1. Chp4 Tutorial: Prob 4.25 Solution
Engineering 25
Chp4 Tutorial: Prob 4.25 Solution
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. The Situation Optimize Distribution Center Location for Customers:     

6. The Final Hand Notes

7. GamePlan Illustrated Set x-value (m-index) Test 31 y-values (n-index)
Increment x-value (m = m+1)
Test 31 new y-values
Repeat

8. The Function File Prob4_25_Dense_Road.m % Bruce Mayer, PE * 21Feb06
% ENGR25 * Problem 4-25
% file = Prob4_25_Dense_Road.m
% Find Optimum Location for Shipping Distribution Center %
% INPUT SECTION - Data For Six Shipping Destinations
xloc = [1,7,8,17,22,27]; % X-CoOrd in Miles
yloc = [28,18,16,2,10,8]; % Y-CoOrd in Miles
V = [3,7,4,5,2,6]; % Shipping Volume in Tons/Week
% Initialize Low-Cost Location & CoOrds Collection Vectors
%% Units for Collection Vectors: [$-Cost, X-miles, Y-Miles]
C_Lo = [1e6, 31, 31];
C_L1 = [1e6, 31, 31];
C_L2 = [1e6, 31, 31];
% CALCULATION SECTION
% vary X&Y for Dist Cntr; calc cost at each location
for m = 1:30 % vary X-CoOrd
for n = 1:30 % vary Y-CoOrd
for k = 1:6 % Calc Dist & Cost for 6 Destinations
d(k) = sqrt((m-xloc(k)).^2 + (n-yloc(k)).^2);
cost(k) = 0.5*d(k)*V(k);
end
% Sum the 6 costs for location (m,n)
Cmn = sum(cost);
% Compare to Previous value for Low-Cost
if Cmn < C_Lo(1) % replace previous if true
C_L2 = C_L1;
C_L1 = C_Lo;
C_Lo = [Cmn, m, n];
% REPORT RESULTS
display('The LOWEST-Cost Option in [$-cost, X-miles, y-miles]')
disp(C_Lo)
display('The 2nd Lowest-Cost Option in [$-cost, X-miles, y-miles]')
disp(C_L1)
display('The 3rd Lowest-Cost Option in [$-cost, X-miles, y-miles]')
disp(C_L2)

9. The Command Session >> run Prob4_25_Dense_Road
The LOWEST-Cost Option in [$-cost, X-miles, y-miles]
The 2nd Lowest-Cost Option in [$-cost, X-miles, y-miles]
The 3rd Lowest-Cost Option in [$-cost, X-miles, y-miles]