Agenda Kuliah Review Case Study Transport Fundamental

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1 Agenda Kuliah Review Case Study Transport Fundamental
Transport Decission Case Study Software Logware  Router Fowler Distributing Company

2 Performance - Average transit time - Transit time variability -
Loss and damage - Other factors including availability, capability, frequency of movement, and various less tangible services

3 Importance of Modes By Products Hauled · Air -- very high -
valued, time sensitive products Truck moderately high valued, time sensitive products. Many finished and semifinished goods Rail low valued products including many raw materials Water very low valued products moved domestically, high valued if moved internationally Pipe generally limited to petroleum products and natural gas

4 Basic Cost Trade-Off Example Finished goods are to be shipped from a plant inventory to a warehouse inventory some distance away. The expected volume to be shipped in a year is 1,200,000 lb. The product is worth $25 per lb. and the plant and carrying costs are 30% per year. Other data are: Transport choice Rate, $/lb. Transit time, days Shipment size, lb. Rail 0.11 25 100,000 Truck 0.20 13 40,000 Air 0.88 1 16,000

5 Include transport rate
Basic cost trade-offs  Transport Selection Analysis Cost Compu- type tation Rail Truck Air Trans- RD .11(1,200,000) .20(1,200,000) .88(1,200,000) portation = $132,000 = $240,000 = $1,056,000 In-transit ICDT 365 [.30(25) [.30(25) [.30(25) inventory 1,200,000(25)]/365 1,200,000(13)]/365 1,200,000(1)]/365 = $616,438 = $320,548 = $24,658 Plant ICQ 2 [.30(25) [.30(25) [.30(25) inventory 100,000]/2 40,000]/2 16,000]/2 = $375,000 = $150,000 = $60,000 Whse IC ' Q 2 [.30(25.11) [.30(25.20) [.30(25.88) inventory 100,000]/2 40,000]/2 16,000]/2 = $376,650 = $151,200 = $62,112 Totals $1,500,088 $ 861,748 $1,706,770 Improved service Include transport rate

6 Can be a weighted index of time and distance
Carrier Routing (Cont’d) Origin Amarillo Oklahoma City Destination Fort Worth A B E I C D G F H J 90 minutes 84 138 348 156 48 132 150 126 120 66 60 Note : All link times are in minutes 90 Can be a weighted index of time and distance

7 Shortest Route Method Step Solved Nodes Directly Connected to Unsolved
Its Closest Unsolved Node Total Cost Involved nth Nearest Minimu m Cost Its Last Connection a 1 A B 90 AB * 2 C 138 AC 90+66=156 3 D 348 E 90+84=174 174 BE F 138+90=228 4 228 CF I 174+84=258 5 =294 258 EI H 228+60=288 6 228+60= 288 FH J =384 7 294 CD G =360 288+48=336 8 =414 384 IJ Shortest Route Method

8 Routing from Multiple Points
This problem is solved by the traditional transportation method of linear programming Plant 1 Requirements = 600 Plant 2 Requirements = 500 Plant 3 Requirements = 300 Supplier A Supply  400 Supplier C Supply  500 Supplier B Supply  700 4 a 7 6 5 9 8 The transportation rate in $ per ton for an optimal routing between supplier A and plant 1 .

9 TRANLP problem setup Solution

10 Konsumen 1 2 3 4 Supply A B 10 8 5 7 C 6 Permintaan gudang

11 Konsumen 1 2 3 4 Supply A B 10 8 5 7 C 6 Permintaan gudang

12 Konsumen 1 2 3 4 Supply A B 10 8 5 6 C 7 Permintaan gudang

13 Konsumen 1 2 3 4 Supply A B 10 8 5 C 7 6 Permintaan gudang

14 Konsumen 1 2 3 4 Supply A B 10 8 5 C 7 6 Gudang

15 Konsumen 1 2 3 4 Supply A B 10 8 5 C 7 6 Gudang

16 Konsumen 1 2 3 4 Supply A B 10 8 5 C 7 6 Permintaan Gudang

17 Practical Guidelines for Good Routing and Scheduling

18 Practical Guidelines for Good Routing and Scheduling
1. Load trucks with stop volumes that are in closest proximity to each other D Depot Stops Depot (b) Better clustering (a) Weak clustering

19 May need to coordinate with sales to achieve clusters
Practical Guidelines for Good Routing and Scheduling 2. Stops on different days should be arranged to produce tight clusters F T D Depot (a) Weak clustering-- routes cross (b) Better clustering Stop May need to coordinate with sales to achieve clusters

20 Practical Guidelines for Good Routing and Scheduling
3. Build routes beginning with the farthest stop from the depot . Savings Method Observation The points that offer the greatest savings when combined on the same route are those that are farthest from the depot and that are closest to each other.

21 Practical Guidelines for Good Routing and Scheduling
4. The stop sequence on a route should form a teardrop pattern (without time windows) don’t cross 5. The most efficient routes are built using the largest vehicles available first 6. Pickups should be mixed into delivery routes rather than assigned to the end of the routes 7. A stop that is greatly removed from a route cluster is a good candidate for an alternate means of delivery 8. Narrow stop time window restrictions should be avoided (relaxed) --> negotiation

22 Savings Method Observation
The points that offer the greatest savings when combined on the same route are those that are farthest from the depot and that are closest to each other. This is a good principle for constructing multiple-stop routes

23 Methods for Routing and Scheduling

24 “Sweep” Method Solution
Geographical region Pickup points 1,000 4,000 2,000 3,000 2,000 3,000 3,000 2,000 Depot 1,000 2,000 2,000 2,000 Stop Volume and Location

25 “Sweep” Method Solution
Sweep direction is arbitrary Route #3 8,000 units Route #1 10,000 units 1,000 4,000 2,000 3,000 2,000 3,000 3,000 2,000 Depot 1,000 2,000 2,000 2,000 Route #2 9,000 units

26 The “Savings” Method for VRP
Stop d A,0 d 0,A A A d 0,A d A,B d 0,B Depot Depot d B,0 B d B,0 B Stop (a) Initial routing (b) Combining two stops on a route Route distance = d +d +d + d Route distance = d +d +d 0,A A,0 0,B B,0 0,A A,B B,0 “Savings” is better than “Sweep” method—has lower average error

27 besarnya saving(penghematan dengan menggabungkan beberapa stop),
The “Savings” Method for VRP Stop d A,0 d 0,A A A d 0,A d A,B d 0,B Depot Depot d B,0 B d B,0 B Stop (a) Initial routing (b) Combining two stops on a route Route distance = d +d +d + d Route distance = d +d +d 0,A A,0 0,B B,0 0,A A,B B,0 besarnya saving(penghematan dengan menggabungkan beberapa stop), S = dOA + dOB - dAB

28 Savings Method Observation
The points that offer the greatest savings when combined on the same route are those that are farthest from the depot and that are closest to each other. This is a good principle for constructing multiple-stop routes

29 Cost Matrix(Distance Matrix)

30 The Saving Matrix Kapasitas angkut 200

31 1,3,4 2 , 9 6,7,11,8 5,10,12,13

32 Route Sequencing in VRP
AM PM 8 9 10 11 12 1 2 3 4 5 6 Route #1 Route #10 Route #6 Truck #1 Route #9 Route #4 Truck #2 Route #5 Route #8 Truck #3 Route #2 Route #7 Truck #4 Route #3 Truck #5 Minimize number of trucks by maximizing number of routes handled by a single truck

33 Freight Consolidation
Inventory Consolidation Pengiriman dilakukan dalam jumlah besar, biaya transportasi turun tapi biaya inventory akan naik. Vehicle Consolidation Pickup and delivery dilakukan oleh kendaraan yang sama. Warehouse Consolidation Mengirim dalam jumlah besar dalam jarak jauh dan mengirim dalam jumlah yang kecil dalam jarak dekat Temporal Consolidation Order dari berbagai konsumen digabung menjadi satu pengiriman sehingga menghindari untuk mengirim setiap order dengan menggunakan kendaraan yang berbeda.

34 Freight Consolidation Analysis
Suppose we have the following orders for the next three days. Consider shipping these orders each day or consolidating them into one shipment. Suppose that we know the transport rates. From: Ft Worth Day 1 Day 2 Day 3 To: Topeka 5,000 lb. 25,000 lb. 18,000 lb. Kansas City 7,000 12,000 21,000 Wichita 42,000 38,000 61,000

35 Freight Consolidation Analysis (Cont’d)
Separate shipments Day 1 Day 2 Rate x volume = cost Topeka 3.42 x 50 = $171.00 1.14 x 250 = $285.00 Kansas City 3.60 x 70 = 1.44 x 120 = Wichita 0.68 x = 285.60 0.68 x 400 a = 272.00 Total $708.60 Total $729.80 a Ship 380 cwt., as if full truckload of 400 cwt. Day 3 Totals 1.36 x 180 = $244.80 $700.80 1.20 x 210 = 676.80 0.68 x 610 = 414.80 972.40 Total $911.60 $2,350.00

36 Freight Consolidation Analysis (Cont’d)
Consolidated shipment Computing transport cost for one combined, three-day shipment Day 3 Rate x volume = cost a Topeka 0.82 x 480 = $393.60 Kansas City 0.86 x = Wichita 0.68 x = 958.80 Total $1,696.40 a 480 = Cheaper, but what about the service effects of holding early orders for a longer time to accumulate larger shipment sizes?

37 Fowler Distributing Company

38 Truk berangkat antara jam 6.30 AM -8.00 AM
 (6 x 60) + 30 = 390

39 Istirahat makan siang boleh dimulai pada jam 11.30
Istirahat makan siang diberi waktu 30 menit

40 Istirahat makan siang hanya sekali , angka 9999 diberikan untuk memastikan bahwa waktu ini tidak akan pernah tercapai

41 kendaraan paling lama berada dalam suatu route adalah 10 jam (tidak termasuk istirahat) di suatu route adalah 10 jam  60 x 10 = 600 Istirahat makan 30 menit  total waktu berada di suatu route = = 630 Kendaraan berangkat paling awal jam 6.30 390 menit setelah midnight Lattest return time= = 1020 menit setelah pukul (midnight)

42 Presentase muatan setelah delivery
Truk berangkat membawa 190 unit Stop Order Presentase muatan setelah delivery 1 30 84,21 2 80 42,11 3 40 21,05 4 0,00 setelah muatannya tinggal berapa persenkah proses pickup boleh dilakukan? karena tidak ada aktivitas pickup dan juga tidak ada batasan maka diasumsikan proses pickup boleh dilakukan meski muatan masih 100%

43 SW

44 Tidak ada rumusan yang pasti tentang waktu yang diperlukan untuk loading dan unloading sehingga nilainya dibiarkan nol, loading dan unloading tidak “hanya’ ditentukan oleh banyaknya atau volume produk yang di load ataupun unload

45 Tidak dibatasi oleh volume produk, angka 9999 hanya untuk memastikan bahwa angka tersebut tidak akan terlewati atau dengan kata lain batasan ini sebenarnya tidak pernah membatasi

46 Nol karena kapasitas truk diasumsikan tidak dibatasi oleh volume produk

47 Fowler Distributing Company
Adakah solusi yang lebih baik? Berapakah jumlah truk yang diperlukan? Apakah ‘fair’ membandingkan solusi sekarang dengan kondisi existing?

48 Fowler Distributing Company
Konsekuensi apakah yang harus ditanggung oleh Roy akibat adanya time windows selain 8.00 am to pm? Apakah yang harus dilakukan oleh Roy?

49 Jumlah truk yang diperlukan bisa diturunkan dengan menegosiasikan time windows
AM PM 8 9 10 11 12 1 2 3 4 5 6 Route #1 Route #10 Route #6 Truck #1 Route #9 Route #4 Truck #2 Route #5 Route #8 Truck #3 Route #2 Route #7 Truck #4 Route #3 Truck #5 Minimize number of trucks by maximizing number of routes handled by a single truck

50 Fowler Distributing Company
Jika ada truk seharga $3500, dengan kapasitas 600, haruskah truk ini dibeli? Jika menggunakan truk ini biaya operasi akan naik $0.05 per miles

51 Fowler Distributing Company
Jika Roy dapat menggunakan jasa tranportasi luar untuk semua pelanggan yang permintaanya kurang dari atau sama dengan 50 dengan biaya $35 per pelanggan. Haruskah pilihan ini diambil?

52 Fowler Distributing Company
Serikat sopir menegoisasikan untuk regular time hanya 7,5 jam sehingga jika harus bekerja lebih dari jam itu maka harus dibayar lembur. Apakah dampaknya terhadap rancangan route dan biaya?

53 Fowler Distributing Company
Roy memiliki rencana untuk memindahkan warehousenya ke lokasi yang lebih sentral yaitu di koordinat x=20 y=25. Apakah ide ini seharusnya dilaksanakan? Jika biaya sewa bangunan sama dengan warehouse yang lama?

54 Fowler Distributing Company
Apakah yang harus diperhatikan jika menggunakan software ini untuk menentukan route pengiriman produk setiap hari? Masalah apakah yang mungkin akan muncul? Bagaimana mengatasi masalah tersebut?