Problem 2, Pg 576

Consider a company that allows back ordering. That is, the company notifies customers that a temporary stock-out exists and that their order will be filled shortly. What conditions might argue for such a policy? What effect does such a policy have on storage costs? Should costs be assigned to stock-outs? Why? How would you make such an assignment? What assumptions are implied by the model in Figure 13.7? Suppose a “loss of goodwill cost”" of \(w\) dollars per unit per day is assigned to each stock-out. Compute the optimal order quantity \(Q^*\) and interpret your model.

Figure 13.7

Figure 13.7

If inventory costs are very high or there is little demand for an item, keeping a minimum inventory that could allow for lock-outs may be justified. This is very common as with carry almost all items would not make sense for companies like Amazon and Home Depot.

Costs should be assigned to stock-outs as there would be delivery costs assigned to order/deliver the item and there is a chance that the customer would go elsewhere to procure the item, i.e. a cost applied to a lost sale or lost revenue.

In Figure 13.7, the quantity appears to go into the negative after some time \(T-t\). This is a state when there is a back order for the item in question. Creating a model for when the inventory goes into this state is structured as follows:

Let:

\[cost\space per\space cycle = d + \frac{wB}{2}t^* + s\frac{(Q-B)(t-t^*)}{2}\]

Noting similar triangles, it can be seen from Figure 13.7 that:

\[\frac{Q-B}{t-t^*}=\frac{B}{t^*} \\ \rightarrow t^* = \frac{tB}{Q}\]

\[cost\space per\space cycle = d + \frac{twB^2}{2Q} + s\frac{(Q-B)(t-\frac{tB}{Q})}{2} \\ = d + \frac{twB^2}{2Q} + \frac{s}{2}(Qt-2tB+\frac{tB^2}{Q})\]

Dividing through by the time between orders \(t\):

\[c = \frac{d}{t} + \frac{wB^2}{2Q} + \frac{s}{2}(Q-2B+\frac{B^2}{Q})\]

It can be seen from Figure 13.7 that the slope r is \(\frac{Q}{t}\). Incorporating this into the equation from above:

\[c = \frac{dr}{Q} + \frac{wB^2}{2Q} + \frac{s}{2}(Q-2B+\frac{B^2}{Q}) \space\space (Eq. 1)\]

Taking the partial derivative with respect to \(B\) and simplifying

\[\frac{\partial c}{\partial B} = \frac{(w+s)B-Qs}{Q} = 0\]

the second derivative with respect to \(B\) is

\[\frac{\partial^2 c}{\partial B^2} = \frac{w+s}{Q}\]

which is always positive, so we will solve for B setting the first derivative to 0.

\[B = \frac{Qs}{w+s} \space\space (Eq. 2)\]

Taking the partial derivative with respect to \(Q\) and simplifying

\[\frac{\partial c}{\partial B} = \frac{sQ^2-B^2w-B^2s-2dr}{2Q^2} = 0\]

the second derivative with respect to \(Q\) is

\[\frac{\partial^2 c }{\partial B^2} = \frac{B^2w + B^2s + 2dr}{Q^3}\]

which is always positive.

Substituting \(B\) from Eq. 2, simplifying and setting the first derivative to 0:

\[\frac{sw}{w+s} - \frac{sqQ^2-2drw+2drs}{2(w+s)Q^2} = 0\]

Solving for \(Q\), which is \(Q^*\), yields:

\[Q^* = \sqrt{2} \sqrt{\frac{rd(s+w)}{ws}}\]

Problem 2, Pg. 584

Find the local minimum value of the function

\[f(x,y) = 3x^3+6xy+7y^2-2x+4y\]

Find the first derivative with respect to \(x\):

\[\frac{\partial f}{\partial x} = 6x+6y-2\]

The second derivative would be \(6\). Setting the first derivative with respect to \(x\) to 0 and solving:

\[6x+6y-2 = 0\]

\[x = \frac{1-3y}{3} \space \space (Eq. 1)\]

Find the first derivative with respect to \(y\):

\[\frac{\partial f}{\partial y} = 6x+14y+4\]

The second derivative would be \(14\). Setting the first derivative with respect to \(y\) to 0, substituting Eq. 1 and solving:

\[2(1-3y)+14y+4 = 0\] \[8y+6 = 0\] \[y = \frac{-3}{4} \space \space (Eq. 2)\] Substituting Eq. 2 into Eq. 1

\[x = \frac{1+\frac{9}{4}}{3} = \frac{13}{12}\]

Therefore, the local minimum would be at the point \((\frac{13}{12}, \frac{-3}{4})\).

Problem 5, Pg. 591

Find the hottest point \((x, y, z)\) along the elliptical orbit

\[4x^2+y^2+4z^2 = 16\] where the temperature function is

\[T(x, y, z) = 8x^2 +4yz-16z+600\]

Using Lagrange multipliers:

\[L(x, y, z, \lambda) = 8x^2 +4yz-16z+600 - \lambda(4x^2+y^2+4z^2 - 16) \space \space (Eq. 1)\]

Taking partial derivatives and setting them to 0:

\[\frac{\partial L}{\partial x} = 16x-8\lambda x = 0\] \[\lambda = 2\]

\[\frac{\partial L}{\partial y} = 4z-2 \lambda y = 0\]

\[z = y\]

\[\frac{\partial L}{\partial z} = 4y - 16 - 8\lambda z = 0\]

\[y = \frac{-4}{3}\]

\[\frac{\partial L}{\partial \lambda} = -4x^2-y^2-4z^2+16 = 0\]

\[x = 4/3\]

Therefore, the hottest point is at \((\frac{4}{3}, -\frac{4}{3}, -\frac{4}{3})\)