Eight of the entries have been deleted from the LINGO output that follows.Use what you know about linear programming to find values for the blanks.
MIN 6 X1 + 7.5 X2 + 10 X3
SUBJECT TO
2)25 X1 + 35 X2 + 30 X3 >= 2400
3)2 X1 + 4 X2 + 8 X3 >= 400
END
LP OPTIMUM FOUND AT STEP 2
OBJECTIVE FUNCTION VALUE
1)612.50000 VARIABLE X1 X2 X3 VALUE −−27.500000 REDUCED COST 1.312500−− ROW2)3) SLACK OR SURPLUS −− DUAL PRICE −.125000−.781250 NO.ITERATIONS= 2
RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ. COEFFICIENT RANGES VARIABLE X1 X2 X3 COEFFICIENT CURRENT 6.0000007.50000010.000000 INCREASE ALLOWAB LE −1.5000005.000000 DECREASE ALLWABLE −2.5000003.571429 RIGHTHAND SIDE RANGES ROW23 CURRENT RHS 2400.000000400.000000 ALLOWABLE INCREASE 1100.000000240.000000 ALLOWABLE DECREASE 900.000000125.714300
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