```! Infinite period newsvendor problem with lost sales, setup cost, and lead time = 1.!(Lost_sales1.lng). Solved by formulating as a Markov decision problem; ! Keywords: Markov decision process, inventory, lost sales; SETS: INVL; ! Possible inventory levels; ORDER; ! Possible order amounts; DD: PD; ! Possible demands & probability; ! Our decision transitions: Given a current inventory level, the next state is the inventory level & order level; DTRAN(INVL, ORDER): CD, X; ! Probabilistic transitions: given I0 and R0, the demand D gives the new inventory level I1 = max(0,I0-D0) + R, so R <= I1 <= I0 + R; PTRAN(INVL,ORDER, INVL)| &2 #LE# &3 #AND# &3 #LE# &1 + &2 - 1 : CP, PR; ! Notation: X(i,r) = joint probability of entering period in with inventory i available to satisfy today's demand and deciding to order an amount r to be delivered tomorrow. PR(i,r,s)= probability of having an amount s available for tomorrow, given there is i available today, and amount r on order for tomorrow. CD(i,r)= cost of making decision t when state is s, CP(i,r,s)= expected cost when nature moves state to u given state i,r PD(d) = probability demand is d-1, for d = 1, 2, ...; ENDSETSDATA: V = 18; ! Revenue per unit sold; P = 2; ! Explicit shortage penalty on each lost sale; H = .7; ! Holding cost/unit per period; C = 3; ! Purchase cost per unit; K = 1; ! Fixed order cost; ! Demand distribution. Lowest corresponds to 0 demand; DD = D0 D1 D2 D3 D4; PD = .01 .90 .04 .03 .02; ORDER = R0..R5; INVL = I0..I7; ! Possible inventory levels. Lowest corresponds to 0 inventory; ENDDATA !--------------------------------------------------; ! The sequence of events each day is: 1) Begin period with an amount i, including order about to arrive, available to satisfy today's demand, 2) We order an amount r for delivery tomorrow, 3) Order scheduled for today arrives, 3) Demand occurs and revenue is realized, 4) Shortage penalties or holding charges are assessed; CALC: ! Compute costs of our decisions or deterministic transitions, note order size is r-1; @FOR(DTRAN(i,r): CD(i,r) = K*(r#GT#1) + C*(r-1)); ! Compute conditional probabilities and expected costs of nature's moves, or Probabilistic transitions, with i = current inventory level, r = order size, s = new inventory level available tomorrow. Note, indexing starts at 1, so index 1 corresponds to inventory or demand = 0; ! No stockout today case; ND = @SIZE(DD); @FOR(PTRAN(i,r,s) | r #LT# s: PR(i,r,s) = @IF(i+r-s #LE# ND,PD(i+r-s),0); CP(i,r,s) = H *(s-r) - V*(i+r-s-1); ); ! Out of stock today case; @FOR(PTRAN(i,r,s) | r #eq# s: PR(i,r,s) = @SUM(DD(u)| u #ge# i: PD(u)); @IFC( PR(i,r,s) #GT# 0: CP(i,r,s) = P*@SUM(DD(u)| u #ge# i: PD(u)*(u-i))/PR(i,r,s) - V*(i-1); @ELSE CP(i,r,s) = 0; ); ); ENDCALC !Minimize the average cost per period; MIN=@SUM(DTRAN( i,r): CD(i,r)*X(i,r)) + @SUM(PTRAN(i,r,s): CP(i,r,s)*X(i,r)*PR(i,r,s)); !The probabilities must sum to 1; @SUM( DTRAN(i,r): X( i,r)) = 1; !Rate at which we exit state i = rate of entry to i; @FOR( INVL(i): @SUM( DTRAN(i,r): X(i,r))= @SUM( PTRAN(i0,r0,s1)|s1 #EQ# i: X(i0,r0)* PR(i0,r0,i)) ); ```