Page 576 Question 2

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.

Page 576 Answer 2

Imagine you’re selling. In fact, imagine you’re selling a lot; the product is flying off the shelves so quickly that you can’t keep up with the demand. So what do you do? Stop selling? Disappoint customers, lose sales and drive them into the arms of a competitor? No. To prevent that from happening, the argument and introduction of backordering becomes very attractive business solution. Backordering is the process of allowing your customers to shop your products even when you don’t have sufficient stock on hand.

Running business with backorders and a reliable supplier works. You’ll save a on carrying costs – no added storage and service costs needed to store inventory – especially if you dropship your backorders, your customers won’t even have to wait for their backorders to reach you. Your supplier will fulfill these for you, further cutting down your customers’ waiting time.
 

There are some real costs associated with not having products on hand when customers want them. Some of the costs are tangible. For instance, Company ABC will have to spend a lot more money on expedited shipping to get parts from its suppliers faster in order to fill the order. Then, it may have to pay its laborers a lot of money in overtime in order to hasten production, and when it finally has the 100,000 units ready to ship, it might have to overnight those items to customers who are anxious about having the widgets in time. In addition, backorder have intangible costs as well. Customers may get irate and buy from a competitor instead.


The model in figure 13.7 is assuming that inventories are allowed to run negative. a situation in which the amount of goods available at a particular time or from a particular place appears to be less than zero, often because of a delay by accounts in recording their movement

alt text

Model formulation:
I will be referencing the above figure
Q= The order quantity
B = the maximum amount of backlog allowed
Q-B remaining units of on-hand inventory satisfies demand in the first sub cycle.
T1 = The first sub-cycle (ADC)
T2= The second sub-cycle (CEF) is T2 during which demand is backordered
The total length of a cycle is T = T1 +T2.
r = number of items required per unit time
s = storage cost per unit time
w = stock out cost per unit time
d= delivery cost per delivery

The on-hand inventory decreases at rate \(\lambda\). and becomes zero in T1 time, Q-B = \(\lambda\)T1.
In the second sub-cycle, the number of backorders increases from 0 to B at rate ?? over a period of length T2 time. Thus:B= \(\lambda\)T2  

and Q = \(\lambda\)(T1 +T2) = \(\lambda\)T.
First compute the average inventory per cycle and then multiply the result by the holding cost w to get the annual holding cost. The average inventory per cycle is equal to:
\[\frac {(Area of triangle ADC)} {T} = \frac {\frac {(Q-B)T_1} {2}} {T}\]

next substitute for \(T_1\) and T, \[Average Inventory per Cycle =\frac { \frac {(Q - B)^2} {\lambda}} {\frac {2Q} {\lambda}}\] \[Average Inventory per Cycle =\frac {(Q - B)^2} {2Q} \]

Before we apply the cost of backordering a unit for time T, we need to first compute the average number of backorders per cycle.
Since all cycles are alike, this means that the average number of outstanding backorders per year is the same as the average per cycle. To get the average backorder cost per year, we multiply the average backorder quantities per year by the backorder cost rate. The average number of backorders per cycle is equal to the area of triangle CEF (the Figure) divided by the length of the cycle T:

\[\frac {Area of triangle CEF} {T} = \frac {\frac {B_2} {2}} {T} \] \[\frac {Area of triangle CEF} {T} = \frac {\frac {B^2} {2 \lambda}} {\frac {Q} {\lambda}} \] \[\frac {Area of triangle CEF} {T} = \frac {B^2} {2Q}\]

Therefore, the average backorder cost is equal to \[w\frac {B^2} {2Q}\] We now combine all the cost components and express the average annual cost of managing inventory as:

\[\frac {(Q - B)^2} {2Q} + w\frac {B^2} {2Q}\]

Now we will add the storage cost, demand, and fixed delivery cost to the above equation, we get:

\[ r\lambda + \frac {d\lambda} {Q} + \frac {s(Q - B)^2} {2Q} + w\frac {B^2} {2Q}\]

\[F(x) = r\lambda + \frac {d\lambda} {Q} + \frac {s(Q - B)^2} {2Q} + w\frac {B^2} {2Q} \qquad Equation.1\]

Now, all we have to do is to obtain the optimal solution, we take the first partial derivatives of F(B,Q) in Equation.1 with respect to Q and B and set them equal to zero. using mathematical tools, \(Q^*\) can be expressed as follow:

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

\[Q^* = Q_e \sqrt {\frac {s+w} {w}}\]

where \(Q_e\) is the optimal solution.

Page 584 Question 2

Find the local minimum value of the function:  

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

Page 584 Answer 2

First, let’s find the derivatives:

with respect to \(x\) \[ \frac{\partial f}{\partial x} = 6x + 6y - 2 \] with respect to \(y\) \[ \frac{\partial f}{\partial y} = 6x + 14y + 4 \]

Now, we will assign the partial dx to zero: \[ \frac{\partial f}{\partial x} = 6x + 6y - 2 \] \[6x + 6y - 2 = 0\] \[ x = \frac {2 - 6y} {6}\] \[ x = \frac {1} {3} - y\]

Then, we will assign the partial dy to zero and substitute x: \[6x + 14y + 4 = 0\] \[6(\frac {1} {3} - y) + 14y + 4=0\] \[2 - 6y + 14y + 4=0\] \[6 - 6y + 14y =0\] \[6 +8y =0\] \[y= \frac{-3} {4}\]

Now we need to find x: \[ x = \frac {1} {3} - y\] \[ x = \frac {1} {3} - \frac{-3} {4}\] \[ x= \frac{13} {12}\]

Therefore, the local minimum value of the function \[f(x,y) = 3x^2 + 6xy + 7y^2 - 2x + 4y \] happens at point: \[(x,y) = (13/12, -3/4)\]

library(scatterplot3d)
x <- seq(-4, 4, length= 500)
y<- seq(-4, 4, length= 500)
fxy <- function(x,y){
  return(3*x^2 +6*x*y + 7*y^2 - 2*x + 4*y) 
}
z<- fxy(x,y)

scatterplot3d(x,y,z,  main="3D f(x,y) = 3x^2+6xy+7y^2-2x+4y ")

library(spatialEco)
## Warning: package 'spatialEco' was built under R version 3.2.5
## spatialEco 0.1-4
## Type se.news() to see new features/changes/bug fixes.
 x <- seq(-4, 4, length= 1000)
 y<- seq(-4, 4, length= 1000)
 fxy <- function(x,y){
   return(3*x^2 +6*x*y + 7*y^2 - 2*x + 4*y) 
 }
 z<- fxy(x,y)
 mean1<- mean(z)
 dev= mean
 (lmm <- local.min.max(z, dev=mean1, add.points=FALSE, main="Local Minima and Maxima") )

## $minima
## [1] -0.06240475
## 
## $maxima
## [1] 248 264
## 
## $devmin
## [1] 85.56658
## 
## $devmax
## [1] 162.4958 178.4958

Page 591: Question #5

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

\[ 4x^2 + y^2+4z^2 = 16\] Where the temperature function is: \[T(x,y,z) = 8x^2 +4yz -16z+600\]

Page 591: Answer #5

let us find the maximum value of T using Lagrange multipliers. let us introduce \(\lambda\). the function for our problem: \[ L(x,y,z, \lambda) = 8x^2 +4yz -16z+600 -\lambda(4x^2+y^2+4x^2-16) \qquad Equation.1\]

The solution methodology is to take partial derivative of equation.1 with respect to x,y,z,\(\lambda\) and set them equal to zero. Next find the critical points and check for which critical point The value of the function is maximum.

\[ \frac{\partial L}{\partial x} = 16x - 8x \lambda\] \[ \frac{\partial L}{\partial y} = 4z - 2\lambda y\] \[ \frac{\partial L}{\partial z} = 4y - 16 - 8z\lambda\] \[ \frac{\partial L}{\partial \lambda} = -(4x^2+y^2+4z^2-16)\]

Next, we will set partial derivatives equal to zero:

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

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

Next, \[ \frac{\partial L}{\partial z} = 4y - 16 - 8z\lambda\] \[ 4y - 16 - 8z\lambda = 0\] \[ 4y - 16 - 8y\lambda = 0\] \[ 4y - 16 - 16y = 0\] \[ - 16 - 12y = 0\] \[ - 16 = 12y \] \[ y= -16 / 12\] \[ y = -4/3\]

We also have z = y. Then z = -4/3.

Then, finally,

\[ \frac{\partial L}{\partial \lambda} = -(4x^2+y^2+4z^2-16)\] \[ -(4x^2+y^2+4z^2-16) = 0\] \[ 4x^2+y^2+4z^2-16 = 0\] \[ 4x^2+5y^2-16 = 0\] \[ 4x^2+5y^2 =16\] \[ x^2 = \frac {16-5y^2} {4} \] \[ x^2 = 4 - \frac {5y^2} {4} \] \[ x^2 = 4 - 20/9 \] \[ x = 4/3 \]

Therefore, the hottest point (x,y,z) is: ( 4/3, -4/3, -4/3).