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Inventory Planning

Inventory Planning. Sumber : Sistem Inventori – Senator Nur Bahagia. Karakteristik. Demand bervariasi dengan pola distribusi diketahui Faktor lain yang mempengaruhi dapat bersifat deterministik, probabilistik atau uncertainty. Contoh Permasalahan.

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Inventory Planning

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  1. Inventory Planning Sumber : SistemInventori – Senator Nur Bahagia Andary A Munita Hanafiah

  2. Karakteristik • Demand bervariasi dengan pola distribusi diketahui • Faktor lain yang mempengaruhi dapat bersifat deterministik, probabilistik atau uncertainty

  3. ContohPermasalahan Pemakaian barang selama satu tahun terakhir ini adalah sebagai berikut:

  4. Contoh Permasalahan Rencanakanpengadaanbaranguntuktahundepanjikadiketahui: • Ongkospemesanan rp.300.000,-/pesan • Hargabarang rp.100.000/unit • Ongkossimpan 20% darihargabarang/unit/ tahun • Lead time 1 bulan • Polapermintaanmendatangsamadenganpolapermintaanmasalaludanberdistribusinormal • Saatinidigudangtersedia 100 unit barangdanpihakdihendakisetiapsaattersedia minimal sebanyak 100 unit barang • Tidakadabarangdalampesanan

  5. Analogi Deterministik KetidakpastianKetidakpastian Probabilistik

  6. Sumber Ketidakpastian Supplier Management User

  7. Dampak Ketidakpastian PerluAdanyaInventoriPengaman Safety Stock Buffer Stock

  8. Safety Stock SAFETY STOCK/SS

  9. Hubungan Tingkat Pelayanandan Z

  10. Contoh Perhitungan SS Diketahui : S = 100 Unit /Bulan t= Lead Time = 1 bulan Tingkat pelayanan 95 % Maka: SS = Z. S  t SS =

  11. Tingkat Pelayanan

  12. Kekurangan Inventori (N)

  13. Contoh Kebutuhanbarangtiaptahunnyaberdistribusi normal dengan rata-rata sebesar 10.000 unit dandeviasistandar 2.000 unit. Jikalead timeuntukmendapatkanbarangsebesar 3 bulan. Berapacadanganpengamandantingkatpelayanannyajikadikehendakikemungkinanterjadinyakekuranganinventoritidakbolehlebihdari 5%.

  14. Solusi • Dari soal di atasmakadapatdiidentifikasikanhal-halsebagaiberikut : • D =…………. unit/tahun • s = …………. unit/tahun • L = …………. tahun •  5% z = 1,65 (lihattabel Normal)

  15. Solusi a. Cadanganpengaman (ss) sebesar : • ss = z s L = ……. unit b. Tingkat Pelayanan Dari tabeldapatdicariuntukz = 1,65 f(z) = 0,1023 (Table) dan(z) = 0,0206 (Table), maka : N = ….. unit  = ……… %

  16. Problem Bagaimana menentukan operating stock dan safety stock ? • Berapa jumlah barang yang akan dipesan untuk setiap kali melakukan pengadaan ( economic order quantity / EOQ) ? • Kapan saat pemesanan dilakukan (reorder point/r)? • Berapa besarnya safety stock(ss) ?

  17. Kebijakan Inventori 1. Besarnya ukuran pemesanan (qo) 2. Saat pemesanan dilakukan (r) 3. Besarnya cadangan pengaman (ss)

  18. Kriteria Kinerja • Tingkat Pelayanan • Ongkos Inventori • Tingkat Pengembalian Investasi

  19. Metoda • Model Sederhana • Model Q • Model P

  20. Model Sederhana Model Sederhana = Model Deterministik + Safety Stock

  21. Q* r* ss PosisiInventori Model Sederhana

  22. Formulasi Model Ot = Ob + Op + Os +Ok Dimana: Ob = Dp Op = AD/Q0 Os = (ss + Qo/2) Ok = .N Ot= Dp +AD/Q0 + h(SS+Q0/2) + .N

  23. Solusi: Ukuran Lot Syarat Ot minimal: Ot/Qo = 0 2 -AD/Qo + h/2 =0 1/2 Q*o = {2AD/h}

  24. Solusi: SS dan Reorder Point • Cadangan Pengaman (SS) ss = z s L • Reorder Point (r) r* = DL + SS

  25. Contoh Kebutuhan barang tiap tahunnya berdistribusi normal dengan rata-rata sebesar 10.000 unit dan deviasi standard 2.000 unit. Untuk mengadakan barang tersebut biasa dipesan dari seorang pemasok dengan ongkos pesan sebesar Rp 1.000.000,-/pesan harga Rp 25.000,-/unit dan lead time 3 bulan. Jika ongkos simpan sebesar 20% dari harga barang/unit/tahun dan kemungkinan terjadinya kekurangan inventori tidak lebih dari 5% dan ongkos kekurangan inventori sebesar Rp 10.000,-/unit.

  26. Contoh Tentukan : 1. Kebijakan inventori yang optimal ! 2. Berapa tingkat pelayanan yang diberikan ? 3. Berapa ongkos inventori selama 1 tahun ?

  27. Solusi 1. Kebijakaninventori a. Ukuran lot Ekonomis Q*o= ……… Unit

  28. Solusi b.Cadanganpengaman (ss) : • untuk = 5% z = 1,65 • ss= ….. unit c. Saattitikpemesanankembali (r*) : • r* = D.L + ss • r* = ……….. unit

  29. Solusi 3. Total ongkosinventori (OT) : OT= …………. juta/ tahun

  30. MODEL Q 1.Ukuran pesanan (Qo) selalukonstan untuksetiap kali melakukanpemesanan 2.Pesanan dilakukanbilapersediaandigudangmencapaitingkat r (reorder point)

  31. Q* r* ss Posisi Inventori

  32. Problem Bagaimana menentukan operating stock dan safety stock ? • Berapa besarnya ukuran pemesanan (Qo) • Berapa besarnya reorder point (r) • Berapa besarnya safety stock (ss) ?

  33. Kriteria Kinerja • Tingkat Pelayanan • Ongkos Inventori

  34. Formulasi Model Ot = Ob + Op + Os +Ok Dimana: Ob = Dp Op = AD/Qo Os = (ss + m) Ok = Cu N

  35. Solusi Syarat Ot minimal: Ot/Qo = 0 2 -AD/Qo + h/2 =0 1/2 Qo* = {2AD/h}

  36. Solusi • Cadangan Pengaman (SS) ss = z s L • Reorder Point (r) r* = DL + SS

  37. MODEL P 1.Pesanan dilakukan menurut suatu interval pemesanan yang tetap (T). 2.Ukuran pesanan (Qo) bergantung pada jumlah barang yang ada pada saat pemesanan dilakukan(r) dan persediaan maximum yang diinginkan(R).

  38. Posisi Inventori

  39. Problem Bagaimanamenentukan operating stock dan safety stock ? • Berapabesarnya interval waktuantarpemesanan • Berapapersediaan maximum (R) • Berapabesarnya safety stock ?

  40. Kriteria Kinerja • Tingkat Pelayanan • Ongkos Inventori

  41. Formulasi Model Ot = Ob + Op + Os +Ok Dimana Ob = Dp Op = A/T Os = (ss + m) Ok = Cu N

  42. Solusi Syarat Ot minimal: Ot/Qo = 0 2 -A/T + hD/2 =0 1/2 T* = {2A/hD}

  43. Solusi • Cadangan Pengaman (SS) ss = z s (L+T) • Persediaan Maximum (R) R* = D(L +T) + SS

  44. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 .1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 .2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 .3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 .4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 .5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 .6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 .7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 .8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 .9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 A table entry is the proportion of the area under the curve from a z of 0 to a positive value of z. To find the area from a z of 0 to a negative z, subtract the tabled value from 1. Tabel Normal

  45. Deviasi Normal Standar Za Prob. Kekurangan a Ordinat f(z) Ekspektasi Parsial Î(z) .00 .5000 .3989 .3989 .05 .4801 .3984 .3744 .10 .4602 .3969 .3509 .15 .4404 .3945 .3284 .20 .4207 .3910 .3069 .25 .4013 .3867 .2863 .30 .3821 .3814 .2668 .35 .3632 .3752 .2481 .40 .3446 .3683 .2304 .45 .3264 .3605 .2137 .50 .3086 .3521 .1978 .55 .2912 .3429 .1828 .60 .2743 .3332 .1687 .65 .2579 .3229 .1554 .70 .2420 .3123 .1429 .75 .2267 .3011 .1312 .80 .2119 .2897 .1202 .85 .1977 .2780 .1100 .90 .1841 .2661 .1004 .95 .1711 .2541 .0916 1.00 .1587 .2420 .0833 1.05 .1469 .2300 .0757 1.10 .1357 .2179 .0686 1.15 .1251 .2059 .0621 1.20 .1151 .1942 .0561 1.25 .1057 .1826 .0506 1.30 .0968 .1714 .0455 1.35 .0886 .1604 .0409 1.40 .0808 .1497 .0367 1.45 .0736 .1394 .0328 1.50 .0669 .1295 .0293 1.55 .0606 .1200 .0261 1.60 .0548 .1109 .0232 1.65 .0495 .1023 .0206 Tabel A

  46. Deviasi Normal Standar Za Prob. Kekurangan a Ordinat f(z) Ekspektasi Parsial Î(z) 1.70 .0446 .0940 .0183 1.75 .0401 .0863 .0162 1.80 .0360 .0790 .0143 1.85 .0322 .0721 .0126 1.90 .0288 .0656 .0111 1.95 .0256 .0596 .0097 2.00 .0228 .0540 .0085 2.05 .0202 .0488 .0074 2.10 .0179 .0440 .0065 2.15 .0158 .0396 .0056 2.20 .0140 .0355 .0049 2.25 .0122 .0317 .0042 2.30 .0107 .0283 .0037 2.35 .0094 .0252 .0032 2.40 .0082 .0224 .0027 2.45 .0071 .0198 .0023 2.50 .0062 .0175 .0020 2.55 .0054 .0154 .0017 2.60 .0047 .0136 .0015 2.65 .0040 .0119 .0012 2.70 .0035 .0104 .0011 2.75 .0030 .0091 .0009 2.80 .0026 .0079 .0008 2.85 .0022 .0069 .0006 2.90 .0019 .0059 .0005 2.95 .0016 .0051 .00045 3.00 .0015 .0044 .00038 3.10 .0010 .0033 .00027 3.20 .0007 .0024 .00018 3.30 .0005 .0017 .00013 3.40 .0004 .0012 .00009 3.50 .0003 .009 .00006 3.60 .0002 .006 .00004 3.80 .0001 .003 .00002 4.00 .00003 .0001 .00001 Tabel A (Lanjutan)

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