1 1000 1000 0 100.0% 9999 0 0 .0% Truck #1 7 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 8 1 1000 950 0 95.0% 9999 0 0 .0% Truck #1 9 1 1000 550 0 55.0% 9999 0 0 .0% Truck #1 10 1 1000 400 0 40.0% 9999 0 0 .0% Truck #1 Total 10000 7925 0 79.3% 99990 0 0 .0%
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. Although the methodology is somewhat elementary, it can be useful in providing some first approximations to good warehouse locations. It allows students to evaluate the financial implications of alternative network designs, and it is intended to be solved with the aid of the COG module provided in the LOGWARE software.
Answers to Questions
(1) Based on information for the current year, is Kansas City the best location for a warehouse? If not, 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, the primary location costs are transportation, both inbound to the warehouse and outbound from it, and the warehouse lease, which varies with the location. The current location serves as a benchmark against which the costs for other locations can be compared. That is, based on information given in the case, the total relevant cost for the Kansas City location is:
Inbound transportation $ 2,162,535
Outbound transportation 4,819,569
Lease $2.75/sq. ft. × 200,000 sq. ft. 550,000 Total relevant cost $ 7,532,104
Using the COG module with inbound transport rates from Phoenix set at $16.73/1163 = $0.014/cwt./mile and from Monterrey set at $9.40/1188 = $0.008/cwt./mile, and the outbound transport rate from the unknown warehouse location set at $0.0235/cwt./mile, the coordinates for the best location are X = 7.61 and Y = 4.51, or approximately Oklahoma City. Total transportation cost for this location would be $6,754,082. The total relevant cost would be:
Transportation $ 6,754,082
Lease $3.25/sq. ft. × 200,000 sq. ft. 650,000
Total $ 7,404,082
The annual cost savings can be projected as $7,532,104 − 7,404,082 = $128,022.
(2) In five years, management expects that Seattle, Los Angeles, and Denver markets to grow by 5 percent, but the remaining markets to decline by 10 percent. Transport costs are expected to be unchanged. Phoenix output will increase by 5 percent, and Monterrey's output will decrease by 10 percent. Would your decision about the location of the warehouse change? If so, how?
A new benchmark for the fifth year can be computed from the data given in Tables 1 and 2. After adjusting the plant and market volumes according to the changes indicated, the fifth year benchmark costs can be computed as follows:
Volume, Rate, Transport
Point cwt. $/cwt. cost, $ 1 64,575 $16.73 $1,080,340 2 108,540 9.40 1,020,276 3 17,850 33.69 601,367 4 33,600 30.43 1,022,448 5 13,125 25.75 337,969 6 8,550 18.32 156,636 7 26,550 25.24 670,122 8 18,900 19.66 371,574 9 37,170 26.52 985,748 10 7,740 26.17 202,556 11 9,630 27.98 269,447 Totals 346,230 $6,718,483
The total 5th-year benchmark relevant cost would be:
Transportation $ 6,718,483
Lease $2.75/sq. ft. × 200,000 sq. ft. 550,000
Total $ 7,268,483
Optimizing the location with the fifth year data gives a location at X = 7.05 and Y = 4.52. The relevant costs for this location are:
Transportation $ 6,464,206
Lease $3.25/sq. ft. × 200,000 sq. ft. 650,000
Total $ 7,114,206
The annualized cost savings would be $7,268,483 − 7,114,206 = $154,277.
The average annual cost savings are (128,022 + 154,277)/2 = $141,150. The simple annual return on investment (moving cost) can be computed as:
ROI = $141, × =
$300, .
150
000 100 471%
Management must now judge whether 47.1 percent annual return is worth the risk of changing warehouse locations.
It is assumed here that the fifth year demand level applies. A revised fifth year benchmark can be recomputed by applying the cost growth factors to the overall fifth year transport costs. That is,
Outbound 1.25×4,617,867 = 5,772,334 Lease 550,000 Total $8,738,042
Running COG shows that the minimum transport cost location would be at coordinates
X = 7.20 and Y = 4.62, which is near the previous location in question two. The shift in location is minimal. The cost for this location is:
Transportation $ 7,939,545 Lease $3.25/sq. ft. × 200,000 sq. ft. 650,000 Total $ 8,589,545
The annualized cost savings would be $8,738,042 − 8,589,545 = $148,497. It can be concluded that:
1. Location is similar to the optimized fifth year location.
2. The increase in possible cost savings further encourages relocation from Kansas City and toward a site near Oklahoma City, OK.
(4) If the center-of-gravity method is used to analyze the data, what are its benefits and
limitations for locating a warehouse?
The center-of-gravity method locates a facility based on transportation costs alone. This is reasonable when only one facility is being located and the general location for it is being sought. Such costs as inventory carrying, production, and warehouse fixed are not included, but they are not particularly relevant to the problem. However, costs such as warehouse storage and handling, 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, although distance may be nonlinear.
The obvious benefits of the method are (1) it is a fast solution methodology; (2) it considers all possible locations (continuous); (3) it is simple to use; (4) its data are readily available; and (5) it gives precise locations through a coordinate system. Some potential limitations are (1) coordinates need to be linear; (2) transportation rates on a per-mile basis are constant; (3) volumes are known and constant for given demand and source points; and (4) locations may be suggested that are not feasible such as in lakes, central cities, or restricted lands.
Inbound 1.15 × 2,100,616 = $2,415,708 Subtotal $8,188,042
Concluding Comments
The analysis in the case seems to suggest a move from Kansas City to a region around Oklahoma City would be advantageous. A return on investment of 47 percent or higher is possible, however management must now seek a particular site in the area whose choice may add or detract from this savings potential. In any case, the COG method has assisted in the selection of good potential locations and testing their sensitivity to changes in costs and volumes.
APPENDIX 1 LOGWARE COG Module Input Data for the Current Year Title: SUPERIOR MEDICAL EQUIPMENT COMPANY
Power factor: .5 Scale factor: 230
Point X-coordinate Y-coordinate Volume Rate
1 3.60 3.90 61,500 0.0140 2 6.90 1.00 120,600 0.0080 3 0.90 9.10 17,000 0.0235 4 1.95 4.20 32,000 0.0235 5 5.60 6.10 12,500 0.0235 6 7.80 3.60 9,500 0.0235 7 10.20 6.90 29,500 0.0235 8 11.30 3.95 21,000 0.0235 9 14.00 6.55 41,300 0.0235 10 12.70 7.80 8,600 0.0235 11 14.30 8.25 10,700 0.0235
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. It also provides an application for the multiple center of gravity location methodology. The MULTICOG module in the LOGWARE software can effectively be applied. The module allows students to quickly evaluate alternatives as to the number of bureaus to use, the bureau locations, and the size of the territory that each bureau should serve.
The case may be assigned as a homework problem, a short case study project, or as a case for class discussion. 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, what data he needs and where to obtain it, and what concerns he should have about changing the existing network design. This would encourage students to think beyond the computational aspects of the problem.
Answers to Questions
(1) Do you think there is any benefit to changing the network of license bureaus in the
Cleveland area? If so, 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. Therefore, a possible approach to the analysis is outlined as follows.
First, establish a benchmark against which changes to the network can be compared. Much of the data for this is given in the case write up. The costs can simply be applied to the size of each bureau and its associated staff. The cost for residents traveling to the bureaus is not known because the bureau territories are not known. However, an estimate can be made of travel costs by solving the problem in MULTICOG for eight bureaus. Since MULTICOG attempts to optimize bureau location, this travel cost is probably understated. The benchmark costs are summarized in Table 1. The location costs for the current operation are estimated to be $1,355,706.
Second, what improvements can be made on the existing eight locations? Besides moving the locations of the bureaus, which results in resizing the facilities and adjusting the staff numbers, there are no obvious improvements to be made. Therefore, the benchmark remains the base for comparison.
Third, it is now necessary to estimate the approximate number of bureaus that are needed to serve the area. Since the costs for a particular network design depend on the size of each bureau, which cannot be known until the problem is solved, an initial assumption must be made. It will be assumed that all bureaus are of the same size. Hence, for 5 bureaus, the average number of residents in each bureau’s territory would be the total number of residents divided by the number of bureaus, or 691,700/5 = 138,340. Rent, staff salaries, and utility expenses can be derived from this estimate. Table 2 is developed to show the bureau size and the number of staff for one to 10 bureaus. Table 3 extends the average costs from these estimates. A reduced number of bureaus, in the range of two, is about right.
Bureau Size, sq. ft. Staff Rent, $ Salaries, $ Utilities, $ Customer Travel, $ 1 1,700 4 37,400 84,000 6,800 2 1,200 4 26,400 84,000 4,800 3 2,000 5 44,000 105,000 8,000 4 1,800 4 39,600 84,000 7,200 5 1,500 4 33,000 84,000 6,000 6 2,200 5 48,400 105,000 8,800 7 2,700 5 59,400 105,000 10,800 8 1,500 5 33,000 105,000 6,000 Totals 14,600 36 321,200 756,000 58,400 220,106*
*Approximated by running MULTICOG at 8 bureaus.
TABLE 2 Average Size and Staff for Various Numbers of Bureaus
(1) No. of bureaus 691,700/(1) Avg. no. of residents per bureau Avg. bureau size, sq. ft. Staff per bureau 1 691,700 4,500 10 2 345,850 3,000 7 3 230,567 2,500 6 4 172,925 2,000 5 5 138,340 2,000 5 6 115,283 2,000 5 7 98,814 1,500 4 8 86,463 1,500 4 9 78,855 1,500 4 10 69,170 1,500 4
TABLE 3 Average Costs by Number of Bureaus No. of bureaus Rent, $ Staff salaries, $ Utilities, $ Resident travel, $ Annual total cost, $ 1 99,000 210,000 18,000 662,319 989,319 2 132,000 294,000 24,000 430,922 880,922⇐⇐⇐⇐ 3 165,000 378,000 30,000 354,239 927,239 4 176,000 420,000 32,000 298,000 926,965 5 220,000 525,000 40,000 278,181 1,063,181 6 264,000 630,000 48,000 249,287 1,191,287 7 231,000 588,000 42,000 237,635 1,098,636 8 264,000 672,000 48,000 220,106 1,204,106 9 297,000 756,000 54,000 206,496 1,313,496 10 330,000 840,000 60,000 198,600 1,428,600
The cost estimates can now be refined around two bureaus. A comparison with the benchmark costs and a return of the initial investment (costs related to changing the network design) are sought. A sample analysis for two bureaus, based on a design provided by MULTICOG, is shown below.
Bureau Resi-dents Size, sq. ft. Staff Rent, $ Staff salaries, $ Utilities, $ Resident travel cost, $ 1 290,200 2,500 6 55,000 126,000 10,000 166,332 2 401,500 3,500 8 77,000 168,000 14,000 264,590 Totals 691,700 6,000 14 132,000 294,000 24,000 430,922
The total annual variable cost is $132,000 + 294,000 + 24,000 + 430,922 = $880,922. There are one-time costs due to staff separation and equipment moves. Compared with the benchmark, 36 − 14 = 22 staff members will be separated for a cost of 22 × $8,000 = $176,000. Equipment movement costs to two bureaus would be 2 × 10,000 = $20,000. Total movement costs would be $176,000 + 20,000 = $196,000.
Annual variable cost savings compared with the benchmark would be $1,355,706 − 880,922 = $474,784. A simple return on investment would be:
ROI = $474, × =
$196, 784
000 100 242%
Similar calculations are carried out for various numbers of bureaus. These results are tabulated in Table 4.
TABLE 4 Cost Savings and Return of Investment for Alternate Network Designs as Produced by MULTICOG No. of bureaus Total size, sq. ft. Total staff Annual variable cost, $ Moving cost, $ Savings, $ Return on invest-ment, % 1 4,500 10 989,319 218,000 366,387 168 2 6,000 14 880,922 196,000 474,784⇐⇐⇐⇐ 242 3 7,500 18 927,239 176,000 428,467 243 4 8,500 21 960,965 160,000 394,741 246⇐⇐⇐⇐ 5 9,500 24 1,029,181 146,000 326,525 223 6 11,500 29 1,157,287 106,000 198,419 187 7 12,000 31 1,200,635 110,000 155,071 141 8 13,000 34 1,272,106 96,000 83,600 87 Benchmark 14,600 36 1,355,706 ---- ---- ---
The maximum annual savings occurs with a network containing two bureaus. However, the maximal return on investment occurs with four bureaus. ROI is selected as the appropriate measure on which to base this economic decision. The details for a design with four bureaus are given in Table 5. The design is shown pictorially in Figure 1 of this note.
TABLE 5 Design Details for a Network with Four Bureaus
Bureau Column grid coordi-nate Row grid
coordi-nate Residents Grid box number assignment
1 3.36 2.74 218,200 1,2,3,4,5,6,7,8,9,10,11,12,13, 14,15,16,17,18,19,20,21,22,23, 24,25,26,27,28,29,30,31,32,33 2 7.00 3.00 168,700 34,35,36,37,38,39,40,41,43,44, 45,46,47,50,51,52,53 3 9.74 5.66 195,000 42,48,49,54,55,56,60,61,62,63, 67,68,69,70,74,75,76,77,81,82, 83,84, 4 10.00 2.00 109,800 57,58,59,64,65,66,71,72,73,78, 79,80
(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. The center-of-gravity approach is appropriate in this problem since there are no capacity limitations on the facilities, locations across a continuous space are desired, and transportation cost (except for some fixed costs) is the primary location variable. The data requirements are relatively straightforward, although they are not always easy to
• The effect of bureau location on the resident's perception of service is not as well known as portrayed in the case. In addition, service may need to be represented by more than location.
• Travel to the bureaus is assumed straight line. However, location in the area is likely to be influenced by a road network. Time may be more important than distance to residents.
• The fixed costs associated with location are not handled directly by the center-of-gravity approach.
• Residents are assumed to travel to the locations within their assigned territories. They may not strictly do this.
• Good facilities may not be available at the indicated location coordinates.
The analysis is particularly weak around the estimate of the resident travel cost. While an exact cost is not likely to be known, Dan should conduct a sensitivity analysis around this cost. He may find that the design does not change a great deal over a wide range of assumed values. If this is the case, he can feel comfortable that his recommendation is fundamentally sound. If not, he should seek to find a more precise value.
(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. Rather, it provides a starting point for further analysis. There are a number of other factors to be considered before the revised network design can be implemented. First, there are site selection factors to be taken into account such as the availability of adequate space near the location coordinates, proximity to good highway linkages, and reasonable neighborhood reactions to this type of operation.
Second, there are political concerns. Reducing the number of locations will result in a releasing some of the staff. Dan may experience some political resistance to this. Since staff is a large expense in the operation, currently about 2/3 of the costs, retention of a larger number of bureaus may be required. Of course, transferring staff to other governmental operations may be a way of dealing with this issue. This assumes their willingness to be relocated, although this is not likely to be a strong issue if relocation were to occur in the same area.
Third, there may be difficulty in demonstrating the economics of network redesign. Although others may appreciate the costs of rent, salaries, and utilities, the cost of resident travel is subject to much interpretation. 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.
FIGURE 1 Four Bureau Locations and Their Territories
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: 2.5
POINT X-COORDINATE Y-COORDINATE VOLUME RATE
1 1 1 4100 .12 2 1 2 6200 .12 3 1 3 7200 .12 4 1 4 10300 .12 5 1 5 200 .12 6 1 6 0 .12 7 1 7 0 .12 8 2 1 7800 .12 9 2 2 8700 .12 10 2 3 9400 .12 11 2 4 11800 .12 12 2 5 100 .12 13 2 6 0 .12 14 2 7 0 .12 15 3 1 8100 .12 16 3 2 10500 .12 17 3 3 15600 .12
Current bureau locations Revised bureau locations Grid column number
Gr id ro w n u m b er
Current bureau locations Revised bureau locations Grid column number
Gr id ro w n u m b er
SUPPLEMENT (Continued)
POINT X-COORDINATE Y-COORDINATE VOLUME RATE
22 4 1 10700 .12 23 4 2 12800 .12 24 4 3 13800 .12 25 4 4 15600 .12 26 4 5 400 .12 27 4 6 0 .12 28 4 7 0 .12 29 5 1 11500 .12 30 5 2 13900 .12 31 5 3 14500 .12 32 5 4 13700 .12 33 5 5 600 .12 34 5 6 0 .12 35 5 7 0 .12 36 6 1 9300 .12 37 6 2 14900 .12 38 6 3 13700 .12 39 6 4 10200 .12 40 6 5 1200 .12 41 6 6 0 .12 42 6 7 0 .12 43 7 1 10100 .12 44 7 2 12600 .12 45 7 3 16700 .12 46 7 4 15800 .12 47 7 5 12400 .12 48 7 6 2600 .12 49 7 7 0 .12 50 8 1 8800 .12 51 8 2 13700 .12 52 8 3 15200 .12 53 8 4 14100 .12 54 8 5 10800 .12 55 8 6 17200 .12 56 8 7 500 .12 57 9 1 5300 .12 58 9 2 16700 .12 59 9 3 13800 .12 60 9 4 11900 .12 61 9 5 13500 .12 62 9 6 18600 .12 63 9 7 12000 .12 64 10 1 5100 .12 65 10 2 17400 .12 66 10 3 10300 .12 67 10 4 9800 .12 68 10 5 10300 .12 69 10 6 15500 .12 70 10 7 11700 .12 71 11 1 7700 .12 72 11 2 9200 .12 73 11 3 7500 .12 74 11 4 8500 .12
POINT X-COORDINATE Y-COORDINATE VOLUME RATE 75 11 5 7800 .12 76 11 6 9900 .12 77 11 7 8700 .12 78 12 1 4300 .12 79 12 2 6700 .12 80 12 3 5800 .12 81 12 4 6800 .12 82 12 5 5400 .12 83 12 6 7100 .12 84 12 7 6400 .12
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. They first should identify the major costs and alternatives that are important to such a design problem. Second, 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. Finally, 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.
Answers to Questions
(1) If you were Carolyn Carter, would you agree with the proposal to build the new
brewery? If you do, what plan for distribution would you suggest?
If growth is uniform over the next five years, Southern can expect that demand for its products will exceed the currently available plant capacity. That is, demand is increasing at the rate of (0 + (595,000 − 403,000)) ÷ 5 = 38,400 barrels per year. Thus, the current annual capacity of 500,000 barrels will be used up in (500,000 − 403,000) ÷ 38,400 = 2.5 years. A major concern is whether it would be profitable to construct the new plant. Rough estimates of its profitability can be made, projecting profits with and without the new plant. We know that 595,000 − 500,000 = 95,000 barrels of beer would not be sold annually in the fifth year if additional capacity is not constructed. This represents a potential average lost revenue of $280/barrel × 0.20 × [(0 + 95,000) ÷ 2.5] = $2,128,000. (The figure of 2.5 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.) From the benchmark1 costs for the current system, as shown in Table 1, Southern is currently producing and distributing 403,000 barrels at a total cost of $60,015,000. This is an average cost per barrel of $60,015,000 ÷ 403,000 = $148.92. The overhead and sales expense is 27 percent, or $280 × 0.27 = $75.60 per barrel. Total costs per barrel are $148.92 + 75.60 = $224.52, which is about 80 percent of the sales dollar. The 20 percent profit margin seems valid. Therefore, the benefit of serving the potentially lost demand with a new brewery can be estimated using on a simple return on investment:
21% or 0.21, 0 $10,000,00 $2,128,000 ROI= =
If management feels that this is an adequate return for such a project, Carolyn should proceed with her analysis. Let's assume that she has this approval.
Next, she may wish to explore the opportunities available by improving upon the existing distribution system without the presence of the new plant. This is an improved
1
A benchmark refers to the costs of producing and distributing demand as currently allocated throughout the network.
benchmark2 and it can be found by solving a transportation-type linear programming problem of the type shown in Table 2. The results given in the TRANLP module of LOGWARE can be summarized in Table 3. This shows that with some slight reallocation of demand among the plants, production and distribution costs can be reduced by $60,015,000 − 59,804,000 = $211,000 annually. 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 28,000 barrels of Columbia's market should be shifted from the Columbia plant to the Montgomery plant. The Columbia plant is a high-cost producer and cost savings are achieved by trading production costs for transportation costs. That is, production costs will be reduced by $400,000 while transportation costs will be increased by $189,000. Note that this $211,000 savings can be realized without any capital investment.
TABLE 1 Benchmark of Production and Transportation Costs ($000s) for Current Demand Market area Brewery of origin Demand in 000s barrels Produc-tion costs Trans-port
costs Total costs
1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Columbia 22 3,190 282 3,472 4 Columbia Columbia 44 6,380 306 6,686 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Montgomery 13 1,781 179 1,960 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Montgomery 26 3,562 355 3,917 9 Jacksonville Montgomery 38 5,206 577 5,783 Total 403 $56,000 $4,015 $60,015
TABLE 2 TRANLP Problem Setup
From\To RICH-MOND RA-LEIGH KNOX-VILLE COL-UMBIA AT-LANTA SAVAN-NAH MONT-GOMRY TALL AHAS JACK-SONVL Supply RICHMD 148.49 150.70 156.38 152.54 155.48 154.64 159.98 164.30 158.84 100 COLMBA 157.54 154.78 157.81 151.96 156.85 154.54 157.93 160.18 157.27 100 MONTGM 156.98 153.35 150.80 149.93 147.20 150.80 143.69 150.65 152.18 300 JACKVL 152.13 149.25 150.48 146.16 148.80 144.54 148.80 144.72 142.68 0 Demand 56 31 22 44 94 13 79 26 38
