6.3 Habitat Suitability Index Problem. We need to formulate this problem to maximize habitat suitability index scores, and require some timber volume to be produced over a 30-year time period. After examining the tables on page 315 in the book, we developed the following problem formulation. The LINDO input file contained the following: max + 102.79 X11 + 103.17 X12 + 102.60 X13 + 97.80 X21 + 98.40 X22 + 97.40 X23 + 92.69 X31 + 95.09 X32 + 96.76 X33 + 91.56 X41 + 93.03 X42 + 92.19 X43 subject to 2) X11 + X12 + X13 = 1 3) X21 + X22 + X23 = 1 4) X31 + X32 + X33 = 1 5) X41 + X42 + X43 = 1 6) 893 X11 + 1020 X21 + 1128.5 X31 + 1197 X41 - H1 = 0 7) 1026 X12 + 1060 X22 + 1128.5 X32 + 1239 X42 - H2 = 0 8) 1064 X13 + 1080 X23 + 1147 X33 + 1260 X43 - H3 = 0 9) H1 >= 1 10) H2 >= 1 11) H3 >= 1 end inte X11 inte X12 inte X13 inte X21 inte X22 inte X23 inte X31 inte X32 inte X33 inte X41 inte X42 inte X43 Here, we are maximizing HSI values. To obtain the objective function value for variable X11, multiply the number of acres in Stand A by the HSI score for Stand A during period 1 (0.541 x 190). Rows 2-4 indicate that each stand will only be entered once during the three time periods, since X11-X43 will take on integer values (as indicated by "inte X11" at the end of the LINDO formulation). Rows 6-7 sum the volume produced during each time period. To obtain the value for X11, multiply the number of acres in Stand A by the volume produced from Stand A during time period 1 (4.7 x 190). H1, H2, and H3 are the total harvest volumes for periods 1, 2, and 3, respectively. Rows 9-11 indicate that the total harvest volume for periods 1-3 must be greater than (or equal to) 1, or that there must be "some" volume produced in each time period. The LINDO output contained the following: LP OPTIMUM FOUND AT STEP 20 OBJECTIVE VALUE = 391.359589 FIX ALL VARS.( 7) WITH RC > 0.220001 NEW INTEGER SOLUTION OF 390.980011 AT BRANCH 0 PIVOT 42 OBJECTIVE FUNCTION VALUE 1) 390.9800 VARIABLE VALUE REDUCED COST X11 1.000000 0.000000 X12 0.000000 0.000000 X13 0.000000 23.697525 X21 0.000000 0.000000 X22 1.000000 1.000000 X23 0.000000 488.940002 X31 0.000000 37682.738281 X32 0.000000 0.000000 X33 1.000000 0.000000 X41 0.000000 0.000000 X42 1.000000 0.000000 X43 0.000000 0.000000 H1 893.000000 0.000000 H2 2299.000000 0.000000 H3 1147.000000 0.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 103.169998 3) 0.000000 98.180000 4) 0.000000 96.760002 5) 0.000000 92.190002 6) 0.000000 0.000000 7) 0.000000 0.000000 8) 0.000000 0.000000 9) 892.000000 0.000000 10) 2298.000000 0.000000 11) 1146.000000 0.000000 12) 0.000000 -0.379997 NO. ITERATIONS= 42 BRANCHES= 0 DETERM.= 1.000E 0 BOUND ON OPTIMUM: 390.9800 ENUMERATION COMPLETE. BRANCHES= 0 PIVOTS= 42 LAST INTEGER SOLUTION IS THE BEST FOUND RE-INSTALLING BEST SOLUTION... After interpreting the LINDO output, we arrived at the following answers to the problem: a) The optimum management schedule is for Stand A to be thinned during period 1, producing 893 cords; Stands B and D during period 2, producing 2299 cords; and Stand C during period 3, producing 1147 cords. Stand A actually would have had a higher HSI score if thinned during period 2, yet one of the stands had to be thinned during period 1 to produce "some" timber volume during this period. b) The total HSI score was 390.98, or a 0.498 HSI score per acre. c) Over the three time periods, the owners could expect to harvest 4339 cords, or an average of 5.5 cords per acre.