Rent cost depends on maximum inventory accumulation

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Rent cost depends on maximum inventory accumulation; holding cost is insensitive to accumulations at other times Given a set of n shipments , motion cost is independent of shipment sizes and times; find sets of shipment times & sizes to minimize holding cost Lower bound on max accumulation = size of largest shipment received, which is minimized when all shipments are equal ( ) ( ) n t D on accumulati max : shipments equal with n t D on accumulati max max max = ( ) n t D c r max Holding cost = Special case ( ) n t D max 2 ( ) n t D max ( ) max t D ( ) t D ( ) t R 0 t 1 t 2 t 3 t 4 t max t time Cum’l items
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LSA: Chapter 3 Lot size problem with variable demand: c r >> c i ( ) n t D v max = Optimal strategy (i) Divide y-axis from 0 to D(t max ) into n equal segments; find times t i for which D(t) equals (i/n) D(t max ) for i = 0,…,n-1 (ii) Dispatch barely enough to cover demand until the following shipment ( ) n t D max 2 ( ) n t D max ( ) max t D ( ) t D ( ) t R 0 t 1 t 2 t 3 t 4 t max t time Cum’l items ( ) + = max max t n c n t D c cost/time f r ( ) ( ) + = max max ' t D n c n t D D c cost/item f r ( ) max max ' t t D D = The total cost (relevant to shipping decision) where Finally, we optimize with regard to n (integer) or LSA: Chapter 3 Lot size problem with variable demand: c r << c i Solution by Newell’s shooting method (1971) P Q T ( ) max t D ( ) t D 0 t 1 t 2 t max t time Cum’l items Build R(t) Choose a point P 1 on the y-axis and move across to T 1 Draw from P 1 a line parallel to the tangent to D(t) at T 1 and draw from T 1 a vertical line. Label the point of intersection P 2 . Repeat to define the curve R(t) If R(t) does not pass through the end point, perturb the position of P 1 until it does P 1 A set of transshipment times ( t 1 , …, t n -1 ) is optimal if each line PQ is parallel to the tangent of D(t) at the receiving time, T ( ) t R
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LSA: Chapter 3 Lot size problem with variable demand: c r << c i Solution by Continuous approximation, Newell 1973 Search for continuous function to yield a set of t i with near minimal cost Works well if D’(t) changes smoothly: D’(t i )D’(t i+1 ) (i) Assume an optimal solution is found; let I i = i th interval between receiving times [ t i -1 ,t i ) , i= 1,2 ,… (ii) Divide the total cost over study period ( ) i i f i area c c cost + = cost otal t cost i i = D(t) D’(t i’ ) 1 i t i t ' i t ( ) 1 i i t t D’(t i’ ) ( ) ( ) ( ) ( ) dt t D t t t D t t area i t t i i i i i i i i ' 1 2 1 ' 2 1 2 1 ' ' 1 = = ( ) ( ) ' 2 1 2 1 ' : i i i i i' t D t t area t point at = One-dispatch motion cost Inter-dispatch Holding cost LSA: Chapter 3 Lot size problem with variable demand: c r << c i ( ) i i i s I t t t t H = , 1 Define a step function H s (t) such that: ( ) ( ) ( ) + = i i t t i s i s f i dt t D t H c t H c cost 1 ' ' 2 Approximate D’ ( t i’ ) by D’ ( t ) : - since D’(t) varies slowly ( ) ( ) ( ) + max 0 ' 2 cost total t s i s f dt t D t H c t H c Replace H s (t) with a smooth function H(t) ( ) ( ) ( ) + max 0 ' 2 cost total t i f dt t D t H c t H c ( ) ( ) 2 1 ' 2 = t D c c t H i f * Note that the integrand is a constant 0 t 1 t 2 t time 45 ( ) t H s
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LSA: Chapter 3 Lot size problem with variable demand: c r << c i 0 t 1 t 2 t time Headways P 1 45 ( ) t H ( ) t H s
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