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Case study of Supply Chain and Operation Management Email: [email protected] This article discusses solution of Inventory and Aggregate Planning based on case in the book of supply chain by S. Chopra and Operation Management by J Heizer

Case 1 Motorola obtains cell phones from its contract manufacturer located in China to supply the U.S. market, which is served from a warehouse located in Memphis, Tennessee. Daily demand at the Memphis warehouse is normally distributed, with a mean of 5,000 and a standard deviation of 4,000. The warehouse aims for a Type I CSL of 99 percent. The company is debating whether to use sea or air transportation from China. Sea transportation results in a lead time of 36 days and costs $0.50 per phone. Air transportation results in a lead time of 4 days and costs $1.50 per phone. Each phone costs $100, and Motorola uses a holding cost of 20 percent. Assume that Motorola takes ownership of the inventory on delivery. Assume that Motorola follows a periodic review policy. Given lot sizes by sea and air, Motorola would have to place order every 20 days using sea transport but could order daily using air transport a. Assume that Motorola follows a periodic review policy. What Order up to level (OUL) and safety inventory should the warehouse aim for when using sea or air transportation? How many days of safety inventory will Motorola carry under each policy? b. How many days of cycle inventory does Motorola carry under each policy? c. Under a periodic review policy, do you recommend sea or air transportation? Answer: Given

Average Daily Demand (D) Minimum Lot Size Q Standard deviation 𝜎𝐷 Holding Cost 20% CSL=99%=Z=Norm.inv Order frequency Unit cost (C) Order cost (S) Lead time (L)

Sea Transport

Air Transport

100.000

5.000

5.000 4000 0.2 2.33 every 20 days

every day

$ 0.5 per phone 36 days

$1.5 per phone 4 days

$100

Safety Stock= Z. 𝝈𝑫 .√𝐋 ROP= Average daily demand x Lead time + Z. 𝝈𝑫 . √𝐋 Sea Transport

Air Transport

Safety Stock =2.33 x 4000 x √36 = 55.920 unit

Safety Stock =2.33 x 4000 x √4 = 18.640 unit

ROP = 5.000 x 36 + Safety stock = 180.000 + 55.920 = 235.920 unit

ROP = 5.000 x 4 + Safety stock = 20.000 + 18.640 = 38.640 unit

Cycle Inventory =

𝑄 2

=

100.000 2

=

Cycle Inventory =

𝑄 2

=

5.000 2

= 2.500

50.000 unit/ 20 days = 2.500 unit /day

unit/day

Total Inventory = 55.920 + 2.500 = 58.420 Total cost = 58.420 x $ 0.5= $ 29.210

Total Inventory = 18.640 + 2.500 = 21.140 Total cost = 21.140 x $ 1.5= $ 31.710

Motorola should use Sea transport because it will give less cost compare to Air transport

Case 2 TopOil, a refiner in Indiana, serves three customers near Nashville, Tennessee, and maintains consignment inventory (owned by TopOil) at each location. Currently, TopOil uses TL transportation to deliver separately to each customer. Each truck costs $800 plus $250 per stop. Thus delivering to each customer separately costs $1050 per truck. TopOil is considering aggregating deliveries to Nashville on a single truck. Demand at the large customer is 60 tons a year, demand at the medium customer is 24 tons per year, and demand at small customer is 8 tons per year. Product cost for TopOil is $10,000 per ton, and it uses a holding cost of 25 percent. Truck capacity is 12 tons. a. What is the annual transportation and holding cost if TopOil ships a full truckload each time customer is running out of stock? How many days of inventory is carried at each customer under this policy? b. What is the optimal delivery policy to each customer if TopOil aggregates shipments to each of the three customers on every truck that goes to Nashville? What is the annual transportation and holding cost? How many days of inventory are carried at each customer under this policy? c.what is the optimal delivery price to each customer if TopOil aggregates each shipments to each of the three customers on every truck that goes to nashville? what is the total annual transportatioin and hlding cost? how many days of inventory are carried at each customer under this policy? Answer: given

Demand (D) Order cost (S) $800+$250 Holding cost as a friction (h) Cost per unit (C) Truck Capacity: 12 tons

Answer point B 2.𝐷.𝑆

EOQ (Q)= √ ℎ.𝐶

Cycle Inventory = 𝑄 2

Order frequency (n)

𝐷 𝑄 Annual holding cost

𝑄 (ℎ). (𝐶) 2

Small Customer 8 tons/year $1050 0.25 $10.000/ton

Small Customer 2.(8).(1050)

Medium Customer 24 tons/year $1050 0.25 $10.000/ton

Medium Customer 2.(24).(1050)

Large Customer 60 ton/year $1050 0.25 $10.000/ton

Large Customer 2.(60).(1050)

𝑄 = √0.25(10.000) =2.5

𝑄 = √ 0.25(10.000) =4.4

𝑄 = √ 0.25(10.000) =7

2.5 = 1.25 2

4.4 = 2.2 2

7 = 3.5 2

8 𝑡𝑜𝑛/𝑦𝑒𝑎𝑟 =3 2.5 𝑡𝑜𝑛

24 𝑡𝑜𝑛/𝑦𝑒𝑎𝑟 =5 4.4 𝑡𝑜𝑛

60 𝑡𝑜𝑛/𝑦𝑒𝑎𝑟 =8 7 𝑡𝑜𝑛

1.25 (0.25).($10.000) = $3125

2.2 (0.25).($10.000) = $5.500

3.5 (0.25).($10.000) = $8.750

Annual order cost

𝐷 (𝑆) 𝑄 Average flow time

𝑄 2𝐷

3.($1050) = $3150

2.5

5.($1050) = $5250

4.4

=0.15/year

=0.09/year

7

=0.05/year

2.(8)

2.(8)

2.(8)

=8/week

=4/week

=2/week $17250

$10850

1.25(0.25)($10.000)+ 3($1050) = $6275

Annual Cost= Cyc.inv.(h)(C) + order freq.(order cost)

8.($1050) = $8400

TC= $34375 Answer point C S=$800, S1=S2=S3=$250 𝑆 ∗=S+S1+S2+S3 =𝑺∗ =$1550

Ʃ 𝑫𝒊(𝒉)(𝑪𝒊) 𝟐𝑆 ∗

n=√

𝟖(𝟎.𝟐𝟓)(𝟏𝟎.𝟎𝟎𝟎)+𝟐𝟒(𝟎.𝟐𝟓)(𝟏𝟎.𝟎𝟎𝟎)+𝟔𝟎(𝟎.𝟐𝟓)(𝟏𝟎.𝟎𝟎𝟎) 𝟐𝑆 ∗

=√

= 𝟖. 𝟔 times/year

Annual order cost = 8.6 x $1550 = $13330

Small Customer 𝐷

Q= 𝑛

8 8.6

= 0.9 ton/order

0.9 2(8)

= 0.05 = 3 weeks

Annual hold cost 0.45.(0.25)($10000)= ℎ(𝐶)𝑄𝑖 $1125 2 Cycl.iv x hold cost TC=$26455

Large Customer

24

60

8.6

= 2.7 ton/order

0.9 = 0.45 2

𝑄

Cycle inv= 2

Avergflowtime 𝑄 2𝐷

Medium Customer

8.6

= 6.9 ton/order

2.7 = 1.35 2 0.9 2(24)

= 0.05

= 3 weeks

$3375

6.9 = 3.45 2 6.9 2(60)

= 0.05

= 3 weeks

$8625

Quantity order = Qsmall+Qmed+Qlarge =0.9+2.7+6.9=10.5 tons/order Quantity order