Chapter 8, Problem 15P

Summary Introduction

To determine: Select the better site using center of gravity.

Introduction: Location is where a firm chooses to site its operations. Location decisions can large effects on expenses and incomes. Location choices are normally quite imperative to both substantial and private companies. The area choice directly affects an operation's expenses and also its capacity to serve clients.

Answer to Problem 15P

The better site using center of gravity is L2.

Explanation of Solution

Given Information:

It is given that L1 and L2 are two possible sites.

Destination	Quantity
D1	900
D2	300
D3	700
D4	600
D5	1,200

Determine the coordinates of each destination using the given map:

Destination	X	Y	Quantity
D1	1	2	900
D2	2	4	300
D3	3	1	700
D4	4	2	600
D5	5	3	1,200

Determine the total quantity

It is calculated by adding all the sum of quantity.

$\begin{array}{lll}\text{Total}\text{Quantity}\hfill & =\hfill & \left[{}_{{\text{Quantity}}_{\text{D4}}+{\text{Quantity}}_{\text{D5}}}^{{\text{Quantity}}_{\text{D1}}+{\text{Quantity}}_{\text{D2}}+{\text{Quantity}}_{\text{D3}}+}\right]\hfill \\ \hfill & =\hfill & [900+300+700+600+1,200]\hfill \\ \hfill & =\hfill & 3,700\hfill \end{array}$

Determine the Center of gravity for the optional location for each distribution center

It is calculated by multiplying each destination value with the quantity of each destination and the result is divided by total quantity.

$\begin{array}{lll}\text{Center}\text{of}{\text{Gravity}}_{\text{X}}\hfill & \text{=}\hfill & \left[\frac{\text{Destination}{\text{1}}_{\text{X}}\text{×}\text{Quantity}}{\text{Total}\text{Quantity}}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{(1\text{×}900)+(2\text{×}300)+(3\text{×}700)+(4\text{×}600)+(5\text{×}1,200)}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{900+600+2,100+2,400+6,000}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{12,000}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & 3.2432\text{or}\text{3}\text{.2}\hfill \\ \text{Center}\text{of}{\text{Gravity}}_{\text{Y}}\hfill & \text{=}\hfill & \left[\frac{\text{Destination}{\text{1}}_{\text{Y}}\text{×}\text{Quantity}}{\text{Total}\text{Quantity}}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{(2\text{×}900)+(4\text{×}300)+(1\text{×}700)+(2\text{×}600)+(3\text{×}1,200)}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{1,800+1,200+700+1,200+3,600}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & \left[\frac{8,500}{3,700}\right]\hfill \\ \hfill & \text{=}\hfill & 2.2972\text{or}\text{2}\text{.3}\hfill \end{array}$

Determine the coordinates for L1 and L2.

The below are the coordinates for L1 and L2: which are found using the map

$\begin{array}{c}{\text{Coordinates}}_{\text{L1}}\approx 2.6,2.4\\ {\text{Coordinates}}_{\text{L2}}\approx 3.5,2.5\end{array}$

Determine the distance between each location and the center of gravity.

It is calculated by adding the difference in X coordinates with the difference in Y coordinates.

$\begin{array}{lll}\text{Distance}{\text{between}}_{\text{Center}\text{of}\text{Gravity\&L1}}\hfill & \text{=}\hfill & \left|{\text{Difference}}_{\text{Coordinate}\text{X}}\right|+\left|{\text{Difference}}_{\text{Coordinate}\text{Y}}\right|\hfill \\ \hfill & \text{=}\hfill & \left|2.6-3.2\right|+\left|2.4-2.3\right|\hfill \\ \hfill & \text{=}\hfill & 0.6+0.1\hfill \\ \hfill & \text{=}\hfill & 0.7\hfill \\ \text{Distance}{\text{between}}_{\text{Center}\text{of}\text{Gravity\&L2}}\hfill & \text{=}\hfill & \left|{\text{Difference}}_{\text{Coordinate}\text{X}}\right|+\left|{\text{Difference}}_{\text{Coordinate}\text{Y}}\right|\hfill \\ \hfill & \text{=}\hfill & \left|3.5-3.2\right|+\left|2.5-2.3\right|\hfill \\ \hfill & \text{=}\hfill & 0.3+0.2\hfill \\ \hfill & \text{=}\hfill & 0.5\hfill \end{array}$

Hence, L2 is closer to the better site using center of gravity