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Basic Problem The basic idea in a transportation problem is that there are sites or sources of product that need to be shipped to destinations. Typically the routes and the amounts shipped on each route must be determined and the goal is to minimize the cost of shipping. The constraints are that you can not ship more from a source than made at that source and you do not want to ship more to a place than needed. Some vocabulary: Source is supply. Factory capacity is supply. Destination is demand. Warehouse requirements can also be called demand.
Some assumptions If production costs are the same at each source, ignore them. If they are not the same include them in the analysis with the shipping cost. From a given site, shipping costs per unit are the same no matter the volume of shipping. We have a balanced problem, or the total amount from sources add up to the total demand at destinations – we change this later.
Let’s do an example to highlight details of the work needed to arrive at the optimal solution. Say the following amounts are made at the various sources: Des Moines 100 (units) Evansville 300 Ft. Lauderdale 300 and say the following sites demand the amounts shown, Albuquerque 300 Boston 200 Cleveland 200 (note the balance of supply and demand).
Say the costs of shipping a unit from each source to each destination is as in the following table:
Now, let’s combine the information into one table and I will explain more about the table on the next slide.
You will notice on the previous slide I added some information and changed how I presented the information from slide 5. First, note I added a supply column. This is keeping track of how much is made at each source. (I initially thought the info should be in a row, but is in a column that has information about the amount supplied in each row.) Second I added a demand row similar to the supply column. The last thing I did was move over the unit shopping cost into a small boxed-off area in each cell. The table sort of looks like a bowling score sheet now. This is done for a reason. We will get into that next.
A solution Ultimately we will have the computer do our work for us. But we want to get some of the logic of the solution process. We will follow the Northwest Corner rule for now (in the context of our problem.) We want to take care of Des Moines and Albuquerque first – the Northwest corner of our table. Note that Des Moines has 100 supplied and Albuquerque needs 300 units. Take all 100 to Albuquerque. Des Moines is done. Drop down to Evansville. It has 300 available and Albuquerque needs 200 more. Drop 200 at Al’s. With the other 100 from Evansville, take them to Boston, since Boston needs 200. Evansville is done, but Boston still needs 100, so ship them from Ft. Lauderdale and since Ft. Lauderdale has some left give the rest to Cleveland.
Since we had a balanced problem the last destination will always get the right amount. You may have noticed that in this method we did not use the cost figures yet. What we have done is pick an initial solution and now we will check for improvements. But first, note the next slide has our solution put into the table we had before.
Now we want to check if this is the best routing. Some routes are not used. We want to ask what would happen if we allocated 1 unit to a route – lets start with DM/B. Here is where it gets kind of complicated to explain, but here we go. If we add a unit to DM/B then E/B will have one too much(because B should only get 200 total) so we will subtract 1 from E/B. But then we have to add 1 to E/A (because E has 300 to supply). But now we have to subtract 1 from DM/A (because A should get 300). Here is a way to see through the “map.” DM/B has none at first. Move vertical or horizontal toward a cell that has a value in it. I went down to E/B (I could have gone to DM/A). Now turn toward the other direction to a cell with a value – here E/A. Now again turn in the other direction to a cell with a value – here DM/A. Then we would turn back to the block from which we started.
Note on the next screen I put a +1 in the block I started in. Then when I went down to E/B I put a –1. Then I went to E/A and put a +1 and finally a –1 in DM/A. Remember we want to see if our initial routing is cheapest. Now we are checking to see if an unused route could be used and give a cheaper cost. We have suggested adding a unit to an unused route and have made adjustments to the other routes.
Now, the cells with a +1 means we would add the cost of shipping a unit on that route because we are contemplating adding on that route and the cells with a –1 means we would subtract the cost on that route because we are contemplating not shipping on that route. Following our path we would have + 4 – 4 + 8 – 5 = + 3. This suggests we would actually add $3 to cost for every unit we ship on the DM/B route by changing the initial solution. SO, do not change the initial solution. But we need to check the other empty cells.
An initial solution start
You noted on the previous slide I started in the empty cell DM/C and went to DM/A and then kept turning until I got back to the start cell. Now put a +1 in the start cell and then a –1 in the next turn cell and then a +1 in the next turn cell, alternating back and forth until you get back to start. Then we add cost of + cells and subtract cost of – cells.
Here are the changes needed: DM/C add -- +3 DM/A minus -- -5 E/A add -- +8 E/B minus -- -4 FL/B add -- +7 FL/C minus -- -5 total change +4 so do not change to this route. You will notice in my list here I started in an unused cell and traced my way through used cells back to the unused cell. If the total change is positive do not make that change. The number obtained can be called the marginal cost of moving one unit of output to that route. Do not use unused cells if they have a positive marginal cost.
What about the E/C route? Initially we did not have it used. Here are the changes needed (do you see the path?): E/C add -- +3 E/B minus -- -4 FL/B add -- +7 FL/C minus -- -5 total change +1 so do not change to this route.
What about the FL/A route? Initially we did not have it used. Here are the changes needed: FL/A add -- +9 FL/B minus -- -7 E/B add -- +4 E/A minus -- -8 total change -2 so change to this route because the marginal cost is negative. This means cost will fall by $2 when a unit is added to this route. The next question is how much to change to this route.
Here I have added the original solution shipped amounts on the affected routes: FL/A add -- +9 0 FL/B minus -- -7 100 E/B add -- +4 100 E/A minus -- -8 200 Note the two routes with a minus sign –> FL/B and E/A. The smaller shipment is 100 from those two. Add this amount to the plus sign routes here, and Subtract this from minus sign routes here. The new shipments are on the next screen.
Now, this is another routing. We have to check all the empty cells again. Some folks would call this a royal pain in the as…., er, a, neck. Now, you would see that the empty cell E/C has a negative marginal cost, and we would make adjustments again just like what we went through to see the final solution on the next page.
QM for windows has a Transportation Problem module. Go to the module and do a file new for each new problem you do. Save the file. • When you input the data dress up the from and to places with the places in the problem. The input would be similar to what I have on screen 6 here. • Output to look at • Transportation shipments has cost and amount to ship from/to each place. • Marginal Costs has costs for routes not used. If a value is zero then more than one set of routes would lead to the same low costs of transportation. • Shipping list is another form of the final solution.
Optimal solution Marginal costs based on optimal solution. Could be different from before because we didn’t have optimal solution before.
Unbalanced problems would be a situation where the supply amounts coming out of warehouses does not equal the demand amounts needed by destinations (retail stores, maybe) QM for windows would handle the problem automatically, just input the data as is. If demand > supply, a dummy source is created in amount to the excess demand. (create an additional row.) If demand < supply, a dummy destination is created in amount to the excess supply. (create an additional column.) The dummy is given a zero cost coefficient and proceed as usual.
Facility location analysis Say we want to expand and build either a new source or destination (or both) . Say for now we want to expand a source. Say we have two choices – Cincinnati or Seattle. The way to do this is include one new source in an analysis with existing sources and get the total costs throughout the system. Then do another analysis for the other new source (taking out the other proposed new source) and get the total cost. From the point of view of distribution costs, the site to chose is the one, that when included in the overall analysis, has the lowest total cost.
10 point assignment Solve the following in QM for Windows but write up the results in a text file and save on your computer. Then cut and paste from that file to a WebCT mail message to me. Also attach QM for windows files you use to complete the work Thanks! Problem 1 use data of problem 10-24 page 443 1. Without using the production cost data (like assuming the production costs are the same across cities) a. State total cost for organization with each addition. b. Which site will it add? c. In context of answer to part b. Does every source ship to every destination? Why or why not? 2. With the production cost data used a. State the total cost for organization for each addition. b. Which site will it add?
