Handling Indivisibilities. Notes for AGEC 622. Bruce McCarl Regents Professor of Agricultural Economics Texas A&M University

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1 Notes for AGEC 6 Bruce McCarl Regents Professor of Agricultural Econoics Texas A&M University

2 All or None -- Integer Prograing McCarl and Spreen Chapter 5 Many investent and probles involve cases where one has to tae all or none of an ite. We cannot build ½ of a plant or buy 3/4 of a achine. We build or or 3 or none but not a fractional part This leads to integer prograing ax s. t. C A W W W C A C W is a noral,continuous LP variable, is an integer variable, is a zero one variable A 3 3 b 0 0 and int eger equals 0 or When probles have only they are called pure integer only they are called pure zero one W and they are called ixed integer other variants exist

3 Logical Conditions -- Integer Prograing Integer prograing also allows logical conditions to be iposed. Suppose I a odeling a bottling plant that runs white il but can run chocolate. If they run any chocolate they encounter cleaning cost of F. Let be aount of chocolate il processed. Then add odel coponent ax C F s. t. M 0 0 equals 0or Note in this coponent M is a big nuber (0 billion) and for >0 this iplies = while if F>0 =0 if =0. So if we run any chocolate il we clean whether it be gallon or one illion. is an indicator variable. 3

4 Integer Prograing Suppose we can buy fro different types of achines and get fro the capacity for the ith tie period ax C F s. t. a i CAP i equals 0 0 for for all all 0 or for In this case if they were utually exclusive we could also add or if buying one eant we ust buy another - =0 or if a achine can only be purchased if we have a iniu volue a LL for all i = i i 0 i all 4

5 Integer Prograing Solution Difficulty All sounds good but probles are hard. Let s explore why Calculus is basis of all continuous optiization but not here because there is no neighborhood around a point in which a derivative can be defined 3 3, 6 6 0& integer Feasible Region for, nonnegative integers 0 Figure 5. Graph of Feasible Integer Points for First LP Proble 8 6 -Axis Axis 5

6 Integer Prograing Solution Difficulty 0 Figure 5. Graph of Feasible Integer Points for First LP Proble 8 6 -Axis 4 Note Axis. Solutions are finite. A line between feasible points does not contain all feasible points 3. Moving between points is not always easy 4. Points are on boundary, interior and not in general at corners 5. Rounding of LP point ay not be bad 6

7 Integer Prograing Solution Difficulty Consider 7 0, &int eger Figure 5. Graph of Feasible Integer Points for Second Integer Proble Here Rounding not good and Moveent between points hard 7

8 Integer Prograing Solution Difficulty Mixed Integer Prograing Feasible Region integer, continuous 3 3, & int eger 0 Figure 5.3 Mixed Integer Feasible Region

9 Integer Prograing Solution Rounding Solving the proble Max s. t , & int eger as an LP yields ==3. Ob=7.68 which can be rounded to ==3, Ob=7. But this ay not always be feasible or optial In this case an obective of 7.6 arises at =4,= (solint.gs) Rounding only wors well if variable values are large 9

10 Integer Prograing Solution -- Branch and Bound Solving Max s. t as an LP yields ==3. ob=7.68, & int eger We can generate related probles that collectively do not exclude integer variables as follows ax.4 ax.4 Prob A 3 6 or Prob B & int eger & int eger Suppose we solveprob A we get x=3.x=3.33 and again generate ore probles the first with x#3 and the other with x 4. Solving these yields an integer solution at x=x=3 ob=7. and another at x=,x=4 ob=6.8. But optiat solution is x=4,x=, ob=7.6 fro proble B above 0

11 Integer Prograing Solution -- Branch and Bound Now since we are axiizing we choose the 7. as the best solution and call it the incubent. But it is not necessarily optial (in fact it is not at all). To verify its optiality we need to go bac and investigate the probles we have not yet solved which still have the potential of having an obective function above our current best (7.). We would then go bac to the right hand proble fro the first setup and eventually find =4,=, Ob=7.6. The above reveals the basic nature of branch and bound. It begins by solving an LP then finds a variable that is not integer and generates probles (creating a branch). It then solves one of these and continues until it finds an integer solution which establishes a bound. We then bactrac and try to eliinate all other possible branches either by finding they cannot have a better obective than the incubent (the bounding step) or are infeasible. This eans interediate solutions are found and we converge to within a tolerance

12 Integer Prograing Solution in ECEL Integer prograing probles can be solved with Excel. To do this one adds a constraint restricting the variables to tae on integer values. This constraint is iposed as one adds any other constraints. Naely a variable is chosen with the inquality iposed as one of the last two choices. This is portrayed in the following screenshot. Naely choosing int restricts the variables to integer values and Bin is chosen to restrict variables to binary or 0- values.

13 Integer Prograing Solution in ECEL Unfortunately the default Excel solver does not reliably get the proper solution for integer prograing probles unless you set soe options Naely when you call up the solver go into the options box. Then chec that you are to assue a linear odel and set the tolerance saller. 3

14 Integer Prograing Solution in ECEL In addition, students should realize that ECEL is not an extraordinarily good choice for the solution of large linear, integer, or nonlinear prograing probles and software lie GAMS ay be better in a professional setting. Excel was taught in this class so as to not require students to learn ultiple and initially unfailiar software pacages One can buy iproved plug-ins by looing on the Internet and purchasing an upgraded solver which has this capability. For exaple, a Google search for the words excel in solver integer turns up a nuber of choices. 4

15 McCarl and Spreen Chapter 6 Integer Prograing Solution -- Knapsac The napsac proble is a faous IP forulation. Suppose a hier selects the ost valuable ites to carry, subect to a weight or capacity liit. Partial ites are not allowed, thus choices are depicted by zero-one variables. The general proble forulation assuing only one of each ite is available is Max v s.t. d = W 0or for all The decision variables indicate whether the th alternative ite is chosen. Each ite is worth v. The obective function gives total value of ites chosen. The capacity used by each is d. The constraint requires total capacity use to be less than or equal to the capacity liit (W). 5

16 Integer Prograing Solution -- Knapsac Suppose an individual is preparing to ove. Assue a truc is available that can hold at ost 50 cubic feet of ites. Suppose there are 0 ites which can be taen and that their naes, volues and values iplicit below. The resultant forulation is Max s.t. 7x 70x x 5x 0x = 0 x 0x or 3 3, x 0x for 4 4 all 5x 5x 5 5 x 5x 6 6 5x 0x 7 7 x 5x 8 8 5x 9 0x 9 0x 0x The optial obective function value equals 8. The solution indicates that all ites except furniture, 7, should be taen. There are peculiarities which should be noted. First, the constraint has 65 units in slac (50-85) and no shadow price. However, for practical purposes the constraint does have a shadow price as the 7 variable would coe into the solution if there were 0 ore units of capacity, but slac is only 65. Further, note that each of the variables has a non-zero reduced cost. This is because bounds have been added. Shadow price and red cost are isleading. 6

17 Warehouse Location McCarl and Spreen Chapter 6 Warehouse location probles involve location of warehouses within a transportation syste so as to iniize overall costs. Basic decision involves tradeoffs between fixed warehouse construction costs and transportation costs. z i Supply point x i Warehouse y Deand Point v Supply point Warehouse Deand Point Supply point i Warehouse Deand Point 7

18 Warehouse Location Min s.t. F CAP V V Ci i i i i i D E i i i Z Z Z i i i S D i 0 0 for alli for all for all for all A V b for all V = 0or, i,, Z i 0 for alli,, Merges Fixed Charge - Capacity and Transportation Proble with transshipents. We consider oving goods fro supply i to deand or fro i to warehouse and then on to deand. 8

19 Warehouse Location V A V B V C A B C A B C A A B B C C Z Z Z Z Min V A, V B, V C (0,) x i,, Z i 0 for all i,, 9

20 Warehouse Location Variable Value Reduced Cost Row Activity Shadow Price V A V B V C A B C A B C A A B 0 0 B C 0 0 C

21 Machinery Selection McCarl and Spreen Chapter 6 The achinery selection proble is a coon investent proble. In this proble one axiizes profits, trading off the annual costs of achinery purchase with the extra profits obtained by having that achinery. A general forulation of this proble is Max s.t. F Cap G i r is a C J A D n i nonnegative integer, 0 b e n r 0 for alli and for alln for allr for all,,and The decision variables are, the integer nuber of units of the th type achinery purchased;, the quantity of the th activity produced using the th achinery alternative. The paraeters of the odel are: F, the annualized fixed cost of the th achinery type; Cap i, the annual capacity of the th achinery type to supply the i th resource; G r, the usage of the r th achinery restriction when purchasing the th achinery type; C, the per unit net profit of ; A i, the per unit use by of the i th capacity resource supplied by purchasing achine ; D n, the per unit usage of fixed resources of the n th type by ; b n, the endowent of the n th resource in the year being odeled; and e r, the endowent of the r th achinery restriction.

22 Machinery Selection Machinery Use Continuous Variables Plow with Tractor Plow with Tractor Plant Disc 8 Harvest with and Plow and Plow and Plow ans Plow Tractor Tractor Tractor Tractor Machinery Acquisition Integer Variables in Period in Period Planter Planter Harvester Harvester Tractor Plow Planter Disc Harvester Crop Sales Input Purchases Obectives (ax) Tractor Capacity in Period Tractor Capacity in Period Plow Capacity in Period Plow Capacity in Period Capacity of Planter Capacity of Disc Capacity of Har- vester Labor Available in Period Plow-Plant Sequencing Plant-Harvest Sequencing Land Available 600 Planters Discs Lin Disc- Planter ield Balance Input Balance - 0

23 Table 6.7. Solution for the Machinery Selection Proble ob = 6,00 Variable Value Reduced Cost Equation Slac Shadow Price Buy Tractor -5,000 Tractor capacity in Period 00 0 Buy Tractor 0 0 Tractor capacity in Period 30 0 Buy Plow 0 0 Tractor capacity in Period Buy Plow -,00 Tractor capacity in Period 0 Buy Planter 0 0 Tractor capactiy in Period Buy Planter Tractor capacity in Period Buy Disc 0 0 Plow capacity in Period Buy Disc 0 Plow capacity in Period 0 0 Buy Harvester 0 0 Plow capacity in Period 00 0 Buy Harvestor 0 Plow capacity in Period 80 0 Plow with Tractor and Plow in Period Planter capacity 0 0 Plow with Tractor and Plow in Period Planter capacity 30 0 Plow with Tractor and Plow in Period Disc 0 0 Plow with Tractor and Plow in Period 0 0 Disc 30 0 Plow with Tractor and Plow in Period Harvester 0 50 Plow with Tractor and Plow in Period Harvester 50 0 Plow with Tractor and Plow in Period 0 0 Labor available in Period 8 0 Plow with Tractor and Plow in Period Labor available in Period 44 0 Plant with Tractor and Planter Labor available in Period Plant with Tractor and Planter Plow Plant Plant with Tractor and Planter Plant Harvester Plant with Tractor and Planter 0 0 Land Harvest with Tractor and Harvester One Planter 0 0 Harvest with Tractor and Harvester One Disc 0 0 Harvest with Tractor and Harvester Planter to Disc 0 0 Harvest with Tractor and Harvester Planter to Disc 0 0 Sell Crop 84,000 0 ield Balance 0.5 Purchase Inputs Input Balance 0 0