DAFTAR PUSTAKA. Al-Salamah, Muhammad & Alsawafy Omar. (2011). Journal Economic Order Quantity for Items with Two Types of Imperfect Quality.

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Vol. IX, No. 1, 12.

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Lampiran 1. Input Data Tahun Anggaran 2014

============================================================

MODEL

:

! Economic order quantity with quantity discounts;

! This model determines the optimal order quantity

for a product that has quantity discounts;

SETS

:

! Each order size range has;

RANGE/1..4/:

B, ! An upper breakpoint;

P, ! A price/unit over this range;

H, ! A holding cost/unit over this range;

EOQ, ! An EOQ using this ranges H and K;

Q, ! An optimal order qty within this range;

AC; ! Average cost/year using this range's Q;

ENDSETS

DATA

:

D = 298; ! The yearly demand;

K = 8500; ! The fixed cost of an order;

IRATE = .0752; ! Yearly interest rate;

!The upper break points, B, and price per unit, P:

Range: 1 2 3 4;

B = 99, 199, 299, 399;

P = 41470, 37323, 36286, 35250;

ENDDATA

! The model;

! Calculate holding cost, H, and EOQ for each

range;

@FOR

( RANGE:

H = IRATE * P;

EOQ = ( 2 * K * D/ H) ^.5;

);

! For the first range, the optimal order quantity

is equal to the EOQ ...;

Q( 1) = EOQ( 1)

! but, if the EOQ is over the first breakpoint,

lower it;

- ( EOQ( 1) - B( 1) + 1) *

( EOQ( 1) #GE# B( 1));

@FOR

( RANGE( J)| J #GT# 1:

! Similarly, for the rest of the ranges, Q = EOQ;

Q( J) = EOQ( J) +

! but, if EOQ is below the lower breakpoint,

raise it up;

( B( J-1) - EOQ( J)) *

( EOQ( J) #LT# B( J - 1))

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! or if EOQ is above the upper breakpoint,

lower it down;

- ( EOQ( J) - B( J) + 1) *

( EOQ( J) #GE# B( J));

);

! Calculate average cost per year, AC,

for each stage;

@FOR

( RANGE: AC = P * D + H * Q/ 2 + K * D/ Q);

! Find the lowest average cost, ACMIN.;

ACMIN =

@MIN

( RANGE: AC);

! Select the Q that gives the lowest AC per year;

! Note: TRUE = 1, FALSE = 0;

QUSE =

@SUM

( RANGE: Q * ( AC #EQ# ACMIN));

END

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Lampiran 1. Output Tahun Anggaran 2014

==================================================================== Feasible solution found.

Total solver iterations: 0

Variable Value D 298.0000 K 8500.000 IRATE 0.7520000E-01 ACMIN 0.1090927E+08 QUSE 299.0000 B( 1) 99.00000 B( 2) 199.0000 B( 3) 299.0000 B( 4) 399.0000 P( 1) 41470.00 P( 2) 37323.00 P( 3) 36286.00 P( 4) 35250.00 H( 1) 3118.544 H( 2) 2806.690 H( 3) 2728.707 H( 4) 2650.800 EOQ( 1) 40.30479 EOQ( 2) 42.48498 EOQ( 3) 43.08778 EOQ( 4) 43.71637 Q( 1) 40.30479 Q( 2) 99.00000 Q( 3) 199.0000 Q( 4) 299.0000 AC( 1) 0.1248375E+08 AC( 2) 0.1128677E+08 AC( 3) 0.1109746E+08 AC( 4) 0.1090927E+08 Row Slack or Surplus 1 0.000000 2 0.000000 3 0.000000 4 0.000000 5 0.000000 6 0.000000 7 0.000000 8 0.000000 9 0.000000 10 0.000000 11 0.000000 12 0.000000 13 0.000000 14 0.000000 15 0.000000 16 0.000000 17 0.000000 18 0.000000 ====================================================================

http://digilib.mercubuana.ac.id/

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Lampiran 3. Input Data Tahun Anggaran 2015

====================================================================

MODEL

:

! Economic order quantity with quantity discounts;

! This model determines the optimal order quantity

for a product that has quantity discounts;

SETS

:

! Each order size range has;

RANGE/1..4/:

B, ! An upper breakpoint;

P, ! A price/unit over this range;

H, ! A holding cost/unit over this range;

EOQ, ! An EOQ using this ranges H and K;

Q, ! An optimal order qty within this range;

AC; ! Average cost/year using this range's Q;

ENDSETS

DATA

:

D = 253; ! The yearly demand;

K = 9754; ! The fixed cost of an order;

IRATE = .0752; ! Yearly interest rate;

!The upper break points, B, and price per unit, P:

Range: 1 2 3

4;

B = 99, 199, 299, 399;

P = 41470, 37323, 36286,35250;

ENDDATA

! The model;

! Calculate holding cost, H, and EOQ for each

range;

@FOR

( RANGE:

H = IRATE * P;

EOQ = ( 2 * K * D/ H) ^.5;

);

! For the first range, the optimal order quantity

is equal to the EOQ ...;

Q( 1) = EOQ( 1)

! but, if the EOQ is over the first breakpoint,

lower it;

- ( EOQ( 1) - B( 1) + 1) *

( EOQ( 1) #GE# B( 1));

@FOR

( RANGE( J)| J #GT# 1:

! Similarly, for the rest of the ranges, Q = EOQ;

Q( J) = EOQ( J) +

! but, if EOQ is below the lower breakpoint,

raise it up;

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( EOQ( J) #LT# B( J - 1))

! or if EOQ is above the upper breakpoint,

lower it down;

- ( EOQ( J) - B( J) + 1) *

( EOQ( J) #GE# B( J));

);

! Calculate average cost per year, AC,

for each stage;

@FOR

( RANGE: AC = P * D + H * Q/ 2 + K * D/ Q);

! Find the lowest average cost, ACMIN.;

ACMIN =

@MIN

( RANGE: AC);

! Select the Q that gives the lowest AC per year;

! Note: TRUE = 1, FALSE = 0;

QUSE =

@SUM

( RANGE: Q * ( AC #EQ# ACMIN));

END

====================================================================

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Lampiran 4. Output Tahun Anggaran 2015

==================================================================== Feasible solution found.

Total solver iterations: 0

Variable Value D 253.0000 K 9754.000 IRATE 0.7520000E-01 ACMIN 9322798. QUSE 299.0000 B( 1) 99.00000 B( 2) 199.0000 B( 3) 299.0000 B( 4) 399.0000 P( 1) 41470.00 P( 2) 37323.00 P( 3) 36286.00 P( 4) 35250.00 H( 1) 3118.544 H( 2) 2806.690 H( 3) 2728.707 H( 4) 2650.800 EOQ( 1) 39.78237 EOQ( 2) 41.93430 EOQ( 3) 42.52929 EOQ( 4) 43.14974 Q( 1) 39.78237 Q( 2) 99.00000 Q( 3) 199.0000 Q( 4) 299.0000 AC( 1) 0.1061597E+08 AC( 2) 9606577. AC( 3) 9464265. AC( 4) 9322798. Row Slack or Surplus 1 0.000000 2 0.000000 3 0.000000 4 0.000000 5 0.000000 6 0.000000 7 0.000000 8 0.000000 9 0.000000 10 0.000000 11 0.000000 12 0.000000 13 0.000000 14 0.000000 15 0.000000 16 0.000000 17 0.000000 18 0.000000 ====================================================================