BAB 6 KESIMPULAN DAN SARAN
6.1. Kesimpulan Berdasarkan
analisis
dan
pembahasan
yang
telah
dilakukan, dapat diambil beberapa kesimpulan yaitu: 1. Dua
model
ukuran
yang
lot
ukuran
dikembangkan
gabungan
lot
produsen
pemesanan
dengan dan
dapat
menggunakan
pemasok
sebagai
digunakan
untuk
menentukan harga jual produk. 2. Pengembangan
2
model
dengan
ukuran
lot
gabungan
tersebut menghasilkan harga jual produk per unit dan total biaya gabungan per periode yang lebih kecil daripada model Banerjee (1986). 6.2. Saran Penelitian
ini
dilakukan
dengan
asumsi
bahwa
proses produksi mempunyai laju yang tetap. Saran untuk penelitian
selanjutnya
adalah
dapat
dilakukan
penelitian lanjutan dengan mempertimbangkan: 1. Proses pemeriksaan produk dilakukan oleh pemasok dan memperhitungkan
biaya
garansi
yang
ditanggung
pemasok jika terjadi kasus produk cacat sampai ke tangan produsen. 2. Dilakukan proses pemeriksaan produk pada produsen dengan menggunakan metode sampling. Selain
itu,
penelitian
lanjutan
untuk
solusi model secara analitik perlu dilakukan.
54
mencari
DAFTAR PUSTAKA
Banerjee, A., 1986, A supplier’s pricing model under a customer's economic purchasing policy, Omega, 14(5), 409-414. Banerjee, A., 2005, Concurrent pricing and lot sizing for make-to-order contract production, International Journal of Production Economics, 94, 189-195. Bintoro, A.G., 2019, Penentuan ukuran lot ekonomis gabungan produsen dan distributor untuk produk yang dijual dengan garansi, Jurnal Teknologi Industri, Vol XIII, No. 2, Hal. 165-179, Universitas Atma Jaya, Yogyakarta Darwish, M., dan Ertogral, K., 2008, The joint economic lot sizing problem: Review and extensions, European Journal of Operational Research, 185, 726-742. Hamid, S., dan Pasandideh, R., 2010, An investigation of vendor-managed inventory application in supply chain: the EOQ model with shortage, International Journal of Advanced Manufacturing Technology, 49, 329-339. Kim, J., Hong, Y., dan Kim, T., 2011, Pricing and ordering policies for price-dependent demand in a supply chain of a single retailer and a single manufacturer, International Journal of Systems Science, 42(1), 81-89. Kim, S.L., dan Ha, D., 2003, A JIT lot-splitting model for supply chain management: Enhancing buyer– supplier linkage, International Journal of Production Economics, 86(1), 1-10. Lin, J.M.H.C., 2005, A buyer-vendor EOQ model with changeable lead-time in supply chain, International Journal of Production Economics, 26, 917-921. Sana, S.S., dan Chaudhuri, K., 2005, A Stochastic EOQ Policy of Cold-Drink-For a Retailer, Vietnam Journal of Mathematics, 33(4), 437-442. 55
Santiyasa, I.W., 2011, Algoritma Newton Raphson Dengan Fungsi Non-Linear, Universitas Udayana, Bali. Siagian, P., 1987, Penelitian Operasional: Teori dan Praktek, UI-Press, Jakarta. Siswanto, 2007, Operations Research, Penerbit Erlangga, Jakarta. Tersine, R.J., 1994, Principles Materials Management, Fourth Hall, New Jersey.
Jilid
Dua,
of Inventory and Edition, Prentice
Wijanarko, W.S., 2012, Penentuan Ukuran Lot Ekonomis Gabungan antara Suplier dan Produsen di PT. Blambangan Foodpackers Indonesia, skripsi di Jurusan Teknik Industri Universitas Atma Jaya Yogyakarta, Yogyakarta. Zimmer, K., 2002, Supply chain coordination with uncertain just-in-time delivery, International Journal of Production Economics, 77, 1-15.
56
LAMPIRAN
57
Lampiran 1 : Perhitungan model Banerjee (1986) Diketahui : D := 12000 P := 48000 h1 := 0.30 h2 := 0.24
S1 := 10 S2 := 500 C2 := 4.5 G := 1.5
Total Biaya Produsen Per Periode Q⋅ h1⋅ C1 D TRC1 D⋅ C1 + ⋅ S1 + 2 Q EOQ Produsen 2⋅ D⋅ S1 Q C1⋅ h1 Total Biaya Pemasok Per Periode D D⋅ Q⋅ h2⋅ C2 TRC2 D⋅ C2 + ⋅ S2 + Q 2⋅ P Jika : a :=
S2 2⋅ D⋅ S1
dan
= 0.559
b :=
h2⋅ C2 2⋅ P
2⋅ D⋅ S1
⋅
h1
h1 Metode Newton-Raphson Iterasi 0 C1 := C2 + G = 6 1
−1
2
2
fC1 := C1 − C2 − a⋅ C1 − b ⋅ C1 −1
ffC1 := 1 −
1 2
⋅ a⋅ C1
2
− G = −1.373
−3
+
1 2
⋅ b ⋅ C1
2
= 0.886
1
−1
2
2
Gross0 := C1 − C2 − a⋅ C1 − b ⋅ C1
= 0.127
58
= 0.01
Iterasi 1 C11 := C1 −
fC1 ffC1
= 7.55 1
−1
2
2
fC11 := C11 − C2 − a⋅ C11 − b ⋅ C11 −1
ffC11 := 1 −
1 2
⋅ a⋅ C11
2
− G = 0.01
−3
1
+
⋅ b ⋅ C11
2
2
= 0.899
1
−1
2
2
Gross1 := C11 − C2 − a⋅ C11 − b ⋅ C11
= 1.51
Iterasi 2 C111 := C11 −
fC11 ffC11
= 7.539 1
−1
2
2
fC111 := C111 − C2 − a⋅ C111 − b ⋅ C111 −1
ffC111:= 1 −
1 2
⋅ a⋅ C111
2
−7
− G = 4.197 × 10
−3
+
1 2
⋅ b ⋅ C111
2
= 0.898
1
−1
2
2
Gross2 := C111 − C2 − a⋅ C111 − b ⋅ C111
= 1.5
Iterasi 3 C1111:= C111 −
fC111 ffC111
= 7.539 1
−1
2
2
fC1111:= C1111 − C2 − a⋅ C1111 − b ⋅ C1111 −1
ffC1111:= 1 −
1 2
⋅ a⋅ C1111
2
− 15
− G = 1.332 × 10
−3
+
1 2
⋅ b ⋅ C1111
2
= 0.898
1
−1
2
2
Gross3 := C1111 − C2 − a⋅ C1111 − b ⋅ C1111 Harga jual produk per unit = $7.539
59
= 1.5
Lampiran 2 : Perhitungan model 1 Diketahui : D := 12000 P := 48000 h1 := 0.30 h2 := 0.24
S1 := 10 S2 := 500 C2 := 4.5 G := 1.5
Total Biaya Produsen Per Periode Q⋅ h1⋅ C1 D TRC1 D⋅ C1 + ⋅ S1 + 2 Q Total Biaya Pemasok Per Periode D⋅ Q⋅ h2⋅ C2 D TRC2 D⋅ C2 + ⋅ S2 + 2⋅ P Q Total Biaya Gabungan Per Periode Q⋅ h1⋅ C1 D D⋅ Q⋅ h2⋅ C2 D + D⋅ C2 + ⋅ S2 + TRCGab D⋅ C1 + ⋅ S1 + 2 Q 2⋅ P Q Ukuran Lot Gabungan [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P QGab ( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) Metode Newton-Raphson Iterasi 0 C1 := C2 + G = 6 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P fC1 := C1 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
−
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
⋅ h2⋅ C2
2⋅ P
− G = −0.233
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) ffC1 :=
2
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2) 2
4⋅ ( C2⋅ D⋅ h2 + C1⋅ P⋅ h1) ⋅
P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C1⋅ P⋅ h1
P ⋅ S2⋅ h1⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
3 2 P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C1⋅ P⋅ h1) ⋅
C2⋅ D⋅ h2 + C1⋅ P⋅ h1
[ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P Gross0 := C1 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
−
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) 2⋅ P
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
60
⋅ h2⋅ C2 = 1.267
2
+ 1 = 0.987
Iterasi 1 C11 := C1 −
fC1 ffC1
= 6.236 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P S2
fC11 := C11 − C2 −
[ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2
−5
− G = 1.79 × 10
2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2) 2
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
ffC11 :=
P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2
4⋅ ( C2⋅ D⋅ h2 + C11⋅ P⋅ h1) ⋅
P ⋅ S2⋅ h1⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
C2⋅ D⋅ h2 + C11⋅ P⋅ h1
3 2 P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C11⋅ P⋅ h1) ⋅
+ 1 = 0.987
2
C2⋅ D⋅ h2 + C11⋅ P⋅ h1
[ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P Gross1 := C11 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2 = 1.5
2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2) Iterasi 2 C111 := C11 −
fC11 ffC11
= 6.236 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
fC111 := C111 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
−
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
⋅ h2⋅ C2
2⋅ P
− 13
− G = 1.035 × 10
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) ffC111:=
2
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2) 2
4⋅ ( C2⋅ D⋅ h2 + C111⋅ P⋅ h1) ⋅
P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C111⋅ P⋅ h1
P ⋅ S2⋅ h1⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
3
2⋅ ( C2⋅ D⋅ h2 + C111⋅ P⋅ h1) ⋅ 2
[ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P Gross2 := C111 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P
−
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) 2⋅ P
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
Harga Jual Produk Per Unit = $6.236
61
P⋅ ( 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C111⋅ P⋅ h1
⋅ h2⋅ C2 = 1.5
2
+ 1 = 0.987
Lampiran 3 : Perhitungan model 2 Diketahui : D := 12000 P := 48000 h1 := 0.30 h2 := 0.24
S1 := 10 S2 := 500 C2 := 4.5 G := 1.5
Cp := 20 Ct := 50
Total Biaya Produsen Per Periode (Dengan Biaya Pemeriksaan) Q⋅ h1⋅ C1 D D + ⋅ Cp TRC1 D⋅ C1 + ⋅ S1 + 2 Q Q Total Biaya Pemasok Per Periode (Dengan Biaya Pengiriman) D⋅ Q⋅ h2⋅ C2 D D + ⋅ Ct TRC2 D⋅ C2 + ⋅ S2 + 2⋅ P Q Q Total Biaya Gabungan Per Periode (Dengan Biaya Pemeriksaan + Biaya Pengiriman) Q⋅ h1⋅ C1 D D D⋅ Q⋅ h2⋅ C2 D ⋅ S1 + + ⋅ Cp + D⋅ C2 + ⋅ S2 + + ⋅ Ct Q 2 Q Q 2⋅ P Q Ukuran Lot Gabungan [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P QGab ( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) TRCGab
D⋅ C1 +
D
Metode Newton-Raphson Iterasi 0 C1 := C2 + G = 6 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P fC1 := C1 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2 −
2⋅ P
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) ffC1 :=
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2) 2
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) 2
4⋅ ( C2⋅ D⋅ h2 + C1⋅ P⋅ h1) ⋅
P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C1⋅ P⋅ h1
−
3 2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C1⋅ P⋅ h1) ⋅
S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
2
P ⋅ S2⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P Gross0 := C1 − C2 −
− G = −0.241
−
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
C2⋅ D⋅ h2 + C1⋅ P⋅ h1 ⋅ h2⋅ C2 −
2⋅ P
2
Ct⋅ P ⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
3
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P ( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C1⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
62
2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C1⋅ P⋅ h1) ⋅
= 1.259
C2⋅ D⋅ h2 + C1⋅ P⋅ h1
2
+ 1 = 0.987
Iterasi 1 C11 := C1 −
fC1 ffC1
= 6.245 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P S2
fC11 := C11 − C2 −
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2
Ct
−
2⋅ P 2
P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2
4⋅ ( C2⋅ D⋅ h2 + C11⋅ P⋅ h1) ⋅
−
C2⋅ D⋅ h2 + C11⋅ P⋅ h1
3 2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C11⋅ P⋅ h1) ⋅
C2⋅ D⋅ h2 + C11⋅ P⋅ h1
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
2
P ⋅ S2⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
Gross1 := C11 − C2 −
− G = 1.937 × 10
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2) C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
ffC11 :=
−5
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2 −
2⋅ P
Ct⋅ P ⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
2
3 2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C11⋅ P⋅ h1) ⋅
Ct
C2⋅ D⋅ h2 + C11⋅ P⋅ h1
+ 1 = 0.987
2
= 1.5
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P ( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C11⋅ h1⋅ P) + ( C2⋅ D⋅ h2) Iterasi 2 C111 := C11 −
fC11 ffC11
= 6.244 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
fC111:= C111 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2 −
2⋅ P 2
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) 2
4⋅ ( C2⋅ D⋅ h2 + C111⋅ P⋅ h1) ⋅
P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C111⋅ P⋅ h1
−
3 2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C111⋅ P⋅ h1) ⋅
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
−
2
P ⋅ S2⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
Gross2 := C111 − C2 −
− 13
− G = 1.221 × 10
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) ffC111:=
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) 2⋅ P
C2⋅ D⋅ h2 + C111⋅ P⋅ h1 ⋅ h2⋅ C2 −
2
Ct⋅ P ⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
3
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P ( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
63
2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C111⋅ P⋅ h1) ⋅
= 1.5
C2⋅ D⋅ h2 + C111⋅ P⋅ h1
2
+ 1 = 0.987
Iterasi 3 C1111:= C111 −
fC111 ffC111
= 6.244 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
fC1111:= C1111 − C2 −
S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C1111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
−
⋅ h2⋅ C2 −
2⋅ P
( C1111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) ffC1111:=
4⋅ ( C2⋅ D⋅ h2 + C1111⋅ P⋅ h1) ⋅
2
P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) C2⋅ D⋅ h2 + C1111⋅ P⋅ h1
−
3 2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C1111P ⋅ ⋅ h1) ⋅
[ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P S2 [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
−
2
P ⋅ S2⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
Gross3 := C1111 − C2 −
− G= 0
( C1111h1 ⋅ ⋅ P) + ( C2⋅ D⋅ h2)
C2⋅ P⋅ h1⋅ h2⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2) 2
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P
( C1111h1 ⋅ ⋅ P) + ( C2⋅ D⋅ h2) 2⋅ P
C2⋅ D⋅ h2 + C1111P ⋅ ⋅ h1 ⋅ h2⋅ C2 −
2
Ct⋅ P ⋅ h1⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
−
3
Ct [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P ( C1111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
( C1111⋅ h1⋅ P) + ( C2⋅ D⋅ h2)
Harga Jual Produk Per Unit = $6.244
64
2 P⋅ ( 2⋅ Cp⋅ D + 2⋅ Ct⋅ D + 2⋅ D⋅ S1 + 2⋅ D⋅ S2)
2⋅ ( C2⋅ D⋅ h2 + C1111⋅ P⋅ h1) ⋅
= 1.5
C2⋅ D⋅ h2 + C1111⋅ P⋅ h1
2
+ 1 = 0.987
Lampiran 4 : Perhitungan ukuran lot optimal dan total biaya Diketahui : D := 12000 P := 48000 h1 := 0.30 h2 := 0.24
S1 := 10 S2 := 500 C2 := 4.5 G := 1.5
Cp := 20 Ct := 50 C1 := 7.539 C11 := 6.236
C111 := 6.244
Banerjee 1986 Ukuran Lot 2⋅ D⋅ S1 Q1 := → 325.753 C1⋅ h1 Total Biaya Produsen Per Periode D Q1⋅ h1⋅ C1 TRC1 := D⋅ C1 + ⋅ S1 + → 91204.755 Q1 2 Total Biaya Pemasok Per Periode D D⋅ Q1⋅ h2⋅ C2 TRC2 := D⋅ C2 + ⋅ S2 + → 72462.853 Q1 2⋅ P Total Biaya Gabungan Per Periode TRCgab1 := TRC1 + TRC2 → 163667.608 Model 1 Ukuran Lot QGab1 :=
[ ( D⋅ S1) + ( D⋅ S2) ] ⋅ 2⋅ P ( C11⋅ h1⋅ P) + ( C2⋅ h2⋅ D)
→ 2391.127
Total Biaya Produsen Per Periode QGab1⋅ h1⋅ C11 D ⋅ S1 + → 77118.846 TRC11 := D⋅ C11 + 2 QGab1 Total Biaya Pemasok Per Periode D⋅ QGab1⋅ h2⋅ C2 D ⋅ S2 + → 56832.079 TRC21 := D⋅ C2 + 2⋅ P QGab1 Total Biaya Gabungan Per Periode TRCgab3 := TRC11 + TRC21 → 133950.925
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Model 2 Ukuran Lot [ ( D⋅ S1) + ( D⋅ S2) + ( D⋅ Cp) + ( D⋅ Ct) ] ⋅ 2⋅ P QGab2 := → 2548.521 ( C111⋅ h1⋅ P) + ( C2⋅ D⋅ h2) Total Biaya Produsen Per Periode D QGab2⋅ h1⋅ C111 D TRC13 := D⋅ C111 + ⋅ S1 + + ⋅ Cp → 77456.203 QGab2 2 QGab2 Total Biaya Pemasok Per Periode D D⋅ QGab2⋅ h2⋅ C2 D TRC23 := D⋅ C2 + ⋅ S2 + + ⋅ Ct → 56933.787 QGab2 2⋅ P QGab2 Total Biaya Gabungan Per Periode TRCgab4 := TRC13 + TRC23 → 134389.991
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