Understanding Sensitivity Analysis in Linear Programming

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dear student in the previous lecture i have discussed about the graphical solutions using a graphical calculator

and also i have discussed about how to use solver in this lecture we can discuss

sensitivity analysis the agenda for this lecture is sensitivity analysis in that if there is any change in in the

coefficient of objective function how that will affect our result so the another name for sensitivity analysis is

post optimality analysis so sensitivity analysis is the study of

how the changes in the coefficient of an optimization model affect the optimal solution there are two way it will

affect how will a change in a coefficient of objective function affect the optimal

solution other one is how will a change in right hand side of a constraint affect the

optimal solution in this lecture we will discuss about the first case

that is if there is if there is any change in coefficient of objective function

how our solution optimal solution get affected

this was the problem what we have used right from the beginning maximize

ten years plus nine d there are four constraint so this problem

will ah i will explain how to do sensitivity analysis what is the center to analysis we are talking about if any

changes in the coefficient of this objective function for example if the 10 may be changed

the tennis is the profit contribution obviously the profit contribution will not be same

sometime the profit contribution will be more sometime the profit contribution will be less

but what will be effect of this profit contribution on our final solution i have brought the same problem here

the optimal solution for the given problem which is on the right hand side s equal to 540 standard bags and d equal

to that is a deluxe bags is equal to 252. so

based on the profit contribution fig figures of ten dollar per standard bag and nine

dollar per deluxe bag suppose we later learned that

a price reduction causes the profit contribution of the standard back fall from

10 dollar to say 8.5 dollar so sensitivity analysis

can be used to determine whether the production schedule calling for 540 standard bags

and 252 deluxe bags is still best if there is a drop in price

our solution that is s equal to 540 and d equal to 252 is the still best or where are this drop

in profit contribution will affect our final result or not that is the purpose

of sensitivity analysis if it is solving a modified linear programming problem

by changing the coefficient of objective function as the new objective function will not

be necessary so obviously you may ask this question if the profit contribution is decreasing

we can simply change that coefficient value in the excel then we can get the final answer but the recall redoing of

the whole process is not required so the management would especially want to consider

how the optimal production quantity quantities should be revised if the profit contribution per

deluxe bag were to drop so that is if the price of the product is dropping

how that will affect our our production quantities there is another situation where

the sensitivity analysis can be done suppose look at the right hand side of the constraint 630 in the cutting and

dying department the total available hours is 630 that is what i have

brought here the another aspect of sensitivity analysis concern changes in the right hand side value of

the constraint in the problem the optimal solution

used all available time in the cutting and dying department and finishing department

you might have recollect from our previous previous lectures so this 630 hours

this constraint is our binding constraint what is the meaning of binding constraint we have utilized all

630 hours this is happening for to the finishing department also that's how i

brought in different colors so what would happen to the optimal solution

and the total profit contribution if that company could obtain additional quantities of either of the resources

so what is the meaning here suppose if the company is trying to bring some additional resources

they say plus one something by bringing this additional resources we need to check whether that will

affect our final solution or that will maximize or that will

affect our objective function whether it is a maximization or minimization type this is another way of doing sensitive

analysis just i am recollecting there are two way to do the sensitivity

analysis one is by changing the coefficient of the objective function

the second one is by changing the value on the right hand side of our constraints

so sensitivity analysis can help determine how much each additional hour

of production time is worth and how many hours can be added before diminishing return certain

so what is the meaning of this is if i add one extra resources whether the adding this extra resources

is worth or not how much it is worth how much it is worth means how that will affect my affect my objective function

and sometime what will happen when you keep on adding the resources instead of it is

ah positively affecting our objective function sometime it will decrease so by

adding more resources will not help for our objective function so that is the case of this diminishing returns but in

this class we discuss about the first case that is if any changes in the

coefficient of the objective function how that will affect our final result let us consider how the changes in the

objective function coefficient might affect the optimal solution the current contribution to the profit

is ten dollar per unit for the standard bag and nine dollar per unit for the deluxe

bag it seems obvious that an increase in the profit

contribution for one of the bags might lead management to increase the production of that bag

and a decrease in the profit contribution for one of the bags might lead

management to decrease the production of that bag so the profit contribution is increases

obviously will company go for manufacturing more bags if it is decreases they will try to reduce the

number of quantities manufactured it is not as obvious however how much the profit contribution

would have to change before the management would want to

change the production quantities so what is the meaning of this we know that already we got the result

that is s equal to five hundred and forty standard bags and deluxe bag is two

fifty two if there is any small change in bag price

right sometime the small change in bag price will not affect your production quantities

but if the bag price is significantly changing so what it is required sometime you have

to change this production quantities also so what we are going to do in this one

how much allowable change is permitted how much how much profit contribution how much how much change in profit

contribution is permitted so that our solution remain changed so

i am going to introduce the term called range of optimality the current optimal solution to this problem call for

producing 540 standard golf bags and 252 deluxe golf bags the range of optimality for

each objective function coefficient provides the range of values over which the

current solution remain optimal i think this picture will explain you clear idea you see this is our coefficient of our

objective function so there is a lower limit for profit contribution that is upper limit for

profit contribution so this much change is permitted

so this change will not affect your final solution that is called range of optimality

even though there is a variation in the coefficient of your objective function that variation will not disturb your

solutions that allowable variation the range of variation is nothing but your range of optimality

the managerial attention should be focused on those objective function coefficient that have narrow

range range of optimality and the coefficient near the end

end point of the range obviously when then when the the range is very narrow

the management should provide more attention for that because immediately your solution need to be

changed because if there is slight variation in the product

the product price that may affect your scheduling so

with this coefficient a small change can necessitate modifying the optimal solution

now i will explain this range of optimality graphically we know that this region is our feasible

region which we have discussed in our previous lectures this point

is our where we got the optimal solution so now we will go back to the problem

the graphical solution of the problem with the slope of objective function line between

slope of line a and b the extreme point three is optimal so

this point is optimal point for the given problem see there are i am writing what is line

a and line b the cutting and dying department is called that line is i am calling it line a

the finishing department corresponding constraint i am calling it as line b ok

so the range of optimality so we have the objective function objective function is

this n ah which is mentioned in the orange color this is 10 s plus 90.

so as long as the slope of the objective function is between the line a and b

right as long as the slope of our objective function is between line and b

that will not affect our result so that will provide your range of optimality as long as the slope of objective

function line is between slope of line a see that which if it is slope of line a is this one which coincides with the

cutting and dying constraint line and the slope of line b which coincides with finishing

constraint line the extreme point 3 with s equal to 540

d equal to 252 will remain optimal so changing an objective function

coefficient for s r d will cause the slope of objective function line to change

if there is any change in the coefficient of s and d that

will affect the slope of that objective function in figure we see that such changes cause

the objective function line to rotate around the extreme point three so if any changes in the coefficient of

objective function that will cause rotating the objective function

however as long as the objective function line stays within the shaded region the

extreme point three will remain optimal what is the shaded region this region

this region rotating the objective function counter clockwise

suppose the objective function is in orange color suppose if you rotate this objective function counterclockwise

the slope to become less negative ok then the slope increases ok so when you rotate this way when the

objective function line rotates counter clockwise because the slope is increased

enough to coincide with the line a so suppose this line

coincide with line a we obtain alternative optimal solutions between extreme points three and four

what will happen if this orange line is coinciding with this red line so what will happen in this region

ok see all the points which are in this region will be your optimal solution so we will get alternate optimal solution

any further counter clockwise rotation of the objective function line will cause

extreme point 3 to be non optimal so beyond this beyond this red line if you rotate this

orange line beyond that so what will what will become then the solution point three will not be the

optimal solution hence the slope of line a right the slope of line a provides an

upper limit for the slope of the objective function line now we will see the other case

rotating the objective function line clockwise ok now the objective objective function

line is going to be rotated clockwise causes the slope to become more negative and slope decreases

when the objective function line rotates clockwise then slope will decrease

enough to coincide with line b what is line b which is in the red color

line b we obtain alternate optimal solution between three

and two so what will happen in this region your objective objective

function line will sit there so when it is coinciding so all these points between two and three will become

your optimal solution you will get multiple optimal solution

any further clockwise rotation of the objective function line will cause extreme point 3 to be non optimal so if

you rotate beyond this so 3 will not be your optimal solution hence the slope of line b provides a

lower limit for the slope of the objective function line so now we got the upper limit and lower limit

of your objective function line so that the point 3 will remain optimal so the extreme point 3 will be optimal solution

as long as slope of line b is less than or equal to

slope of line objective function line is less than or equal to slope of line a

what it says that if the slope of your objective function line

is between slope of line b and a your optimal solution will remain same

that will not disturb your optimal solution in given figure which is given on the right hand side we see that

the equation for line a line a is the this one line a that is the cutting and

dying constraint line is like this

what is that line because it is equal to written seven upon ten years plus one d equal to

six hundred and thirty so i want to know the slope of this equation

so i am going to write rearrange we know that very very gen y equal to m x plus b where m is the slope

here y is nothing but your d number of deluxe bags so i am going to write this this equation

to bring in this form y equal to m x plus b so that i can find out what is the slope

so when you simplify d ok you will get minus seven upon ten s

plus six hundred and thirty so the minus seven upon ten says the slope of your line a

the 630 is the intercept of line a on d d axis

ok similarly we will find the slope of line b

in figure three in figure we see that the equation for line b is this one one s plus

two upon three d equal to seven hundred and eight so two upon

three d i am simplifying when i brought on the right hand side minus s plus seven zero eight

so when you simplify this d equal to minus three by two s plus one thousand sixty two now this is your slope of your

line b now i am going to write our objective function in the generic form

so p equal to c s s plus c d d c s means coefficient of standard back c d means coefficient of deluxe back

so when i simplify c d equal to minus c s e s plus p whole divided by

d so if i further simplify i get d equal to minus c s by c d s

plus p by c d we hold the profit contribution for the deluxe bag

fixed at at initial value c d equal to 9 we know that it is 10 s

plus 9 d so i am going to fix

this value of c d equal to nine we know that minus three by two less than or equal to minus c s by c d

so instead of c d i have substituted the value nine less than or equal to minus seven by ten

so when i simplify suppose i take only on the left hand side when i simplify this minus three by two

less than or equal to minus c s by nine so when minus get cancelled so the value of c s what i am getting is

thirteen point five similarly if i take the right hand side inequality

minus cs by nine less than or equal to minus seven upon ten so i am getting the cs value six point

three so what we got i have got the coefficient value for the standard bag

between 6.3 to 13.5 so what this number says that

as long as the coefficient of standard bag is within this range that will not disturb our final answer

there is a final solution so that will not affect our optimality so how we are interpreting

in the original problem the standard back had a profit contribution of 10 dollar

so the resulting optimal solution was s equal to 540 d equal to 252

so the range of optimality for c s tells for the given problem

with other coefficients unchanged what is other coefficient the c d

c cd is unchanged thats what we have substituted cd equal to 9 the profit contribution for the standard

bag can be anywhere between 6.3 dollar and 13.5 dollar

and the production quantities of 540 standard bags and 252 deluxe bags will remain optimal

that is the interpretation of in the what i the slope which i have written in the previous slide

even though the production quantities will not change the total profit contribution

that is the value of objective function will change the value of objective function will

change due to change in profit contribution of standard backs so what is that our objective function z

equal to 10 s plus 90. so the value of this

coefficient of standard biax even though it is goes up to 13.5 and 6.3

our solution which we got it s equal to 540 d equal to 252 will remain same that is

the interpretation of this range of optimality so range of optimality for standard bags

similarly we can find the range of optimality for the deluxe bag also that is what we have done it here

the same equations so here c s by c d we know that c s by

c d minus so the c s value i have substituted equal to 10 because our

objective function is 10 s plus 9 d

so i am going to find out what is the allowable range for the coefficient of deluxe bags

by assuming that the coefficient of standard bag is 10. so the 10 i have substituted

and then i have readjusted on the left hand side so i got 6.67 is lower limit

and the upper limit is fourteen point two nine so what this number

implies that as long as the coefficient of deluxe bags this one

goes up to six point six seven and fourteen point two nine our solution that is our

s equal to 540 d equal to 252 remain same now how to see this range of optimality in our

solver output so i have brought the output of the solver in the variable cell you see that

the final value that is our answer this is objective function coefficient 10 and 9

so here the allowable increase is 3.5 that is 13.5 allowed decrease

three point seven similarly for the deluxe bag the

currently it is nine it can be increased ah nine plus 5.2

and decreased by 9 minus 2.3 so this is the place where you can get

the range of optimality so the range of optimality is providing

kind of your flexibility in the price changes even though the price changes we need

not bother about that because that will not affect our our optimal solution now i will show this

output in our excel sheet now this problem which we have solved in the

previous class we got the answer is 540 and 252 when i go to

this answer report you have to go to sensitivity report look at this location this location that

is final value is our answer there is a objective function there is

allowable increase and allowable decrease so this says the range of optimality

now we can see the a special case in range of optimality in cases where the rotation of the

objective function line about an optimal extreme point causes the objective function line is become

vertical see this green color saves our given given our

objective function sometime if the objective function is vertical there will be either no upper limit or

no lower limit for the slope as it appears in the form of expression

so we know that minus three by two less than or equal to minus c s by c d less than or equal to minus seven by ten

to show this special situation can occur suppose the objective function for the problem is eighteen c s plus

nine d say this is the new objective function in this case

the the extreme point two so now this point two provides the optimal solution

so rotating this objective function line counter clockwise counter clockwise

this way rotating in counter clockwise around the extreme point 2 provides an upper limit for the slope

when the objective function line coincides with the line b so rotating the objective function line

counter clockwise around extreme point 2 provides an upper limit for the slope when the objective function line

coincides with the line b o this is our line b we showed previously

the slope of line b is minus three upon two so the upper limit for the slope of the objective function line must be

minus three by two there is upper limit there is no problem however rotating the objective function

in clockwise suppose if you rotate this objective function this objective function clockwise

clockwise result in the slope becoming more and more negative approaching a value of

minus infinity as the objective function line become vertical in this case the slope of the objective

function has no lower limit using the upper limit of minus 3 upon 2 we can write minus c s by c d less than or

equal to minus 3 by 2 that is the slope of our objective function following the previous procedure of

holding c d constant and the original value that is c d equal to nine

so we have minus c s by nine less than or equal to minus three by two

so when you simplify this we are getting c s is thirteen point five in reviewing this figure we note that

the extreme point two remains optimal for all values of c s about thirteen point five

thus we obtain the following range of optimality for cs at extreme point so

the val the lower limit for this coefficient of standard bag is 13.5 and there is no upper limit this is the

special case of optimality when the objective function is vertical now there may be another situation

simultaneous changes what is the meaning of simultaneous changes

coefficient of standard bags and coefficient of deluxe bags so simply

compute the slope of the objective function minus c s by c d for the new coefficient

values if this ratio is greater than or equal to the lower

limit on the slope of the objective function r

less than or equal to the upper limit then the changes made will not cause a change in optimal solution

if there is a simultaneous change you have to find the ratio of minus cs by cd

if that ratio is within the limit so our final result will not change so consider changes in both of the

objective function coefficients for the problem which we are discussing suppose the profit contribution

for the standard bag is increased to 13 and the profit contribution per deluxe bag is simultaneously reduced to eight

now both the things are happening together recall that changes for optimality for c

s and c d both computed in one at a time manner the way we have arrived c s the range of

optimality because we have kept the coefficient of one

for example c s is kept constant we found the answer for the d similarly then d kept constant so we got

the answer for yes but now there is a simultaneous changes

so c s is 6.3 to 13.5 now c d is 6.67 to 14.29

so for these ranges of optimality we can conclude that

changing either cs to 13 dollar rcd to 8 but not both

any one changes can take place would not cause change in optimal solution s equal to 540

d equal to 252 even though there are two changes are taking place

but we are considering only one change at a time

if there is any one change taking place at a time both the

changes are within the limit that will not affect our our optimal solution

but we cannot conclude from the ranges of optimality that changing both coefficients

simultaneously would not result in change in the objective function

so what will happen if there is simultaneous changes so that will affect our

range of optimality simultaneous changes we showed that the extreme point 3 remains optimal as long as

ok minus 3 upon 2 less than or equal to minus c s by c d minus 7 by 7 upon 10. so i brought in the

the number line 0 minus so as long as this ratio is within this limit

that will not disturb our optimal solution if the c s is changed to 13

and simultaneously c d is changed to 8 the new objective function slope will will be given by minus c s by c d so we

will get minus thirteen upon eight so minus one point six two five so this minus one

point six two by uh minus 1.625 will go on this side so what will happen if both the things

are changes are taking place simultaneously so you will not get the optimal suit that is you will get

optimal solution the solution which you are already are having that will be disturbed

because it is going beyond the ranges yeah minus one point six two five because minus

one point sixty five is less than the lower limit of minus 3 by 2 the current solution

s equal to 540 and d equal to 252 will now will no longer be optimal so in this case by resolving the problem with

the c s equal to thirteen and c d equal to eight we will find the extreme point two is the new optimal solution now this

is two is the new optimal solution so what we are learning from here if there is a simultaneous changes then we have

to resolve the problem so sensitivity analysis what we have discussed so far is any one value changes at a time

not both if both the coefficients are changing at a time we have to resolve the problem

range of optimality simultaneous changes looking at the ranges of optimality we conclude that

changing either cs to 13 or cd to 18 but not both would not cause change in the optimal

solution but in a recomputing the slope of the objective function with simultaneous

changes for both c s and c d we

saw that the optimal solution did change so what we are learning if there is a simultaneous changes

our optimal solution will change so the result emphasize this fact that the range of optimality by itself

can only be used to draw a conclusion about changes made to one objective function coefficient at a

time if both the coefficients are changing simultaneously the range of optimality the answer which

we got will not be valid

now we will summarize what we have learnt in this class we have learnt

how will a change in the coefficient of objective function affect the optimal solution then we have learnt a new term

called range of optimality then we have seen if there is a simultaneous changes of coefficient of objective function

how that will affect our optimal solution ok student the next class

we will learn how a change in right side value of constraint affect the optimal solution

so far we have seen if any changes in the coefficient of objective function how data will affect

our optimal solution but in the next lecture if the right hand side of the constraint

is changing how that will affect our optimal solution

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