Chapter 11, Problem 56AP

(a)

Summary Introduction

Interpretation: A company is considering where to locate its cafeteria to service six buildings. The location of the buildings and the fraction of the company’s employees working at these locations are given below. The optimal location of the cafeteria to minimize the weighted rectilinear distance to all the buildings needs to be determined.

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Concept Introduction: When the differences between the $x\text{ and }y$ coordinates are added, then it gives the Recti-linear distance between two points along an orthogonal path.

Answer to Problem 56AP

The optimal location of the cafeteria to minimize the weighted rectilinear distance to all the buildings has been determined as $\left(4,\left[3,6\right]\right)$ .

Explanation of Solution

Given information: The locations of the buildings and the fraction of company’s employees working at these locations are as given below:

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

In order to find out the optimal location of the cafeteria to minimize the weighted rectilinear distance to all the buildings, the optimal coordinates a and b needs to be derived.

Optimal coordinate a:

The fraction of workforce needs to be multiplied by 12 for normalization.

Building	a	Weight	Cumulative Weight
B	1	1	1
A	2	1	2
C	3	2	4
E	4	3	7
D	5	3	10
F	10	2	12

By dividing the total cumulative value by 12 and finding out the first location where the cumulative weight exceeds this value, following is calculated:

The first time where the cumulative weight exceeds the value $\frac{12}{2}=6$ , is at $a=4$ .

Optimal Coordinate b:

The fraction of workforce needs to be multiplied by 12 for normalization.

Building	a	Weight	Cumulative Weight
B	0	1	1
E	2	3	4
C	3	2	6
A	6	1	7
F	7	2	9
D	9	3	12

By dividing the total cumulative value by 12 and finding out the first location where the cumulative weight exceeds this value, following is calculated:

The first time where the cumulative weight exceeds the value $\frac{12}{2}=6$ , is at $b=6$ . However, at $b=3$ there is an exact value of 6. Hence the optimal value of coordinate $b$ is in closed interval $\left[3,6\right]$ .

Therefore the optimal location of the cafeteria to minimize the weighted rectilinear distance to all the buildings is $\left(4,\left[3,6\right]\right)$ .

(b)

Summary Introduction

Interpretation: A company is considering where to locate its cafeteria to service six buildings. The locations of the buildings and the fraction of the company’s employees working at these locations are given below. The optimal location of the cafeteria to minimize the maximum rectilinear distance to all the buildings needs to be determined.

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Concept Introduction: When the differences between the $x\text{ and }y$ coordinates are added, then it gives the Recti-linear distance between two points along an orthogonal path.

Answer to Problem 56AP

All the points on line connecting $\left(2,7\right)\text{ and }\left(6.5,2.5\right)$ provides the optimal location of the cafeteria.

Explanation of Solution

Given information: The locations of the buildings and the fraction of company’s employees working at these locations are as given below:

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Following is the calculation for finding out the optimal location of the cafeteria to minimize the maximum rectilinear distance to all the buildings:

  $c_1,c_2,c_3,c_4\text{ and }{c}_5$ are being defined as shown below:

  $\begin{array}{l}{c}_1=\min (a_1+b_1)\\ c_2=\max (a_1+b_1)\\ c_3=\min (-a_1+b_1)\\ c_4=\max (-a_1+b_1)\\ c_5=\max (c_2-c_1,c_4-c_3)\end{array}$

Assuming $x_1=\frac{(c_1-c_3)}{2}$

Recitlinear distance $(a+b)$

A

2

6

8

B

1

0

1

C

3

3

6

D

5

9

14

E

4

2

6

F

10

7

17

Substituting the obtained values:

  $\begin{array}{l}{c}_1=\min (a_1+b_1)\\ \text{ =min(8,1,6,14,6,17)}\\ \text{ = 1}\\ c_2=\max (a_1+b_1)\\ \text{ =}\max (8,1,6,14,6,17)\\ \text{ = 17}\end{array}$

  $\begin{array}{l}{c}_3=\min (-a_1+b_1)\\ \text{ =min(}4,-1,0,4,-2,-3)\\ \text{ = }-3\end{array}$

  $\begin{array}{l}{c}_4=\max (-a_1+b_1)\\ \text{ = 4}\end{array}$

  $\begin{array}{l}{c}_5=\max (c_2-c_1,c_4-c_3)\\ \text{ =}\max (17-1,4+3)\\ \text{ =max(16,7)}\\ \text{ = 16}\end{array}$

Thus

  $\begin{array}{l}{x}_1=\frac{(c_1-c_3)}{2}=\frac{(1+3)}{2}=2\\ y_1=\frac{(c_1+c_3+c_5)}{2}=\frac{(1-3+16)}{2}=7\\ x_2=\frac{(c_2-c_4)}{2}=\frac{(17-4)}{2}=6.5\\ y_2=\frac{(c_2+c_4-c_5)}{2}=\frac{(17+4-16)}{2}=2.5\end{array}$

Hence all the points on line connecting $\left(2,7\right)\text{ and }\left(6.5,2.5\right)$ provides the optimal location of the cafeteria.

(c)

Summary Introduction

Interpretation: A company is considering where to locate its cafeteria to service six buildings. The locations of the buildings and the fraction of the company’s employees working at these locations are given below. The gravity solution needs to be determined.

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Concept Introduction: In a gravity modeling, locations are determined that minimizes the cost of transportation of raw materials from the supplier as well as manufactured goods to the market.

Answer to Problem 56AP

The optimal solution derived through gravity method is $\left(4.67,4.91\right)$

Explanation of Solution

Given information: The locations of the buildings and the fraction of company’s employees working at these locations are as given below:

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

A company might be considering to provide a new facility in a region, for various reasons, like increasing the capacity of production of a particular existing product, introduction of a new product, or finding new markets in order to have business on their existing or new products. The smallest productive entity wherein a single commodity is manufactured is called a facility. In order to identify appropriate geographic locations in a region, gravity location models are used.

In the gravity solution method it is assumed that there is a linear growth in the transportation costs, with the quantity that is being shipped. In this case all the distances between two points are being calculated as the geometric distance on a plane.

Using gravity modeling, the optimal solution can be obtained by the following formula:

  $x=\frac{{\displaystyle \sum _{i=1}^n{w}_i{a}_i}}{{\displaystyle \sum _{i=1}^n{w}_i}}y=\frac{{\displaystyle \sum _{i=1}^n{w}_i{b}_i}}{{\displaystyle \sum _{i=1}^n{w}_i}}$

Following tabulation gives the values of coordinates multiplied by their respective weights:

a	b	Weight (w)	wa	wb
2	1	1	2	6
1	1	1	1	0
3	2	2	6	6
5	3	3	15	27
4	3	3	12	6
10	2	2	20	14
		Sum=12	Sum=56	Sum=59

Substituting the summation values of w, wa and wb in the formula:

  $x=\frac{{\displaystyle \sum _{i=1}^n{w}_i{a}_i}}{{\displaystyle \sum _{i=1}^n{w}_i}}=\frac{56}{12}=4.67$

  $y=\frac{{\displaystyle \sum _{i=1}^n{w}_i{b}_i}}{{\displaystyle \sum _{i=1}^n{w}_i}}=\frac{59}{12}=4.91$

Through gravity method the optimal solution arrived at is $\left(4.67,4.91\right)$

(d)

Summary Introduction

Interpretation: A company is considering where to locate its cafeteria to service six buildings. The locations of the buildings and the fraction of the company’s employees working at these locations are given below. Supposing that the cafeteria must be located in one of the buildings; the particular building, minimizing the weighted rectilinear distance to all the buildings, needs to be identified.

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Concept Introduction: When the differences between the $x\text{ and }y$ coordinates are added, then it gives the Recti-linear distance between two points along an orthogonal path.

Answer to Problem 56AP

The weighted rectilinear distance for C and E are minimum and hence the optimal solution will be either C or E.

Explanation of Solution

Given information: The locations of the buildings and the fraction of company’s employees working at these locations are as given below:

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

The tabulation showing the coordinates along with their weights for various buildings is as shown below:

Building	a	b	Weights
A	2	6	1
B	1	0	1
C	3	3	2
D	5	9	3
E	4	2	3
F	10	7	2

Below mentioned formula gives the rectilinear distance of a building from other buildings:

  $\begin{array}{l}f\left(x,y\right)={\displaystyle \sum _{i=1}^n{w}_i}\left(\left|x-a_i\right|+\left|y-b_i\right|\right)\\ \text{ =1(1+6)+2(1+3)+3(3+3)+3(2+4)+2(8+1)}\\ \text{ = }69\end{array}$

Similarly for all the other buildings the weighted rectilinear distance is being calculated as follows:

Therefore weighted rectilinear distance of building B from other buildings

  $\begin{array}{l}f\left(x,y\right)=\text{(1+6)+2(2+3)+3(4+9)+3(3+2)+2(9+7)}\\ \text{ = 103}\end{array}$

Weighted rectilinear distance of building C from other buildings

  $\begin{array}{l}f\left(x,y\right)=\text{(1+3)+(2+3)+3(2+6)+3(1+1)+2(7+4)}\\ \text{ = 61}\end{array}$

Weighted rectilinear distance of building D from other buildings

  $\begin{array}{l}f\left(x,y\right)=\text{(3+3)+(4+9)+2(2+6)+3(1+7)+2(5+2)}\\ \text{ = 73}\end{array}$

Weighted rectilinear distance of building E from other buildings

  $\begin{array}{l}f\left(x,y\right)=\text{(2+4)+(3+2)+2(1+1)+3(1+7)+2(6+5)}\\ \text{ = 61}\end{array}$

Weighted rectilinear distance of building F from other buildings

  $\begin{array}{l}f\left(x,y\right)=\text{(8+1)+(9+7)+2(7+4)+3(5+2)+2(6+5)}\\ \text{ = 101}\end{array}$

From the results the weighted rectilinear distance for C and E are minimum and hence the optimal solution will be either C or E.

(e)

Summary Introduction

Interpretation: A company is considering where to locate its cafeteria to service six buildings. The locations of the buildings and the fraction of the company’s employees working at these locations are given below. Supposing that the cafeteria must be located in one of the buildings; using Euclidean distance the particular building needs to be identified.

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Concept Introduction: The simple measurement of distance in a straight line between two facilities is called Euclidean Distance.

Answer to Problem 56AP

The optimal solution can be considered as building C, as the Euclidean distance is least for the same.

Explanation of Solution

Given information: The locations of the buildings and the fraction of company’s employees working at these locations are as given below:

Building	$a_i$	$b_i$	Fraction of Workforce
A	2	6	$\frac{1}{12}$
B	1	0	$\frac{1}{12}$
C	3	3	$\frac{1}{6}$
D	5	9	$\frac{1}{4}$
E	4	2	$\frac{1}{4}$
F	10	7	$\frac{1}{6}$

Following is the formula for weighted Euclidean distance:

  $f\left(x,y\right)={\displaystyle \sum _{i=1}^n{w}_i\sqrt{{\left(x-a_1\right)}^2+{\left(y-b_i\right)}^2}}$

For building A, the weighted Euclidean distance from other buildings can be calculated as follows:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{1^2+6^2}+2\sqrt{1^2+3^2}+3\sqrt{3^2+3^2}+3\sqrt{2^2+4^2}+2\sqrt{8^2+1^2}\\ \text{ = 55}\text{.67}\end{array}$

For building B:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{1^2+6^2}+2\sqrt{2^2+3^2}+3\sqrt{4^2+9^2}+3\sqrt{3^2+2^2}+2\sqrt{9^2+7^2}\\ \text{ = 76}\text{.46}\end{array}$

For building C:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{1^2+3^2}+\sqrt{2^2+3^2}+3\sqrt{2^2+6^2}+3\sqrt{1^2+1^2}+2\sqrt{7^2+4^2}\\ \text{ = 46}\text{.11}\end{array}$

For building D:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{3^2+3^2}+\sqrt{4^2+9^2}+2\sqrt{2^2+6^2}+3\sqrt{1^2+7^2}+2\sqrt{5^2+2^2}\\ \text{ = 58}\text{.72}\end{array}$

For building E:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{4^2+2^2}+\sqrt{2^2+3^2}+2\sqrt{1^2+1^2}+3\sqrt{7^2+1^2}+2\sqrt{5^2+6^2}\\ \text{ = 47}\text{.74}\end{array}$

For building F:

  $\begin{array}{l}f\left(x,y\right)=\sqrt{1^2+8^2}+\sqrt{7^2+9^2}+2\sqrt{4^2+7^2}+3\sqrt{2^2+5^2}+2\sqrt{5^2+6^2}\\ \text{ = 76}\text{.17}\end{array}$

From the results, Euclidean distance is least for building C and building C is the optimal solution.