El Paso Manufacturing Co

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El Paso Manufacturing Co

El Paso Manufacturing Co. (EPMC)

Let: z  number of assemblies to make

x1  number of frames to manufacture x2  number of shafts to manufacture

Objective: Maximize z

LP: Max z [number of assemblies] s.t. z  x1 [frame required for each assembly] z  x2 [shaft required for each assembly]

0.8x1  0.5x2  48000 [forging machine hours]

0.1x1  7200 [drilling machine hours]

0.3x1  14400 [milling machine hours]

0.5x1  0.3x2  24000 [grinding machine hours]

0.2x2  12000 [lathe machine hours] z, x1, x2  0 [nonnegativity] bubba - EPMC PROBLEM DATA IN EQUATION STYLE Maximize 1 Z Subject to FRAME REQD 1 Z - 1 X1 <= 0 SHAFT REQD 1 Z - 1 X2 <= 0 FORGING 0.8 Z + 0.5 X1 <= 48000 DRILLING 0.1 Z <= 7200 MILLING 0.3 Z <= 14400 GRINDING 0.5 Z + 0.3 X1 <= 24000 LATHE 0.2 Z <= 12000 0 <= Z 0 <= X1 0 <= X2 bubba - EPMC OPTIMAL SOLUTION - DETAILED REPORT Variable Value Cost Red. cost Status

1 Z 30000.0000 1.0000 0.0000 Basic 2 X1 30000.0000 0.0000 0.0000 Basic 3 X2 30000.0000 0.0000 0.0000 Basic

Slack Variables 4 FRAME REQD 0.0000 0.0000 -0.3750 Lower bound 5 SHAFT REQD 0.0000 0.0000 0.0000 Lower bound 6 FORGING 9000.0000 0.0000 0.0000 Basic 7 DRILLING 4200.0000 0.0000 0.0000 Basic 8 MILLING 5400.0000 0.0000 0.0000 Basic 9 GRINDING 0.0000 0.0000 -1.2500 Lower bound 10 LATHE 6000.0000 0.0000 0.0000 Basic

Objective Function Value = 30000 bubba - EPMC OPTIMAL SOLUTION - DETAILED REPORT Constraint Type RHS Slack Shadow price

1 FRAME REQD <= 0.0000 0.0000 0.3750 2 SHAFT REQD <= 0.0000 0.0000 0.0000 3 FORGING <= 48000.0000 9000.0000 0.0000 4 DRILLING <= 7200.0000 4200.0000 0.0000 5 MILLING <= 14400.0000 5400.0000 0.0000 6 GRINDING <= 24000.0000 0.0000 1.2500 7 LATHE <= 12000.0000 6000.0000 0.0000

Objective Function Value = 30000

Grinding Machine constraint is tight (slack = 0; shadow price = 1.25). EPMC should first consider increasing the number of grinding machines. Alternative Formulation

Let: x  number of frames to manufacture 1 x2  number of shafts to manufacture

Objective: Maximize x1 (or x2 )

LP:

Max x1 [number of assemblies ( number of frames)] s.t. x1  x2 [1 frame & 1 shaft required for each assembly]

0.8x1  0.5x2  48000 [forging machine hours]

0.1x1  7200 [drilling machine hours]

0.3x1  14400 [milling machine hours]

0.5x1  0.3x2  24000 [grinding machine hours]

0.2x2  12000 [lathe machine hours] x1, x2  0 [nonnegativity] bubba - EPMC PROBLEM DATA IN EQUATION STYLE Maximize 1 X1 Subject to FR EQ SHFT 1 X1 - 1 X2 = 0 FORGING 0.8 X1 + 0.5 X2 <= 48000 DRILLING 0.1 X1 <= 7200 MILLING 0.3 X1 <= 14400 GRINDING 0.5 X1 + 0.3 X2 <= 24000 LATHE 0.2 X2 <= 12000 0 <= X1 0 <= X2 bubba - EPMC OPTIMAL SOLUTION - DETAILED REPORT Variable Value Cost Red. cost Status

1 X1 30000.0000 1.0000 0.0000 Basic 2 X2 30000.0000 0.0000 0.0000 Basic

Slack Variables 4 FORGING 9000.0000 0.0000 0.0000 Basic 5 DRILLING 4200.0000 0.0000 0.0000 Basic 6 MILLING 5400.0000 0.0000 0.0000 Basic 7 GRINDING 0.0000 0.0000 -1.2500 Lower bound 8 LATHE 6000.0000 0.0000 0.0000 Basic

Objective Function Value = 30000 bubba - EPMC OPTIMAL SOLUTION - DETAILED REPORT Constraint Type RHS Slack Shadow price

1 FR EQ SHFT = 0.0000 0.0000 0.3750 2 FORGING <= 48000.0000 9000.0000 0.0000 3 DRILLING <= 7200.0000 4200.0000 0.0000 4 MILLING <= 14400.0000 5400.0000 0.0000 5 GRINDING <= 24000.0000 0.0000 1.2500 6 LATHE <= 12000.0000 6000.0000 0.0000

Objective Function Value = 30000

Grinding Machine constraint is tight (slack = 0; shadow price = 1.25). EPMC should first consider increasing the number of grinding machines.

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