1. Route Planning Texas Transfer Corp (TTC) Case 1

2. Linear programming Example: Woodcarving, Inc. Manufactures two types of wooden toys  Soldiers sell for $27 and cost $24 to produce  Trains sell for $21 and cost $19 to produce Requires two types of skilled labor  Finishing (100 available hours each week)  Carpentry (80 available hours each week)  Soldiers require 2 hours of finishing labor and 1 hour of carpentry labor  Trains require 1 hour of finishing labor and 1 hour of carpentry labor Demand for trains is unlimited, but no more than 40 soldiers are bought each week Objective: maximize weekly profit

3. Example: Woodcarving, Inc. (continued) Decision Variables: X=number of soldiers to produce each week Y=number of trains to produce each week Objective Function: Maximize the total profit of soldier and train sales Max 3X+2Y Sell PriceProduction cost Soldier$27$24 Train$21$19 → Profit $3 → Profit $2

4. Example: Woodcarving, Inc. (continued) Constraints:  Labor hour constraints: Total finishing hours must be  100 hours: 2X+Y  100 (1) Total carpentry hours must be  80 hours: X+Y  80 (2)  Demand constraint: do not produce more than 40 soldiers: X  40 (3) FinishingCarpentry Soldier21 Train11 Total available hours10080

5. Linear programming problem Decision variables What decision must we make (e.g. production levels)? Objective function What is our measure of success? What are we trying to maximize (e.g. profits) or minimize (e.g. cost)? Constraints What limits our success (e.g. available resources)? Relationships (equations) are linear

6. Feasible Region Graphical solution X  40 2X + Y  100 X + Y  80 100 0 0 Y = # trains X = # soldiers X=0, Y=80, profit=160 X=20, Y=60, profit=180 (optimal solution) X=40, Y=20, profit=160 X=40, Y=0, profit=120

7. Use Excel Solver: Data-Solver Objective Function Decision Variables Constraints

8. The solution

9. Easy rider toys Easy Rider Toys (ERT) manufactures and markets toy vehicles, including cars and trucks Currently ERT has10,000 cars and 12,000 trucks in inventory. Inventories can be sold in two different bundles:  The Racer Set consists of 7 cars and 2 trucks, and is sold for $34.99.  The Construction Set consists of 3 cars and 12 trucks, and is sold for $43.99. How many sets of each bundle ERT should produce to maximize the profit?

10. Easy rider formulation X = # of Racer Sets to produce Y = # of Construction Sets to produce Problem formulation: CarsTrucksPrice Racer Set72$34.99 Construction Set312$43.99 Available Inventory10,00012,000

11. Transportation example Philadelphia and Washington D.C. are supply nodes Buffalo, Pittsburgh, and Harrisburg are demand nodes Total supply = total demand Washington D.C. 150 Harrisburg 100 $30 Philadelphia 100 $50 $20 $25 Buffalo 50 Pittsburgh 100 $30$15 $70 1 2 3 4 5

12. Graphical representation of network flow problems (5,10)(4,9) (3,6) (1,3) (4,8) 4 6 2 5 4 Arc (i, j) Cost of sending one unit from i to j Minimum and Maximum allowable flows on arc ( i,j ) i j

13. Translating a graph to an LP One decision variable Xi,j for each arc (i,j) (the amount of flow to send on that arc) Two constraints for each arc (i,j) Xi,j  upper bound for flow on arc (i,j) Xi,j  lower bound for flow on arc (i,j) One constraint for each node (flow-balance contr.) flow into node i ═ flow out of node i

14. LP formulation Objective function: Minimize the overall flow cost

15. Decision variables One decision variable for every arc  Interpretation: How much flow (i.e. how many cars) should we send along that arc? Washington D.C. 150 Harrisburg 100 Philadelphia 100Buffalo 50 Pittsburgh 100 1 2 3 4 5 X23 X13X34 X35 X45X54 X12

16. Objective function Minimize total transportation costs 70 × X12 + 30 × X13 + 50 × X23 + 20 × X34 + 25 × X35 + 30 × X45 + 15 × X54 Interpretation: We will ship at the lowest cost possible while still satisfying demand.

17. Constraints One constraint for every node  Outflow = Inflow Philadelphia: X12 + X13 = 100 Washington: X23 = 150 + X12 Harrisburg: X13 + X23 = 100 + X34 + X35 Buffalo: X34 + X54 = 50 + X45 Pittsburgh: X45 + X35 = 100 + X54

18. The shortest path problem Texas Transfer Corp. is an express package delivery company Services 10 cities in Texas Must plan fastest routes

19. The shortest path problem The flow limits on origin and destination arcs are (1,1) The flow limits on all other arcs are (0,1) Assign a flow of 1 along an arc if it is used in the route; otherwise the flow is 0. Minimize the overall flow cost (distance).

20. Decision variables One decision variable for each arc in the network Example: X 12: A flow from City 1 to City 2 X 12 is either 0 or 1

21. User input variables Oi = 1 if the origin is city i, 0 otherwise Example: O 1 = 1, O 2 = … = O 10 = 0 Dj = 1 if the destination is city j, 0 otherwise Example: D 10 = 1, D 1 = … = D 9 = 0

22. Travel time City 1 → City 2 122 / 51 + 0 = 2.39 22 Distance / Avg. speed + Delay

23. 23 Min 2.39 X 12 + 6.08 X 13 + … + 3.41 X 10_9 Excel function: =SUMPRODUCT(, ) Objective function Column

24. Constraints Flow-balance constraints CityConstraint (1) Amarillo O1 + X21 + X31 = D1 + X12 + X13 (2) LubbockO2 + X12 + X42 + X52 = D2 + X21 + X24 + X25 (3) Fort WorthO3 + X13 + X53 + X83 + X93 = D3 + X31 + X35 + X38 + X39 (4) El PasoO4 + X24 + X54 + X64 = D4 + X42 + X45 + X46 (5) AbileneO5 + X25 + X35 + X45 + X65 = D5 + X52 + X53 + X54 + X56 (6) San AngeloO6 + X46 + X56 + X76 = D6 + X64 + X65 + X67 (7) San AntonioO7 + X67 + X87 + X97 + X10_7 = D7 + X76 + X78 + X79 + X7_10 (8) AustinO8 + X38 + X78 = D8 + X83 + X87 (9) HoustonO9 + X39 + X79 + X10_9 = D9 + X93 + X97 + X9_10 (10) Corpus ChristiO10 + X7_10 + X9_10 = D10 + X10_7 + X10_9

25. Flow-balance constraints Example: City 1 O1 + X21 + X31 = D1 + X12 + X13 Excel Function: SUMIF O1 + SUMIF(,, ) D1 + SUMIF(,, ) 25 Column1 1

26. Other constraints We also include non-negativity constraints. We can ignore the upper bounds on the decision variables (e.g. X21≤1) since they are redundant. 26

27. Results Shortest route: 1→3→9→10 Travel time: 13.81 hours 27