Business Logistics/

Supply Chain Management

Planning, Organizing, and Controlling the Supply Chain

Fifth Edition

Instructors Manual

Ronald H. Ballou Weatherhead School of Management

Case Western Reserve University

ii

CONTENTS Preface iii Chapter 1 Business Logistics/Supply ChainA Vital Subject 1

2 Logistics/Supply Chain Strategy and Planning

2

3 The Logistics/Supply Chain Product... 4 4 Logistics/Supply Chain Customer Service.. 9 5 Order Processing and Information Systems. 13 6 Transport Fundamentals.. 14 7 Transport Decisions.

Fowler Distributing Company.. Metrohealth Medical Center. Orion Foods, Inc............... R & T Wholesalers...

17 35 41 48 52

8 Forecasting Supply Chain Requirements. World Oil.. Metro Hospital .

65 84 88

9 Inventory Policy Decisions.. Complete Hardware Supply, Inc... American Lighting Products. American Red Cross: Blood Services..

94 121 124 131

10 Purchasing and Supply Scheduling Decisions. Industrial Distributors, Inc

134 144

11 The Storage and Handling System... 147 12 Storage and Handling Decisions.. 148 13 Facility Location Decisions.

Superior Medical Equipment Company.... Ohio Auto & Drivers License Bureau. Southern Brewery

162 186 190 198

14 The Logistics Planning Process... Usemore Soap Company... Essen USA...

204 208 217

15 Logistics/Supply Chain Organization. 229 16 Logistics/Supply Chain

Control... 230

iii

PREFACE This instructor's guide provides answers to the more quantitatively oriented problems at the end of the textbook chapters. If the questions or problems are for discussion or they involve a substantial amount of individual judgment, they have not been included. Solutions to the cases and exercises in the text are also included. These generally require computer assistance for solution. With the text, you are provided with a collection of software programs, called LOGWARE, that assist in the solution of the problems, cases, and exercises in the text. The LOGWARE software along with a users manual is available for downloading from the Prentice Hall website or this book. The users manual is in Microsoft Word or Acrobat .pdf formats. This software, along with the users manual, may be freely reproduced and distributed to your classes without requiring permission from the copyright holder. This permission is granted as long as the use of the software is for educational purposes. If you encounter difficulty with the software, direct questions to Professor Ronald H. Ballou Weatherhead School of Management Case Western Reserve University Cleveland, Ohio 44106 Tel: (216) 368-3808 Fax: (216) 368-6250 E-mail: Ronald.Ballou@CASE.edu Web site: www.prenhall.com/ballou

1

CHAPTER 1 BUSINESS LOGISTICS/SUPPLY CHAINA VITAL SUBJECT

12 (a) This problem introduces the student to the evaluation of alternate channels of

production and distribution. To know whether domestic or foreign production is least expensive, the total of production and distribution costs must be computed from the source point to the marketplace. Two alternatives are suggested, and they can be compared as follows.

Production at Houston: Total cost = Production cost at Houston + Transportation and storage costs = $8/shirt100,000 shirts + $5/cwt. 1,000 cwt. = $805,000/year Production at Taiwan: Total cost = Production cost in Taiwan + Transportation and storage costs from Taiwan to Chicago + Import duty + Raw material transportation cost from Houston to Taiwan = $4/shirt100,000 shirts + $6/cwt. 1,000 cwt. + $0.5/shirt100,000 shirts + $2/cwt. 1,000 cwt. = $458,000/year Producing in Taiwan would appear to be the least expensive. (b) Other factors to consider before a final decision is made might be: (i) How reliable would international transportation be compared with domestic

transportation? (ii) What is the business climate in Taiwan such that costs might change in favor of

Houston as a production point? (iii) How likely is it that the needed transportation and storage will be available? (iv) If the market were to expand, would there be adequate production capacity

available to support the increased demand?

2

CHAPTER 2 LOGISTICS/SUPPLY CHAIN STRATEGY AND PLANNING

13 The purpose of this exercise is to allow the student, in an elementary way, to examine the tradeoffs between transportation and inventory-related costs when an incentive transportation rate is offered. Whether the incentive rate should be implemented depends on the shipment size corresponding to the minimum of the sum of transportation, inven-tory, and order processing costs. These costs are determined for various shipping quantities that might be selected to cover the range of shipment sizes implied in the problem. Table 2-1 gives a summary of the costs to Monarch for various shipment sizes. From Monarch's point of view, the incentive rate would be beneficial. Shipment sizes should be approximately doubled so that the 40,000 lb. minimum is achieved. It is important to note that the individual cost elements are not necessarily at a minimum at low shipment sizes, whereas order-processing costs are low at high shipment sizes. They are in cost conflict with each other. Transportation costs are low at high shipment sizes, but exact costs depend on the minimum volume for which the rate is quoted. In preparation for a broader planning perspective to be considered later in the text, the student might be asked what the place of the supplier is in this decision. How does he affect the decision, and how is he affected by it? This will focus the student's attention on the broader issues of the physical distribution channel.

3

TABLE 2-1 Evaluation of Alternative Shipment Sizes for the Monarch Electric Company Current Proposed Type of cost

57 motors or

10,000 lb.

114 motors or

20,000 lb.

171 motors or

30,000 lb.

228 motors or

40,000 lb.

285 motors or

50,000 lb. Transportation RD

98,750 = $78,750

58,750 = $43,750

58,750 = $43,750

38,750 = $26,250a

38,750 = $26,250

Inventory carryingb ICQ/2

0.2520057/2 = $1,425a

0.25200114/2 = $2,850

0.25200171/2 = $4,275

0.25200228/2 = $5,700

0.25200285/2 = $7,125

Order processingc DS/Q

5,00015/57 = $1,316

5,00015/114 = $658

5,00015/171 = $439

5,00015/228 = $329

5,00015/285 = $263a

Handling HD

0.308,750 = $2,625

0.308,750 = $2,625

0.308,750 = $2,625

0.308,750 = $2,625

0.308,750 = $2,625

Total $84,116 $49,883 $51,089 $34,904a $36,263 a Minimum values. b Students should be informed that average inventory can be approximated by one half the shipment size. c Demand D has been converted to units per year. LEGEND R = transportation rate, $/cwt. D = annual demand, cwt. I = inventory carrying cost, %/year. C = cost of a motor, $/motor. Q = shipment size in motors, where Q/2 represents the average number of motors maintained in inventory. S = order processing costs, $/order. H = handling costs, $/cwt.

4

CHAPTER 3 THE LOGISTICS/SUPPLY CHAIN PRODUCT

3 The 80-20 principle applies to sales and items where 80 percent of the dollar volume is generated from 20 percent of the product items. While this ratio rarely holds exactly in practice, the concept does. We can apply it to these data by ranking the products by sales, and the percentage that the cumulative sales represent of the total. The following table shows the calculations.

The 80-20 rule cannot be applied exactly, since the cumulative percent of items does not break at precisely 20 percent. However, we might decide that only products 08776 and 12121 should be ordered directly from vendors. The important principle derived from the 80-20 rule is that not every item is of equal importance to the firm, and that dif-ferent channels of distribution can be used to handle them. The 80-20 rule gives some rational basis for deciding which products should be shipped directly from vendors and which are more economically handled through a system of warehouses. 6 (a) Reading the ground transport rates for the appropriate zone as determined by zip code

and the weight of 27 lb. (rounding upward of 26.5 lb.) gives the following total cost table for the four shipments.

Product

code

Dollar sales

Cumulative

sales

Cumulative sales as

% of total

Cumulative items as

% of total 08776 $71,000 $ 71,000 18.2 8.3 12121 63,000 134,000 34.3 16.7 10732 56,000 190,000 48.6 25.0 11693 51,000 241,000 61.6 33.3 10614 46,000 287,000 73.4 41.7 12077 27,000 314,000 80.3 50.0 07071 22,000 336,000 85.9 58.3 10542 18,000 336,000 90.5 66.7 06692 14,000 354,000 94.1 75.0 09721 10,000 368,000 96.7 83.3 14217 9,000 378,000 98.9 91.7 11007 4,000 391,000 100.0 100.0

Total $391,000

5

(b) The transport rate structure is reasonably fair, since ground rates generally follow

distance and size of shipment. These are the factors most directly affecting transport costs. They are not fair in the sense that customers within a zone are all charged the same rate, regardless of their distance from the shipment origin point. However, all customers may benefit from lower overall rates due to this simplified zone-rate structure.

10 (a) This is a delivered pricing scheme where the seller includes the transport charges in

the product price. The seller makes the transport arrangements. (b) The seller prices the product at the origin, but prepays any freight charges; however,

the buyer owns the goods in transit. (c) This is a delivered pricing scheme where the freight charges are included in the

product price, however the freight charges are then deducted from the invoice, and the seller owns the goods in transit.

(d) The seller initially pays the freight charges, but they are then collected from the buyer

by adding them to the invoice. The buyer owns the goods in transit, since the pricing is f.o.b. origin.

(e) The price is f.o.b. origin. The buyer pays the freight charges and owns the goods in

transit. Regardless of the price policy, the customer will ultimately pay all costs. If a firm does not consider outbound freight charges, the design of the distribution system will be different than if it does. Since pricing policy is an arbitrary decision, it can be argued that transport charges should be considered in decision making, whether the supplying firm directly incurs them or not. 11 This shows how Pareto's law (80-20 principle) is useful in estimating inventory levels when a portion of the product line is to be held in inventory. An empirical function that approximates the 80-20 curve is used to estimate the level of sales for each product to be held in inventory. According to Equation 3-2, the constant A is determined as follows.

To zip code

Catalog price

UPS zone

Transport costa

Total cost

11107 $99.95 2 $ 7.37 $107.32 42117 99.95 5 10.46 110.41 74001 99.95 6 13.17 113.12 59615 99.95 8 18.29 118.24

a Use 27 lb.

6

A X YY X

( ) . ( . ). .

.1 0 25 1 750 75 0 25

0125

The 80-20 type curve according to Equation 3-1 is:

Y AA X

XX

( ) ( . )

.1 1 0125

0125

This formula can be used to estimate the cumulative sales from the cumulative item proportion. For example, item 1 is 0.05 of the total number of items (20) so that:

Y ( . )( . )

. ..1 0125 0 05

0125 0 050 321

Of the $2,600,000 in total annual warehouse sales, item 1 should account for 0.3212,600,000 = $835,714. By applying this formula to all items, the following inventory investment table can be developed which shows sales by item. The average inventory investment by item is found by dividing the turnover ratio into the item sales. The sum of the average inventory value for each item gives a total projected inventory of $380,000.

Inventory Investment Table

Pro-duct

Cumulative item pro-portion, X

Cumulative

sales, Y

Projected item sales

Turnover

ratio

Average inventory

value 1 0.05 $ 835,714 $ 835,714 8 $104,464 2 0.10 1,300,000 464,286 8 58,036 3 0.15 1,595,454 295,454 8 36,932 4 0.20 1,800,000 204,546 8 25,568 5 0.25 1,950,000 150,000 6 25,000 6 0.30 2,064,705 114,706 6 19,118 7 B 0.35 2,155,263 90,558 6 15,093 8 0.40 2,228,571 73,308 6 12,218 9 0.45 2,289,130 60,559 6 10,093

10 0.50 2,340,000 50,870 6 8,478 11 0.55 2,383,333 43,333 4 10,833 12 0.60 2,420,689 37,356 4 9,339 13 0.65 2,453,226 32,537 4 8,134 14 0.70 2,481,818 28,592 4 7,148 15 C 0.75 2,507,142 25,324 4 6,331 16 0.80 2,529,719 22,587 4 5,647 17 0.85 2,550,000 20,271 4 5,068 18 0.90 2,568,293 18,293 4 4,473 19 0.95 2,584,884 16,591 4 4,148 20 1.00 2,600,000 15,116 4 3,779

Total $380,000

A

7

12 This problem involves the application of Equations 3-1 and 3-2. We can develop an 80-20 curve based on 30 percent of the items accounting for 70 percent of sales. That is,

A X YY X

( ) . ( . ). .

.1 0 30 1 0 700 70 0 30

0 225

Therefore, the sales estimating equation is:

Y XX

( . )

.1 0 2250 225

By applying this estimating curve, we can find the sales of A and B items. For example, 20 percent of the items, or 0.220 = 4 items, will be A items with a cumulative proportion of sales of:

YA ( . )( . )

. ..1 0 225 0 20

0 225 0 200 5765

and 3,000,0000.5765 = 1,729,412. The A+B item proportion will be:

YA B ( . )( . )

. ..1 0 225 0 50

0 225 0 500 8448

and 3,000,0000.8448 = 2,534,400. The product group B sales will A+B sales less A sales, or 2,534,400 1,729,412 = $804,988. The product group C will be the remaining sales, but these are not of particular interest in this problem. The average inventories for A and B products are found by dividing the estimated sales by the turnover ratio. That is,

A: 1,729,412/9 = 192,157 B: 804,988/5 = 160,988 Total inventory 353,155 cases

The total cubic footage required for this inventory would be 353,1551.5 = 529,732 cu. ft. The total square footage for products A and B is divided by the stacking height. That is, 529,731/16 = 33,108 sq. ft.

8

13 This problem is an application of Equations 3-1 and 3-2. We first determine the constant A. That is,

A X YY X

( ) . ( . ). .

.1 0 20 1 0 650 65 0 20

0156

and

0 75 1 01560156

. ( . ).

X

X

Solving algebraically for X, we have:

X AxYA Y

x 10156 0 75

1 0156 0 750 288. .

. ..

That is, about 29% of the items (0.2885,000 = 1,440 items) produce 75% of the sales. 14 The price would be the sum of all costs plus an increment for profit to place the automotive component in the hands of the customer. This would be 25+10+5+8+5+transportation cost, or 53+T. Based on the varying transportation cost, the following price schedule can be developed.

Quantity Price per unit Discount 1 to 1,000 units 53+5=$58 0 1,001 to 2,000 units 53+4.00=57 1.7%a >2,000 units 53+3.00=56 3.5%

a [(58 - 57)/58][100]=1.7%

9

CHAPTER 4 LOGISTICS/SUPPLY CHAIN CUSTOMER SERVICE

6 (a) This company is fortunate to be able to estimate the sales level that can be achieved at

various levels of distribution service. Because of this, the company should seek to maximize the difference between sales and costs. These differences are summarized as follows.

Percent of orders delivered within 1 day Contribution to 50 60 70 80 90 95 100 profit -1.8 2.0 3.5 4.0 3.4 2.8 -2.0 The company should strive to make deliveries within 1 day 80 percent of the time for a maximum contribution to profit. (b) If a competing company sets its delivery time so that more than 80 percent of the

orders are delivered in 1 day and all other factors that attract customers are the same, the company will lose customers to its competitor, as the sales curve will have shifted downward. Cleanco should adjust its service level once again to the point where the profit contribution is maximized. Of course, there is no guarantee that the previous level of profits can be achieved unless the costs of supplying the service can correspondingly be reduced.

7 (a) This problem solution requires some understanding of experimental design and

statistical inference, which are not specifically discussed in the text. Alert the students to this.

The first task is to determine the increase in sales that can be attributed to the change in the service policy. To determine if there is a significant change in the control group, we set up the following hypothesis test.

z X XsN

sN

2 1

22

2

12

1

2 2

224 18561102

79102

3936 48 6118

394. .

.

Now, referring to a normal distribution table in Appendix A of the text, there is a

significant difference at the 0.01 level in the sales associated with the control group. That is, some factors other than the service policy alone are causing sales to increase.

Next, we analyze the test group in the same manner.

10

z

2 295 1 34257656

33556

9535 924 2 004

10 72 2

, ,, ,

.

This change is also significant at the 0.01 level. The average increase in sales for the control group is 224/185 = 1.21, or 21%.

The average sales increase in the test group is 2295/1342 = 1.71, or 71%. If we believe that 21% of the 71% increase in the test group is due to factors other than service policy, then 71 21 = 50% was the true service effect. Therefore, for each sales unit, an incremental increase in profit of (0.4095)(0.50) = $19 can be realized. Since the cost of the service improvement is $2, the benefit exceeds the cost. The service improvement should be continued.

Note: If the students are not well versed in statistical methodology, you may wish to instruct them to consider the before and after differences in the mean values of both groups as significant. The solution will be the same.

(b) The use of the before-after-with-control-group experimental design is a methodology

that has been used for some time, especially in marketing research studies. The outstanding feature of the design is that the use of the control group helps to isolate the effect of the single service variable. On the other hand, there are a number of potential problems with the methodology:

The sales distributions may not be normal. The time that it takes for diffusing the information that a service change has taken

place may distort the results. The products in the control group may not be mutually exclusive from those in the

test group. The method only shows the effect of a single step change in service and does not

develop a sales-service relationship. It may not always be practical to introduce service changes into on-going

operations to test the effect. 8 (a) The optimum service level is set at that point where the change in gross profit equals

the change in cost. The change in gross profit: P = Trading margin Sales response rate Annual sales = 1.000.0015100,000 = $150 per year per 1% change in the service level The change in cost: C = Annual carrying cost Standard product cost z

11

Demand standard deviation for order cycle = 0.3010.00400z Now, set P = C and solve for z. 150 = 1200z z = 0.125 From the tabulated changes in service level with those changes in z, the service level should be set between 96-97%. (b) The weakest link in this analysis is estimating the effect that a change in service will

have on revenue. This implies that a sales-service relationship is known. 9 The methodology is essentially the same as that in question 7, except that we are asked to find X instead of Y. That is, P = 0.750.001580,000 = 90 and C = 0.251,000500z = 1250z Then, P= C 90 = 1250z z = 0.072 From the normal distribution (see Appendix A), the z for an area under the curve of 93% is 1.48, and for 92%, z is 1.41. Since the difference of 1.48 1.41 = 0.07, we can conclude that the in-stock probability should be set at 92-93%. Of course, the change in z is found by taking the difference in z values for 1% differences in the area values under the normal distribution curve for a wide range of area percentages. 10 Apply Taguchis concept of the loss function. First, estimate the loss per item if the target level of service is not met. We know the profit per item as follows.

12

Sales price $5.95 Cost of item -4.25 Other costs -0.30 Profit per item $1.40

Since one-half of the sales are lost, the opportunity loss per item would be

/item70.0$880(1/2)(880) $1.40lossy Opportunit

Next, find k in the loss function.

L k y mkk

k

( ). ( ). ( )

.

2

20 70 10 50 70 25

0 03

Finally, the point where the marginal supply cost equals the marginal sales loss is

%67.1)03.0(2

10.02

)5( k

By

%67.6567.1 y The retailer should not allow the out-of-stock percentage to deviate more than 1.67%, and should not allow the out-of-stock level to fall below 1.67 + 5 = 6.67%.

Profit per item Sales lost

Current sales

Target %

out-of-stock % at point where sales are lost

13

CHAPTER 5 ORDER PROCESSING AND INFORMATION SYSTEMS

All questions in this chapter require individual judgment and response. No answers are offered.

14

CHAPTER 6 TRANSPORT FUNDAMENTALS

14 The maximum that the power company can pay for coal at its power plant location in Missouri is dictated by competition. Therefore, the landed cost at the power plant of coal production costs plus transportation costs cannot exceed $20 per ton. Since western coal costs $17 per ton at the mine, the maximum worth of transportation is $20 $17 = $3 per ton. However, if the grade of coal is equal to the coal from the western mines, eastern coal can be landed in Missouri for $18 per ton. In light of this competitive source, transportation from the western mines is worth only $18 $17 = $1 per ton. 15 Prior to transport deregulation, it was illegal for a carrier to charge shippers less for the longer haul than for the shorter haul under similar conditions when the shorter haul was contained within the longer one. To be fair, the practice probably should be continued. If competitive conditions do not permit an increase in the rate to Z, then all rates that exceed $1 per cwt. on a line between X and Z should not exceed $1 per cwt. Therefore, the rate to Z is blanketed back to Y so that the rate to Y is $1 per cwt. By blanketing the rate to Z on intervening points, no intervening point is discriminated against in terms of rates. 16 (a) From text Table 6-4, the item number for place mats is 4745-00. For 2,500 lb., the

classification is 100 since 2,500 lb. is less than the minimum weight of 20,000 lb. for a truckload shipment. From text Table 6-5, the rate for a shipment 2,000 lb. is 8727/cwt. The shipping charges are $87.27 25 cwt. = $2,181.75.

(b) This is an LTL shipment with a classification of 100, item number 4980-00 in text

Table 6-4. From Table 6-5, the minimum charge is 9351 and the rate for a

15

Try (1): Rate is $5.65/cwt. 5.65 270 = $1,525.50. Try (2): Rate is $3.87/cwt. 3.87 300 = $1,161.00 Lowest cost Try (3): Rate is $3.70/cwt. 3.70 360 = $1,332.00 (d) The shipment is a truckload classification (2070-00) of 65. The rate at 30,000 lb. is

$4.21/cwt. The charges are 4.21 300 = $1,263.00. (e) Classification of this product is 55 (4860-00) for a truckload of 24,000 lb. Check the

break weight according to Equation 6-1.

Break weight = 3.87 30,0005.65

lb. 20 549, Since current shipping weight of 24,000 lb. exceeds the break weight, ship as if 30,000 lb. Hence, 3.87 300 = $1,161.00. Now, discount the charges by 40 percent. That is, $1,161 (1 0.40) = $696.60 21 The question involves evaluating two alternatives. The first is to compute the transport charges as if there are three separate shipments. The next is to see if a stop-off privilege offers any cost reduction. The comparison is shown below.

With stop-off Ship direct to B and split deliver thereafter. Rate, Stop-off Loading/unloading Route $/cwt. charge Charges 25,000 A to B $1.20 $ 300.00 Direct shipment 40,000 B to D 2.20 880.00 Stop-off @ C $25.00 25.00 Stop-off @ D 25.00 25.00 Total charges $1,230.00

Separate shipments Rate, Stop-off Loading/unloading Route $/cwt. charge Charges 22,000 A to D $3.20 --- $704.00 3,000 A to C 2.50 --- 75.00 15,000 B to C 1.50 --- 225.00 Total charges $1,004.00

16

Split deliver at all stops. Rate, Stop-off Loading/unloading Route $/cwt charge Charges 40,000 A to D 3.20 1,280.00 Stop-off @ B 25.00 25.00 Stop-off @ C 25.00 25.00 Stop-off @ D 25.00 25.00 Total charges $1,335.00 Other combinations may be tried. In this case, there appears to be no advantage to using the stop-off privilege.

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 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.

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. Service type

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) = $56,438

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 $5,026,438 a C refers to price less transport cost per unit.

18

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 Q DS IC* / ( , )( ) / . ( ) . 2 2 9 600 100 0 3 50 357 8 cwt. 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 DS IC* / ( , )( ) / . ( ) . 2 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: cwt. 5.50)365/600,9(5.128.1)( dzsr LT For B: Q* ( , )( ) / . ( . ) . 2 9 600 100 0 3 6180 3218 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. TABLE 7-2 Tabulation of Computational Steps for the Shortest Route Method

Applied to Transcontinental Trucking Company Problem

19

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:

20

5 This problem can be used effectively as an in-class exercise. Although the problem might be solved using a combination of the shortest route method to find the optimum path between stops and then a traveling salesman method to sequence the stops, it is intended that students will use their cognitive skills to find a good solution. The class should be divided into teams and given a limited amount of time to find a solution. They should be provided with a transparency of the map and asked to draw their solution on it. The instructor can then show the class each solution with the total distance achieved. From the least-distance solutions, the instructor may ask the teams to explain the logic of their solution process. Finally, the instructor may explore with the class how this and similar problems might be treated with the aid of a computer. Although the question asks the student to use cognitive skills to find a good route, a route can be found with the aid of the ROUTER software in LOGWARE. The general approach is to first find the route in ROUTER without regard to the rectilinear distances of the road network. Because this may produce an infeasible solution, specific travel distances are added to the database to represent actual distances traveled or to block infeasible paths from occurring. A reasonable routing plan is shown in Figure 7-1 and the ROUTER database that generates it is given in Figure 7-2. The total distance for the route is 9.05 miles and at a speed of 20 miles per hour, the route time is approximately 30 minutes.

Origin

Destination

Number of carriers

Cleveland Letterkenny 150 S. Charleston Letterkenny 150 Cleveland Fort Hood 50 San Jose Fort Hood 50 Cleveland Fort Riley 100 San Jose Fort Carson 100 Cleveland Fort Benning 100

0.5 1.0 1.5 2.0

0 5

00.5 1.0 1.5 2.0

0 5

01719

2021

0

0.5 1.0 1.5 2.0

0 5

00.5 1.0 1.5 2.0

0 5

00.5 1.0 1.5 2.0

0 5

00.5 1.0 1.5 2.0

0 5

01719

2021

0

21

FIGURE 7-2 Input Data for ROUTER for School Bus Routing Problem

PARAMETERS AND LABELS Problem label School Bus Routing Exercise Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NW DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate 0.14 Vertical coordinate 0.45 Earliest starting time (min) - 0 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 20 After how many clock hours will overtime begin - 9999 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 1 Vertical scaling factor - 1 Maximum TIME allowed on a route (hours) - 9999 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 0 To begin after - 9999 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999

22

--STOP DATA

NO

STOP DESCRIPTION

TY

LOAD WGHT

VOL. CUBE

HCRD

VCRD

ZN

LOADTIME

BEG1

END1

BEG2

END2

1 Stop 1 D 1 0 0.14 0.80 0 0 0 9999 9999 9999 2 Stop 2 D 1 0 0.14 1.14 0 0 0 9999 9999 9999 3 Stop 3 D 1 0 0.14 1.31 0 0 0 9999 9999 9999 4 Stop 4 D 1 0 0.35 1.31 0 0 0 9999 9999 9999 5 Stop 5/22 D 1 0 0.52 0.61 0 0 0 9999 9999 9999 6 Stop 6 D 1 0 0.58 1.31 0 0 0 9999 9999 9999 7 Stop 7 D 1 0 0.80 1.31 0 0 0 9999 9999 9999 8 Stop 8 D 1 0 1.03 0.61 0 0 0 9999 9999 9999 9 Stop 9 D 1 0 1.03 0.96 0 0 0 9999 9999 9999

10 Stop 10 D 1 0 1.03 1.31 0 0 0 9999 9999 9999 11 Stop 11 D 1 0 1.36 1.31 0 0 0 9999 9999 9999 12 Stop 12 D 1 0 1.48 1.31 0 0 0 9999 9999 9999 13 Stop 13 D 1 0 1.80 1.31 0 0 0 9999 9999 9999 14 Stop 14 D 1 0 1.87 1.31 0 0 0 9999 9999 9999 15 Stop 15 D 1 0 1.84 0.61 0 0 0 9999 9999 9999 16 Stop 16 D 1 0 1.95 0.61 0 0 0 9999 9999 9999 17 Stop 17 D 1 0 1.29 0.10 0 0 0 9999 9999 9999 18 Stop 18 D 1 0 1.26 0.61 0 0 0 9999 9999 9999 19 Stop 19 D 1 0 1.15 0.10 0 0 0 9999 9999 9999 20 Stop 20 D 1 0 0.69 0.23 0 0 0 9999 9999 9999 21 Stop 21 D 1 0 0.14 0.26 0 0 0 9999 9999 9999

VEHICLE DATA

--CAPACITY-- --VEHICLE-- --DRIVER--

NO.

VEHICLE DESCRIPTION

TP

NO

WGHT

CUBE

FIXED COST

PER MI COST

FIXED COST

PER HR COST

OVER TIME COST

1 Bus 1 1 9999 9999 0 0 0 0 0

SPECIFIED STOP-TO-STOP DISTANCES

NO

STOP NO.

STOP DESCRIPTION

STOP NO.

STOP DESCRIPTION

DISTANCEIN MILES

1 14 Stop 14 16 Stop 16 0.78 2 14 Stop 14 15 Stop 15 0.90 3 15 Stop 15 17 Stop 17 1.06 4 16 Stop 16 17 Stop 17 1.18 5 18 Stop 18 9 Stop 9 0.58 6 19 Stop 19 8 Stop 8 0.76 7 19 Stop 19 20 Stop 20 0.59 8 19 Stop 19 5/22 Stops5&22 1.14 9 19 Stop 19 18 Stop 18 0.53

10 9 Stop 9 20 Stop 20 1.08 11 9 Stop 9 19 Stop 19 1.11 12 9 Stop 9 21 Stop 21 1.69 13 5/22 Stops 5&22 1 Stop 1 0.56 14 5/22 Stops 5&22 21 Stop 21 1.05 15 5/22 Stops 5&22 20 Stop 20 1.14 16 5/22 Stops 5&22 9 Stop 9 0.97 17 20 Stop 20 21 Stop 21 0.84 18 20 Stop 20 0 School 1.03 19 20 Stop 20 5/22 Stops 5&22 0.55

23

20 17 Stop 17 0 School 2.43 21 0 School 5/22 Stops 5&22 1.37 22 2 Stop 2 5/22 Stops 5&22 1.03

6 Strategy 1 is to stay at motel M2 and serve the two routes on separate days. Using the ROUTESEQ module in LOGWARE gives us the sequence of stops and the coordinate distance. The routes originating at M2 would be: Route Stop sequence Distancea 1 8,6,1,4,2,3,5,7,9 95.55 mi. 2 10,13,14,17,18,16,12,15,11 86.45 182.00 mi. a Includes map scaling factor The total cost of this strategy would be: Motel 3 nights @ 49.00 $147.00 Travel 182 miles @ $.30/mi. 54.60 Total $201.60 Strategy 2 is a mixed strategy involving staying at motels closest to the center of the stops clusters. The route sequences from different motels are: Route Stop sequence Distance 1 4,2,3,5,7,9,8,6,2 98.50 mi. 2 18,17,13,14,10,11,15,12,16 80.30 178.80 mi. The total cost of this strategy is:

Strategy 2 appears to be most economical. 7 (a) Since distances are asymmetrical, we cannot use the geographically based traveling

salesman method in LOGWARE. Rather, we use a similar module in STORM that allows such asymmetrical matrices, or the problem is small enough to be solved by inspection. For this problem, the minimal cost stop sequence would be:

Motel M1 1st night $ 40.00 M1 2nd night 40.00 M1 3rd night 45.00 Travela 214.80 mi. @ 0.30/mi. 64.44 Total $189.44 a178.80 + 36 = 214.80

24

BakeryStop 5Stop 3Stop 4Stop 2Stop 1Bakery

with a tour time of 130 minutes. (b) Loading/unloading times may be added to the travel times to a stop. The problem

may then be solved as in part a. (c) The travel times between stop 3 and all other nodes are increased by 50%. The

remaining times are left unchanged. Optimizing on this matrix shows no change in the stop sequence. However, the tour time increases to 147.50 minutes.

8 This may be solved by using the ROUTER module in LOGWARE. The screen set up for this is as follows.

25

FIGURE 7-3 Input Data for ROUTER for Sima Donuts

Making a run with ROUTER will give the route design.

--PARAMETERS AND LABELS Problem label - Sima Donuts Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NE DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate - 2084 Vertical coordinate - 7260 Earliest starting time (min) - 180 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 45 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 0.363 Vertical scaling factor - 0.363 Maximum TIME allowed on a route (hours) - 40 Maximum DISTANCE allowed on a route (miles) - 1400 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 60 To begin after - 1200 Duration of 3rd break (minutes) - 60 To begin after - 2160 Duration of 4th break (minutes) - 60 To begin after - 2640

--STOP DATA NO

STOP DESCRIPTION

TY

LOAD WGHT

VOL. CUBE

HCRD

VCRD

ZN

LOAD TIME

BEG1

END1

BEG2

END2

1 Tampa FL D 20 0 1147 8197 0 15 360 1440 1800 2880 2 Clearwater FL P 14 0 1206 8203 0 45 360 1440 1800 2880 3 Daytona Beach F D 18 0 1052 7791 0 45 360 1440 1800 2880 4 Ft Lauderdale FL D 3 0 557 8282 0 45 180 1440 1800 2880 5 N Miami FL D 5 0 527 8341 0 45 360 1440 1800 2880 6 Oakland Park FL P 4 0 565 8273 0 45 180 1440 1800 2880 7 Orlando FL D 3 0 1031 7954 0 45 180 1440 1800 2880 8 St Petersburg FL P 3 0 1159 8224 0 45 180 1440 1800 2880 9 Tallahassee FL D 3 0 1716 7877 0 15 600 1440 1800 2880

10 W Palm Beach F D 3 0 607 8166 0 45 360 1440 1800 2880 11 Puerto Rico D 4 0 527 8351 0 45 360 1440 1800 2880

--VEHICLE DATA -CAPACITY-- --VEHICLE-- --DRIVER--

NO.

VEHICLE DESCRIPTION

TP

NO

WGHT

CUBE

FIXED COST

PER MI COST

FIXED COST

PER HR COST

OVER TIME COST

1 Truck #1-20 1 3 20 9999 0 1.30 0 0 0 2 Truck #2-25 2 1 25 9999 0 1.30 0 0 0 3 Truck #3-30 3 1 30 9999 0 1.30 0 0 0

26

FIGURE 7-4 Graphical Display of Route Design for Sima Donuts The route design involves 3 routes for a total distance of 3,830 miles, a cost of $4,978.71, and a total time of 100.4 hours. The route details are as follows:

Route #1 with 20-pallet truck Depot Start time 3:00AM of day 1 Daytona Beach Deliver 18 pallets Clearwater Pickup 14 pallets Depot Return time 5:48AM of day 2

Route #2 with 20-pallet truck Depot Start time 3:00AM of day 1 Orlando Deliver 3 pallets W Palm Beach Deliver 3 pallets Ft Lauderdale Deliver 3 pallets N Miami Deliver 5 pallets Miami-Puerto R. Deliver 4 pallets Depot Return time 4:43PM of day 2

Pickup Pickup

27

Route #3 with 30-pallet truck Depot Start time 4:13AM of day 1 Tallahassee Deliver 3 pallets Tampa Deliver 20 pallets St Petersburg Pickup 3 pallets Oakland Park Pickup 4 pallets Depot Return time 4:03PM of day 2 9 Given sailing times and dates when deliveries are to be made, loadings need to be accomplished no later than the following dates: To: A B C D From: 1 16 40 1 2 69 25 5 The problem can be expressed as a transportation problem of linear programming. There will be 6 initial states [(1,1), (2,5), (1,16), (2,25), (1,40), and (2,69)] and 6 terminal states [(D,10), (C,15), (A,36), (B,39), (C,52), and (A,86)]. The linear program is structured as shown in Figure 7-4. Using a transportation solution method, we determine one of the optimum solutions. There are several. The solution is read by starting with the slack on initial loading state 1. This tells us to next select the cell of terminal state 1. In turn, this defines initial state 3 and hence terminal state 3. And so it goes until we reach the terminal state slack column. This procedure is repeated until all initial state slacks are exhausted. Our solution shows two routings. The first is (1,1)(D,10)-(1,16)(A,36)(2,69)(A,86). The second is (2,5)(C,15)(2,25)(B,39)(1,40)(C,52). Two ships are needed.

28

FIGURE 7-5 Transportation Matrix Setup and Solution for the Queens Lines Tanker Scheduling Problem

10 This is a problem of freight consolidation brought about by holding orders so they can be shipped with orders from subsequent periods. The penalty associated with holding the orders is a lost sales cost. (i) Orders shipped as received

Weight Rate = Cost Haysa 10,000 0.0519 = $519.00 Manhattan 14,000 0.0519 = 726.00 Salina 13,000 0.0408 = 530.00 Great Bend 10,000 0.0498 = 498.00 Transportation $2,274.00 Lost sales .00 Total $2,274.00

a Ship 8,000 lb. as if 10,000 lb. Average period cost is $2,274.00

Loading points and dates Load date Discharge date

1 1

2 5

1 16

2 25

1 40

2 69

Slack

Rim re-striction

D 10

100 XXa

100 XX

1 1

1 1 1 10 1

C 15

100 XX

100 XX

1

1 1

1 1 10 1

A 36

100 XX

100 XX

100 XX

100 XX

1 1 1

10 1

B 39

100 XX

100 XX

100 XX

100 XX

1 1

1 10 1

C 52

100 XX

100 XX

100 XX

100 XX

100 XX

1 10 1

1

A 86

100 XX

100 XX

100 XX

100 XX

100 XX

100 XX

10 1

1

Slack

10 1

10 1

10 10 10 10 10 4

6

Rim re- striction

1

1

1

1

1

1

6

a XX inadmissible cells given a high cost

29

(ii) Consolidate first period orders with second period orders.

Weight Rate = Cost Hays 16,000 0.0519 = $830.40 Manhattana 40,000 0.0222 = 888.00 Salina 26,000 0.0342 = 889.20 Great Bend 10,000 0.0498 = 498.00 Transportation $3,105.60 Lost sales 1,050.00 Total $4,155.60

a Ship 28,000 lb. as if 40,000 lb. The lost sales cost is 1,000 cases $1.05 = $1,050.00 to hold one group of orders for 2 weeks. Average cost per period is $4,155.60/2 = $2,077.80. (iii) Hold all orders until the third period.

Weight Rate = Cost Hays 24,000 0.0426 = $1,022.40 Manhattan 42,000 0.0222 = 932.40 Salinaa 40,000 0.0246 = 984.00 Great Bend 15,000 0.0498 = 747.00 Transportation $3,685.80 Lost sales 3,150.00 Total $6,835.80

a Ship as if 40,000 lb. Lost sales Hold 1st period orders for 2 periods 1,0001.05.2 = $ 2,100 Hold 2nd period sales for 1 period 1,0001.05 = 1,050 $ 3,150 Average period cost is $6,835.80/3 = $2,278.60 Summary Ship immediately $ 2,274.00 Hold orders 1 period 2,077.80 Optimum Hold orders 2 periods 2,278.60 11 Routes are built by placing the trips end-to-end throughout the day from 4AM until 11PM, respecting the times that a warehouse can receive a shipment. This is a 19-hour block of time per day, or there are 95 hours per week per truck in which a truck may

30

operate. If there were no delivery time restrictions on warehouses and trips could be placed end-to-end for a truck without any slack at the end of the day, the absolute minimum number of trucks can be found multiplying the number of trips by the route time and then dividing the total by the 95 hours allowed per week. That is,

Warehouse location

(1) Number of

trips

(2) Total time

per trip, hr.

(3)=(1)(2) Total time,

hr. Flint 43 1.25 53.75 Alpena 5 10.50 52.50 Saginaw 8 2.25 18.00 Lansing 21 3.75 78.75 Mt. Pleasant 12 5.50 66.00 W. Branch 5 6.00 30.00 Pontiac 43 2.75 118.25 Traverse City 6 10.50 63.00 Petoskey 5 11.75 58.75 Total 539.00

For 539 trip hours, 539/95 = 5.67 rounded to six trucks needed per week. Now, it is necessary to adjust for the problem constraints. A good schedule can be found by following a few simple rules that can be developed by examining the data. First, begin the day with a trip where the driving time to a warehouse is just long enough for the truck to arrive at the warehouse just after it opens. One-half the driving time should exceed 6:30 4:00 = 2:30, or 2 hr. Trips to Alpena, Traverse City, and Petoskey qualify. Second, use the short trips at the end of the day to avoid slack time. Third, allocate the trips to the days using the longest ones first. Make sure that the total trip time for a day does not exceed 19 hours. For a minimum of six trucks, the following feasible schedule can be developed by inspection. Day 1 Day 2 Day 3 Day 4 Day 5 Truck 1 Petoskey 11.75

W Branch 6.00 Flint 1.25 Total =19.00 hr.

Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 hr.

Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00

Petoskey 11.75 5 Flint 6.25 Total =18.00 hr.

Petoskey 11.75 5 Flint 6.25 Total =18.00 hr.

Truck 2 T. City 10.50 2 Lansing 7.50 Total =18.00 hr.

T. City 10.50 2 Lansing 7.50 Total =18.00 hr.

T. City 10.50 2 Lansing 7.50 Total =18.00

T. City 10.50 2 Lansing 7.50 Total =18.00 hr.

T. City 10.50 2 Lansing 7.50 Total =18.00 hr.

Truck 3 T. City 10.50 2 Lansing 7.50 Total =18.00 hr.

Alpena 10.50 2 Lansing 7.50 Total =18.00 hr.

Alpena 10.50 2 Lansing 7.50 Total =18.00 hr.

Alpena 10.50 2 Lansing 7.50 Total =18.00 hr.

Alpena 10.50 2 Lansing 7.50 Total =18.00 hr.

Truck 4 Alpena 10.50 Lansing 3.75 3 Flint 3.75 Total =18.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00

Truck 5 M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 4 Pontiac 11.00 Flint 1.25 Total =17.75

M Pleasant 5.50 4 Pontiac 11.00 Total =16.50 hr.

M Pleasant 5.50 4 Pontiac 11.00 Total = 16.50

Truck 6 M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

M Pleasant 5.50 3 Pontiac 8.25 2 Flint 2.50 Total =16.25 hr.

M Pleasant 5.50 6 Saginaw 13.50 Total =19.00 hr.

W Branch 6.00 2 Saginaw 4.50 Total =10.50 hr.

W Branch 6.00 10 Flint 12.50 Total =18.50 hr.

31

Although this schedule meets the requirements of the problem, it might be improved by better balancing the workload across the trucks and the days. 12 (a) A sweep method solution is shown on the following figure. Five trucks are needed

with a total route distance of (30+29+39+44+19.5)10 = 1,615 miles.

(b) The sweep method is a fast and relatively simple method for finding a solution to

rather complex vehicle routing problems. Solutions can be found graphically without the aid of a computer. However, there are some limitations. Namely,

The method is heuristic and has an average error of about 10 to 15 percent. This

error is likely to be low if the problem contains many points and the weight of each point is small relative to the capacity of the vehicle.

The method does not handle timing issues well, such as time windows. Too many trucks may be used in the route design.

13 This problem may be solved with the aid of ROUTER in LOGWARE. The model input data may be formatted as shown in Figure 7-6.

5 3

0

64

5Warehouse 5

33

4 3 3 4

25

43

4

32 3 1

4

0 2 4 6 8 10 12 14 16 18 20 22 24 26Miles x 10

2

4

Milesx 10

20

18

16

14

12

10

8

2

6

2

Route #1 Load 19

Route #2 Load 20

Route #3 Load 17

Route #4 Load `8

Route #5 Load 9

32

(a) The solution from ROUTER shows that four routes are needed with a minimum total distance of 492 miles. The route design is shown graphically in Figure 7-7. A summary for these routes is given in following partial output report.

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 1.2 1.0 .3 .0 .4 08:59AM 10:12AM 3 29 .00 2 8.9 6.6 1.3 1.0 1.1 08:32AM 05:25PM 19 199 .00 3 6.2 3.7 1.4 1.0 .9 08:42AM 02:54PM 14 112 .00 4 7.5 5.1 1.5 1.0 1.4 08:30AM 04:02PM 12 152 .00 Total 23.8 16.4 4.4 3.0 3.8 48 492 .00 (b) Note that route #1 is short and that a driver and a station wagon would be used for a

route that takes 1.2 hours to complete. By attaching route #1 to route #3, the same driver and station wagon may be used, and the constraints of the problems are still met. The refilled station wagon can leave the depot by 3:30-3:45PM and still meet the customers time windows and return to the depot by 6PM. Thus, only three drivers and station wagons are actually needed for this problem.

FIGURE 7-6 Input Data for ROUTER for Medic Drugs

--PARAMETERS AND LABELS Problem label - Medic Drugs Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - SW DEPOT DATA Depot description - Pharmacy Located in zone - 0 Horizontal coordinate - 13.7 Vertical coordinate - 21.2 Earliest starting time (min) - 480 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 30 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 4.6 Vertical scaling factor - 4.6 Maximum TIME allowed on a route (hours) - 168 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999

--STOP DATA NO

STOP DESCRIPTION

TY

LOAD WGHT

VOL. CUBE

HCRD

VCRD

ZN

LOAD TIME

BEG1

END1

BEG2

END2

1 Covington House D 1 0 23.40 12.90 0 2 540 1020 9999 9999 2 Cuyahoga Falls D 9 0 13.40 13.40 0 18 540 1020 9999 9999 3 Elyria D 1 0 6.30 16.80 0 5 540 1020 9999 9999

33

46 Westbay D 6 0 8.40 18.00 0 10 630 690 9999 9999 47 Westhaven D 2 0 8.50 18.10 0 5 540 1020 9999 9999 48 Broadfiels Mnr D 6 0 18.20 22.90 0 2 540 1020 9999 9999

--VEHICLE DATA -CAPACITY-- --VEHICLE-- --DRIVER--

NO.

VEHICLE DESCRIPTION

TP

NO

WGHT

CUBE

FIXED COST

PER MI COST

FIXED COST

PER HR COST

OVER TIME COST

1 Station wagon 1 50 63 9999 0 0 0 0 0

FIGURE 7-7 Graphical Solution for Medic Drugs

14 There is no exact answer to this problem nor is one intended. Several approaches might be taken to this problem. We could apply the savings method or the sweep method to solve the routing problem for each day of the week, given the current demand patterns. However, we can see that there is much overlap in the locations of the customers by delivery day of the week. We might encourage orders to be placed so that deliveries form tight clusters by working with the sales department and the customers. Perhaps some incentives could be provided to help discipline the order patterns. The orders should form a general pattern as shown below. Currently, the volume for Thursday exceeds the available truck capacity of 45 caskets. Maybe the farthest stops could be handled by a for-hire service rather than acquiring another truck for such little usage.

34

Monday Tuesday

Wednesday Thursday

Friday

Depot

It appears that the truck capacity is about right, given that some slack capacity is likely to be needed. Once the pattern orders are established, either as currently given or as may be revised, apply principles numbers 1, 3, 4, 5, and 7.

35

FOWLER DISTRIBUTING COMPANY Teaching Note

Fowler Distributing Company involves a problem of routing and scheduling a fleet of trucks to serve customers for beer and wine products on a daily basis. Determining an efficient design for making daily deliveries when stop sequence, number of trucks and their sizes, warehouse location, and time window restrictions are variables is the objective of this exercise. For the most part, students should consider that they are to prepare a consulting report to management about this problem. The several questions provided at the end of the case study should help direct the analysis. The ROUTER model is used to cost out and to optimize the various design scenarios. Run results are tabulated in Table 1.

Q1. The current design may not be as efficient as it might be. Therefore, our first task is establish a profile of the current design and then to plan the routes so as to minimize the total miles driven and the number of trucks needed to serve the customers, subject to the restrictions of truck capacity, time windows, total time on the route, etc. Running the current route design in ROUTER establishes a benchmark as shown in Figure 1. The design requires 334 miles of travel with 5 trucks. The daily routing cost is $764.62. Next, optimizing this problem, given no change in data or restrictions, gives the revised benchmark design as shown in Figure 2. Now, costs are reduced to $731.31 per day for a total routing distance of 311 miles. Six trucks are needed. There appears to be a substantial change in the design of the routes that can achieve a ((334 311)/334)100 = 7% reduction in the total miles traveled to make the deliveries. The saving in cost is ($764.62 731.31)250 = $8,327.50 per year. This is not a large savings, but it can be achieved without changing the restriction on deliveries or incurring additional investment. Q2. Time window restrictions often force route design into stop sequences that are not very economical, that is, routes cross themselves. This is the case for the revised benchmark design as shown in Figure 2. To determine the impact of the time windows, we can make an optimizing ROUTER run where there are no time window restrictions. The daily time windows are all set 8:00AM to 5:00PM. The optimized routes are shown in Figure 3. The route mileage is 270 miles per day for five trucks. The total cost is $673.64 per day. From the revised benchmark, this is an additional cost reduction of ($731.31 673.64) 250 = $14,417.50 per year.

36

FIGURE 1 Current Route Design

The questions for management are: How restrictive are the time windows? Can deliveries be made outside of the account's "open hours", such as by giving the driver the key to a safe storage area? Can management offer a small incentive to widen the time window when it otherwise would not be convenient for the account? Relaxing such time windows is often one of the important sources for cost savings in routing problems. Q3. When there are no truck capacity restrictions on a routing problem, the most efficient route design would be to use one large truck to serve all accounts. Therefore, we would expect that trucks of larger capacity would reduce the total distance traveled. A ROUTER optimizing run was made with 600 case capacity trucks to find out. All other conditions were set at the current design.

37

FIGURE 2 Optimized Current Route Design Compared with

the optimized current design, the potential savings is ($731.31 722.98) 250 = $2,082.50 per year. Two 600-case trucks would be needed, along with 3 of the smaller trucks. Although there is a positive savings associated with using larger trucks, the savings seems quite small and perhaps not worth switching to some larger trucks at this time. We should note that these savings are a result of comparing full costs, which include truck depreciation. Although a present value analysis would be appropriate here, we would need some additional information which might include (1) the proportion of truck costs allocated to equipment depreciation, (2) the life of the trucks, (3) the estimated salvage value of the trucks, and (4) and the required rate of return on investments of this type.

38

FIGURE 3 Route Design with Relaxed Time Window Constraints

Q4. From the incremental costs in the detailed report of the optimized current design, we can see that account 14 costs $39.05 and account 19 costs $35.04 to make the deliveries. Since the volume of account 19 is 90 cases and exceeds the 50 case capacity of an outside transport service, account 19 cannot be considered for alternate delivery. If it only costs $35.00 for an outside delivery service to handle account 14, then a cost savings can be realized. Dropping this account and redesigning the routes shows that only 5 trucks are needed for a route cost of $690.17 and route miles of 286. The total cost for handling all accounts would be $690.17 + 35.00 = $725.17. Compared with the optimized current design, this is an annual savings of ($731.31 725.17) 250 = $1,535.00. Economically, Roy should use the outside transportation. However, losing direct control over the deliveries and possible adverse reactions from route salesmen may give him second thoughts about it. Q5. Two ROUTER runs were made here in order to determine the effect of a shorter workday before overtime begins. If no overtime is allowed, then six trucks are required for a total daily routing cost of $754.77. If some overtime is permitted, then the route cost can be reduced to $733.28 with 6 trucks required. Compared with the optimized current design, Roy Fowler would seem to be putting himself at a disadvantage by not wanting to pay overtime.

Q6. Moving the warehouse to a more central location would have the appeal of being closer to all stops and would result in shorter routes. Testing this shows that total route distance can be reduced to 305 miles with a daily cost of $716.91. However, in order to meet the time window and other constraints, an extra truck is needed. There is a potential annual savings of ($731.31 716.91) 250 = $3,600.00. Since it costs $15,000 to make

39

the move, a simple return on investment would be (3,600/15,000) 100 = 24%. At this ROI, the move should be seriously considered. Q7. Routing and scheduling beer trucks from a central depot is quite similar to delivery problems found in manufacturing, retail, and service industries. Several examples are listed below. Serving branch banks by making pickups and deliveries of canceled checks,

supplies, and other paper work from a central domicile point for the trucks Making Federal Express pickups and deliveries around an airport Dispatching school buses around a school for student pick up and drop off Dispatching service personnel for various service companies such as electric,

water, gas, and telephone Making retail deliveries from a warehouse Making deliveries from a plant location to warehouses and then picking up

supplies from vendors on the return trip. Making deliveries (and pickups) of medical records to doctor's offices and other

locations from a central record-keeping location such as a hospital Summary Roy Fowler should consider optimizing his route design. This can be done at no investment to him, and he can gain about a 7% reduction in operating costs on the average. A substantial benefit can be realized by widening the time windows. He should especially see if those time windows that cause routes to overlap themselves can be widened. The potential for doing this is up to $14,418 per year. He should explore the possibility and reasonableness of serving accounts 14 and 19 by an outside transport service. Finally, he should consider relaxing his rigid policy of not wanting to pay overtime, especially if the union is successful in negotiating a 7-hour workday. There seems to be little opportunity for reducing cost by increasing truck capacity or relocating the warehouse. Although there are operating cost reductions available, they do not seem sufficient to justify a change in truck size unless Roy can make good use of the trucks in other ways.

40

TABLE 1 Results of Various ROUTER Runs Run type Miles Cost Trucks Comments Current design 334 $764.62 5 Cost out current design Optimized current

311

731.31

6

Improves on current route design

No time windows restrictions

270

673.64

5

Time windows opened

Trucks at 600-case capacity

299

722.98

5

Two routes require larger trucks

7 hr. work day - no overtime

332

754.77

6

Route time allowed to be no longer than 8.0 hours

7 hr. work day - with overtime

311

733.28

6

Route time allowed to exceed 8.0 hours

New warehouse location

305

716.91

7

Warehouse moved to more central location

41

METROHEALTH MEDICAL CENTER Teaching Note

Strategy MetroHealth Medical Centers mission is to provide transportation to and from its facility for those patients that cannot afford the expense or are unable to use other modes of transportation. MMC currently has a fleet of vans and drivers to serve this patient population. In addition, some transportation can be provided as part of the hospitals contract ambulance service. The primary service provided by Transportation Services is to pick patients up and bring them to MMC for their scheduled appointments and then to return them to their origin points. Careful management of the fleet is needed through vehicle assignment to patients, routing the vans, and assigning some of the patients to the ambulance service for transportation. It is the purpose of this case to show how a basic routing and scheduling model can be used to assist in answering the typical questions surrounding fleet utilization in a service setting. The ROUTER module in LOGWARE is used as the analytical tool. This case does require some insight on the part of the student to deal with the complexities of this problem. Considering the amount of data provided and the richness of the issues involved, this case is probably best assigned as a project rather than as a homework exercise.

Answer to Questions (1) Which vans from the current fleet should be used? To what extent should

subcontracting be used? This problem is complex, considering that it is quite dynamic. Although the patient pick up list appears to be fixed for a particular day, changes take place in the form of cancellations and occasional additions. Patient appointment times are met a high percentage of the time, but not always. Patients to be returned to their origins may find alternate means of transportation so that 100 percent of the patients picked up may not have to be returned. For simplicity, it will be assumed that the patient pick up list is fixed for a particular day and appointment times are rigid. In addition, the returns are not directly considered in designing the routes. Rather, they are expected to be seeded into the pick up routes. This need not be a serious limitation as long as van capacity (number of pick-up patients to the number of available seats) is not highly utilized on a pickup route and van utilization for pick ups drops as appointments are concentrated in the morning hours and return times are shifted toward the afternoon hours, as shown in Figure 1.

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Appointment times Return times FIGURE 1 Appointment Time and Return Time Distributions A typical days patient list can be submitted to ROUTER. Coordinate points for zip code areas are scaled from the map in Figure 2 of the case study. Appointment times are represented as time windows with a 2-hour gap for pick up. A large number of 15-passenger and 6-passenger vans are assumed available to start the ROUTER model. Since ROUTER is a basic vehicle routing and scheduling model, it simply assigns patients to vans and gives a sequence to pick them up. It assumes that a single van leaves and returns to MMC only once in a day. Analysis must account for this, since vans make several out and back trips per day. It is not possible to accurately recreate the current routing patterns for the vans from the data given. However, optimized routing and scheduling can be found through the application of ROUTER. First, the typical daily appointment list is solved with ROUTER using a 2-hour time window where the ending window time is the appointment time. An average speed of 27.5 miles per hour, variable costs of $0.11 per mile1, and a large number of 15-passenger and 6-passenger vans are used. A complete database is shown in the appendix of this note. Using the appointment list for the typical day as representative with all 56 patients, ROUTER shows that 6 vans are needed, where one is a 15-passenger type (route 1). This is routing where no patient is transported by the contract transport service. The routing represents a total of 329 miles driven at a variable cost of 329 mi. $0.11/mi. = $36.19. The routes are placed end-to-end to cover the full day and then ranked, first by the number of patients transported throughout the day on a van and second by the total miles driven, as shown in Table 1. A reasonable procedure for allocating routes to vans is to assign routes with the largest number of patients first to the largest vans. This is because

1Per mile variable cost is determined as repair cost/30,000 mi. per yr. = $1,000/30,000 = $0.03 per mi. plus fuel cost at $1.00 per gallon/average miles per gallon = 1/13 = $0.08 per mi. for a total of 0.08 + 0.03 = $0.11 per mile.

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the ambulance service charges on a per patient basis, and these routes would be most costly to subcontract. If the number of patients is tied for a van, rank first the van with the shortest total route distance. TABLE 1 Route Design for Representative Patient Appointments with

No Transport Subcontracting Van no.

Start from MMC

Return to MMC

Patients on route

Route distance

Route time

1 6:07AM 6:28AM 1 7 mi. 0.3 hr. 1 6:52AM 8:28AM 81 22 1.6 1 8:55AM 10:30AM 4 33 1.6 1 11:04AM 11:31AM 1 9 0.4 1 11:41AM 1:27PM 5 35 1.8 2 7:24AM 9:05AM 5 32 1.7 2 10:59AM 11:57AM 5 13 1.0 3 7:41AM 9:16AM 5 29 1.6 3 10:45AM 12:27PM 5 33 1.7 4 7:37AM 9:12AM 5 30 1.6 4 11:52AM 1:34PM 5 33 1.7 5 7:24AM 8:44AM 4 25 1.3 6 7:43AM 9:03AM 3 28 1.6

56 329 mi. 17.6 hr. 1A 15-passenger van needed. The cost of this initial route configuration can be summarized as follows:

Annual cost of 1 15-passenger van $ 5,750 1 Annual cost of 5 6-passenger vans 28,500 2 Annual repair cost of 6 vans 6,000Annual cost for 6 drivers 141,000 Variable cost 8,686 3 Total cost $189,936

1$23,000/4 yr. = $5,750 2($19,000/4 yr.) 6 = $28,500 3329 mi. x $0.11/mi. 240 days/yr. = $8,686 The question now is whether substituting contract carriage for some of the routes will significantly lower vehicle and driver costs while only slightly increasing variable costs. Since the routes are ranked in Table 1, a reasonable rule for dropping routes would be to start from the bottom of the route list in Table 1 and work upward. This will determine the number of vans needed and the patients to be transported by each mode. These results are shown in Table 2. An unlimited number of subcontracting trips is assumed. TABLE 2 Optimal Number of Vans

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Vans

Driver

Vehicle

Variable (gas + repair)

Subcon-tracting

Total 6 $141,000 $34,250 $14,686 $ 0 $189,936 5 117,500 29,500 12,946 6,235 1 166,181 4 94,000 20,000 11,286 14,549 139,835 3 70,500 15,250 8,623 35,333 129,706 2 47,000 10,500 5,986 56,117 119,603 1 23,500 5,750 3,798 76,901 109,9490 0 0 0 116,390 116,390

1Annual subcontracting cost is $8.66 per round trip 240 days per year 3 patients = $6,235.20 (2) How many subcontracted trips should MMC negotiate and at what price? One 15-passenger van appears to be the optimal number, but this requires 37 patients 20 days per month = 740 subcontracting trips per month. This is more than the 500 allowed. If the 500-trip limit (500/20 days per mo. = 25 patients per day) is to be respected, approximately one additional van will be needed. On the other hand, the cost of the additional van is the annual cost of a van + the annual drivers salary + the gas and repair cost = $19,000/4 + 23,500 + 1,1882 = $29,438/yr. to transport 10 patients per day. At this rate, MMC could afford to pay $29,438/240/10 = $12.27 per round trip (find 10 patients for van 2 in Table 1). Possibly the subcontractor would be willing to offer this additional transport for a price between the $8.66 and $12.27 per trip. It would be in Macs interest to do this. Renegotiating all trips at the 750-trip level is another possibility. MMC can afford to pay a subcontracting cost of up to $189,936 - 31,0483 = $158,888/yr. This would require transporting 37 patients per day at a maximal average trip cost of $158,888/240/37 = $17.89. Since $17.89 is the optimal worth of subcontracting, MMC probably can negotiate a rate much less than this while still encouraging the subcontractor to provide service at the higher level. (3) MMC is considering using all 6-passenger vans. Would this be a good decision? Developing daily routes and assigning them to 6-passenger vans is the same as for Table 1. This initial allocation without subcontracting is shown in Table 3. Considering the subcontracting trip limit of 500 patients per month (25 patients per day), two vans are needed in the MMC fleet. This number of vans is determined by counting the patients from the bottom of Table 3 until approximately 25 patients are found, considering increments of full vans and noting the number of vans remaining at the top of the list. (Assume 27, an extra 2 patients can be accommodated by the subcontractor at no additional cost.) The variable cost for two vans is shown below. Variable cost type

Two 6-passenger

One 15- & one 6-passenger van

245 miles x $0.11/mile 240 days/yr. = $1,188. 3Driver + vehicle +variable costs for one van, or 19,000/4 + 23,500 + 2,798 = $31,048.

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vans Annual cost of 2 drivers 47,000 47,000 Vehicle mileage + repair cost 5,9601 5,986 Subcontracting cost 56,1172 56,117 Total $109,077 $109,103 1150 mi. $0.11/mi. 240 days/yr. + 2,000 = $3,960. 227 patients $8.66/trip 240 days/yr. = $56,117. In contrast, the 15-passenger and 6-passenger van configuration compared with the two 6-passenger van configuration saves $109,103 109,077 = $26 per year. The initial outlay is less for the smaller van by $4,000. From purely an economic standpoint, using just 6-passenger vans would appear to be the best choice, however there may be customer service reasons for having the larger van available to meet peak demand needs.

TABLE 3 Route Design for 6-Passenger Vans Only Van no.

Start from MMC

Return to MMC

Patients on route

Route distance

Route time

1 6:07AM 6:28AM 1 7 mi. 0.3 hr. 1 6:52AM 7:37AM 3 12 0.8 1 7:37AM 9:12AM 5 30 1.6 1 11:04AM 11:31AM 1 9 0.4 1 11:52AM 1:34PM 5 33 1.7 2 7:22AM 8:19AM 5 13 1.0 2 8:55AM 10:30AM 4 33 1.6 2 10:59AM 11:57AM 5 13 1.0 3 7:41AM 9:16AM 5 29 1.6 3 10:45AM 12:27PM 5 33 1.7 4 7:24AM 9:05AM 5 32 1.7 4 11:41AM 1:27PM 5 35 1.8 5 7:24AM 8:44AM 4 25 1.3 6 7:43AM 9:03AM 3 28 1.3

56 332 mi. 17.8 hr.

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APPENDIX A Database for ROUTER

--PARAMETERS AND LABELS-- Problem label - METROHEALTH MEDICAL HOSPITAL Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - SW DEPOT DATA Depot description - Medical Center Located in zone - 0 Horizontal coordinate - 7.40 Vertical coordinate - 6.13 Earliest starting time (min) - 360 Latest return time (min) - 960 Default vehicle speed (miles per hour) - 28 After how many clock hours will overtime begin - 68.0 GENERAL DATA Percent of vehicle in use before allowing pickups - 100 Horizontal scaling factor - 3.102 Vertical scaling factor - 2.820 Maximum TIME allowed on a route (hours) - 2.0 Maximum DISTANCE allowed on a route (miles) - 9999.0 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 6.00 Variable time per stop by weight - 0.00 By cube - 0.00 BREAK TIMES Duration of 1st break (minutes) - 0 To begin after - 9999 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999

--STOP DATA-- STOP LOAD VOLUME -COORDINATES- LOAD ---TIME WINDOWS---- NO. DESCRIPTION---- TY WGHT CUBE HCRD VCRD ZN TIME BEG1 END1 BEG2 END2 1 Baker P 1 0 6.50 8.90 0 0 390 510 9999 9999 2 Boyd P 1 0 6.20 9.00 0 0 390 510 9999 9999 3 Carver P 1 0 11.00 5.00 0 0 420 540 9999 9999 4 Ivey P 1 0 10.00 6.00 0 0 420 540 9999 9999 5 Rashed P 1 0 10.00 10.00 0 0 420 540 9999 9999 6 Walsh P 1 0 3.50 5.00 0 0 420 540 9999 9999 7 Johnson P 1 0 8.00 5.50 0 0 450 570 9999 9999 8 Burgess P 1 0 9.00 7.50 0 0 465 585 9999 9999 9 Delgado P 1 0 8.20 7.20 0 0 465 585 9999 9999 10 Fairrow P 1 0 8.60 7.80 0 0 480 600 9999 9999 11 Middlebrooks P 1 0 8.00 5.40 0 0 480 600 9999 9999 12 Suech P 1 0 5.00 6.50 0 0 480 600 9999 9999 13 Lawson P 1 0 6.30 8.80 0 0 510 630 9999 9999 14 Reed P 1 0 9.05 7.30 0 0 510 630 9999 9999 15 Bongiovanni P 1 0 9.00 5.80 0 0 525 645 9999 9999 16 Miller P 1 0 7.80 5.40 0 0 540 660 9999 9999 17 Talley P 1 0 6.50 10.20 0 0 570 690 9999 9999 18 Williams P 1 0 7.50 7.00 0 0 585 705 9999 9999 19 Dumas P 1 0 7.20 5.90 0 0 630 750 9999 9999 20 Taylor P 1 0 9.80 7.00 0 0 645 765 9999 9999 21 Barker P 1 0 7.20 6.00 0 0 660 780 9999 9999 22 Lhota P 1 0 8.00 6.20 0 0 660 780 9999 9999 23 Manco P 1 0 10.10 9.00 0 0 660 780 9999 9999 24 Webb P 1 0 9.70 7.00 0 0 660 780 9999 9999 25 Wilson P 1 0 9.00 5.80 0 0 660 780 9999 9999 26 Arrington P 1 0 10.20 5.90 0 0 720 840 9999 9999 27 Staunton P 1 0 6.40 8.80 0 0 720 840 9999 9999 28 Wall P 1 0 9.50 5.10 0 0 720 840 9999 9999 29 Williams P 1 0 8.20 7.20 0 0 720 840 9999 9999 30 Caruso P 1 0 7.00 5.00 0 0 375 495 9999 9999 31 West P 1 0 5.60 6.00 0 0 390 510 9999 9999 32 Amaro P 1 0 6.00 6.60 0 0 420 540 9999 9999 33 Brown P 1 0 7.20 4.20 0 0 420 540 9999 9999 34 Ciesicki P 1 0 6.50 3.00 0 0 420 540 9999 9999 35 Pinkevich P 1 0 6.60 5.10 0 0 420 540 9999 9999 36 Staufer P 1 0 5.80 6.00 0 0 420 540 9999 9999 37 Winterich P 1 0 7.30 4.80 0 0 420 540 9999 9999 38 Brown P 1 0 4.00 5.00 0 0 435 555 9999 9999 39 Ball P 1 0 6.20 6.30 0 0 450 610 9999 9999 40 Lanza P 1 0 6.20 5.90 0 0 450 610 9999 9999 41 Mayernik P 1 0 7.10 7.00 0 0 450 610 9999 9999 42 Suech P 1 0 4.00 6.50 0 0 465 625 9999 9999 43 Heffner P 1 0 5.20 4.50 0 0 480 640 9999 9999 44 Jarrell P 1 0 4.20 6.80 0 0 480 640 9999 9999

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45 Piatak P 1 0 7.00 3.50 0 0 480 640 9999 9999 46 Swaysland P 1 0 6.10 6.70 0 0 480 640 9999 9999 47 Baer P 1 0 3.80 4.90 0 0 540 700 9999 9999 48 Wills P 1 0 4.70 6.10 0 0 540 700 9999 9999 49 Fauber P 1 0 5.20 6.80 0 0 660 780 9999 9999 50 Mullins P 1 0 7.20 6.30 0 0 660 780 9999 9999 51 Pack P 1 0 6.20 5.00 0 0 660 780 9999 9999 52 Westerfield P 1 0 5.90 6.30 0 0 675 795 9999 9999 53 Lisiewski P 1 0 5.00 5.50 0 0 690 810 9999 9999 54 McPherson P 1 0 4.20 6.20 0 0 720 840 9999 9999 55 Mykytuk P 1 0 6.00 6.50 0 0 750 870 9999 9999 56 Gutschmidt P 1 0 5.50 5.90 0 0 780 900 9999 9999

--VEHICLE DATA-- ---VEHICLE--- ----DRIVER---- OVER VEHICLE --CAPACITY--- FIXED PER MI FIXED PER HR TIME NO. DESCRIPTION---- TP NO WEIGHT --CUBE ---COST --COST ---COST --COST --COST 1 15-pasngr vehcl 1 20 15 99 0.00 0.11 0.00 0.00 0.00 2 6-pasngr vehcl 2 20 6 99 0.00 0.11 0.00 0.00 0.00

48

ORION FOODS, INC. Teaching Note

Strategy The purpose of this case study is to allow students to analyze a distribution problem where determining the optimal paths through a network is central to the problem solution. The shortest route methodology applies, and the ROUTE module in the LOGWARE software can effectively be used. This module permits students to quickly find the shortest distance paths from any starting point to all other points in the network. A prepared data file of the network is available for use under the ROUTE module. The case has several dimensions that allow it to be used as a homework assignment, a short case study project, or as a basis for classroom discussion. It is a simple problem in network design highlighting transportation routing issues, however additional factors such as inventory consolidation and investment costs allow for an enriched analysis. Note: A database for this problem is available in the ROUTE module of the LOGWARE software. Answers to Questions (1) Can Anita improve upon the current distribution operations? It should be obvious to Anita that the current distribution system may not be performing at optimum. Allowing carriers to decide the routes to use when they are being paid on a mileage basis is like asking a fox to watch the chicken coop. She really cannot expect that carriers will be motivated to seek out optimal routing patterns. Therefore, she should determine the best routes between regional and field warehouses and insist that carriers invoice according to the mileages along these specified routes. Using the map provided in Figure 1 of the case study and the current assignments of field warehouses to regional warehouses, she can develop a database for the ROUTE module in LOGWARE. The database is shown in the Supplement to this note. Running ROUTE will give the optimal routes from which she can develop transportation costs, as shown in Table 1 below. Compared with the current cost level, a savings of $652,274 630,140 = $22,134 per year can be realized. This savings does not require any investment; however, it will be magnified by the growth in demand in the next five years.

49

(2) Is there any benefit to expanding the warehouse at Burns, OR? Finding the optimal distances when serving all field warehouses from Burns or Fresno gives a slightly different allocation of field warehouses to regional warehouses than is currently the case. That is,

Field warehouse

If served from Burns

If served from Fresno

Los Angeles 806 mi. 219 mi.* Phoenix 973 588* Salt Lake City 536* 815 San Francisco 555 183* Portland 293* 757 Butte 676* 1,120 Seattle 467* 925

*Indicates optimal assignment. This shows that Salt Lake City would be better served out of Burns rather than Fresno. Burns currently is near its capacity limit, so to assign Salt Lake City's volume to it would require expansion. In the short term, 35,000/8 = 4,375 cwt. of inventory capacity is needed. However, (43,000 + 5,000 + 56,000)/8 = 13,000 cwt. of the 15,000 cwt. of available capacity is currently being used. At minimum, an additional increment of capacity is required at a cost of $300,000. Reassignment of Salt Lake City to Burns would save 815 536 = 279 miles per trip. On 116.7 trips per year, the annual savings would be 1.30 279 116.7 = $42,327. The simple return on investment (ROI) would be:

ROI $42,$300,

.327000

100 141%

TABLE 1 Optimal Transportation Costs Under Current Distribution System Design

Regional warehouse

Field warehouse

Optimal route miles

Average number of

tripsa

Transport

cost, $ Fresno Los Angeles 219 366.7 104,399b Fresno Phoenix 588 200.0 152,880 Fresno Salt Lake City 815 116.7 123,644 Fresno San Francisco 183 280.0 66,612 Burns Portland 293 143.3 54,583 Burns Butte 676 16.7 14,676 Burns Seattle 467 186.7 113,346 Totals 3,241 630,140 a Determined from the warehouse throughput divided by the average shipment size. e.g., 110,000 cwt./300 cwt. = 366.7 trips. b $1.30 per mi. 219 mi. 366.7 trips = $104,399.

50

If the anticipated growth in demand is realized, the number trips to Salt Lake City would increase to 56,000/300 = 186.7. The projected savings in the fifth year would be 1.30 279 186.7 = $67,716. The average annual savings would be (42,327 + 67,716)/2 = $55,022. The average annual ROI is:

ROI $55,$300,

.022000

100 18 3%

Anita must now compare this return to other worthy investments in the firm to see if this opportunity is worth the risk. (3) Is there any merit to consolidating the regional warehousing operation at Reno, NV? If Reno were to replace the Burns and Fresno warehouses, an