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ADVANCED
OPERATIONS
RESEARCH
By: -
Hakeem–Ur–Rehman
IQTM–PU 1
RA O
LINEAR PROGRAMMING USING EXCEL SOLVER
LINEAR PROGRAMMING USING EXCEL SOLVER
2
Excel Lingo
On the toolbar at the bottom of the screen, click on:
 Start  All Programs  Microsoft Office  Microsoft Office Excel 2007
 Spreadsheet: A two-dimensional array of rectangles.
 Cell: Each rectangle in excel (Four types of information can be typed into a cell:
Number, Fraction, Function, and Text.) It is identified by its column and row
location on the spreadsheet, which are designated by letter and numbers,
respectively (i.e. cell A1).
 SUMPRODUCT: A function that first multiplies the numbers in n consecutive cells
(i.e. A1 through E1) by the numbers in another set of n consecutive cells (i.e. A5
through E5), respectively, then takes the sum of the n number of products (i.e.
A1*A5 + B1*B5 + C1*C5 + D1*D5 + E1*E5), and finally deposits that sum in the
cell you have selected (i.e. F1).
 Cell reference: Lets you repeat patterns of information between cells, which
occurs a selected cell refers to information typed in another cell.
 Absolute reference: A cell that always refers to the originally referred cell; if
the location of the selected cell changes, the referred cell will not change. It
includes a “$” sign before the cell’s column (i.e. $A1), row (i.e. A$1), or both
(i.e. $A$1).
 Relative reference: A cell that initially refers to the originally selected cell; if
the location of the selected cell changes, the referred cell will change and the
location of the new referred cell will reflect the location change of the selected
cell. It omits the “$” sign (i.e. A1).
LINEAR PROGRAMMING USING EXCEL SOLVER
3
How to activate Solver:
LINEAR PROGRAMMING USING EXCEL SOLVER
4
Solver can find a solution to:
 Systems of equations
 Inequalities
 Optimization problems
 Linear programs***
 Integer programs
 Nonlinear programs
EXAMPLE
5
XYZ manufacturing company has a division that produces two models of
grates, model–A and model–B. To produce each model–A grate requires ‘3’ g.
of cast iron and ‘6’ minutes of labor. To produce each model–B grate requires
‘4’ g. of cast iron and ‘3’ minutes of labor. The profit for each model–A grate
is Rs.2 and the profit for each model–B grate is Rs.1.50. One thousand g. of
cast iron and 20 hours of labor are available for grate production each day.
Because of an excess inventory of model–A grates, Company’s manager has
decided to limit the production of model–A grates to no more than 180 grates
per day.
Solve the given LP problem and perform sensitivity analysis.
LP MODEL: Let X1 and X2 be the number of model–A and model–B grates
respectively.
The complete LP model is as follow:
Maximum: Z = 2X1 + 1.5X2  2X1 + (3/2)X2
Subject to:
3X1 + 4X2 ≤ 1000 (Cast Iron Constraint)
6X1 + 3X2 ≤ 1200 (Labor Hour Constraint)
X1 ≤ 180 (Production limit of Model-A Constraint)
X1, X2 ≥ 0
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – I: ENTER THE DATA & FUNCTION
Cell I8: Enter:
=SUMPRODUCT($G$6:$H$6,G8:H8)
Drag to cells G11:H11
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS
8
The “Solver Parameters” dialog box:
WINDOWS
 “Set Target Cell” window:
Identifies the cell that Solver
will use to record the optimal
z-value for the problem.
 “By Changing Cells”
window: Identifies the cells
that Solver will use to record
the optimal solution for the
decision variables.
 “Subject to the
Constraints” window:
Identifies the non-negativity
constraints and the constraints
given by the problem.
Buttons
 “Options” button: Identifies
the type of optimization
problem; remember to check
off the “Assume Linear Model”
option.
 “Add” button: Used to insert
the constraints; identified
constraints are displayed in the
“Subject to the Constraints”
window.
 “Solve” button: Used to
determine the optimal value
for the objective z and the
decision variables.
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
With the CURSOR in the
“Set Target Cell Box”: Click
on Cell “I8”
SET TARGET CELL:
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
LEAVE THE BUTTON
FOR Max
HIGHLIGHTED
EQUAL TO:
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
WITH THE CURSOR IN THE
“BY CHANGING CELLS
BOX”: HIGHLIGHT CELLS
“G6” & “H6”
BY CHANGING
CELLS:
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
SUBJECT TO THE CONSTRAINTS:
 In the “Solver
Parameters” dialog
box, click on the “Add”
button.
 Fill in the “Cell
Reference” and
“Constraint” windows
by clicking on the
changing cells and the
function cells.
 Click on the “OK”
button after adding
each constraint.
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
 With the cursor in the cell reference box: highlight cells “I9 through
I11”. Leave the direction as “≤”. With the cursor in the constraint
box: : highlight cells “K9 through K11”.
 If more constraints were to be added, click “Add” and follow the
same procedure.
SUBJECT TO THE CONSTRAINTS (Cont…):
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
OPTIONS:
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
SOLVE:
LINEAR PROGRAMMING USING EXCEL SOLVER
STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
REPORT:
LINEAR PROGRAMMING USING EXCEL SOLVER
Analyzing the Excel Spreadsheet
LINEAR PROGRAMMING USING EXCEL SOLVER
THE ANSWER REPROT
LINEAR PROGRAMMING USING EXCEL SOLVER
THE SENSITIVITY REPROT
Range of Optimality
 Changing the profit coefficient of the objective function
 Will the original optimal solution still be optimal?
 Range of Optimality?
 Profit coefficient for X1
 2, range of optimality (2 + 1, 2 – 0.875) = (3, 1.125)
 Profit coefficient for X2
 1.5, range of optimality (1.5 + 1.167, 1.5 – 0.5) = (2.667, 1)
LINEAR PROGRAMMING USING EXCEL SOLVER
THE SENSITIVITY REPROT
Changing the RHS – CAST IRONS
 Binding Constraints
 3X1 + 4X2 ≤ 1000 (Cast Irons Constraint)
 3(120) + 4 (160) = 1000
 Suppose we increase one gram Cast Iron, what’s the impact on
the optimal profit?
 The unit change in the objective function is the shadow price
of the resource.
 Shadow price of Cast Iron Gram = 0.2
 Range of Feasibility: (1000 + 600, 1000 – 300) = (1600, 700)
LINEAR PROGRAMMING USING EXCEL SOLVER
THE SENSITIVITY REPROT
Changing the RHS – LABOUR HOUR
 Binding Constraints
 6X1 + 3X2 ≤ 1200 (Labor Hours Const.)
 6(120) + 3(160) = 1200
 Suppose we increase one Labour hour, what’s the impact on the
optimal profit?
 The unit change in the objective function is the shadow price
of the resource.
 Shadow price of Labour Hour = 0.23333
 Range of Feasibility: (1200 + 225, 1000 – 450) = (1425, 550)
LINEAR PROGRAMMING USING EXCEL SOLVER
THE SENSITIVITY REPROT
Changing the RHS – LABOUR HOUR
 NON–Binding Constraints
 X1 ≤ 180 (Model-A Production Cont.)
 Optimum: 120 + 0 = 120 (Model–A Grates)
 We have 60 excessive Model–A Grates (slack)
 Increasing the Grates?
 Decreasing the Grates?
 Shadow price of Model–A = 0
 Range of Feasibility: (180 + ∞, 180 – 60) = (∞, 120)
EXAMPLE: PRODUCTION SCHEDULING
23
Cool-bike Industries manufactures boys and girls bicycles in both 20-inch and 26-inch models. Each week it
must produce at least 200 girl models and 200 boy models. The following table gives the unit profit and the
number of minutes required for production and assembly for each model.
X1 = Number of 20-inch girls bicycles produced this week; X2 = Number of 20-inch boys bicycles
produced this week; X3 = Number of 26-inch girls bicycles produced this week; X4 = Number of 26-inch
boys bicycles produced this week
MAX 27X1 + 32X2 + 38X3 + 51X4
S.T.
X1 + X3  200 (Min girls models)
X2 + X4  200 (Min boys models)
12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes)
6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes)
2X1 + 2X2  500 (20-inch tires)
2X3 + 2X4  800 (26-inch tires)
All X's  0
Bicycle Unit Profit Production Minutes Assembly Minutes
20-inches girls $27 12 6
20-inches boys $32 12 9
26-inches girls $38 9 12
26-inches boys $51 9 18
The Production and assembly areas run two (eight-hour) shifts per day, five days per week. This week there
are 500 tires available for 20-inch models and 800 tires available for 26-inch models. Determine Cool-bike’s
optimal schedule for the week. What profit will it realize for the week?
EXAMPLE: PRODUCTION SCHEDULING (Cont…)
24
X1 = Number of 20-inch girls bicycles produced this week; X2 = Number of 20-inch boys bicycles
produced this week; X3 = Number of 26-inch girls bicycles produced this week; X4 = Number of 26-inch
boys bicycles produced this week
MAX 27X1 + 32X2 + 38X3 + 51X4
S.T.
X1 + X3  200 (Min girls models)
X2 + X4  200 (Min boys models)
12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes)
6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes)
2X1 + 2X2  500 (20-inch tires)
2X3 + 2X4  800 (26-inch tires)
All X's  0
EXAMPLE: PRODUCTION SCHEDULING (Cont…)
25
MAX 27X1 + 32X2 + 38X3 + 51X4
S.T.
X1 + X3  200 (Min girls models)
X2 + X4  200 (Min boys models)
12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes)
6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes)
2X1 + 2X2  500 (20-inch tires)
2X3 + 2X4  800 (26-inch tires)
All X's  0
EXAMPLE: PRODUCTION SCHEDULING (Cont…)
26
MAX 27X1 + 32X2 + 38X3 + 51X4
S.T.
X1 + X3  200 (Min girls models)
X2 + X4  200 (Min boys models)
12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes)
6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes)
2X1 + 2X2  500 (20-inch tires)
2X3 + 2X4  800 (26-inch tires)
All X's  0
QUESTIONS
27

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4. linear programming using excel solver

  • 2. LINEAR PROGRAMMING USING EXCEL SOLVER 2 Excel Lingo On the toolbar at the bottom of the screen, click on:  Start  All Programs  Microsoft Office  Microsoft Office Excel 2007  Spreadsheet: A two-dimensional array of rectangles.  Cell: Each rectangle in excel (Four types of information can be typed into a cell: Number, Fraction, Function, and Text.) It is identified by its column and row location on the spreadsheet, which are designated by letter and numbers, respectively (i.e. cell A1).  SUMPRODUCT: A function that first multiplies the numbers in n consecutive cells (i.e. A1 through E1) by the numbers in another set of n consecutive cells (i.e. A5 through E5), respectively, then takes the sum of the n number of products (i.e. A1*A5 + B1*B5 + C1*C5 + D1*D5 + E1*E5), and finally deposits that sum in the cell you have selected (i.e. F1).  Cell reference: Lets you repeat patterns of information between cells, which occurs a selected cell refers to information typed in another cell.  Absolute reference: A cell that always refers to the originally referred cell; if the location of the selected cell changes, the referred cell will not change. It includes a “$” sign before the cell’s column (i.e. $A1), row (i.e. A$1), or both (i.e. $A$1).  Relative reference: A cell that initially refers to the originally selected cell; if the location of the selected cell changes, the referred cell will change and the location of the new referred cell will reflect the location change of the selected cell. It omits the “$” sign (i.e. A1).
  • 3. LINEAR PROGRAMMING USING EXCEL SOLVER 3 How to activate Solver:
  • 4. LINEAR PROGRAMMING USING EXCEL SOLVER 4 Solver can find a solution to:  Systems of equations  Inequalities  Optimization problems  Linear programs***  Integer programs  Nonlinear programs
  • 5. EXAMPLE 5 XYZ manufacturing company has a division that produces two models of grates, model–A and model–B. To produce each model–A grate requires ‘3’ g. of cast iron and ‘6’ minutes of labor. To produce each model–B grate requires ‘4’ g. of cast iron and ‘3’ minutes of labor. The profit for each model–A grate is Rs.2 and the profit for each model–B grate is Rs.1.50. One thousand g. of cast iron and 20 hours of labor are available for grate production each day. Because of an excess inventory of model–A grates, Company’s manager has decided to limit the production of model–A grates to no more than 180 grates per day. Solve the given LP problem and perform sensitivity analysis. LP MODEL: Let X1 and X2 be the number of model–A and model–B grates respectively. The complete LP model is as follow: Maximum: Z = 2X1 + 1.5X2  2X1 + (3/2)X2 Subject to: 3X1 + 4X2 ≤ 1000 (Cast Iron Constraint) 6X1 + 3X2 ≤ 1200 (Labor Hour Constraint) X1 ≤ 180 (Production limit of Model-A Constraint) X1, X2 ≥ 0
  • 6. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – I: ENTER THE DATA & FUNCTION Cell I8: Enter: =SUMPRODUCT($G$6:$H$6,G8:H8) Drag to cells G11:H11
  • 7. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS
  • 8. 8 The “Solver Parameters” dialog box: WINDOWS  “Set Target Cell” window: Identifies the cell that Solver will use to record the optimal z-value for the problem.  “By Changing Cells” window: Identifies the cells that Solver will use to record the optimal solution for the decision variables.  “Subject to the Constraints” window: Identifies the non-negativity constraints and the constraints given by the problem. Buttons  “Options” button: Identifies the type of optimization problem; remember to check off the “Assume Linear Model” option.  “Add” button: Used to insert the constraints; identified constraints are displayed in the “Subject to the Constraints” window.  “Solve” button: Used to determine the optimal value for the objective z and the decision variables. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)
  • 9. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) With the CURSOR in the “Set Target Cell Box”: Click on Cell “I8” SET TARGET CELL:
  • 10. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) LEAVE THE BUTTON FOR Max HIGHLIGHTED EQUAL TO:
  • 11. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) WITH THE CURSOR IN THE “BY CHANGING CELLS BOX”: HIGHLIGHT CELLS “G6” & “H6” BY CHANGING CELLS:
  • 12. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) SUBJECT TO THE CONSTRAINTS:  In the “Solver Parameters” dialog box, click on the “Add” button.  Fill in the “Cell Reference” and “Constraint” windows by clicking on the changing cells and the function cells.  Click on the “OK” button after adding each constraint.
  • 13. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…)  With the cursor in the cell reference box: highlight cells “I9 through I11”. Leave the direction as “≤”. With the cursor in the constraint box: : highlight cells “K9 through K11”.  If more constraints were to be added, click “Add” and follow the same procedure. SUBJECT TO THE CONSTRAINTS (Cont…):
  • 14. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) OPTIONS:
  • 15. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) SOLVE:
  • 16. LINEAR PROGRAMMING USING EXCEL SOLVER STEP – II: RECORD THE SOLVER PARAMETERS (Cont…) REPORT:
  • 17. LINEAR PROGRAMMING USING EXCEL SOLVER Analyzing the Excel Spreadsheet
  • 18. LINEAR PROGRAMMING USING EXCEL SOLVER THE ANSWER REPROT
  • 19. LINEAR PROGRAMMING USING EXCEL SOLVER THE SENSITIVITY REPROT Range of Optimality  Changing the profit coefficient of the objective function  Will the original optimal solution still be optimal?  Range of Optimality?  Profit coefficient for X1  2, range of optimality (2 + 1, 2 – 0.875) = (3, 1.125)  Profit coefficient for X2  1.5, range of optimality (1.5 + 1.167, 1.5 – 0.5) = (2.667, 1)
  • 20. LINEAR PROGRAMMING USING EXCEL SOLVER THE SENSITIVITY REPROT Changing the RHS – CAST IRONS  Binding Constraints  3X1 + 4X2 ≤ 1000 (Cast Irons Constraint)  3(120) + 4 (160) = 1000  Suppose we increase one gram Cast Iron, what’s the impact on the optimal profit?  The unit change in the objective function is the shadow price of the resource.  Shadow price of Cast Iron Gram = 0.2  Range of Feasibility: (1000 + 600, 1000 – 300) = (1600, 700)
  • 21. LINEAR PROGRAMMING USING EXCEL SOLVER THE SENSITIVITY REPROT Changing the RHS – LABOUR HOUR  Binding Constraints  6X1 + 3X2 ≤ 1200 (Labor Hours Const.)  6(120) + 3(160) = 1200  Suppose we increase one Labour hour, what’s the impact on the optimal profit?  The unit change in the objective function is the shadow price of the resource.  Shadow price of Labour Hour = 0.23333  Range of Feasibility: (1200 + 225, 1000 – 450) = (1425, 550)
  • 22. LINEAR PROGRAMMING USING EXCEL SOLVER THE SENSITIVITY REPROT Changing the RHS – LABOUR HOUR  NON–Binding Constraints  X1 ≤ 180 (Model-A Production Cont.)  Optimum: 120 + 0 = 120 (Model–A Grates)  We have 60 excessive Model–A Grates (slack)  Increasing the Grates?  Decreasing the Grates?  Shadow price of Model–A = 0  Range of Feasibility: (180 + ∞, 180 – 60) = (∞, 120)
  • 23. EXAMPLE: PRODUCTION SCHEDULING 23 Cool-bike Industries manufactures boys and girls bicycles in both 20-inch and 26-inch models. Each week it must produce at least 200 girl models and 200 boy models. The following table gives the unit profit and the number of minutes required for production and assembly for each model. X1 = Number of 20-inch girls bicycles produced this week; X2 = Number of 20-inch boys bicycles produced this week; X3 = Number of 26-inch girls bicycles produced this week; X4 = Number of 26-inch boys bicycles produced this week MAX 27X1 + 32X2 + 38X3 + 51X4 S.T. X1 + X3  200 (Min girls models) X2 + X4  200 (Min boys models) 12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes) 2X1 + 2X2  500 (20-inch tires) 2X3 + 2X4  800 (26-inch tires) All X's  0 Bicycle Unit Profit Production Minutes Assembly Minutes 20-inches girls $27 12 6 20-inches boys $32 12 9 26-inches girls $38 9 12 26-inches boys $51 9 18 The Production and assembly areas run two (eight-hour) shifts per day, five days per week. This week there are 500 tires available for 20-inch models and 800 tires available for 26-inch models. Determine Cool-bike’s optimal schedule for the week. What profit will it realize for the week?
  • 24. EXAMPLE: PRODUCTION SCHEDULING (Cont…) 24 X1 = Number of 20-inch girls bicycles produced this week; X2 = Number of 20-inch boys bicycles produced this week; X3 = Number of 26-inch girls bicycles produced this week; X4 = Number of 26-inch boys bicycles produced this week MAX 27X1 + 32X2 + 38X3 + 51X4 S.T. X1 + X3  200 (Min girls models) X2 + X4  200 (Min boys models) 12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes) 2X1 + 2X2  500 (20-inch tires) 2X3 + 2X4  800 (26-inch tires) All X's  0
  • 25. EXAMPLE: PRODUCTION SCHEDULING (Cont…) 25 MAX 27X1 + 32X2 + 38X3 + 51X4 S.T. X1 + X3  200 (Min girls models) X2 + X4  200 (Min boys models) 12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes) 2X1 + 2X2  500 (20-inch tires) 2X3 + 2X4  800 (26-inch tires) All X's  0
  • 26. EXAMPLE: PRODUCTION SCHEDULING (Cont…) 26 MAX 27X1 + 32X2 + 38X3 + 51X4 S.T. X1 + X3  200 (Min girls models) X2 + X4  200 (Min boys models) 12X1 + 12X2 + 9X3 + 9X4  4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4  4800 (Assembly minutes) 2X1 + 2X2  500 (20-inch tires) 2X3 + 2X4  800 (26-inch tires) All X's  0