17 CHAPTER 7 TRANSPORT DECISIONS 1 Selecting a mode of transportation requires balancing the direct cost of transportation with the indirect costs of both vendor and buyer inventories plus the in-transit inventory costs. The differences in transport mode performance affect these inventory levels, and, therefore, the costs for maintaining them, as well as affect the time that the goods are in transit. We wish to compare these four cost factors for each mode choice as shown in Table 7-1 of the manual. The symbols used are: R = transportation rate, $/unit D = annual demand, units C = item value at buyers inventory, $ C = item value at vendors inventory, $ T = time in transit, days Q = Shipping quantity, units Rail has the lowest total cost. 2 As in question 1, this problem is one of balancing transport costs with the indirect costs associated with inventories. However, in this case we must account for the variability in transit time as it affects the warehouse inventories. We can develop the following decision table. TABLE 7-1 An Evaluation of the Transport Alternatives for the Wagner Company Cost type Method Rail Piggyback Truck Transport RD 2550,000 = $1,250,000 4450,000 = $2,200,000 8850,000 = $4,400,000 In-transit inventorya ICDt/365 0.2547550,000 (16/365) = $260,274 0.2545650,000 (10/365) = $156,164 0.2541250,000 (4/365) = $564,384 Wagers inventorya ICQ/2 0.25475(10,000/2) = $593,750 0.25456(7,000/2) = $399,000 0.25412(5,000/2) = $257,500 Electronics inventory ICQ/2 0.25500(10,00/2) = $625,000 0.25500(7,000/2) = $437,500 0.25500(5,000/2) = $312,500 Total $2,729,024 $3,192,664 $5,534,384 a C refers to price less transport cost per unit. 18 Service type Cost type Method A B Transport RD 129,600 = $115,200 11.809,600 =$114,048 In-transit inventory ICDt/365 0.20509,600 (4/365) = $1,052 0.20509,600 (5/365) = $1,315 Plant inventory ICQ*/2 0.3050(321.8/2) = $2,684 0.3050(357.8/2) = $2,684 Warehouse inventory ICQ*/2 + ICr 0.3062(321.3/2) + 0.306250.5 = $3,927 0.3061.80(321.8/2) + 0.3061.8060.6 = $4,107 Total $122,863 $122,154 Recall that QDSIC * /( ,)() / . ().=22 9 600 10003 503578cwt. for the plant, assuming the order cost is the same at plant and warehouse. However, for the warehouse, we must account for safety stock (r) and for the transportation cost in the value of the product. Therefore, For A: QDSIC * /( ,)() / . ().=22 9 600 10003 623213 cwt. and for z = 1.28 for an area under the normal distribution of 0.90, the safety stock is: cwt. 5 .50)365/600, 9(5 . 128. 1)(=dzsr LT For B: Q*( ,)()/ . (.).=2 9 600 10003 61803218 cwt. and cwt. 6 .60)365/600, 9(8 . 128. 1=r Service B appears to be slightly less expensive. 3 The shortest route method can be applied to this problem. The computational table is shown in Table 7-2. The shortest route is defined by tracing the links from the destination node. They are shown in Table 7-2 as A D F G for a total distance of 980 miles. 19 TABLE 7-2 Tabulation of Computational Steps for the Shortest Route Method Applied to Transcontinental Trucking Company Problem Step Solved nodes directly connected to unsolved nodes Its closest connected unsolved node Total time involved nth nearest node Its minimum time Its last connectiona 1 A B 186 mi. B 186 mi. AB A D 276 2 A D 276 D 276 AD* B C 186+110= 296 3 B C 186+110= 296 C 296 BC D C 276+ 58= 334 D F 276+300= 576 4 C E 296+241= 537 E 537 CE C F 296+350= 646 D F 276+300= 576 5 C F 296+350= 646 E G 537+479=1016 D F 276+300= 576 F 576 DF* 6 E G 537+479=1016 F G 576+404= 980 G 980 FG* a Asterisk indicates the shortest route 4 In this actual problem, the U.S. Army used the transportation method of linear programming to solve its allocation problem. The problem can be set up in matrix form as follows: Origin Destination Cleve- land South Charleston San Jose Demand Letterkenny 150 150 100 150 800 300 Fort Hood 325 50 350 300 50 100 Fort Riley 275 100 325 350 100 Fort Carson 375 400 275 100 100 Fort Benning 300 100 250 0 450 100 Supply 400 150 150 The cell values shown in bold represent the number of personnel carriers to be moved between origin and destination points for minimum transportation costs of $153,750. An alternative solution at the same cost would be: