Solutions for Chapter 6 Solutions To Problems From Chapter 6 6.1 a) The Greedy Heuristic Solution is x11=60, x21=40, x22=84, x23=21, x33=56, and x34=69, at a total cost of $438,940. b) The optimal solution one obtains from Solver is : x14=60, x21=61, x22=84, x31=39, x33=77, and x34=9 at a total cost of $381,880. The spreadsheet appears below. Solution to Problem 6.1 Variables x11 Values x12 60 Objective Coeff x13 x14 x21 x22 0 0 60 40 250 420 380 280 1280 1 1 x23 84 x24 0 990 1440 x31 0 x32 x33 x34 39 0 56 69 1520 1550 1420 1660 1730 Operator Value Min 438940 RHS st Constraint 1 1 1 Constraint 2 1 1 1 1 Constraint 3 1 Constraint 4 1 Constraint 5 1 1 Constraint 6 1 1 1 1 1 1 Constraint 7 1 1 1 1 1 1 = 120 60 = 124 145 = 164 125 = 139 100 = 84 84 = 56 77 = 129 69 c) The percentage error from using the greedy heuristic is approximately 15%. 6.2 By using very large costs for the routes that are eliminated and resolving the problem, one finds the new optimal solution is: x11=39, x14=21, x21=61, x22=84, x33=77, x34=78. The total cost of the solution is 387,730, and the percentage error difference is 30%. The spreadsheet solution appears below. Solution to Problem 6.2 Variables Values Objective Coeff x11 x12 39 x13 0 250 1.00E+30 0 x14 21 x21 61 x22 x23 84 x24 0 0 x31 x32 0 0 x33 77 x34 Operator Value RHS 48 380 280 1280 990 1.00E+30 1520 1.00E+30 1420 1660 1730 Min 387730 st Constraint 1 1 1 1 1 Constraint 2 1 1 1 1 Constraint 3 Constraint 4 Constraint 5 Constraint 6 Constraint 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 127 1 = 60 60 = 145 145 = 125 125 = 100 100 = 84 84 = 77 77 = 69 69 Production and Operations Analysis, Fourth Edition 6.3 The network for this case looks essentially the same as Figure 6.2, except that we add a transshipment node between the plants and warehouses, and all flow goes through the transshipment node in Oklahoma City. If we identify the nodes as follows: 1: Sunnyvale, 2: Dublin, 3: Bangkok, 4: Oklahoma City, 5:Amarillo, 6:Teaneck, 7: Chicago, and 8: Sioux Falls, we obtain the following balance of flow equations: x14 x15 x16 x17 – x18 = – 45 node 1: node 2: x24 x25 x26 x27 x28 120 node 3: x34 x35 x36 x37 x38 95 node 4: x14 + x24 +x34 x43 x46 x47 x48 = 0. node 5: x15 + x25 + x35 + x45 = 80 node 6: x16 + x26 + x36 + x46 = 78 node 7: x17 + x27 + x37 + x47 = 47 node 8: x18 + x28 + x38 + x48 = 55. The spreadsheet with the solution is: Solution to Problem 6.3 Variables x14 x15 x16 x17 x25 x26 x35 x36 x47 x49 x57 x58 x59 x67 x68 x69 Operator RHS Value Values 45 0 Obj Funct 76 ## -1 -1 0 0 0 120 82 13 20 170 250 1440 1200 1660 1600 110 0 35 0 95 180 195 0 0 78 55 35 245 145 0 Min 3E+05 st Node 1 -1 -1 Node 2 -1 -1 Node 3 Node 4 Node 5 Node 6 Node 7 -1 -1 1 -1 1 1 1 1 1 1 -1 -1 -1 -1 1 -1 1 1 Node 8 1 -1 1 1 Node 9 -1 1 1 1 <= -45 -45 <= -120 -120 <= -95 -95 <= 25 25 <= 47 47 <= -0 0 <= 55 80 <= 78 78 <= 55 55 Note that the objective function value is 294,410. The addition of the transshipment node provides only small savings over the system without the transshipment point. The savings are $3390, or about 1.1%. 6.4 a) We solve the problem in units of 100 cars. Total production=100, total requirements = 84. It follows that this is an unbalanced problem supply exceeding demand. To find the greedy heuristic solution, add a dummy column with high cost. Letting Flint, Fresno, and Monterrey correspond to sources 1, 2, and 3 and Phoenix, Davenport, Columbia, and the Dummy column corresponding to sinks 1, 2, 3, and 4, the greedy heuristic yields x 12=28, 128 Solutions for Chapter 6 x13=15, x21=26, and x32=15 at a total cost of 1126 (thousands). This is exactly the same cost as was obtained in the chapter for the optimal solution. This shows that this problem has multiple optimal solutions. See spreadsheet below. Solution to Problem 6.4 b) Variables Values Obj Funct x11 x12 x13 x21 x22 x23 x31 x32 x33 Operator 0 13 30 26 0 0 0 15 12 8 17 7 14 21 18 22 1 1 1 1 1 1 Value RHS 0 31 min 1126 st Cons1 Cons2 Cons3 Cons4 1 1 1 Cons5 1 Cons6 1 < 43 43 < 26 26 1< 15 31 = 26 26 = 28 28 1= 30 30 1 1 1 1 1 c) 0%. d) Replacing the Monterrey/Columbia cost with a high number and resolving the problem gives exactly the same cost as part b). This is to be expected since x 33 = 0 in the original optimal solution. 6.5 The spreadsheet solution for this problem appears below. Interpret the node designations as follows: 1=Flint, 2=Fresno, 3=Monterrey, 4=Sante Fe, 5=Jefferson City, 6=Phoenix, 7=Davenport, 8=Columbia. The value of the objective function at the optimal solution is 1052 (thousands). Solution to Problem 6. 5 Variables x14 x15 x24 x25 x34 x35 x46 x47 x48 x56 x57 x58 Operator Values 0 43 26 0 15 0 26 0 15 0 28 15 Obj Funct 8 6 6 9 9 14 3 8 10 5 5 9 -1 -1 -1 -1 Value RHS Min 1052 >= -43 -43 >= -26 -26 >= -15 -31 >= -7.6E-10 0 >= -4.3E-11 0 >= 26 26 >= 28 28 >= 30 30 st Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 Node 7 Node 8 -1 1 1 1 -1 1 1 -1 -1 -1 1 -1 1 -1 -1 1 1 1 1 129 1 Production and Operations Analysis, Fourth Edition 6.6 The solution appears in the spreadsheet below. Solution to Problem 6. 6 Variables x13 Values x14 x15 x16 0 0 5000 0 Obj Funct 182 375 x23 285 460 -1 -1 x24 x25 3500 3500 77 290 -1 -1 x26 x34 x35 0 0 0 x36 x45 x46 3500 245 575 275 Operator 0 0 3500 125 380 90 110 Min Value RHS 4407000 st Node 1 -1 -1 Node 2 Node 3 1 -1 1 Node 4 -1 1 1 Node 5 -1 -1 1 1 -1 1 Node 6 6.7 -1 1 1 1 -1 1 1 1 = -5000 -5000 = -7000.0001 -7000 = 0.0001415 0 = -7.074E-05 0 = 3499.9999 3500 = 8500.0001 8500 The solution appears in the spreadsheet below. Solution of Problem 6. 7 Variables Values Obj Funct x13 x14 x15 0 x16 x23 0 0 3000 182 375 285 460 -1 -1 x24 x25 3000 4000 77 290 x26 x34 0 x35 x36 x45 x46 Operator 0 0 3000 0 0 4000 245 575 275 125 380 90 110 Min Value RHS 3586000.1 st Node 1 -1 -1 Node 2 Node 3 Node 4 Node 5 Node 6 -1 1 -1 -1 -1 1 1 -1 1 1 -1 1 1 1 -1 -1 1 1 130 -1 1 1 1 >= -3000.0003 -5000 >= -6999.9999 -7000 >= 1.116E-08 >= 0 0 0 >= 3000 3000 >= 7000.0002 7000 Solutions for Chapter 6 6.8. The solution appears in the spreadsheet below. Solution of Problem 6. 8 Variables x13 x14 x15 0 Values Obj Funct st Node 1 Node 2 Node 3 Node 4 Node 5 Node 6 6.9 x23 x24 x25 x26 x34 x35 0 1500 3500 2000 5000 182 375 -1 x16 -1 285 460 -1 -1 77 0 0 2000 290 245 575 275 -1 1 0 -1 -1 -1 1 1 0 125 380 Operator -1 90 110 -1 1 1 -1 1 1 -1 1 1 Value RHS 0 5000 -1 1 1 x36 x45 x46 1 1 Min 4441500 = -5000 -5000 = -7000 -7000 = 0 0 = 0 0 = 3500 3500 = 8500 8500 Although the problem asks to compute average sales based on forecasts, it makes more sense to determine the net demand after taking out on order and in transit stock. Doing so yields the following Net Demand Forecasts: Week 0 Net Demand 0 1 0 2 0 3 25 4 12 5 0 6 19 7 85 8 33 The average of these values based on six periods of data is 29. (If one uses the original data one obtains 33.25 here.) Using this as the value of in the EOQ formula gives the ___________ EOQ as 2*40*29/.25 = 96. (with the larger value of onc obtains 103). Ordering in lots of 96 gives the following DRP Profile: Weeks: Sales Forecasts: In Transit: On Order: EOQ placed: EOQ arrives: Projected Balance: 0 1 2 3 4 5 6 7 8 22 40 35 60 12 0 19 85 33 40 96 51 18 26 96 26 44 96 35 131 96 71 59 59 Production and Operations Analysis, Fourth Edition 6.10 Again, we will use the net demand data as shown in the solution to problem 6.9. Using the Silver Meal Algorithm with K=40, h=.25 gives the following solution (starting in period 3): y3 = 56, y7=118, and all other values zero. The resulting DRP Profile is: Weeks: 0 Sales Forecasts: In Transit: On Order: Order placed: Order arrives: Projected Balance: 1 2 3 4 5 6 7 8 22 40 35 60 12 0 19 85 33 0 118 33 0 26 56 26 44 118 35 56 31 19 19 6.11 In order to incorporate a safety stock of 30 units for each period, we just increase the net demand by 30 units in the first period. Taking into account receipts and on hand, this gives a requirements vector of (0, 0, 21, 60, 12, 0, 19, 85, 33). The Silver Meal solution for this set of requirements and K=40 and h=0.25 is y = (0, 0, 93, 0, 0, 0, 137, 0, 0). 6.12 The ROP method assumes a stationary demand pattern, but allows randomness, while DRP allows for demand variation. The DRP method is conceptually the same as MRP applied to distribution rather than production. The discussion in Chapter 7 comparing MRP and traditional ROP methods applies here as well. 6.13 Consider Example 6.4 assuming that the truck capacity is 250 loaves instead of 300 loaves. Since (1,2) is the first pair in the ranking consider linking (1,2) without violating the constraint. Since the total demand at these two locations is 247, include this link. Clearly it would be impossible to include any other locations on this route. The next pair in the ranking which doesn't include either location 1 or 2 is (3,5). These two locations can be linked since the total demand in these locations is 26 + 110 = 136. Location 4 cannot be included on this route since that brings the demand on this route to 136 + 140 = 276 > 25. Hence location 4 would require a separate route. This solution thus requires three distinct routes: (1,2), (3,5) and 4. 6.14 The distance matrix for this case is 0 1 2 3 4 1 2 3 4 5 45 35 10 30 15 15 10 35 25 20 30 25 35 20 20 The resulting values of sij are: 132 Solutions for Chapter 6 s12 s13 s14 s15 s23 = = = = = 70 60 20 50 50 s24 s25 s34 s35 s45 = = = = = 20 30 20 40 20 It turns out that the ranking of the pairs for this case is exactly the same as the ranking when using the Euclidean metric, if one chooses to break ties to be consistent. Hence the resulting solution is the same as the solution of Example 7.11. 6.15 Although this is not marked as a computer problem, the calculations involved with adding only two additional locations turns out to be tedious. You may wish to recommend that a computer be used to assist with these calculations. The first step required in solving this problem is to complete the cost matrix. This means computing the Euclidean distances from all locations to location 6 (ci6 for i = 0 to 5) and location 7 (ci7 for i = 0 to 6). The two new columns that result are: 6 0 1 2 3 4 5 6 7 17.0 18.2 19.3 8.2 10.0 8.2 23.2 28.2 32.5 21.4 18.1 7.6 14.2 The next step is to compute the new sij values (11 in all). They are s16 s26 s36 s46 s56 = = = = = 32.3 28.1 31.2 14.1 31.2 s17 s27 s37 s47 s57 S67 = = = = = = 28.5 21.1 24.2 12.2 38.0 26.0 The next setp is to rank the 21 pairs of locations in decreasing order of the sij values. The ranking one obtains is: (1,2), (1,3), (2,3), (5,7), (1,5) (1,6), (3,6), (5,6), (3,5), (1,7) (2,6), (2,5), (6,7), (3,7), (2,7) 133 Production and Operations Analysis, Fourth Edition (4,6), (1,4), (3,4), (4,5), (2,4), (4,7). The final step is to obtain the routes by considering the pairs in sequence and including new locations until the capacity constraint is exceeded. We assume the same capacity constraint as in Example 6.4 which is 300 loaves. As before 1, 2, and 3 are on the first route. This results in a requirement of 273 for this route. This route is then closed. The next route starts with 5 and 7 with a req't. of 236. This route is then closed. The final route consists of locations 4 and 6. While tedious, this example does illustrate one point: that by adding only two additional locations, the complexity of the calculations goes up considerably, in fact by a factor of about 2. 6.16 The first step here is to compute the sij values. There will be a total of ten values to compute. One obtains: s12 s13 s14 s15 s23 = = = = = 60 48 10 35 90 s24 s25 s34 s35 s45 = = = = = 27 63 3 43 15 Next, these pairs are ranked in decreasing order giving: (2,3), (2,5), (1,2), (1,3), (3,5), (1,5), (2,4), (4,5), (1,4), (3,4) Now we construct the routes considering each pair in sequence and checking at each assignment that we do not violate the 50 mile constraint. The first route consists of locations 2 and 3 at a total distance of 18 miles. Next we include 5 which requires an additional 20 miles, bringing this route up to 38 miles total. The pairings would recommend including location 1 next, but the distance separating 5 and 1 is 15 miles thus violating the constraint. In fact, this route must be closed at this point. The next route would start with the next pair of unassigned routes, which are 1 and 4, which would constitute the second route. 6.17 The matrix one obtains of the straight-line distances separating locations is: 0 1 2 3 4 1 14.9 2 26.9 12.1 3 30.2 16.1 8.6 The savings terms are: 134 4 37.0 30.1 28.2 36.7 5 12.2 8.6 18.0 24.0 25.5 Solutions for Chapter 6 s12 s13 s14 s15 s23 = = = = = 29.7 29.0 21.8 18.5 41.0 s24 s25 s34 s35 s45 = = = = = 33.8 30.5 37.1 33.8 40.6 This results in the following ranking of the pairs: (2,3), (4,5), (3,4), (2,4), (3,5), (2,5), (1,2), (1,3), (1,4), (1,5) First locations 2 and 3 are combined to be on the same route. The req't. of this route is 1,050. Assuming a 1200 gallon capacity this means this route is closed. Locations 1, 4 and 5 can now be included in the second route. The total requirement of this second route is 1100 gallons. 6.18 Postponement refers to delaying the final configuration of a product as long as possible. The main advantage is that by delaying, new information becomes available, thus allowing a better match between products and demand. As a simple example, consider packing for trip. By delaying the decision of exactly what clothes to pack as long as possible, one can check the weather at the destination, and only pack what's suitable for the expected weather conditions. 6.19 In this example, Lincoln and Jaguar would want to delay putting in the engines as long as possible to see which engine type was preferred by consumers before committing to a particular mix. While it would probably not be cost effective here, the two companies could build local configuration facilities that would allow putting in the engines on a custom order basis. 6.20 Too many suppliers can cause logistics complications. When problems arise, identifying which supplier is responsible could be difficult. Also, more personnel are required to manage relationships with suppliers. There are also risks associated with too few suppliers. For example, if there is only one supplier for a critical part or resource, and something happens to that supplier, disaster could result. Many years ago, a single Japanese company supplied all the resin used in semiconductor production worldwide. When this company experienced a fire, semiconductor production was seriously affected. 6.21 In some cases, it is easier and more efficient for large firms to outsource specific functions, even when some of those functions are duplicated in-house. One notable example is the development of the original IBM PC. The project was initiated by IBM in the early 80’s somewhat secretly. To keep this project separate from the company’s more traditional business, much of the PC was outsourced (Intel produced the microprocessor and the operating system was licensed from Microsoft.) In retrospect, one would imagine that IBM was very sorry it did not develop these key components in-house. Both Microsoft and Intel grew to be giants, largely as a result of IBM’s decision. 6.22 The bullwhip refers to the increase in variability as one moves up the supply chain. It has 135 Production and Operations Analysis, Fourth Edition been observed in practice in several contexts, and can result in highly variable demand patterns on the factory even for items in which customer demand is fairly stable. The idea behind the term "bullwhip" is that a small, but well timed, flick of the wrist results in a very effecive and highly charged snap at the end of the whip. 6.23 Eliminating the middleman should reduce the bullwhip effect, since there are fewer intervening layers in the system. However, there are some circumstances where middlemen are not the cause of the problem. For example, one cause of high variability is “phantom” orders. This is a result of shortage gaming, discussed in the section. Another situation in which there might be high variability of orders is when demand is very hard to predict, such as is the case for fashion items. 6.24 The causes listed for the bullwhip effect include: 1. Demand forecast updating 2. Order batching 3. Price fluctuations 4. Shortage gaming The recommended remedies are: 1. Information sharing 2. Channel alignment 3. Price stabilization 4. Disincentives for shortage gaming. It is unclear if any of these remedies could be implemented and what effect they might have within existing supply chain distribution structures. 6.25 The popularity of EDI can be attributed primarily to: 1) sunk cost of investment in EDI systems, both in terms of dollars and time, and 2) improved security afforded by EDI over web-based transactions systems. For new entrants to the market, web based systems are likely to be less expensive and easier to implement, by avoiding the complex EDI protocols. 6.26 E-tailing has grown into a multibillion dollar business in only a few years. E-tailing web sites require a much lower level of investment, so entry into the marketplace is much easier. However, the other side of the coin is that unless these web sites can distinguish themselves in some way, there is a high risk of failure. For example, Amazon.com was one of the first e-tailers, and continues to be very successful. But at this point, it would be difficult for new web-based book sellers to break into the market. 6.27 Risk pooling refers the decrease of relative variation in multilevel inventory systems in which distribution centers are used to pool the risk at demand points. The risk pooling effect is a consequence of the fact that adding independent random variables, variances rather than standard deviations, are additive. 136 Solutions for Chapter 6 6.28 If a system gets big enough, (say over 1,000 stores), the number of DC’s required to serve these stores can get large as well. By inserting an additional layer between the DC’s and the factory, one can do risk pooling from the DC’s. Usually, only very large scale systems would benefit from such a structure. 6.29 The two factors that determine the risk pooling benefit include the variance of sales and the number of stores served by the DC. That is, risk pooling has a greater effect when the variance of sales is high, and when there is a larger number of stores (or more generally, demand points). 6.30 Certainly China should emerge as one of the most important new markets in coming years. Not only is China the most populous country in the world, but investments of billions of dollars by Japan and the United States, among others, are being poured into China. As the standard of living improves, and (hopefully) the government allows the economic system to migrate towards free enterprise, the need and desire among Chinese for Western goods will increase as well. The second most populous country in the world, India, also has the potential for being a significant market. Another emerging market is Latin America. However, progress in both Latin America, and to a greater extent, Africa, is hampered by political instability. 6.31 In the opinion of this writer, this trend is probably not going to decline. Even so, factors that could slow or reverse the trend include: 1) world-wide economic depression leading to the demise of these firms. 2) war 3) political problems in the emerging new markets discussed in the answer to problem 6.30. 6.32 Manufacturing has always migrated to places where wage rates are lower. This occurred in the United States in the 1940’s and 1950’s when textile and clothing manufacturing moved from the northeast to the southeast. Today, much of the movement is offshore to Asia and Latin America. Some writers suggest that this is a poor choice. That is, they claim that there is too much emphasis placed on direct labor costs, and not enough on other factors, such as stability and communication between manufacturing and design. This migration would probably halt if wage rates equalized worldwide. (This is unlikely to occur any time soon, however.) 6.33 Supply chains cannot react as fast when long distances separate manufacturing facilities from other parts of the corporation. Supply chains are also hampered by less developed infrastructures. Poor roads, airports, and port facilities can significantly impair the flow of goods. Finally, implementation of such initiatives as JIT purchasing can be impossible when suppliers and customers are separated by long distances. 137