First constraint is acreage constraint: so we must calculate
acres per ton of alfalfa raised.
4 tons/acre => 1/4=.25 acres per ton.
The second constraint is in acre-feet: so we must calculate
acre-feet per ton of AR.
.25 acres/ton * 3 acre-feet per acre = .75 acre-feet per ton.
MAX Y1 + 1.42699 Y2 + 0.37998 Y4 + 0.1236 Y3 + 0.00444 Y5 SUBJECT TO 2)Y1 - 1.425 X21 - 0.1234 X31 - 0.3793 X41 - 0.00443 X51 + X12 + X13 + X14 + X15 = 2 3)Y2 - 0.6998 X12 - 0.08647 X32 - 0.2662 X42 -0.0031 X52 + X21 + X23 + X24 + X25 = 5 4)Y3 - 8.078 X13 - 11.55 X23 - 3.073 X43 - 0.03586 X53 + X31 + X32 + X34 + X35 = 0 5)Y4 - 2.627 X14 - 3.754 X24 - 0.325 X34 - 0.01163 X54 + X41 + X42 + X43 + X45 = 3 6) Y5 - 224.7 X15 - 320.7 X25 - 27.76 X35 - 85.51X 45 + X51 + X52 + X53 + X54 = 0 7) Y3 >= 8 8) Y5 >= 1280 END LP OPTIMUM FOUND AT STEP 3 OBJECTIVE FUNCTION VALUE 1) 10.269500 VARIABLE VALUE REDUCED COST Y1 2.000000 .000000 Y2 .000000 .000590 Y4 .000000 .000302 X21 .000000 .002580 X31 .000000 .000200 X41 .000000 .000982 X51 .000000 .000017 X12 .000000 .000979 X13 .000000 .001559 X14 .000000 .000998 X15 .000000 .000708 X32 .000000 .000157 X42 .000000 .000261 X52 .000000 .000022 X23 1.811665 .000000 X24 3.188335 .000000 X25 .000000 .001354 Y3 20.924730 .000000 X43 .000000 .000460 X53 .000000 .000015 X34 .000000 .000008 X35 .000000 .000145 X54 .000000 .000025 X45 14.969010 .000000 Y5 1280.000000 .000000 ROW SLACK OR SURPLUS DUAL PRICES 2) .000000 1.000000 3) .000000 1.427580 4) .000000 .123600 5) .000000 .380282 6) .000000 .004447 7) 12.924730 .000000 8) .000000 -.000007 NO. ITERATIONS= 3 RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE Y1 1.000000 .001621 .000708 Y2 1.426990 .000590 INFINITY Y4 .379980 .000302 INFINITY X21 .000000 .002580 INFINITY X31 .000000 .000200 INFINITY X41 .000000 .000982 INFINITY X51 .000000 .000017 INFINITY X12 .000000 .000979 INFINITY X13 .000000 .001559 INFINITY X14 .000000 .000998 INFINITY X15 .000000 .000708 INFINITY X32 .000000 .000157 INFINITY X42 .000000 .000261 INFINITY X52 .000000 .000022 INFINITY X23 .000000 .000095 .000590 X24 .000000 .000978 .000095 X25 .000000 .001354 INFINITY Y3 .123600 .000088 .000051 X43 .000000 .000460 INFINITY X53 .000000 .000015 INFINITY X34 .000000 .000008 INFINITY X35 .000000 .000145 INFINITY X54 .000000 .000025 INFINITY X45 .000000 .000618 .000269 Y5 .004440 .000007 INFINITY RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 2 2.000000 INFINITY 2.000000 3 5.000000 INFINITY 1.119024 4 .000000 INFINITY 12.924730 5 3.000000 11.969010 4.200818 6 .000000 1023.470000 359.211900 7 8.000000 12.924730 INFINITY 8 1280.000000 359.211900 1023.470000
OBJECTIVE FUNCTION VALUE 1) 10.269500 VARIABLE VALUE REDUCED COST Y1 2.000000 .000000 X23 1.811665 .000000 X24 3.188335 .000000 Y3 20.924730 .000000 X45 14.969010 .000000 Y5 1280.000000 .000000 ROW SLACK OR SURPLUS DUAL PRICES 2) .000000 1.000000 3) .000000 1.427580 4) .000000 .123600 5) .000000 .380282 6) .000000 .004447 8) .000000 -.000007The optimal solution calls for exchanging 1.812M pounds into Francs, 3.188M pounds into Marks, and 14.969M Marks into Yens. The final holding will consist of 2M Dollars, 20.92M Francs and 1280M Yens. The total dollar value of this bundle is 10.2695$.