Home Work Chapter 1, 2, 3, 4, 5, 6, 7: Business Logistics/Supply Chain Management Ronald H. Ballou Selected questions, Excel sheet attached for some calculations.

1. Home Work Chapter 1 to 7
Book: Business Logistics/Supply Chain Management Ronald H. Ballou
Excel sheet:
Logistics
management.xlsx
Student Name: Shaheen Sardar
Course Name: Logistics Management
Department: Industrial and Management Engineering, Hanyang University, South Korea.
Home Work 1
Chapter 1: Business Logistics/Supply Chain
Question 12: Suppose that a manufacturer of men's shirts can produce a dress shirt in its Houston, Texas, plant for
$8 per shirt (including the cost of raw materials). Chicago is a major market for 100,000 shirts per year. The shirt is
priced at $15 at Houston plant. Transportation and storage charges from Houston to Chicago amount to $5 per
hundredweight (cwt.). Each packaged shirt weighs 1 pound.
As an alternative, the company can have the shirts produced in Taiwan for $4 per shirt (including the cost of
raw materials). The raw materials, weighing about 1 pound per shirt, would be shipped from Houston to Taiwan at a
cost of $2 per cwt. When the shirts are completed, they are to be shipped directly to Chicago at a transportation and
storage cost of $6 per cwt. An import duty of $0.50 per shirt is assessed.
(a) From a logistics-production cost standpoint, should the shirts be produced in Taiwan?
(b) What additional considerations, other than economic ones, might be considered before making a final decision?
Solution:
Plant Location
cwt Material
Transport Cost
(Houston to
Taiwan)
Unit
Production
Cost
cwt Product
Transportation/
storage charges
(Houston to Chicago)
cwt Product
Transportation
/storage charges
(Taiwan to Chicago)
Unit Import
Duty
Houston plant
(USA)
NO $8 $5 NO NO
Outsourcing to
Taiwan (Asia)
$2 $4 NO $6 $0.50
Note: hundredweight (cwt) = 100 pounds weight
Each packed shirt weight = 1 pound
Raw material weight per shirt = 1 pound
Unit Material Transport Cost = Raw Material Density/CWT
Unit Product Transport Cost = Product Density/CWT
Example: $2/100 = $0.02
Plant Location
Unit Material
Transport Cost
(Houston to
Taiwan)
Unit
Production
Cost
Unit Product
Transportation/
storage charges
(Houston to Chicago)
Unit Product
Transportation
/storage charges
(Taiwan to Chicago)
Unit Import
Duty
Houston plant
(USA)
NO $8 $0.05 NO NO
Outsourcing to
Taiwan (Asia)
$0.02 $4 NO $0.06 $0.50

2. Following formulas are used.
 Total Material Transport Cost = Product Volume * Unit Material Transport Cost
 Total Product Transport Cost = Product Volume* Unit Product Transport Cost
 Total Production Cost = Product Volume * Unit Production Cost
 Total Import Duty = Product Volume * Unit Import Duty
 Total Cost = Total Material Transport Cost + Total Product Transport Cost + Total Production Cost + Total
Import Duty
 Total Price = Product Volume * Shirt Price/unit
 Total Profit = Total Price - Total Cost
Plant Location
Product
Volume
Total
Production
Cost
Total
Material
Transport
Cost
Unit
Product
Transport
Cost
Total
Import
Duty
Total
Cost
Shirt
price/
unit
Total price Profit
Houston plant
(USA)
100000 $800,000 0 $5,000 0 $805,000 $15 $1,500,000 $695,000
Outsourcing to
Taiwan (Asia)
100000 $400,000 $2,000 $6,000 $50,000 $458,000 $15 $1,500,000 $1,042,000
From a logistics-production cost standpoint, the shirts should be produced in Taiwan. There is a cost for raw
materials from Houston to Taiwan, but still is cheaper than the Houston plant when other costs are combined.
Following additional considerations might be considered before making a final decision.
 How long the shirts would be stored (holding cost) from plant-to-truck-to-final destination.
 Order processing cost for suppliers needs to be done strategically (to have least cost of shipping to Chicago).
 How order is transported at least cost, most efficiently, and within the time allotted.
 Supplier availability in Taiwan.
 Capacity availability in Taiwan.
 Risk of late delivery in Taiwan.
 Quality and reliability issues in Taiwan.
 Effective inventory lot-sizing in Taiwan.
 Other strategic considerations in USA as well as in Taiwan.
Home Work 2
Chapter 2: Logistics/Supply Chain Strategy and Planning
Question 13: The traffic manager of the Monarch Electric Company has just received a rate reduction offer from a
trucking company for the shipment of fractional horsepower motors to the company's field warehouse. The proposal
is a rate of $3 per hundredweight (cwt.) if a minimum of 40,000 pounds is moved in each shipment. Currently,
shipments of 20,000 pounds or more are moved at a rate of $5 per cwt. If the shipment size falls below 20000
pounds, a rate of $9 per cwt. applies.
To help the traffic manager make a decision, the following information has been gathered:
Annual demand on warehouse 5,000 motors a year
Warehouse replenishment orders 43 orders a year
Weight of each motor, crated 175 lb per motor
Standard cost of motor in warehouse $200 per motor
Stock replenishment order handling cost $15 per order
Inventory carrying cost as percentage of average value of inventory on hand for a year 25% per year
Handling cost at warehouse $ 0.30 per cwt.
Warehouse space unlimited
Should the company implement the new rate?

3. Solution:
Circumstance 1: Rate for shipment weight ≥ 40000 = $3 per cwt. (New proposal)
Circumstance 2: Rate for shipment weight ≥ 20000 = $5 per cwt. (Present)
Circumstance 3: Rate for shipment weight < 20000 = $9 per cwt. (Present)
Shipment size is more than 20000 or minimum 40000, so we will not investigate Circumstance 3.
Weight of total motors (annual requirement) = 5000 motors/year * 175 lb. /motor = 875000 lb. / year = 8750
cwt/year (i.e. 1 cwt = 100 lb.)
Cost for Circumstance 1: (For new proposal)
Trucking cost = $3/cwt * 8750 cwt / year = $ 26250/year
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟𝑠 =
875000 𝑙𝑏./𝑦𝑒𝑎𝑟
40000 𝑙𝑏.
= 21 𝑜𝑟𝑑𝑒𝑟𝑠
Ordering cost = 21 orders * $ 15 /order = $ 315
𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 =
40000 𝑙𝑏
175 𝑙𝑏
×
1
2
× $200 × 0.25 = $ 𝟓𝟕𝟏𝟓
Handling cost at warehouse = $0.3/cwt *8750 cwt / year = $ 2625
Total Cost = $ 26250 + $ 315 + $ 5715 + $ 2625 = $ 34905
Cost for Circumstance 2: (For present situation)
Trucking cost = $5/cwt * 8750 cwt / year = $ 43750
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟𝑠 =
875000 𝑙𝑏./𝑦𝑒𝑎𝑟
20000 𝑙𝑏.
= 43 𝑜𝑟𝑑𝑒𝑟𝑠
Ordering cost = 43 orders * $ 15 /order = $ 645
𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 =
20000 𝑙𝑏
175 𝑙𝑏
×
1
2
× $200 × 0.25 = $ 𝟐𝟖𝟓𝟕
Handling cost at warehouse = $0.3/cwt *8750 cwt/ year = $ 2625
Total Cost = $ 43750 + $ 645 + $ 2857 + $ 2625 = 49877
Circumstance 1 Total Cost: (For new proposal) = $ 34905
Circumstance 2 Total Cost: (For present situation) = $ 49877
Yes, the company should implement the new rate.
Home Work 3
Chapter 3: Logistics/Supply Chain Product
Question 11: Davis Steel Distributors is planning to set up an additional warehouse in its distribution network.
Analysis of item-sales data in its other warehouses shows that 25% of the items represent 75% of the sales volume.
The company also has an inventory policy that varies with the items in the warehouse. That is, the first 20% of the

4. items are the A items and are to be stocked with turnover ratio of 8. The next 30% of the items, or B items, are to
have turnover ratio of 6. The remaining C items are to have a turnover ratio of 4. There are to be 20 products held at
the warehouse with sales forecasted to be $2.6 million annually. What dollar value of the average inventory would
you estimate for the warehouse?
Solution:
𝑌 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑎𝑙𝑒𝑠
𝑋 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠
𝐴 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑
𝑌 = 0.75
𝑋 = 0.25
𝐴 = ?
𝐴 =
𝑋(1 − 𝑌)
𝑌 − 𝑋
=
0.25(1 − 0.75)
0.75 − 0.25
= 0.125
Number of items = N = 20
Sales = $2.6 million = $2600000
𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐼𝑡𝑒𝑚 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = 𝑋 = 𝐼𝑡𝑒𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 ×
1
𝑁
Cumulative sales proportion is
𝑌 =
(1 + 𝐴)𝑋
𝐴 + 𝑋
Projected items sales = difference between cumulative sales for successive items
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚

5. Alternative Method
Cumulative sales proportion is
𝑌 =
(1 + 𝐴)𝑋
𝐴 + 𝑋
=
(1 + 0.125)𝑋
0.125 + 𝑋
𝑌𝐴 =
(1 + 0.125)0.20
0.125 + 0.20
= 0.69
Cumulative sales (𝑌𝐴) =$2600000*0.69 = 1,800,000
Projected Item Sales (𝑌𝐴) = $1,800,000
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐴 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚
=
$1,800,000
8
= $225,000
𝑌𝐴+𝐵 =
(1 + 0.125)0.50
0.125 + 0.50
= 0.90
Cumulative sales (𝑌𝐴+𝐵) =$2,600,000*0.90 = $2,340,000
The product group B sales will be A+B sales less A sales

6. Projected Item Sales (𝑌𝐵) = 𝑌𝐴+𝐵 − 𝑌𝐴 = $2,340,000- $1,800,000 = $540,000
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐵 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚
=
$540,000
6
= $90,000
𝑌𝐴+𝐵+𝐶 =
(1 + 0.125)1.00
0.125 + 1.00
= 1.00
Cumulative sales (𝑌𝐴+𝐵+𝐶) =$2,600,000*1.00 = 2,600,000
The product group C sales will be A+B+C sales less A+B sales
Projected Item Sales (𝑌𝐶) = 𝑌𝐴+𝐵+𝐶 − 𝑌𝐴+𝐵 = 2,600,000-$2,340,000= $260,000
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐶 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚
=
$260,000
4
= $65,000
Dollar value of the average inventory for the warehouse is
= 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐴 + 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐵 + 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 𝐶
= $225,000 + $90,000 + $65,000 = $380,000
Question 12: Beta Products is planning to add another warehouse. Ten products from the entire line are to be stored
in the new warehouse. These products will be the A and B items. All C items are to be served out of the plant.
Forecasts of annual sales that are expected in the region of the new facility are 3 million cases (A, B, and C items).
Historical data show that 30 percent of the items account for 70 percent of the sales. The first 20 percent of the entire
line are designated as A items, the next 30 percent as B items, and the remaining 50% as C items. Inventory turnover
ratios in the new warehouse are projected to be 9 for A items and 5 for B items. Each inventory item, on the average,
requires 1.5 cubic feet of space. Product is stacked 16 feet high in the warehouse.
What effective storage space is needed in square feet excluding aisle, office, and other space requirements?
Solution:
All C items are to be served out of the plant. So, we will consider only A and B.
𝑌 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑎𝑙𝑒𝑠
𝑋 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠
𝐴 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑
𝑌 = 0.70
𝑋 = 0.30
𝐴 = ?
𝐴 =
𝑋(1 − 𝑌)
𝑌 − 𝑋
=
0.30(1 − 0.70)
0.70 − 0.30
= 0.225
Number of items = N = 10
Sales = $3 million = $3000000
𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝐼𝑡𝑒𝑚 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = 𝑋 = 𝐼𝑡𝑒𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 ×
1
𝑁

7. Cumulative sales proportion is
𝑌 =
(1 + 𝐴)𝑋
𝐴 + 𝑋
Projected items sales = difference between cumulative sales for successive items
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝑙𝑒𝑣𝑒𝑙 =
𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝐼𝑡𝑒𝑚 𝑆𝑎𝑙𝑒𝑠
𝑇𝑢𝑟𝑛𝑜𝑣𝑒𝑟 𝑟𝑎𝑡𝑖𝑜 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚
Space requirement = 1.5 cubic feet
Stack height requirement = 16 feet
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑡𝑜𝑟𝑎𝑔𝑒 𝑠𝑝𝑎𝑐𝑒 =
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑖𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 × 𝑠𝑝𝑎𝑐𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡
𝑆𝑡𝑎𝑐𝑘 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡
=
$353171 × 1.5 𝑓𝑒𝑒𝑡3
16 𝑓𝑒𝑒𝑡
Effective storage space required = 33110 𝑓𝑒𝑒𝑡2
Question 13: An analysis of the product line items in the retail stores of the Save-More Drug chain shows that 20%
of the items stocked account for 65% of the dollar sales. A typical store carries 5,000 items. The items accounting
for the top 75% of the sales are replenished from warehouse stocks. The remainder is shipped directly to stores from
manufacturers or jobbers. How many items are represented in the top 75% of sales?
Solution:
𝑌 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑎𝑙𝑒𝑠
𝑋 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠
𝐴 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑
𝑌 =
(1 + 𝐴)𝑋
𝐴 + 𝑋
(1)
𝑌 = 0.65
𝑋 = 0.20

8. 𝐴 = ?
From equation (1), we have
𝐴 =
𝑋(1 − 𝑌)
𝑌 − 𝑋
(2)
𝐴 =
𝑋(1 − 𝑌)
𝑌 − 𝑋
=
0.20(1 − 0.65)
0.65 − 0.20
= 0.156
Now
𝐴 = 0.156
𝑌 = 0.75
𝑋 = ?
From equation (2), we have
𝑋 =
𝐴 × 𝑌
1 − 𝑌 + 𝐴
𝑋 =
𝐴 × 𝑌
1 − 𝑌 + 𝐴
=
0.156 × 0.75
1 − 0.75 + 0.156
=
0.117
0.406
= 0.29
29% of the items represent the top 75% of sales
Total number of items = 5,000
Number of items that represent 75% of sales = 0.29*5000 = 1450
Question 14: The cost associated with producing, distributing, and selling a domestically produced automotive
component to Honda in Japan can be summarized as follows:
Cost type Cost per Unit, $
Purchased materials 25
Manufacturing labor 10
Overhead 5
Transportation Varies by shipment size
Sales 8
Profit 5
Transportation costs vary as follows. If the purchase (shipping) quantity is 1000 units or less, the
transportation cost is $5 per unit. For more than 1000 units but less than or equal to 2000 units, the transportation
cost is $4 per unit. For more than 2000 units, the transportation cost is $3 per unit.
Construct a price schedule, assuming the vendor would like to pass the transportation economies on to the
customer. Indicate the discount percentage the customer will receive through buying at various quantities.
Solution:
Case 1: Quantity is 1000 units or less
Total cost = $25 + $10 +$5 + $5 + $8 + $5 = $58
Case 2: Quantity is more than 1000 units but less than or equal to 2000 units
Total cost = $25 + $10 +$5 + $4 + $8 + $5 = $57
Discount rate between Case 1 and Case 2:

9. 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 𝑟𝑎𝑡𝑒 =
58 − 57
58
× 100 = 0.017 = 1.7 %
Case 3: Quantity is more than 2000 units
Total cost = $25 + $10 +$5 + $3+ $8 + $5 = $56
Discount rate between Case 1 and Case 3:
𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 𝑟𝑎𝑡𝑒 =
58 − 56
58
× 100 = 0.034 = 3.4 %
Discount rate between Case 2 and Case 3:
𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 𝑟𝑎𝑡𝑒 =
57 − 56
57
× 100 = 0.018 = 1.8 %
Home Work 4
Chapter 4: Logistics/Supply Chain Customer Service
Question 6: The Cleanco Chemical Company sells cleaning compounds (dishwashing powders, floor cleaners,
nonpetroleum lubricants) in a keenly competitive environment to restaurants, hospitals and schools. Delivery time
on orders determines whether a sale can be made. The distribution system can be designed to provide different
average levels of delivery time through the number and location of warehousing points, stocking levels and order
processing procedures. The psychical distribution manager has made the following estimates of how service affects
sales and the cost of providing service levels:
Percentage of orders delivered in one day
50 60 70 80 90 95 100
Estimated annual sales (millions of $) 4.0 8.0 10.0 11.0 11.5 11.8 12.0
Cost of distribution (millions of $) 5.8 6.0 6.5 7.0 8.1 9.0 14.0
(a) What level of service should the company offer?
(b) What effect would competition likely have on the service level decision?
Solution (a): Profit = Estimated annual sales – cost of distribution
Percentage of orders delivered in one day
50 60 70 80 90 95 100
Estimated annual sales (millions of $) 4.0 8.0 10.0 11.0 11.5 11.8 12.0
Cost of distribution (millions of $) 5.8 6 6.5 7 8.1 9 14
Profit (millions of $) -1.8 2 3.5 4 3.4 2.8 -2
Profit is $4 million at 80% service level. Company maximizes the profit when it serves at 80% service level.
Company should offer 80% of service level.
Solution (b): Suppose the product’s price and quality is the same in all competing companies. If a company A
increases its service level, other competitor companies will also increase their service levels. Based on competition,
the company A will change its service level, resulting in change in its profit (otherwise loss of customers).

10. Question 7: Five years ago, Norton Valves, Inc., introduced and publicized a program under which 56 items in its
hydraulic valve line would be made available on a 24-hour-delievery basis, instead of normal 1-to 12-week delivery
period. Quick order processing, stocking to anticipated demand, and using premium transportation services when
necessary were elements of the 24-hour delivery program. Sales history was recorded for the five years before the
service change as well as for a five-year period after the change. Because only a portion of the product family was
subject to the service improvement, the remaining products (102 items) served as a control group. Statistics for one
of the test product groups showing the before and after annual units sales levels are given as follows:
Sales before service change Sales after service change
Product Family 5-Year Average Standard Deviation 5-Year Average Standard Deviation
Test group 1342 335 2295 576
Control group 185 61 224 76
 Standard Deviation: For the individual sales
 Test group: Products in family with 24-hour delivery
 Control group: Products in family with 1-to 12-week delivery
The average value of products in this family was $95 per unit. The incremental cost for the improved service was $2
per unit, but the company did not intend to pass along the costs as a price increase. Instead it hoped that additional
sales volume would more than offset the added costs. The profit margin on sales at the time was 40 percent.
(a) Should the company continue the premium service policy?
(b) Appraise the methodology as a way of accuracy determining the sales-service effect.
Solution: (a)
 Before: (1342+185)*95*0.40 = $58026
 After: (2295+224)*95*0.40 – (2295*2) = $91132
 Company should continue the premium service policy.
Solution: (b) We appraise the methodology as a way of accurately determining the sales-service effect.
Question 8: A food company is attempting to set the customer service level (in-stock probability in its warehouse)
for a particular product line item. Annual sales for the item are 100,000 boxes, or 3846 boxes biweekly. The product
cost in inventory is $10, to which $1 is added as profit margin. Stock replenishment is every two weeks, and the
demand during this time is assumed to be normally distributed with a standard deviation of 400 boxes. Inventory
carrying costs are 30% per year of item value. Management estimates that a 0.15% change in total revenue would
occur for each 1% change in the in-stock probability.
(a) Based on this information, find the optimum in-stock probability for the item.
(b) What is the weakest link in this methodology? Why?
Solution: (a) Optimum service level is the point where the change of cost equals to the change of profit (∆p = ∆c)
∆p = Trading margin*sales response rate *annual sales
∆p = $1*0.0015*100000
∆p = $150 per year (per 1% change in the service level)
∆c = annual carrying cost * standard product cost * ∆z*demand standard deviation over replenishment lead time
∆c = 0.30*$10*∆z*400
∆c = $ 1200 ∆z
∆p = ∆c
$150 = $1200*$∆z

11. ∆z =
150
1200
= 0.125
For the change in z found in a normal distribution table, the optimum in-stock probability during the lead time is
about 96-97%.
Solution: (b) The weakest link in methodology is that ∆p is assumed as constant for all service levels, in fact, in
most cases ∆p is a decreasing function of service level. We do not have the correct data about interrelation of service
level and sales. (Weakest link in this analysis is estimating the effect that a change in service will have on revenue).
Question 9: An item in the product line for the food company discussed in question 8 has the following
characteristics:
Sales response rate = 0.15% change in revenue for a 1% change in the service level
Trading margin = $0.75 per case
Annual sales through the warehouse = 80,000 cases
Annual carrying cost = 25 %
Standard product cost = $10
Demand standard deviation = 500 cases per 1 week lead time
Lead time = 1 week
Find the optimum service level for this item.
Solution: Optimum service level is the point where the change of cost equals to the change of profit (∆p = ∆c)
∆p = Trading margin*sales response rate *annual sales
∆p = $0.75*0.0015*80000
∆p = $90 per year (per 1% change in the service level)
∆c = annual carrying cost * standard product cost *∆z*demand standard deviation over replenishment lead time
∆c = 0.25*$10*∆z*500
∆c = $ 1250 ∆z
∆p = ∆c
$90 = $1250*∆z
∆z =
90
1250
= 0.072
For the change in z found in a normal distribution table, the optimum service level for this item is about 92-93%.
Question 10: A retailer has targeted a shelf item to be out of stock only 5% of the time (m). Customers have come
to expect this level of product availability, so much so, that when the out of stock percentage increases, customer
seek substitutes and lost sales occur. From market research studies, the retailer has determined that when the out of
stock probability increases to the 10% level (y), sales and profit drop to one-half of those at target level. Decreasing
the out of stock percentage from the target level seems to have little impact on sales, but it does increase inventory
carrying cost substantially. The following data have been collected on the item:
Price $5.95
Cost of item 4.25
Other expenses associated with stocking the item @0.30
Annual item sold @ 95% in-stock 880

12. The retailer estimates that every one percentage point that the in-stock probability is allowed to vary from the target
level, the unit cost of supplying the item decreases according to C = (1.00-0.10) (i.e. y-m), where C is the cost per
unit, y is the out-of-stock percentage, and m is the target out-of-stock percentage
How much variability from the target stocking percentage should the retailer allow?
Solution:
𝐿 = 𝑘 (𝑦 − 𝑚)2
0.7 = 𝑘 (10 − 5)2
𝑘 = 0.028
𝑦 − 5 =
0.10
2 × 0.028
= 1.79
𝑦 = 6.79
The retailer should not allow the out of stock percentage to deviate more than 1.79% and should not allow the out of
stock level to fall below 1.79 + 5 = 6.79%.
Home Work 5
Chapter 5: Order Processing and Information Systems
Question 5: A logistics manager for a television producer in South Korea has been given the responsibility for
setting up a logistics information system for his company. How would you answer his question below?
(a) What types of information do I want from the information system? Where would I obtain the information?
(b) Which item in information database should I retain in the computer for easy access? How should I handle the
remainder?
(c) What types of decision problems would the information system help me address?
(d) What models for data analysis would be most useful in dealing with these problems?
Answer (a): We can get amount of sales, times, sales location, order data, inventory rate, transport cost, etc. from
customers, company, books, and company files.
Answer (b): Logistics manager decides the criteria for judgment when he makes the decision. For example:
 How important is the data stored on computer database?
 What data is not required to keep on computer due to computer memory problems?
 Is it necessary to respond quickly?
 How frequent efforts are required for data management?
Answer (c): Information system helps the company to solve logistics problems.
Answer (d): Mathematical and statistical models.
Question 6: For the following companies, suggest the types of data that should be collected to plan and control their
supply chain:
(a) A hospital
(b) A city government
(c) A tire manufacturer
(d) A retailer of general merchandise
(e) An ore-mining company
Answer (a): A hospital = Number of customers, rooms, patients, doctors, etc.
Answer (b): A city government = Number of customers, employees, etc.
Answer (c): A tire manufacturer = Number of orders, sales, location, etc.

13. Answer (d): A retailer of general merchandise = Information of items, inventory, etc.
Answer (e): An ore-mining company = Expected demand, logistics, amount of minerals, etc.
Home Work 6
Chapter 6: Transport Fundamentals
Question 14: A power company in Missouri can buy coal for its generating plants from western mines in
Utah or from eastern mines in Pennsylvania. The maximum purchase price for coal of $20 per ton at the Missouri
plant is set according to the price of competing energy forms. The cost to mine coal in the West is $17 per ton and
in the East is $15 per ton. Transportation cost from the eastern mines is $3 per ton. What is the value of
transportation from the western mines?
Solution:
The cost to mine coal + transportation cost should be ≤ $20 per ton.
For East:
Cost to mine coal in the East = $15
Transportation cost in the East = $3
Total cost in the East = $15 + $3 = $18
For West:
Cost to mine coal in the West = $17
Transportation cost in the West = $x
Total cost in the West = $17 + $x
The maximum purchase price for coal is decided by competition.
Now, total cost in the West must be as follow:
$17 + $x ≤ $18
Transportation cost in the West must be as follow:
$x ≤ $18 - $17
$x ≤ $1
Therefore, Transportation cost in the West must be ≤ $1, if we want to purchase coal from West.
Question 15: Shipments for a certain product originate at point X and are to be sent to points Y and Z. Y is
an intermediate point to X and Z. The rate of Y is $1.20 per cwt., but due to competitive condition at Z, the rate of Z
is $1.00 per cwt. (hundredweight). Apply the principle of blanketing back, and explain how it eliminates rate
discrimination.
Solution:
Blanketing is a form of rate discrimination, but the benefits of rate simplification for both carriers and shippers
outweigh the disadvantages.
X----------------Y--------------------Z
X to Y = $1.20 Y to Z = $1.00
Cost 1 Cost 2
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

14. is $1 per cwt. By blanketing the rate to Z on intervening points, no intervening point is discriminated against in
terms of rates.
Question 16: Using Table 6-4, 6-5, and 6-6, determine the freight charges for the following shipments:
a) A 2500-lb shipment of paper place mats with printed advertising moving from New York to Los Angeles.
b) A 150-lb shipment of rubber displays for advertising purposes moving from New York to Providence, RI.
c) A 27000-lb shipment of emery cloth in packages moving from Louisville, Kentucky, to Chicago, Illinois. Note:
For any product classification number below 50, use 50 in Table 6-6.
d) A 30000-lb shipment of cat accessories at a density of 10 lb. per cubic foot moving between Louisville,
Kentucky, and Chicago, Illinois.
e) A 24000-lb shipment of advertising circulars not printed on newsprint moving between Louisville, Kentucky, and
Chicago, Illinois. A rate discount of 40 percent is offered.
Solution (a): From Table 6-4, the item number for place mats is 4745-00. Classification is 100, and minimum
weight is 20000 lb. The 2500 lb. is less than minimum weight of 20000 lb. for a truckload shipment. From the Table
6-5, for Los Angeles, the rate for shipment of ≥ 20000 is 87.27 cents/cwt. The shipping charges are as follows:
25 cwt * $87.27 = $2181.75
Solution (b): From Table 6-4, the item number for rubber displays for advertising purposes is 4980-00.
Classification is 100, and minimum weight is 20000 lb. The 150 lb. is less than minimum weight of 20000 lb. for a
truckload shipment. From the Table 6-5, for Providence, RI, the rate for minimum shipment is 93.51 cents/cwt. The
shipping charges are as follows:
1.5cwt * $93.51 = $140.265
The rate for < 500 lb. is 54.01 cents/cwt. The shipping charges using the < 500 lb. are as follows:
$54.01 * 1.5 cwt = $81.02
$81.02 is less than $140.265. We should pay the minimum charges.
Solution (c): From Table 6-4, the item number for emery cloth in packages is 2055-00. Classification is 55 for Less
Than Truckload (LTL), and 37.5 for Truckload. Minimum weight is 36000 lb. There are following three options
(Using Table 6-6):
 Ship LTL at class 55 and 27000 lb. shipment. ($5.65/cwt*270 = $1525.50)
 Ship at class 55 and 30000 lb. rate. ($3.87/cwt*300 = $1161) = Lowest cost
 Ship at class 37.5 and 36000 lb. rate. ($3.70/cwt*360 = $1332.50)
Solution (d): From Table 6-4, the item number for cat accessories is 2070-00. Classification is 65 (see 2070-07: 10
but less than 12). Using Table 6-6, the rate at 30000 lb. is $4.21/cwt, and shipping charges are as follows:
4.21*300 = $1263
Solution (e): Using Table 6-4, not printed on newsprint are classified 55 (4860-00) for a Truck load of 24000 lb.
Calculate break weight as follows.
𝐵𝑟𝑒𝑎𝑘 𝑤𝑒𝑖𝑔ℎ𝑡 =
𝑅𝑎𝑡𝑒 𝑁𝑒𝑥𝑡 × 𝑊𝑒𝑖𝑔ℎ𝑡 𝑁𝑒𝑥𝑡
𝑅𝑎𝑡𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡
=
3.87 × 30000
5.65
= 20549 𝑙𝑏.
Since current shipping weight of 24000 lb. exceeds the break weight, ship as if 30000 lb. Hence,
3.87*300 = $1161. Now, discount the charges by 40 percent as follows:
$1161 * (1-0.40) = $696.60
Question 21: A traffic manager has two options in scheduling a truck to make multiple pickups and
deliveries. The pick-up delivery problem is shown pictorially in Figure 6-11. The traffic manager can ship the
accumulated volumes as single shipments between the designated points or can use the stop-off privilege at $25 per
stop for any or all portions of the trip. If the traffic manager wishes to minimize shipping costs, which alternative
should be chosen? Assume that the final destination point incurs the stop-off change?

15. Solution:
In Figure 6.11, there are two stops (i.e. A and B) where we can load the total 40,000 lb. quantity in such a way that:
 A = (25000 lb. for C and D)
 B = (15000 lb. for C)
In Figure 6.11, there are two stops (i.e. C and D) where we can deliver the total 40,000 lb. quantity in such a way
that:
 C = (18000 lb. from A or B)
 D = (22000 lb. from A)
According to Figure 6.11, alternative 1 is without stop-off.
Alternative 1: Separate shipment (without stop-off privilege)
Loading/unloading Route Rate $/cwt.
cwt Rate
$/cwt.
Stop-off
charge
Charges
22000 A→D $3.20 $0.032 - $704
3000 (i.e. 25000-22000) A→C $2.50 $0.025 - $75
15000 B→C $1.50 $0.015 - $225
Total charges $1,004
Alternative 2: With stop-off privilege
Loading/unloading Route Rate $/cwt.
cwt Rate
$/cwt.
Stop-off
charge
Charges
25000 A→B $1.20 $0.012 - $300
40000 (i.e. 25000+15000) B→D $2.20 $0.022 - $880
Stop-off @ C - - $25 $25
Stop-off @ D - - $25 $25
Total charges $1,230
Alternative 3: With stop-off privilege

16. Loading/unloading Route Rate $/cwt.
cwt Rate
$/cwt.
Stop-off
charge
Charges
40000 A→D $3.20 $0.032 - $1,280
Stop-off @ B - - $25 $25
Stop-off @ C - - $25 $25
Stop-off @ D - - $25 $25
Total charges $1,355
Various other alternatives can be checked. In the present cases, there appears to be no advantage of stop-off
privilege.
Home Work 7
Chapter 7: Transport Decisions
Question 1: The Wagner Company supplies electric motors to Electronic Distributors, Inc.on a delivered-price
basis. Wagner has the responsibility for providing transportation. The traffic manager has three transportation
service choices for delivery----rail, piggyback, and truck. He has compiled the following information.
Transport Mode Transit Time, Days Rate, $/Unit Shipment Size, Units
Rail 16 25.00 10,000
Piggyback 10 44.00 7,000
Truck 4 88.00 5,000
Electronic Distributors purchases 50,000 units per year at a delivered contract price of $500 per unit. Inventory-
carrying cost for both companies is 25 percent per year. Which mode of transportation should Wagner select?
Solution:
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.
We use following formulas:
 Transport cost = transportation rate * annual demand
 Item value at buyer's inventory = price per unit
 Item value at vendor's inventory = price per unit – transport cost per unit (e.g. 500– 25 = 475 for rail)
 In-transit inventory cost= Inventory carrying cost * item value at vendor's inventory* annual demand *
(time in transit /365)
 Inventory cost of Wagner (supplier) = Inventory carrying cost * item value at vendor's inventory*
(shipping quantity/2)
 Inventory cost of Electronic Distributors (Buyer) = Inventory carrying cost * item value at buyer's inventory*
(Shipping quantity /2)
Rail Piggyback truck
Transportation rate 25 44 88
Shipping quantity, units 10000 7000 5000
Transit Time, Days 16 10 4
Transport cost 1250000 2200000 4400000

17. In-transit inventory cost 260274 156164 56438
Inventory cost of Wagner (supplier) 593750 399000 257500
Inventory cost of Electronic Distributors (Buyer) 625000 437500 312500
Total Cost 2729024 3192664 5026438
Mode of transportation Rail Piggyback truck
Rail has the lowest total cost.
Question 2: Suppose two truck services are being considered for deliveries from a company plant to one of its
warehouses. Service B is cheaper but slower and less reliable than service A. The following information has been
assembled:
Demand (known) 9600 cwt./year
Order cost $100/order
Production price, f.o.b. source $50/cwt
Shipping quantity As per EOQ
In-transit carrying cost 20%/year
Inventory carrying cost 30%/year
In-stock probability during the lead time 30%
Out-of-stock costs Unknown
Selling days 365 days/year
Service A Service B
Transit time (LT) 4 days 5 days
Variability (std. dev., sLT) 1.5 days 1.8 days
Rate $12.00/cwt $11.80/cwt
A reorder point control method of inventory control is used at the warehouse. From the point of view of the
inventory in the warehouse, which truck service should be selected? (Note: Refer to Chapter 9 on inventory
management for discussion of the reorder point method of inventory control. Hint: The standard deviation of the
demand-during-lead-time distribution is s’ = d(sLT), where sLT is assumed the transit time variability and d is the
daily demand rate.)
Solution:
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.

18. Now, we use following formulas:
 Transport cost = transportation rate * annual demand
 In-transit inventory cost = Inventory carrying cost * Production price * annual demand * (time in transit/365)
 Inventory cost of Plant = In-transit carrying cost * Production price * (shipping quantity for plant/2)
 Inventory cost of Warehouse = Inventory carrying cost * Production price * (Shipping quantity /2) + Inventory
carrying cost * Production price * Safety stock (r).
Annual demand 9600
Inventory-carrying cost 0.30
Production price 50
In-transit carrying cost 0.20
Shipping quantity for plant 357.8
A B
Transportation rate 12 11.8
Shipping quantity, units 321.3 321.8
Transit Time, Days 4 5
Safety stock (r) 50.5 60.6
Transport cost 115200 113280
In-transit inventory cost 1052 1315
Inventory cost of Plant 2684 2684
Inventory cost of Warehouse 3927 4107
Total Cost 122863 121385
Service B appears to be less expensive.

19. Question 3: The Transcontinental Trucking Company wishes to route a shipment from Buffalo to Duluth over
major highways. Because time and distance are closely related, the company dispatcher would like to find the
shortest route. A schematic network of the major highway links and mileage between city pairs is shown below (i.e.
Figure 7-20 in Book). Find the shortest route through the network by using the shortest route method.
Solution:
Using the shortest route method, results are show in table below. The shortest route is defined by tracing the links
from the destination node. They are shown as A→D → F→ G for a total distance of 980 miles.
Question 4: The U.S. Army Materiel Command was preparing final arrangements to move its M113 Full-
Tracked Armored Personnel Carrier from various subcontractors’ manufacturing facilities to intermediate storage
facilities at Letterkenny, Pennsylvania, for those units destined for Europe and for several army bases within the

20. United States. The production schedule for December plus the units on hand at the plant and requirements for
December are shown as follows:
Production schedule for December
Cleveland, OH 150 units plus 250
South Charleston, WV 150
San Jose, CA 150
Requirements for December
U.S. Army, Europe via Letterkenny, PA 300 units
Fort Hood, TX 100
Fort Riley, KS 100
Fort Carson, CO 100
Fort Benning, GA 100
Figure 7-21 (see Book) shows the location of the supply and demand points as well as the per-unit transportation
costs from supply to demand points. Find the least-costly delivery plan for December to meet the requirements, but
do not exceed the production schedule requirements.
Solution:
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:
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
as follows:
Question 5: The Evansville Local School District provides bus transportation for its elementary
schoolchildren. One bus has been assigned to the neighborhood, as defined in Figure 7-22 (see Book). Each year
there is a new roster of children, and the location of the stops for them can be plotted on a map. Sequencing the stops

21. determines the time and distance required to complete the bus tour. Using your best cognitive skills, design the
shortest bus tour possible given the following conditions:
 Only one bus is to be used.
 The bus starts at the elementary school and returns to it.
 Each stop is to be visited.
 Children may be picked up or dropped off from either side of the street.
 A pickup or dropoff at a corner may be made from either adjacent street.
 No U-turns are permitted.
 The bus has adequate capacity to transport all the children on the route.
Use a ruler or linear grid to determine the total distance for the bus tour.
Solution:
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 distance are added to the database to represent
actual distances traveled or to block infeasible paths from occurring. A reasonable routing plan is shown in the
Figure below:
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. The ROUTER database that generates it is given below.

23. Question 6: Dan Pupp is a jewelry salesperson who calls on store accounts in the Midwest. One of his
territories is shown in Figure 7-23 (see Book). His mode of operation is to arrive at the territory the night before
making his calls to stay at one of the local motels. He covers the region in two days and leaves the morning of the
third day. Since he pays his own expenses, he would like to minimize his total costs for serving the accounts.
Accounts 1 to 9 are covered the first day, and the remainder are visited the second day. He would like to compare
two strategies:
Strategy 1: Stay at Motel M2 all three nights at $49.00 per night.
Strategy 2: Stay at Motel M1 to visit accounts 1 to 9, staying there two nights at $40.00 per night. Then, stay at
Motel M3 one night at $45.00 per night to visit accounts 10 to 18. After visiting accounts 1 to 9, he returns to M1 to
stay overnight before moving to M3. He then stays overnight at M3 before moving on the next morning. The distance
between M1 and M3 is 36 miles.
Disregard any distance that he travels to and from the territory. Dan figures mileage costs at $0.30 per mile.
Which strategy seems best for Dan?
Solution:
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 as
follows:

24. Question 7: A bakery delivers daily to five large retail stores in a defined territory. The driver person for the
bakery loads goods at the bakery, makes deliveries to the retail stores, and returns to the bakery. A diagram for the
territory is shown in Figure 7-24 (see Book). The associated network travel times in minutes are:
To  Bakery 1 2 3 4 5
From
Bakery 0 24 50 38 55 20
1 22 0 32 23 45 18
2 47 35 0 15 21 60
3 39 27 17 0 14 25
4 57 42 18 16 0 42
5 21 16 57 21 41 0
(a) What is the best routing sequence for the delivery truck?
(b) If loading or unloading times are significant, how might they be included in the analysis?
(c) Retail store 3 is located in such a densely populated urban area that the travel times to and from this point may
increase by as much as 50 percent, depending on the time of the day. Travel times for other points remain relatively
unchanged. Would the solution in part (a) be sensitive to such variations?
Solution:
(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 as follows:
Bakery →Stop 5 → Stop 3→ Stop 4→ Stop 2→Stop 1→Bakery (with a tour time of 130 minutes)
(b) Loading/unloading times may be added to the travel times to a stop. Problem may then be solved as in part a.
(c) Travel times between stop 3 and all other nodes are increased by 50 percent. Remaining times are left
unchanged. Optimizing on this matrix shows no change in the stop sequence. However, tour time increases to
147.50 minutes.

25. Question 8: Sima Donuts supplies its retail outlets with the ingredients for making fresh donuts. A central
warehouse from which trucks are dispatched is located in Atlanta. Trucks can leave the Atlanta warehouse as early
as 3 A.M. to make pallet-load deliveries to the Florida market and make returns at any time. The trucks may also
pick up empty containers and supplies from vendors in the general area. Pickups are allowed only after all deliveries
on a route are made. A simple linear grid is placed over the Georgia-Florida area and grid coordinates found for
warehouse, retail, and vendor sites. The 0, 0 coordinates are in the northeast corner. For example, the Atlanta
warehouse is located at X = 2084, Y = 7260. The scaling factor on the map, including a road circuitry factor, is
0.363. The total time on a route may be 40 hours and total route distance may be up to 1400 miles. Team drivers are
used so that no overnight breaks are required, but one-hour rest breaks are allowed at 12-noon and 8 P.M. each day.
Average driving speed is taken to be 45 miles per hour. Additional data about the stops are as follows:
No Stop location Stop
type
Volume,
Pallets
X-
Coordinate
Y-
Coordinate
Loading/ Unloading
Time (Min.)
Time Windows
Open Close
1 Tampa, FL Delivery 20 1147 8197 15 6:00 A.M. 12:00 A.M.a
2 Clearwater, FL Pickup 14 1206 8203 45 6:00 A.M. 12:00 A.M.
3 Daytona Beach, FL Delivery 18 1052 7791 45 6:00 A.M. 12:00 A.M.
4 Fort Lauderdale, FL Delivery 3 557 8282 45 3:00 A.M. 12:00 A.M.
5 North Miami, FL Delivery 5 527 8341 45 6:00 A.M. 12:00 A.M.
6 Oakland Park, FL Pickup 4 565 8273 45 3:00 A.M. 12:00 A.M.
7 Orlando, FL Delivery 3 1031 7954 45 3:00 A.M. 12:00 A.M.
8 St Petersburg, FL Pickup 3 1159 8224 45 3:00 A.M. 12:00 A.M.
9 Tallahassee, FL Delivery 3 1716 7877 45 10:00 A.M. 12:00 A.M.
10 West Palm Beach, FL Delivery 3 607 8166 45 6:00 A.M. 12:00 A.M.
11 Miami-Puerto Rico Delivery 4 527 8351 45 6:00 A.M. 12:00 A.M.
Total 80
a
Midnight
There are three trucks with a 20-pallet capacity, one with a 25-pallet capacity, and one with a 30-pallet capacity. The
cost for drivers and truck is $1.30 per mile. Design the routes for this set of deliveries and pickups. Which trucks are
to be assigned to which routes? What is the dispatch plan? What is the cost for dispatch?
Solution:
This may be solved by using the ROUTER module in LOGWARE. The screen set up for this is as follows:

26. Making a run with ROUTER, we have route deign as given in the following Figure:

27. Question 9: Queens Lines operates a fleet of tanker ships to transport crude oil worldwide. One scheduling
problem concerns the movement of oil from Middle East ports to four European ports in England, France, and
Belgium. The sailing time in days between ports is
European Discharge ports
Middle east ports A B C D
1 20 18 12 9
2 17 14 10 8
Within the next three months, deliveries are to be made according to the following schedule:
From loading port 1 2 1 2 1 2
At discharge port D C A B C A
Day 19 15 36 39 52 86
Assume that ships are available to start anywhere and can end up at any port.
How many ships are needed to meet the schedule, and how should they be deployed? (Hint: Requires solving the
linear programming transportation problem).
Solution:
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

28. 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 below:
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.
Question 10: The Maxim Packing Company is considering a freight consolidation program for serving the
Kansas market. The program would involve the small-volume customers located at Hays, Manhattan, Salina, and
Great Bend. The proposal is to hold all orders from these areas for several weeks in order to realize lower
transportation charges. Assume all orders now shipped LTL directly from Fort Worth, Texas, to their destination in
Kansas. Average biweekly orders from the Kansas territory are as follows:
Hays 200 cases
Manhattan 350
Salina 325
Great Bend 125
The average case weighs 40 pounds. Orders could be shipped in the biweekly period that they are received, held and
shipped after two biweekly order periods, or held and shipped after three biweekly order periods. The potential loss
of sales has been estimated at $1.05 per case for each additional biweekly period that orders are held. Transportation
rates to Kansas are given in Table 7-7. Should the program be implemented? If so, how long should orders be held
before shipment?

29. Table 7-7: Trucks rate between Fort Worth Texas, and selected destination points in Kansas
From Fort
Worth to
Rates ($/cwt.)
Truck AQa
≥ 10000 lb. ≥ 20000 lb. ≥ 40000 lb.
Hays 12.78 5.19 4.26 3.06
Manhattan 12.78 5.19 4.26 2.22
Salina 10.26 4.08 3.42 2.46
Great Bend 12.27 4.98 4.08 2.94
a
Any quantity less than 10000 lb.
Solution:
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.

30. Question 11: The Sunshine Bottling Company bottles soft drinks that it distributes to retail outlets from nine
warehouses in the Michigan area. A single bottling plant is located in Flint, Michigan. The product is shipped from
plant to the nine warehouses in full truckload quantities. The typical plat-to-warehouse movement is to transport a
trailer of palletized soft drink to the warehouse, drop off the loading trailer, and bring a trailer of empty pallets back
to the plant. The unloading and hitching of the trailer at the warehouse takes 15 minutes. Since the routes are
traveled frequently, travel times, unloading times, and break times are known with a greater deal of certainty. The
number of trips needed to meet demand and the route times for a typical week are as follows (see Table below).
Distance,
Miles
Number of
Weekly Trips
Driving Time,
HR.a
Unloading
Time, hr.
Break/Lunch
time, hr.
Total Route
Time, hr.
Flint 20 43 1.00 0.25 0 1.25
Alpena 350 5 9.00 0.25 1.25 10.50
Saginaw 80 8 2.00 0.25 0 2.25
Lansing 118 21 3.25 0.25 0.25 3.75
Mt. Pleasant 185 12 4.50 0.25 0.75 5.50
W. Branch 210 5 5.00 0.25 0.75 6.00
Pontiac 90 43 2.50 0.25 0 2.75
Traverse City 376 6 9.00 0.25 1.25 10.50
Petoskey 428 5 10.00 0.25 1.50 11.75
a
Round trip time
It is desirable to schedule trucks to leave the plant at or after 4:00 A.M. and to return no later than 11:00 P.M. the
same day. Unloading can take place only when the warehouse is open, which is from 6:30 A.M. to 11:00 P.M.
By sequencing the routes to maximize truck utilization, determine the minimum number of trucks that is needed to
serve all the routes. The company was using 10 trucks.
Solution:

31. Routes are built by placing the trips end-to-end throughout the day from 4 a.m. until 11 p.m., 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 operate. If there were no delivery time restrictions on warehouses and trips could be 29
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 as follows:
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.
Although this schedule meets the requirements of the problem, it might be improved by better balancing the
workload across the trucks and the days.

32. Question 12: The Nockem Dead Casket Company supplies funeral homes with caskets throughout the state of
California. The funeral homes for a particular warehouse territory are located as shown on the map in Figure 7-25
(see Book).
(a) Suppose that the funeral home locations (•) and associated number of caskets for each funeral home represents a
single, daily dispatch. If the company has six trucks with capacities of 20 caskets each, develop a routing plan using
the “sweep” method. (Use a counterclockwise sweep with a due north start.) Place your design on the map. How
many trucks are actually used and what is the total travel distance for the route design? You may scale distances
from the diagram.
(b) Appraise the sweep method as a good method for truck routing and scheduling.
Solution:
(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.
Question 13: As an adjunct to its retail business, Medic Drugs fills prescriptions for outlying nursing homes,
extended cares facilities, rehabilitation centers, and retirement homes. Part of this service is to deliver the
prescription order to the customer site. Station wagons having a capacity of 63 cartons are used for delivery.
Customer locations are geocoded by a linear grid overlay with a map-scaling factor of 4.6 per coordinate unit.
Customer data for a typical delivery day is given in the Table below (see full Table 7-8 in Book). The 0, 0
coordinates on the grid are in the southwest corner.

33. Deliveries may begin as early as 8 A.M. (drivers leave depot) and drivers are to return to the base pharmacy by 6
P.M. Average driving speed is 30 miles per hour. Drivers are allowed a one-hour lunch break after 12 noon. Most
customers may receive their deliveries between 9 A.M. and 5 P.M., although there are a few exceptions. The base
pharmacy is located at X (horizontal coordinate) = 13.7, Y (vertical coordinate) = 21.2. If a driver returns early to
the base pharmacy, the station wagon may be reloaded and sent on a second route.
(a) Design a dispatch route plan that will minimize the total distance traveled.
(b) Can any routes be assigned to the same station wagon to reduce the total number of drivers and vehicles needed
to service the customers? If not, is there anything that might be done to accomplish this?
Solution:
This problem may be solved with the aid of ROUTER in LOGWARE. The model input data may be formatted as
shown in the Figure below.
Stop Volume X Y Unloading Time windows
Customer location type (cartons) coordinate Coordinate time (min) Open Close
Covington House D 1 23.4 12.9 2 9am 5pm
Cuyahoga Falls D 9 13.4 13.4 18 9am 5pm
Elyria D 1 6.3 16.8 5 9am 5pm
⁞ ⁞ ⁞ ⁞ ⁞ ⁞ ⁞ ⁞

34. (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 below.
A summary for these routes is given in following partial output report.
(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 3:30-3:45 p.m. and still meet the
customer’s time windows and return to the depot by 6 p.m. Thus, only three drivers and station wagons are actually
needed for this problem.

35. Question 14: The Nockem Dead Casket Company sells and distributes caskets to funeral homes in the
Columbus, Ohio, region. Funeral homes place orders at a warehouse (X = 7.2, Y = 8.4) for delivery throughout the
week. The funeral home locations and the weekdays for delivery are given in Figure 7-26 (see Book). Number of
caskets and the funeral home coordinates are given in Table 7-9. Deliveries are made using one 18-casket truck and
one 27-casket truck. Trucks leave the warehouse to make deliveries and return the same day.
Using the Principle for Good Routing and Scheduling, develop a good routing plan for the company. Be creative.
Solution:
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.
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.