上海海事大学物流学原理第七章

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 buyer's inventory, $ C' = item value at vendor's inventory, $ T = time in transit, days Q = Shipping quantity, units

Rail has the lowest total cost.

TABLE 7-1 An Evaluation of the Transport Alternatives for the Wagner

Company

Method Rail R×D 25×50,000

= $1,250,000

In-transit 0.25×475×50,000

a

inventory I×C’×D×t/365 ×(16/365)

= $260,274

Wager’s 0.25×475×(10,000/2)

a

inventory = $593,750 I×C’×Q/2 Electronic’s 0.25×500×(10,00/2) inventory = $625,000 I×C×Q/2 Total $2,729,024 aC’ refers to price less transport cost per unit. Cost type Transport

Piggyback Truck 44×50,000 88×50,000 = $2,200,000 = $4,400,000 0.25×456×50,000 0.25×412×50,000 ×(10/365) ×(4/365) = $156,164 = $564,384

0.25×456×(7,000/2) 0.25×412×(5,000/2) = $399,000 = $257,500

0.25×500×(7,000/2) 0.25×500×(5,000/2) = $437,500 = $312,500 $3,192,664 $5,534,384

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.

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Cost type

Transport

In-transit inventory Plant

inventory

Warehouse inventory Total

Service type

A

12×9,600 R×D = $115,200 0.20×50×9,600 I×C×D×t/365 ×(4/365)

= $1,052

I×C×Q*/2 0.30×50×(321.8/2)

= $2,684

0.30×62×(321.3/2) I×C’×Q*/2 + 0.30×62×50.5 + I×C’×r = $3,927 $122,863

Method

B

11.80×9,600 =$114,048 0.20×50×9,600 ×(5/365) = $1,315

0.30×50×(357.8/2) = $2,684

0.30×61.80×(321.8/2) + 0.30×61.80×60.6 = $4,107 $122,154

Recall that Q*=2DS/IC=2(9,600)(100)/0.3(50)=357.8cwt. 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:

Q*=2DS/IC=2(9,600)(100)/0.3(62)=3213 cwt. .

and for z = 1.28 for an area under the normal distribution of 0.90, the safety stock is:

r=zsLT(d)=1.28×1.5×(9,600/365)=50.5 cwt.

For B:

Q*=2(9,600)(100)/0.3(6180.)=3218 cwt. . and

r=1.28×1.8×(9,600/365)=60.6 cwt.

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.

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TABLE 7-2 Tabulation of Computational Steps for the Shortest Route Method

Applied to Transcontinental Trucking Company Problem

Solved nodes

Its closest directly

connected connected to

th

n nearest Its minimum Total time unsolved Its last unsolved

involved node connectiona nodes time Step node

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 Cleve- South San Destination land Charleston Jose Demand 100 800 150 300 Letterkenny 150 150 350 300 325 100 Fort Hood 50 50 325 350 275 100 Fort Riley 100 375 400 275 100 Fort Carson 100 250 450 300 100 Fort Benning 100 0 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:

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