Excel solver

or - understanding excel solver - ndc

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UNDERSTANDING EXCEL SOLVERSuper Swim

Super Swim manufactures two styles of men’s swimming costumes, shorts (S) and briefs (B). How many pairs of shorts and briefs should be produced each week to maximise profits given the following constraints:

The profit per shorts is Rs. 3.00, compared with Rs. 4.50 per briefs. Briefs use0.5 metres of material; shorts use 0.4 metres.300 metres of material are available.

It requires 1 hour to manufacture a pair of shorts and 2 hours for a pair of briefs. 900 labour hours are available. (NB. shorts/trousers etc a pair means one piece)There is unlimited demand for shorts but total demand for briefs is 375 per week.

Each pair of shorts has an insignia logo, and 600 logos are in stock.


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Revision - LP Formulation

Steps in formulating a Linear Programming (LP) model

1. Understand the problem.

2. Identify the decision variables.

3. State the objective function as a linear combination of the decision variables.

4. State the constraints: ¶ upper or lower bounds on the decision variables, including non-negativity constraints.

¶ linear combinations of the decision variables.


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Variables: number of briefs = B, number of shorts = S.Objective function:maximize ( Rs.4.50 x B ) + ( Rs.3.00 x S ) = P 4.5B + 3S = P

LHS RHSVARIABLES OBJECTIVE

Constraints:Material: ( 0.5 x B ) + ( 0.4 x S ) <= 300 yards. 0.5B + 0.4S <= 300Labour: ( 2 x B ) + ( 1 x S ) <= 900 hrs. 2B + S <= 900Logos: ( 1 x S ) <= 600 S <= 600

LHS RHSUTILISATION CAPACITY

Demand: ( 1 x B ) <= 375 units. B <= 375LHS RHS

PRODUCED DEMANDED

Non-Negativity: B, S >= 0

Algebraic Formulation

(Refer to file 08_SuperSwim.xls)


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Graphical LP SolutionConvert the inequalities to equations, find the intercepts on the X and Y axes for each constraint and draw them. Since they are all upper bound constraints (<=) the feasible region will be as shown.

The objective function line is tangent to the feasible region at the intersection of labour and material constraints whereS = 500, B = 200 and profitP = 3x500 + 4.5x200 = Rs.2400.

Material0.5B + 0.4S = 300

Logos S = 600

Demand B = 375

Labour 2B + S = 900

Objective Function

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EXCEL SOLVERExcel solver is a computer programme for solving LP, IPand NLP problems. It comes with Microsoft Excel as an Add In.

Click on Tools to see if Solver is present. If not, click on

Add-Ins in the Tools menu and select the Solver Add-In.

Excel was developed for Microsoft by Frontline Systems.

http://www.frontsys.com/xlhelp.htm.


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Layout

Design the layout of the model data for clarity. Use descriptive names and labels for variables, constraints and bounds.

Decision Variables: Assign each to a cell and name it.

Constraints: Enter each RHS value in a cell. Enter the formula for each LHS

in a cell. Use Contiguous Ranges. Combine similar constraints ( < =, =, >= ).

Enter the Objective Function as a formula in a cell.

Check for ErrorsFormulasConsistency (e.g. Units, Scaling)

(Refer to file 08_SuperSwim.xls)

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Solver Worksheet Layout

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Solver Worksheet - Initial Conditions

Explanations in slides 9, 10. (Refer to file: 08_SuperSwim.xls)


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Solver Worksheet ExplanationsFirst note the labels used to identify cells. They clearly identify the contents. When solver creates reports it adds names to cells by taking the label on the left and the one directly above. If appropriate labels are used its very easy to read the reports.

Starting values for the decision variables are arbitrary but should be of the same order of magnitude as the

likely final values. These can usually be estimated from the constraints. Do not use zero as a starting value.

The RHS and LHS values of the constraints. RHS values are the bounds of the problem. They represent

resource availability. LHS values are calculated according to the constraint inequalities and their final values appear corresponding to the optimal solution. They represent utilisation of resources.

C D E F G28 Briefs Shorts29 Quantity 100 10030 Used Available31 90 300 0.5 0.432 300 900 2 133 100 600 0 134 Supplied Maximum35 100 375 1 03637 Profit/unit 4.5 33839 Maximum 450 300

Profit750

Cells E31:F35 are the coefficients of the

decision variables in the constraint inequalities. Cell C31 contains the formula SUMPRODUCT($E$29:$F$29,E31:F31) i.e. (E29 x E31)+(F29 x F31) which calculates the LHS of the material constraint; the quantity used. The $ sign indicates absolute cell references. Absolute references are used to make formula entries easier. Since the constraint LHS coefficients are all multiplied by the decision variables, it is only necessary to enter the formula in Cell C31 and then copy and paste it into C32, C33, C35. (Click on Help for more information about Absolute Cell References and the SUMPRODUCT formula). Cells E39 and F39 contain the formulas (E29 x E37) and (F29 x F37) i.e. the profits generated individually by Briefs and Shorts. The value of the Objective Function appears in cell G39 = SUM(E39:F39). At first all the cells with formulas show the initial set up values. After Solver calculates the model they show the optimal results. E29, F29 show the product mix; C31:C33, C35 show resources used; G39 shows the profit.

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Run Solver1. In the worksheet click on the final result cell to place the cursor there (G39 in this example). Then click on Tools - Solver and you will get:

If the window, Set Target Cell, does not already show $G$39, click on cell G39 in the worksheet and it will be entered in the window.


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2. Select Max for a maximising problem or Min for a minimising problem.

3. Click in the By Changing Cells window then click and drag the cursor across the decision variables cells E29 and F29 :

4. Click on the Add button :


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5. Click in the Cell Reference window then click in cell C31 and drag to C33. Leave the <= sign as it is because the constraint LHS are all <= the RHS. Click in the Constraint window then click in cell D31 and drag to D33:

6. Click on Add and repeat for cells C35 and D35. 7. Click on Add and enter cells E29 and F29 into Cell Reference in the same way as above. In the centre window select >= then click in Constraint and enter 0. This enters the non negativity condition for the decision variables.


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8. Click OK in the Add Constraints dialog box and you will get:

All the model parameters are now entered in solver.


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9. Click on Options and tick the windows Assume Linear Model and Assume Non-Negative. Leave the other parameters as they are. Click on Help for explanations.


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10. Click on OK to go back. Check the entries for errors and correct if any. Click on Solve. If the model is correct and there are no entry errors you will get:

Click on Help for information about this dialog box.

11. In the Reports window click on Answer, Sensitivity and Limits and click OK. The three reports will be created and Solver will close. The reports can be viewed by clicking on the tabs at the bottom of the screen.


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12. It is always a good idea to save the model in case Solver gets reset or used for another problem in the same worksheet. Reopen Solver, click Options and Save Model. A number of cells will be highlighted at the place where the cursor is currently situated. Count them and select the same number of cells anywhere in your worksheet (see previous slide which has the model saved. The title Model is a label. Solver has saved the values in cells H30:H35). Click OK and go back to the Solver Parameters box. Click on Reset All and Solver will be cleared. Click Options and Load Model. Select the cells where you have saved the model and click OK in the dialog box. Go back to Solver Parameters and you will see that all the data have been restored.


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SENSITIVITY ANALYSIS1. Typically many of the parameters of a linear programming model are only estimates of quantities (e.g. unit profits) that cannot be determined precisely in the beginning. Sensitivity analysis reveals how accurate each of these estimates needs to be to avoid obtaining an erroneous optimal solution. It pinpoints the sensitive parameters that need extra care to refine their estimates because even small changes in their values can affect the optimal solution.

2. If conditions change after the study has been completed (a common occurrence) sensitivity analysis gives indications (without solving the model again) whether the resulting change in a parameter of a model changes the optimal solution.

3. When certain parameters of a model represent managerial policy decisions sensitivity analysis provides guidelines regarding altering the impact of these decisions.

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The Answer Report shows us the optimal value of the objective function, the decision variable values at the optimum and the status of the constraints,whether binding or non binding.

Binding constraints - the optimal solution lies on binding constraints. Multiple binding constraints means that the optimal solution lies at their intersection i.e. a vertex.

Slack measures utilisation of resources.

Slack = 0 : the resource is exhausted, the constraint is binding, the optimal vertex includes the constraint.Slack 0 : spare resource is available, constraint is not binding


Logos S = 600

Demand B = 375

Labour 2B + S = 900

Objective Function

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The Answer Report

Graphical Solution to Super Swim


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ConstraintsCell Name Cell Value Formula Status Slack

$C$31 Material Used 300 $C$31<=$D$31 Binding 0$C$32 Labour Used 900 $C$32<=$D$32 Binding 0$C$33 Logos Used 500 $C$33<=$D$33 Not Binding 100$C$35 Demand Supplied 200 $C$35<=$D$35 Not Binding 175$E$29 Quantity Briefs 200 $E$29>=0 Not Binding 200$F$29 Quantity Shorts 500 $F$29>=0 Not Binding 500

From the constraints part of the answer report we can see, for example, that:

… Getting additional logos would not generate any profit because we already have 100 logos in stock at the optimum solution.… Advertising briefs to stimulate demand at the current price would be a waste of money because the optimum production is much less than the demand.… Profit is constrained by material and labour availability.… Briefs and shorts are not binding because we can make more if resources are available.


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The Sensitivity ReportThe Sensitivity Report allows you to see:

… over what range and under what conditions the components of a solution remain unchanged.

… how sensitive a solution is to changes in the data, and to get an insight into how changes may affect optimum values.

What is the Optimal Vertex?

The Optimal Vertex is the intersection of constraints at which the Optimal Solution is found.


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From the constraints part of the sensitivity report we have:

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$31 Material Used 300 5 300 15 52.5$C$32 Labour Used 900 1 900 131.25 60$C$33 Logos Used 500 0 600 1E+30 100$C$35 Demand Supplied 200 0 375 1E+30 175

We can identify binding and non binding constraints by seeing in which of them the final value is respectively equal to the RHS and less than the RHS. Hence material and labour are binding but logos and demand are not. The allowable increase and decrease for binding constraints is the range over which the RHS can be changed while still retaining the same optimal vertex i.e. the optimal vertex will remain at the intersection of the same constraints. However the value of the objective function and the product mix will change.


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Effect of increasing RHS of thematerial constraint by the allowable limit of 15 i.e. making15 m. more material available.The optimal vertex is still on the intersection of material and labour, but notice that logos have become binding as well.S = 600, B = 150 and profitP = 3x600 + 4.5x150 = Rs.2475.We can see that supplying any more material will move this constraint line away from the feasible region leaving logos and labour binding.

Labour 2B + S = 900


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Material0.5B + 0.4S = 247.5

Effect of decreasing RHS of thematerial constraint by the allowable limit of 52.5 - making52.5m less material available.The optimal vertex is still on the intersection of material and labour but demand has alsobecome binding.S = 150, B = 375 and profitP = 3x150 + 4.5x375 = Rs.2137.5Reducing material supply any more will drive the constraint down, squeezing the feasible region further. Material and demand only will be binding.


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Effect of Changing Binding Constraint RHS for >= and <= ConstraintsGraphical Visualisation

Optimal Solution for the above model:

Maximise 5x + y = Z subject to 15x + 8y <= 960, 5x + 4y >= 360, y <= 80, x, y >= 0

Allowable RHS limits:UB: +120, - 200. LB: +66.7, - 40.… if we change the RHS of the binding constraints within allowable limits, the optimal vertex will still remain on their intersection. … the value Z of the objective function and the optimal quantities of the decisionvariables (x, y) will change.

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Upper Bound15x + 8y = 960Lower Bound5x + 4y = 360Z = 270

Non Bindingy = 80

(48,30)


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Constraint <=, increase RHS.

… for constraints of <= type, increasing the RHS means loosening the bound. 15x + 8y <= 1020 (960 + 60)… the feasible region expands, the value of the objective function increases.… the optimal vertex remains on the intersection of bindingconstraints. Optimal values of the decision variables change.

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Non Bindingy = 80

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Constraint <=, decrease RHS.

… for constraints of <= type, decreasing the RHS means tightening the bound. 15x + 4y <= 840 (960 - 120)… the feasible region contracts, the value of the objective function decreases.… the optimal vertex remains on the intersection of binding constraints. Optimal values of the decision variables change.

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Non Bindingy = 80

(24,60)


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Constraint >=, decrease RHS.

… for constraints of >= type, decreasing the RHS means loosening the bound. 5x + 4y <= 320 (360 - 40)… the feasible region expands, the value of the objective function increases.… the optimal vertex remains on the intersection of binding constraints. Optimal values of the decision variables change.

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Non Bindingy = 80

(64,0)


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Constraint >=, increase RHS.

… for constraints of >= type, increasing the RHS means tightening the bound. 5x + 4y <= 400 (360 + 40)… the feasible region contracts, the value of the objective function decreases.… the optimal vertex remains on the intersection of binding constraints. Optimal values of the decision variables change.

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Non Bindingy = 80

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Value of a Scarce Resource

Unit valueof scarce =resource

Change in value of objective function

Corresponding allowable change in resource

Referring to slides 22 - 24, we see that when the RHS of the material constraint changes from 315 to 247.5, the value of the objective function changes from 2475 to 2137.5. Hence the unit value of material is:

(2475 - 2137.5) / (315 = 247.5) = 337.5 / 67.5 = 5

i.e. Rs. 5 per metre. Similarly the unit value of labour is Re. 1 per hour.


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Shadow PriceThe sensitivity report shows that the binding constraints, material and labour have shadow prices. The shadow price is nothing but the unit value of a scarce resource, one that has been exhausted. Because material and labour do not have any available capacity we cannot manufacture more products. If we could make capacity available we could make more products and hence increase the value of the objective i.e. increase profit.

The Shadow Price of a binding constraint is the amount by which the value of the objective function changes if the constraint changes by one unit.

The Shadow Price of a binding constraint only remains valid within the range of Allowable Increase and Decrease for that constraint.


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For the non binding constraints the value 1E +30 is exponential notation for10 raised to the power 30, effectively infinity. As long as the original optimalsolution is true, increasing the constraint RHS of logos and demand will nothave any effect on it. In other words, since there is already unused capacity, no amount of additional capacity will be of any use. The allowable decrease is limited. Below the allowable decrease the constraints will have an effect on the current optimal solution.

For a non binding constraint either the allowable increase or the allowabledecrease will be equal to the slack because either adding / subtracting theslack to / from the constraint RHS will make it binding.


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Example: Only 285 metres of material is found to be useable. How will thisaffect the profit for the week?

New optimal profit = Old optimal profit - (shadow price x material lost)= 2400 - ( 5 x [300 - 285] ) = 2400 - 75 = Rs. 2325.

Example: How much can management afford to pay for 100 hours of overtime?

Addition to optimal profit = shadow price x hours = 1 x 100 = Rs.100.

It can afford to pay a part of the additional profit as overtime wages - subjectto negotiation. However, longer working hours may not yield the sameproductivity.

Figures don’t always tell the whole story!


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The sensitivity report does not tell us what will happen if the coefficients of thedecision variables in the constraints change. If a constraint coefficient changesthe slope of the constraint changes. One or more feasible region vertices willchange. The problem will have to be re-solved.

Shadow Price in Maximisation and Minimisation Problems

A positive shadow price increases the value of the objective function.A negative shadow price decreases the value of the objective function.

For maximisation problems a positive shadow price is beneficial.For minimisation problems a negative shadow price is beneficial.


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Decision Variables and Objective Function CoefficientsFrom the adjustable cells part of the sensitivity report we have:

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$E$29 Quantity Briefs 200 0 4.5 1.5 0.75$F$29 Quantity Shorts 500 0 3 0.6 0.75

The allowable increase and decrease for the for the objective function coefficients tells us how much the objective function coefficients can change before the optimal solution vertex changes. If an objective function coefficient changes, the slope of the O.F. line changes, but it will still be tangential to the feasible region at the current optimal vertex if the change is within the allowable limit.


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Changing a decision variable coefficient in the objective function within allowable limits:… changes the value of the objective function, reduced costs and shadow prices.… does not change the product mix.… does not change the shape of the feasible region.

If the sensitivity report shows zero allowable increase and decrease for anobjective function coefficient it is an indication of multiple optimal solutions.It implies that the objective function line coincides with an edge of the feasible region.


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Coefft. of briefs = 4.5, allowable increase = 1.5. As the briefscoefficient increases the O.F. line will turn towards the labour line but still be tangential at the current optimal vertex. Value of the O.F. will change. When the coefft. = 6 the two lines will have the same slope. There will be multiple optimal solutions. Beyond 6 the O.F. line will be tangential at the intersection of labour and demand and there will be a new optimal solution vertex and a new product mix.


Logos S = 600

Demand B = 375

Labour 2B + S = 900

Objective Function

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New vertex


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Examples:

1. If the contribution by shorts increases by Rs. 0.50 to Rs 3.50. What would happen?

No change in the product mix: briefs = 200, shorts = 500 ( Why? )

Change in weekly profit = Rs.0.50 x 500 shorts/week = Rs. 250 Total profit = 2400 + 250 = Rs. 2,650.

Note the range: What could you say about a Re. 1 increase?

2. The contribution of briefs decreases by $0.75 to $3.75. What is the new mix? What is the change in weekly profit?

No change in the product mix: briefs = 200, shorts = 500. ( Why? )

Change in weekly profit = Rs.0.75 x 200 briefs/week = Rs.150Total profit = 2400 - 150 = Rs. 2,250.


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Reduced CostReduced cost is associated with each decision variable. If a decision variable appears in the final optimal solution its reduced cost will be zero. It shows that there is equilibrium between output (unit profit) and input (unit cost of used resources).

If a decision variable is zero and non basic (does not appear in the final optimal solution) it will have a reduced cost. (There is one exception which will be mentioned later.)

If a non basic decision variable is forced into the final solution, the value of the objective function will worsen. For example, in a production mix problem if the quantity of a product, say P, is shown as zero at the optimal solution, we should not make it. If we insist on making it, it will consume resources that other products need and worsen the value of the objective function.

Reduced cost is the change in the value of the objective function per unitchange in the value of the non basic variable introduced into the mix.


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Alternatively, we can look at the reduced cost as the amount by which the contribution (coefficient) of the decision variable must be improved before it will have a beneficial effect on the optimal solution.

A worse solution for a maximisation problem is a lower value of the objective function. Hence reduced costs in maximisation problems are negative.

A worse solution for a minimisation problem is a higher value of the objective function. Hence reduced costs in minimisation problems are positive.


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Amended Model to Demonstrate Reduced Cost

Let the profit per unit for briefs = Re. 1 and for shorts = Rs. 10.Let the material required per unit for shorts = 1 metre.Keep all other parameters as before.The optimal solution is: 0 briefs, 300 shorts, maximum profit Rs. 3000.The sensitivity report is:

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$E$58 Quantity Briefs 0 -4 1 4 1E+30$F$58 Quantity Shorts 300 0 10 1E+30 8

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$60 Material Used 300 10 300 300 300$C$61 Labour Used 300 0 900 1E+30 600$C$62 Logos Used 300 0 600 1E+30 300$C$64 Demand Supplied 0 0 375 1E+30 375


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The reduced cost of - 4 for briefs means that if we now produce briefs the profit will decrease by Rs. 4 per brief.

Another way of looking at the situation is via material usage, which is the only binding constraint in the new scenario. Briefs contribute Re.1 each. But each pair of briefs uses 0.5m of material and so for every 2 briefs we lose one pair of shorts which requires 1m of material. Shorts contribute Rs. 10. Hence the marginal loss is:

10 - 2 = Rs.8 for 2 briefs or Rs. 4 per brief

which is the reduced cost.


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Multiple OptimaWe saw earlier that one indication of multiple optima was zero allowable increase and decrease to an objective function coefficient.

A second indication of multiple optima is if a decision variable is zero at the optimal solution i.e. non basic and its reduced cost is zero. This means that if the variable is brought into the solution it will not have any effect on the value of the objective function. Therefore there is an alternative solution.


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Changing Multiple Parameters Simultaneously

The 100% RuleSo far we have only considered changes in one parameter at a time. To see what happens if we change more than one we use the 100% rule.

The rule applies to objective coefficients, constraint coefficients and constraint right hand sides. If the sum of percentages is greater than 100% we have to re-solve the problem. The next slide shows an example.

If several parameters are changed simultaneously, the current solution will remain optimal if the sum of the changes expressed as percentages of the allowable changes is not more than 100 percent.


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Suppose material availability reduces from 300 to 290 metres and labour hours increase from 900 to 1000, what can we infer? The sensitivity report shows: Constraints

Final Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease

$C$31 Material Used 300 5 300 15 52.5$C$32 Labour Used 900 1 900 131.25 60$C$33 Logos Used 500 0 600 1E+30 100$C$35 Demand Supplied 200 0 375 1E+30 175

The ratio for material is (10/52.5) x 100 = 19.05% (loss / allowable decrease)The ratio for labour is (100/131.25) x 100 = 76.19% (gain / allowable increase) The sum of percentages = 95.25% which is less than 100% Therefore we can calculate the change in the objective function as Re.1/hr x 100hrs (gain)- Rs.5/m x 10m (loss) = 100 - 50 = Rs.50 and its new value will be Rs.2400 + Rs.50 = Rs.2450. However, we need to re-solve to get the new product mix, which comes out as 366.67 briefs and 266.67 shorts.


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Example - Evaluating a New ProductProduct design has suggested a new line of briefs that require 1 metre of material and 2 hours of labour. Profit will be Rs.6 per brief. Available resources remain the same as before. Should it be introduced?

Remember that the new product will take away resources from the existing products so we are talking about allowable reductions. Referring to the sensitivity report we have (material: 1/52.5 = 0.019 + labour: 2/60 = 0.033). Total 0.052 < 1.000. (We can work with ratios or percentages).

Since the 100% rule applies we have:

Loss from 1 new brief = Rs.5/m x 1m + Rs.1/hr x 2 hrs = Rs.7.

Since the loss of Rs.7 is greater than the profit of Rs.6 the product should not be introduced.

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