$3\mathrm{.}2$ THE SILVER$-$MEAL HEURISTIC The Silver$-$Meal heuristic provides an alternative method for lot sizing. This heuristic is a forward method that requires determining the average cost per period as a function of the number of periods the current order is to span, and stopping the computation when this function first increases. Define C$(7)$ as the average holding and setup cost per period if the current order spans the next T periods. As above, let $(1,...,$n$)$ be the requirements over the n$-$period horizon. Consider period $1,$ then we just incur the order cost K$.$ Hence, C$(1)=$ K$.$ If we order enough in period $1$ to satisfy the demand in both periods $1$ and $2,$ then we must hold r$2$ for one period. Hence, C$(2)=($K $+$ h$*$r$2)/2.$ Similarly, if we order enough in period $1$ to satisfy the demand in periods $1,2,$ and $3,$ then we must hold r$2$ for one period, and r$3$ for two periods. C$(3)=($K$+1*$h$*$r$2+2*$h$*$r$3)/3,$ and, in general, C$($j$)=($K$+$h$*$r$2+2*$h$*$r$3+...+($j$-1)*$h$*$r$)/$j$.$ Once C$($j$)>$ C$($j$-1),$ we stop and set y$1=$r$1+2+...+-1,$ and begin the process again starting at period j$.$ y represents the size of the order $($lot size$).$ EXAMPLE A machine shop uses the Silver$-$Meal heuristic to schedule production lot sizes for computer casings. Over the next five weeks the demands for the casings are r$=(18,30,42,5,20).$ The holding cost is $$2$ per case per week, and the production setup cost is $$80.$ Find the recommended lot sizing. Solution Starting in period $1$ C$(1)=80$ C$(2)=(80+(2)(30))/2=70$ C$(3)=(80+(2)(30)+(2)(2)(42))/3=102\mathrm{.}67.$ Therefore, stop because C$(3)>$ C$(2).$ And, we set the first order lot as y$1=$r$1+$r$2=18+30=48.$ Notice that this order covers the first two requirements. Starting in period $3$: C$(1)=80$ C$(2)=(80+(2)(5))/2=45,$ C$(3)=(80+(2)(5)+(2)(2)(20))/3=56\mathrm{.}67.$ Then, stop. Set y$3=$r$3+$r$4=42+5=47.$ Because period $5$ is the final period in the horizon, we do not need to start the process again. We set y$5=$r$5=20.$ Hence, the Silver$-$Meal heuristic results in the policy y$=(48,0,47,0,20).$ Homework Exercise #$4$ Consider the Valve Casing Assembly from the Harmon Music Company example. Determine the planned order release using the Silver Meal Heuristic. Recall that the Time$-$Phased Net Requirements are the ones shown in the table below. Week $45667$ Gross Requirements $7891011121314151617424232122611245147638424232122611245147638424232122611245147638$ Net Requirements Time$-$Phased Net Requirements Hint: the time$-$phased net requirements represent the r vector used in the explanation of the Silver$-$Meal Heuristic. That is$,$ in this problem: r$=(42,42,32,12,26,112,45,14,76,38)$