ÐÏࡱá>þÿ  þÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ! The yearly demand; K = 90; ! The fixed cost of an order; IRATE = .2; ! Yearly interest rate; !The upper break points, B, and price per unit, P: Range: 1 2 3 4; B = 10000, 20000, 40000, 60000; P = .35225, .34525, .34175, .33825; 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)) ! 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 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 = 40000;þÿÿÿýÿÿÿþÿÿÿþÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿRoot EntryÿÿÿÿÿÿÿÿqCONTENTSÿÿÿÿÿÿÿÿÿÿÿÿqÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ þÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿRoot Entryÿÿÿÿÿÿÿÿ*0_šîÏ»òÀð^àmûeþÄ Contentsÿÿÿÿÿÿÿÿÿÿÿÿå ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿþÿÿÿýÿÿÿþÿÿÿ þÿÿÿ  ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ  !"#$%&'()*+þÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ ( B( J-1) - EOQ( J)) * \par ( EOQ( J) #LT# B( J - 1)) \par \cf3 ! or if EOQ is above the upper breakpoint, \par lower it down;\cf2 \par - ( EOQ( J) - B( J) + 1) * \par ( EOQ( J) #GE# B( J)); \par ); \par \par \cf3 ! Calculate average cost per year, AC, \par for each stage;\cf2 \par \cf1 @FOR\cf2 ( RANGE: AC = P * D + H * Q/ 2 + K * D/ Q); \par \par \cf3 ! Find the lowest average cost, ACMIN.;\cf2 \par ACMIN = \cf1 @MIN\cf2 ( RANGE: AC); \par \par \cf3 ! Select the Q that gives the lowest AC per year;\cf2 \par \cf3 ! Note: TRUE = 1, FALSE = 0;\cf2 \par QUSE = \cf1 @SUM\cf2 ( RANGE: Q * ( AC #EQ# ACMIN)); \par \par \cf1 END\cf2 \par \par } ì‹{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Courier New;}} {\colortbl ;\red0\green0\blue255;\red0\green0\blue0;\red0\green175\blue0;} \viewkind4\uc1\pard\cf1\f0\fs20 MODEL\cf2 : \par \cf3 ! Economic order quantity with quantity discounts;\cf2 \par \cf3 ! This model determines the optimal order quantity \par for a product that has quantity discounts;\cf2 \par \cf1 SETS\cf2 : \par \cf3 ! Each order size range has;\cf2 \par RANGE/1..4/: \par B, \cf3 ! An upper breakpoint;\cf2 \par P, \cf3 ! A price/unit over this range;\cf2 \par H, \cf3 ! A holding cost/unit over this range;\cf2 \par EOQ, \cf3 ! An EOQ using this ranges H and K;\cf2 \par Q, \cf3 ! An optimal order qty within this range;\cf2 \par AC; \cf3 ! Average cost/year using this range's Q;\cf2 \par \cf1 ENDSETS\cf2 \par \par \cf1 DATA\cf2 : \par D = 40000; \cf3 ! The yearly demand;\cf2 \par K = 90; \cf3 ! The fixed cost of an order;\cf2 \par IRATE = .2; \cf3 ! Yearly interest rate;\cf2 \par \cf3 !The upper break points, B, and price per unit, P: \par Range: 1 2 3 4;\cf2 \par B = 10000, 20000, 40000, 60000; \par P = .35225, .34525, .34175, .33825; \par \cf1 ENDDATA\cf2 \par \par \cf3 ! The model;\cf2 \par \cf3 ! Calculate holding cost, H, and EOQ for each \par range;\cf2 \par \cf1 @FOR\cf2 ( RANGE: \par H = IRATE * P; \par EOQ = ( 2 * K * D/ H) ^.5; \par ); \par \par \cf3 ! For the first range, the optimal order quantity \par is equal to the EOQ ...;\cf2 \par Q( 1) = EOQ( 1) \par \cf3 ! but, if the EOQ is over the first breakpoint, \par lower it;\cf2 \par - ( EOQ( 1) - B( 1) + 1) * \par ( EOQ( 1) #GE# B( 1)); \par \par \cf1 @FOR\cf2 ( RANGE( J)| J #GT# 1: \par \cf3 ! Similarly, for the rest of the ranges, Q = EOQ;\cf2 \par Q( J) = EOQ( J) + \par \cf3 ! but, if EOQ is below the lower breakpoint, \par raise it up;\cf2 \par