Single-Period Stochastic Models 61 Newsvendor Models: Optimality of the Base-Stock Policy A newsvendor problem is a single-period stochastic inventory problem First assume that there is no initial inventory, ie, I 0 = 0 Let c be the cost to buy each unit; h be the holding cost for each unit of inventory left over at the end of the period; be the shortage cost for each unit of unsatisfied demand in the period; D be the demand of the period such that D is a continuous random variable ~ F; v() be the expected cost of the period for order uantity The objective is to minimize the total cost of the period, ie, min v() = c + he[d] + + E[D] + (1) As shown in the second set of notes, v() is a differentiable #1 first-order necessary condition, which is also sufficient, gives convex function in The c + hf() F c () = 0, () ie, F( * c ) = h (3) The minimum cost is achieved at ordering * that satisfies (3) Suppose that I 0 * If we order any uantity y, the expected total cost per period = v(i 0 +y) + cy v( * ) Therefore, it is optimal not to order if I 0 * Now consider 0 < I 0 < * Let us compare the two policies of ordering * I 0 and ordering y * I 0 The expected total cost of ordering *I 0 = v( * ) c * + c( * I 0 ) = v( * ) c I 0, while the expected total cost ordering y = v(i 0 +y) c(i 0 +y) + cy = v(i 0 +y) ci 0 v( * ) c I 0 It is optimal to order *I 0 The result shows that the otpimal policy in (3) is of base-stock type: If I 0 *, order nothing; otherwise, order * -I 0 Example 611 Consider the standard Newsvendor problem with c = $1/unit, h = $3/unit, #1 de[ D] de[ D] c c F( ); F ( ), where F ( ) = 1- F( ) d d 1
and = $/unit; and D ~ uniform[0, ] Find the optimal order uantity Sol First of all, it is intuitively clear that the optimal order uantity * is within [0, ] Then there are different ways to compute E[-D] + and E[D-] + Let f be the density function of D A standard way based on the form of [-D] + and the definition of density function is that E[D] + = 0 [ x] f( x) dx = 0 ( x) f( x) dx + [ x] f( x) dx = 00 Another way based on conditional probability is that E[D] + = P( > D)E[(D) + > D] = = 00 One way or another, E[D] + = P(D > )E[(D) + D > ] = = ( ) 00 3 Therefore, v() = + ( ) 00 + 00 The first-order condition is that 3 ( ) v'( ) 1 0 3 1, ie, * = 5 0 Consider a variant of the minimum cost problem Let D be the demand of a particular newspaper at a news stand such that D ~ F and E(D) = Each piece of newspaper is bought at c and sold at p There is no salvage value of a piece of newspaper at the end of a day Let r() be the expected profit if the newsstand orders pieces Find * that maximizes r() max r() = pe[min(d, )] c (4) We are going to express (4) in a form similar to (1): First ignore the order uantity Suppose every unit of demand is satisfied Then the expected profit is (p-c)e(d) = (p-c) This is an over estimate Some demands are unsatisfied when is smaller than D; some items are bought but are left over as scrap when is larger than D The expected profit loss in the former case is (p-c)e(d-) + and the expected profit loss in the latter case is ce(d-d) + Thus, max r() = (p-c) - [(p-c)e(d-) + + ce(-d) + ] (5) The derivation can be more analytical Note that
min (D, ) = D + max(d, ) = D [D ] +, and (6) = D + [ D] + [D ] + (7) r() = pe[min(d, )] c = pe[d] pe[d-] + ce[d] ce[ D] + + ce [D ] + = (p c) [(p c)e[d ] + + ce[ D] + ] As the first term is constant, to maximize r() is the same as minimizing v() = (p-c)e[d-] + + ce[-d] +, which is in the same form of (1) without the fixed ordering cost From (3), the optimal order uantity is given by F( * p c ) = (8) p Remark 611 There are euivalent forms of r(p) other than the one given above Instead of (6), min (D, ) = D + max(d, ) = max(0, D) = [D] + r() = pe[min(d, )] c = p p[d] + c, which leads to the same result as (8) Example 61 Consider a similar context as Example 611, except that each unit of item is sold for $5, and the objective is to maximize the expected cost Sol We are going to use the euivalent expression of r() shown in Remark 611 r() = pe[min(d, )] c he[d] + E[D] + = (pc) (p+h)e[d] + E[D] + With E[D] + and E[D] + found in Example 611, r() = (pc) (p+h) 00 ( ) 00, and r'() = (pc) (p+h) ( ) + 3
The first-order condition r'(p) = 0 gives (pc) + = ( ph ) ; ie, * = p c ph = 51 53 = 06, which gives * = 60 6 Fixed-Cost Models: Optimality of the (s, S) Policy Consider the same cost structure of c, h,, D, and v() leading to (1), plus the additional setup cost K for placing an order (ie, > 0) The objective is still to minimize the total cost in the period In this case, what should the optimal inventory policy be? In the newsvendor model without any setup cost, the base-stock policy is always to order up to * if I 0 < *, and order nothing otherwise However, with the setup cost, the optimal ordering policy can be different Let S * be the optimal on-hand stock in this case, no matter how S * is found If I 0 is only slightly less than S *, ie, the benefit of having on-hand S * over I 0 is not too big, it may not be worthwhile to pay the setup cost K (and additional variable cost on ordering) to change the on-hand inventory to S * The following analysis expresses the above idea rigorous and derivates the optimal inventory policy First let v() = c + he[-d] + + E[D-] + be the sum of the variable cost in ordering, the expected inventory holding cost, and the expected shortage cost Recall that v() is a convex function with a well-defined minimum Let S * be the value of that minimizes v() Let s * < S * such that v( * ) = K+v(S * ) Refer to Figure 1 for the definitions of s * and S * We claim that the optimal inventory (production) policy is of (s, S) type, ie, order S * -I 0, if I 0 < s *, order nothing, if I 0 s * v(s * ) v() v(s * ) K s * S * Figure 1 The Definition of s * and S * 4
I 0 > S * To establish the claim, consider four cases: (i) I 0 < s *, (ii) I 0 = s *, (iii) s * < I 0 S *, and (iv) Case (i) I 0 < s * : Ordering incurs the cost K+c(S * -I 0 ) for the order The on-hand inventory becomes S *, and the expected inventory holding and shortage costs will be v(s * )-cs * The deduction of cs * is necessary because v(s * ) includes the variable cost to buy S * items, which is fact is extra Thus, the expected total cost of ordering is K+c(S * -I 0 )+v(s * )-cs * = K+v(S * )-ci 0 By a similar argument, the expected total cost of not ordering is v(i 0 )-ci 0 Since I 0 < s *, v(i 0 ) > v(s * ) = K+v(S * ) The (s, S) policy is optimal in this case Case (ii) I 0 = s * : From the argument, ordering or not makes no difference The (s, S) policy is optimal in this case Case (iii) s * < I 0 S * : The two expressions of the expected total cost remain valid, ie, the expected total cost for is K+v(S * )-ci 0 = v(s * )-ci 0 and the expected total cost for not ordering is v(i 0 )-ci 0 For I 0 in the given range, v(i 0 ) < v(s * ) It is better not to order, and the (s, S) policy is optimal in this case Case (iv) I 0 > S * : Clearly there is no point to order and the (s, S) policy is optimal Example 61 Consider the same context as Example 611 Now there is a fixed cost K = $5 per order Find the order uantity * that minimizes the expected cost Sol v() = c + he[d] + + E[D] + In the context of the problem, v() = c + h ( ) 00 00 = 00 By construction, v'(s * ) = 0, ie, S * = 0, and v(s * ) = (0) ()(0) 00 90 s * is defined by the value of < S * such that v(s * ) = K+v(S * ) = 95 * * ( s ) s 00 95, ie, s * = 0 00 = 58579 The optimal policy is to order 0I 0 if I 0 < 58579, and do nothing otherwise 5