Introduction to Operations Research, Volume 1-- This classic, field-defining text is the market leader in Operations Research -- and it's now updated and expanded to keep professionals a step ahead -- Features 25 new detailed, hands-on case studies added to the end of problem sections -- plus an expanded look at project planning and control with PERT/CPM -- A new, software-packed CD-ROM contains Excel files for examples in related chapters, numerous Excel templates, plus LINDO and LINGO files, along with MPL/CPLEX Software and MPL/CPLEX files, each showing worked-out examples |
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Page 155
X2 3x1 + 2x2 = 18 8 Z = 3x1 + 5x2 ( 1 ) -2x2 = 13—2 = 3 ( s ) + 5 ( ) = 374 2x2 122
= 3 ( 2 ) + 5 ( 6 ) = 36 AZ = = 1 = y2 6 ( 2,6 ) x1 = 4 4 FIGURE 4.8 This graph
shows that the shadow price is yi = for resource 2 for the Wyndor Glass Co.
problem .
X2 3x1 + 2x2 = 18 8 Z = 3x1 + 5x2 ( 1 ) -2x2 = 13—2 = 3 ( s ) + 5 ( ) = 374 2x2 122
= 3 ( 2 ) + 5 ( 6 ) = 36 AZ = = 1 = y2 6 ( 2,6 ) x1 = 4 4 FIGURE 4.8 This graph
shows that the shadow price is yi = for resource 2 for the Wyndor Glass Co.
problem .
Page 181
( b ) Use graphical analysis to find the shadow prices for the resources . ( c )
Determine how many additional units of resource 1 would be needed to increase
the optimal value of Z by 15 . 4.7-5 . Consider the following problem . Maximize Z
= x1 ...
( b ) Use graphical analysis to find the shadow prices for the resources . ( c )
Determine how many additional units of resource 1 would be needed to increase
the optimal value of Z by 15 . 4.7-5 . Consider the following problem . Maximize Z
= x1 ...
Page 182
Use this information to identify the shadow price for each resource , the allowable
range to stay optimal for each objective function coefficient , and the allowable
range to stay feasible for each right - hand side . and xi 20 , x2 = 0 , X3 = 0 .
Use this information to identify the shadow price for each resource , the allowable
range to stay optimal for each objective function coefficient , and the allowable
range to stay feasible for each right - hand side . and xi 20 , x2 = 0 , X3 = 0 .
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activity additional algorithm alternative amount analysis apply assignment assumed basic variable begin BF solution calculate called changes coefficients column complete Consider constraints Construct corresponding cost CPF solution customers decision demand described determine developed distribution entering equations estimated example expected feasible FIGURE final flow formulation given gives hour identify illustrate increase indicates initial inventory iteration linear programming machine Maximize mean million Minimize month needed node objective function obtained operations optimal optimal solution original parameter path payoff plant player possible presented Prob probability problem procedure profit programming problem queueing respectively resulting shown shows side simplex method solution solve step strategy Table tableau tion transportation unit waiting weeks