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Metode Transportasi merupakan suatu metode yang digunakan untuk mengatur distribusi dari sumber-sumber yang menyediakan produk yang sama ke tempat tempat yang membutuhkan secara optimal deng…Full description
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Riset OperasiFull description
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ykjtrt
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Membahas mengenai: pengertian singkat dan menyeluruh dari penelitian operasional II, analisis jaringan, perencanaan & pengendalian proyek dengan CPM – PERT, programa dinamis, teori permainan…Full description
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riset operasiFull description
riset operasiFull description
MAKALAH “ MATA KULIAH : OPERATION RESEARCH” METODE TRANSPORTASI
P R OD IMA N A J E ME N F EU NI V E RS I T A SP E MB AN GU NA NP A NC ABU DI ME D A N2 0 1 7
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KATA PENGANTAR
Puji syukur kehadirat Allah SWT atas limpahan rahmat dan anugrah dari-Nya kami dapat menyelesaikan makalah tentang”MET!E TRANSPRT TRANSPRTAS"” ini# Sh$la%at dan salam sem$ga senantiasa ter&urahkan ter &urahkan kepada junjungan 'esar kita( Na'i Muhammad SAW yang telah menunjukkan kepada kita semua jalan yang lurus 'erupa ajaran agama islam yang sempurna dan menjadi anugrah ter'esar 'agi seluruh alam semesta# Kami sangat 'ersyukur karena dapat menyelesaikan makalah ini yang menjadi tugas# !isamping itu( kami mengu&apkan 'anyak terimakasih kepada semua pihak yang telah mem'antu kami selama pem'uatan makalan ini 'erlangsung sehingga dapat terealisasikanlah makalah ini# !emikian yang dapat kami sampaikan( sem$ga makalah ini dapat 'erman)aat 'agi para pem'a&a# Kami mengharapkan kritik dan saran terhadap makalah ini agar kedepannya kedepannya dapat kami per'aiki# Karena kami sadar( makalah yang kami 'uat ini masih 'anyak terdapat kekurangannya#
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BAB I PENDAHULUAN 1.1 Latar Belakang Permasalahan transportasi merupakan permasalahan yang sering terjadi dalam kehidupan sehari-hari. Transportasi merupakan bentuk khusus dari program linear yang digunakan untuk memecahkan masalah yang berhubungan dengan transportasi (pengangkutan) untuk meminimalkan biaya, jarak tempuh dan sebagainya sehingga dapat memaksimalkan laba/keuntungan yang diperoleh. Metode transportasi merupakan suatu metode yang digunakan untuk mengatur distribusi dari beberapa sumber ke tempat-tempat yang membutuhkan barang. Pendistribusian barang harus diatur sedemikian rupa, karena ada perbedaan jarak atau biaya dari sumber-sumber yang ada dan tempat-tempat yang yang membutuhkan barang. Tujuan dari masalah transportasi adalah untuk menentukan jumlah yang optimal dari barang yang akan diangkut dari berbagai sumber ke berbagai tujuan sehingga biaya transportasi total minimum. Jumlah yang tersedia di sumber-sumber dapat memenuhi jumlah yang diperlukan pada tujuan sesuai dengan kebutuhan/permintaannya. Dalam transportasi terdapat variabel yang perlu diminimalkan, variabel ini dapat berupa biaya, waktu atau keamanan pengirimannya. Dalam prakteknya parameternya dari transportasi yang berupa nilai permintaan dan persediaan tidak selalu dapat diketahui dengan pasti dan tidak selalu stabil. Ketidakpastian ini meliputi kurangnya informasi yang tepat dan sebagainya. Sebuah cara yang sering digunakan untuk menyatakan ketidaktepatan ini adalah bilangan fuzzy. Bilangan fuzzy dapat disajikan dalam bentuk trapezoidal. Mengambil keputusan adalah bagian dari kehidupan sehari-hari. Nyatanya, permasalahan yang terjadi sering terjadi perbedaan tujuan, kurangnya informasi.
1.2. Perumusan Masalah 1. Apa yang di maksud dengan metode transportasi? transportasi? 2. Bagaimana menyelesaikan masalah transportasi dengan menggunakan metode stepping stone?
1.3. Tujuan Penelitian Tujuan yang ingin dicapai dalam pembuatan sistem ini adalah 1. Mengimplementasikan Metode Me tode !ussel Apro"imation dan Metode #ogel Apro"imation untuk menyelesaikan masalah transportasi. 2. $ntuk membandingkan %aktu proses& jumlah iterasi dan total biaya minimal dari Metode !ussel Apro"imation dan Metode #ogel Apro"imation.
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BABII PEMBAHASAN 2!1! PENGERTIANMETODETRANSPORTASI M&$& T%#)-%$#-" adalah suatu met$de yang digunakan untuk mengatur distri'usi
dari sum'er 4 sum'er yang menyediakan pr$duk 4 pr$duk yang sama di tempat- tempat yang mem'ut mem 'utuhk uhkan an se&a se&ara ra $pt $ptima imal# l# Al Al$ka $kasi si pr$ pr$duk duk ini har harus us dia diatur tur sed sedemi emikia kian n rup rupaa kar karena ena terdapat per'edaan 'iaya transp$rtasi 5al$kasi6 dari suatu sum'er ke 'e'erapa tujuan yang 'er'eda 4 'eda dan dari 'e'erapa sum'er ke suatu tujuan juga 'er'eda 'er'eda 4 'eda# Ada tiga ma&am met$de dalam met$de transp$rtasi7
1# 1!
M&$& Stepping Stone M&$& Stepping
2# 2.
M&$& M" (Modified Distribution)
8# 3!
M&$&A M&$ &AM M *&'*&'- Approximation Approximation Method Method 7
Pada sesi ini hanya akan di'ahas mengenai met$de transp$rtasi dengan met$de stepping st$ne( sedangkan met$de M!" dan 9AM 9AM akan di'ahas pada sesi s esi tulisan yang lain# la in#
2!2 M&$& Stepping M&$& Stepping Stone
Met$de ini dalam meru'ah al$kasi pr$duk untuk mendapatkan al$kasi pr$duksi yang $ptimal menggunakan &ara trial and error atau atau &$'a 4 &$'a# Walaupun meru'ah al$kasi dengan &ara &$'a- &$'a( namun ada syarat yang harus diperhatikan yaitu dengan melihat pengurangan 'iaya per unit yang le'ih 'esar dari pada penam'ahan 'iaya per unitnya# ntuk memperm mem permuda udah h pen penjela jelasan san(( 'eri 'erikut kut ini aka akan n di' di'eri erikan kan se' se'uah uah &$n &$nt$h t$h## Sua Suatu tu per perusa usahaa haan n mempunyai tiga pa'rik di W( *( # !engan kapasitas pr$duksi tiap 'ulan masing- masing :; t$n( <; t$n( dan =; t$n> dan mempunyai tiga gudang penjualan di A( 0( ? dengan ke'utuhan tiap 'ulan masing- masing =; t$n( 11; 11; t$n( dan @; t$n# t$n# 0iaya pengangkutan setiap t$n pr$duk dari pa'rik W( *( *( ke gudang A( 0( ? adalah se'agai s e'agai 'erikut7
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Tentukan Te ntukan al$kasi hasil pr$duksi dari pa'rik 4 pa'rik terse'ut ke gudang 4 gudang penjualan dengan 'iaya pengangkutan terendah#
'olusi
1.1
Penyusunan tabel alokasi
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ij adalah 'anyaknya al$kasi dari sum'er 5pa'rik6 i ke tujuan 5gudang6 j# Nilai ijinilah yang akan kita &ari# 1!2
P%-&+% #'(#-"
Ped$man pr$sedur al$kasi tahap pertama adalah ped$man sudut 'arat laut (North West Corner Rule) yaitu Rule) yaitu pengal$kasian sejumlah maksimum pr$duk mulai dari sudut kiri atas 5 116 dengan melihat kapasitas pa'rik dan ke'utuhan gudang#
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0iaya Pengangkutan untuk al$kasi tahap pertama se'esar B =; 52;6 C @; 5=6 C <; 52;6 C 1; 51;6 C @; 51:6 B 82<;# 1!3
M&%+8#. #'(#-" -&#%# trial and error
Peru'ahan 'isa dari k$tak terdekat atau 'isa juga pada k$tak yang tidak 'erdekatan dengan melihat pengurangan 'iaya per unit yang le'ih 'esar dari pada penam'ahan 'iaya per unit# Misalnya akan di&$'a peru'ahan dari k$tak WA ke k$tak *A artinya =; t$n ke'utuhan gudang A akan dikirim dari pa'rik * dan 'uikan dari pa'rik W# Peru'ahan al$kasi pr$duk dari dua k$tak terse'ut akan mengaki'atkan 'eru'ahnya al$kasi pr$duk k$tak lainnya yang terkait 5k$tak *0 dan k$tak W06# ntuk itu se'elum dilakukan peru'ahan perlu dilihat penam'ahan dan pengurangan 'iaya 'iaya transp$rtasi per unitnya se'agai 'erikut7
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0iaya Pengangkutan untuk al$kasi tahap pertama se'esar B :; 5=6 C =; 51=6 C 1; 52;6 C 1; 51;6 C @; 51:6 B 22<;# Penam'ahan 'iaya7 dari W ke ? B 3 dari ke 0 B 1; C 13
Pengurangan 'iaya 7 dari W ke 0 B = dari ke ? B 1:C 2@
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0iaya Pengangkutan untuk per'aikan kedua se'esar B =; 5=6 C @; 53;6 C =; 51=6 C 1; 52;6 C =; 51;6 B 2;2;# Penam'ahan 'iaya7 dari W ke 0 B =
Pengurangan 'iaya 7 dari * ke 0 B 2;
dari * ke ? B 1; C
dari W ke ? B 3 C
1=
23
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0iaya Pengangkutan untuk per'aikan ketiga se'esar B <; 5=6 C 8; 536 C =; 51=6 C 1; 51;6 C =; 51;6 B 13:; 5'iaya pengangkutan terendah6 Sehingga al$kasi pr$duksi dengan 'iaya terendah adalah7
•
:; unit pr$duksi dari pa'rik W dial$kasikan ke gudang 0 se'anyak <; unit dan ke gudang ? se'anyak 8; unit#
•
<; unit pr$duksi dari pa'rik * dial$kasikan ke gudang A se'anyak =; unit dan ke gudang ? se'anyak 1; unit#
•
=; unit pr$duksi dari pa'rik dial$kasikan ke gudang 0 se'anyak =; unit#
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Padaa sesi ini han Pad hanya ya aka akan n di' di'aha ahass men mengen genai ai met met$de $de tran transp$ sp$rtas rtasii den dengan gan met met$de $de 9AM( sedangkan met$de Stepping Stone dan Stone dan M!" sudah di'ahas pada sesi tulisan se'elumnya# M&$& AM
Teknik Te knik pengerjaan pada met$de ini 'er'eda dengan dua met$de se'elumnya s e'elumnya yaitu met$de transp$rtasi Stepping Stone dan Stone dan M!" dimana untuk mendapatkan s$lusi yang $ptimal dilakukan 'erulang-ulang sampai k$ndisi $ptimal terse'ut terpenuhi# Sedangkan pada met$da 9A 9AM ini( sekali kita menentukan al$kasi pada satu &ell maka al$kasi terse'ut tidak 'eru'ah lagi# ntuk mempermudah penjelasan( penjelasan( kita gunakan &$nt$h yang sama seperti pada met$de transp$rtasi se'elumnya# Suatu perusahaan mempunyai pa'rik W( W( *( dengan kapasitas pr$duksi tiap 'ulan masingmasing :; t$n( <; t$n( dan =; t$n#> dan mempunyai 8 gudang penjualan di A( 0( ? dengan ke'utuhan tiap 'ulan masing-masing =; t$n( 11; t$n( dan @; t$n# 0ia ya pengangkutan setiap t$n pr$duk dari pa'rik W( *( *( ke gudang A( 0( ? adalah se'agai s e'agai 'erikut7 +angkah 4 langkah pengerjaan7 1# Susun Susunlah lah ke'utuha ke'utuhan( n( kapasitas kapasitas masing-masin masing-masing g sum'er dan dan 'iaya pengang pengangkutan kutan ke dalam matriks transp$rtasi
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8# Pili Pilihlah hlah 1 nilai per'edaanper'edaan- per'edaan yang ter'esar diantara semua nilai per'edaaan pada p ada k$l$m dan 'aris# 0aris mempunyai nilai per'edaan per 'edaan ter'esar yaitu :# 0ila nilai nilai per'edaan per'edaan 'iaya ada 2 yang 'esarnya sama( maka maka pilihlah pilihlah 'aris atau k$l$m yang mempunyai 'iaya terendah# terendah# @# "silah pada salah satu segi empat yang termasuk termasuk dalam k$l$m atau 'aris 'aris terp te rpil ilih ih(( ya yait itu u pa pada da se segi gi em empat pat ya yang ng me memp mpun unyai yai 'i 'iay ayaa te tere rend ndah ah## "si "sika kan n se'anyak mungkin yang 'isa dilakukan#
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Karena 0 mempunyai per'edaan ter'esar yaitu 1=( maka isilah se'anyak mungkin yang 'isa diangkut pada k$l$m 0 yang mempunyai 'iaya terendah#
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Baris ( mempunyai perbedaan terbesar yaitu 12 dan langkah selanjutnya
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adalah sebagai berikut
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Met$de ini dalam meru'ah al$kasi pr$duk untuk mendapatkan al$kasi pr$duksi yang $ptimal menggunakan suatu indeks per'aikan yang 'erdasarkan pada nilai 'aris dan nilai k$l$m# ?ara untuk penentuan nilai 'aris dan nilai k$l$m menggunakan persamaan7
Ped$man Ped$m an pr$sed pr$sedur ur al$kas al$kasii tahap pertama mengggunakan mengggunakan pr$sedur ped$man sudut 'arat laut (No (North rth West Cor Corner ner rul rule). e). n ntu tuk k me met$ t$de de M M!" !" ad adaa sy syara aratt ya yang ng ha haru russ di dipe penu nuhi hi(( yaitu 8#)#()# ($#( $&%"-" .#%+- -#,# &)*#) 8#)#()# 8#%"- "$#,8#. 8#)#()#
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'olusi
1. Isilah tabel tabel pertama pertama dari sudut sudut kiri atas
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4
Nilai 'aris yang lain dan nilai semua k$l$m ditentukan 'erdasarkan persamaan
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4
K$tak yang merupakan titik t$lak peru'ahan adalah adalah k$tak yang indeksnya 'ertanda
negati) dan angkanya 'esar# !alam &$nt$h ternyata yang memenuhi syarat adalah k$tak *A dengan nilai -2;#
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indeks yan! ne!ati&.
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