236
Section Mathematics 5.2: Sensitivity analysis using Lindo
(Revised October, 2003)
Simplex Solution to Winco Problem (Winston, pages 231-233)
MAX
New Variables coming into
problem
Z = 4 X1 + 6 X2 + 7 X3 + 8 X4
SUBJECT TO:
2)
Slack
X1 +
X2 +
X3 +
X4 =
950
3)
X4 >=
400
4)
2 X1 + 3 X2 + 4 X3 + 7 X4 <=
4600
3 X1 + 4 X2 + 5 X3 + 6 X4 <=
5000
Artificial
S4
5)
Excess
S5
A2
e3
A3
and X1, X2, X3, X4 all nonnegative.
Note: Numbering of constraints conforms to the rows Lindo uses.
Initial Tableau
Basic Eq.
Var No
Z
1
Z
X1
X2
X3
X4
e3
A2
A3
S4
S5
1
-1M
-4
-1M
-6
-1M
-7
-2M
-8
1M
+0
0
0
0
RHS
0
-1350M
A2
2
0
1
1
1
1
0
1
0
0
0
950
A3
3
0
0
0
0
1
-1
0
1
0
0
400
S4
4
0
2
3
4
7
0
0
0
1
0
4600
S5
5
0
3
4
5
6
0
0
0
0
1
5000
Z
X1
X2
X3
X4
e3
A2
A3
S4
S5
RHS
Final Tableau
Basic Eq.
Var No
Z
1
1
1
0
0
-0
2
1M
+3
1M
-2
1
0
6650
X2
2
0
2
1
0
0
-3
4
3
-1
0
400
X4
3
0
0
0
0
1
-1
0
1
0
0
400
X3
4
0
-1
0
1
0
4
-3
-4
1
0
150
S5
5
0
0
0
0
0
-2
-1
2
-1
1
250
The Winco problem Interpreting the Lindo Output
page 2
Lindo output - Winco problem, page 233
OBJECTIVE FUNCTION VALUE
1) 6650.000
VARIABLE
VALUE
X1
0.000000
X2
400.000000
X3
150.000000
X4
400.000000
Comment:
These reduced costs are the
Z-row coeff's of X1...X4 in
the final tableau.
REDUCED COST
1.000000
0.000000
0.000000
0.000000
The rules below apply for a MAX
problem.
"Dual Prices" are the Shadow Prices for the
constraints.
ROW
SLACK OR SURPLUS
Constraint
Type
Look in Z-row
of
final tableau
DUAL PRICES
2)
0.000000
3.000000
`='
(A2 coeff when
M=0)
3)
0.000000
-2.000000
>
-(coeff of e3)
=
+(coeff of A3
with M=0)
4)
0.000000
1.000000
<
+(coeff of S4)
0.000000
<
+(coeff of S5)
5) 250.000000
NO. ITERATIONS= 3
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE
X1
X2
X3
X4
CURRENT
COEF
4.000000
6.000000
7.000000
8.000000
ALLOWABLE
INCREASE
1.000000
0.666667
1.000000
2.000000
RIGHTHAND SIDE RANGES
ROW
CURRENT
ALLOWABLE
RHS
INCREASE
2
950.000000
50.000000
3
400.000000
37.500000**
4
4600.000000
250.000000
5
5000.000000
INFINITY
ALLOWABLE
DECREASE
INFINITY
0.500000
0.500000
INFINITY
(We answered yes to Lindos
question about wanting the Range
Analysis. This is the output we
got.)
ALLOWABLE
DECREASE
100.000000
125.000000
150.000000
250.000000
*** Winston, page 233, has an error here.
Lets look more closely at how sensitive the optimal solution of the Winco problem is to the changes of
the RHS of for example the constraint in row 5, viz.:
5) 3 X1 + 4 X2 + 5 X3 + 6 X4 <=
The Lindo output tells us several things about this constraint (row 5):
5000
The Winco problem Interpreting the Lindo Output
page 3
the optimal solution has the slack variable s5= 250, so constraint (5) is not binding. Wed expect
the shadow price to be zero. Is it?
The shadow price for a constraint is the number in the DUAL PRICE column. For constraint
5, the DUAL PRICE is 0.
Note that the DUAL PRICE for constraint (3) is -2 (a negative value). This is what we expected:
Constraint (2) requires X4 >= 400; and it is binding at the optimal solution.
The shadow price is the amount z improves for each increase by 1 unit in the RHS of a
constraint (as long as the RHS stays within the allowable range). In this case, increasing
the RHS of constraint (3) makes the feasible region smaller -- its harder to be feasible -- so the
improvement cannot be positive; it has to be negative or zero. Since constraint(3) is binding at
the original optimum, we expect that increasing RHS(3) will force us to a truly less-good value of
Z ; so were not surprised to see a nonzero (and so negative) shadow price there.
Back to our analysis of Constraint (5):
The basic set at the optimal solution consists of X2, X3, X4, and S5. Why is that? We have four
constraints, so a basic set has four members. These four variables are the only four that are
nonzero at the optimum. Anything nonzero must be part of the basic set. (Note the reverse can
fail: A basic variable could be zero at the optimum.)
The line in the
ROW
5
RIGHTHAND SIDE RANGES
CURRENT
RHS
5000.000000
ALLOWABLE
INCREASE
INFINITY
ALLOWABLE
DECREASE
250.000000
tells us that the current basic set (X2, X3, X4, and S5) will remain optimal so long as
4750 = 5000 -250 < RHS (5) < 5000 + INFINITY = INFINITY.
That is, so long as 4750 < RHS(5); and all other data stays the same!!
Remember: even if the RHS(5) were to change within this range, the values of the basic variables which
give the optimal solution would change. For example, if RHS(5) were to change to 4800, the optimal
solution would no longer be X1=0 (nonbasic), X2= 400 , X3= 150, and X4= 400. What we do know is
that X2, X3, X4, and S5 would be the only variables that could possibly be nonzero at the optimal
solution. Does Lindo tell us what the values of these variables will be? No, it does not.
************
What happens if RHS(5) were to go outside this range for example, if the RHS were to drop to
4000, or to 4500? Then the guarantee that X2, X3, X4, and S5 is the basic set no longer applies. A
different basic set would (probably) be needed to get the optimal solution. But WHAT BASIC SET??
The Winco problem Interpreting the Lindo Output
page 4
Lindo can tell us that, with its PARA (for parametric) command, which is discussed in section 5.4 of
the Winston text. We will not discuss it in the course this year. (An Appendix on this topic is available
from your instructor if you want to pursue this further on your own.)
5.2 and 5.3: Decision making using the Lindo output
Winco Problem: version A
The Winco scenario was introduced on page 232 this way. The company is planning a production run
during which it will make units of four products. The information given can be summarized as follows:
Resources consumed in making one item of a product line
Resource
Product 1
Product 2
Product 3
Product 4
Amt Available
Raw material
2 units
3 units
4 units
7 units
4600 units
Hrs of labour
3
4
5
6
5000
$4
$6
$7
$8
Sales price/unit
In addition we were told the company planned to produce exactly 950 units in total, of which at least 400
had to be of product 4. The stated GOAL was to maximize sales revenue . Setting X1, ... X4 to be the
numbers of units of products 1, .. 4 to be produced led to the problem we discussed, namely
MAX
Z
SUBJECT
2)
3)
4)
2
5)
3
and X1,
= 4 X1 + 6 X2 + 7 X3 + 8 X4
TO:
X1 +
X2 +
X3 +
X4 =
X4 >=
X1 + 3 X2 + 4 X3 + 7 X4 <=
X1 + 4 X2 + 5 X3 + 6 X4 <=
X2, X3, X4 all nonnegative.
950
400
4600
5000
The LINDO output with the optimal solution was seen earlier. The optimal production schedule has
X1 = 0; X2 = 400; X3 = 150; and X4 = 400. It gives an optimal Z-value of $ 6,650.
Some Questions
(see Ex. 3, page 235; Example 4 parts (a) & (b), page 237; and Example 5, page 246.)
In each situation (separately) use the output to decide if you have sufficient information at present to
determine if (i) the optimal production schedule will change, and if so, how; and (ii) if the optimal z-value
will change, and if so, how.
a)
Winco decides to raise the price of product 2 by 50 cents a unit. (Ex 3a, p 235)
b)
Winco decides to raise the price of product 1 by 60 cents a unit. (Ex 3b p 235)
c)
Winco decides to lower the price of product 3 by 60 cents a unit. (Ex 3c, p235)
d)
Winco is required to produce a total of 980 units (not 950). (Ex 4a, p 237)
e)
(i) A check of inventory shows that only 4500 units of raw material are available, not the 4600 we
thought. (ii) Later we learn that water damage made an additional 100 units of raw material
unavailable for use, leaving only 4400 usable units. (Ex 4b, p 237)
f)
Returning to the original scenario, Winco has a chance to buy more raw material. Would it like to,
and if so, what would it be willing to pay? And how much would it want at that price?
g)
Winco has a chance to hire more labour. Would it like to, and if so, what would it be willing to
The Winco problem Interpreting the Lindo Output
page 5
pay? And how much would it want at that price?
THINK ABOUT THESE QUESTIONS A BIT BEFORE YOU READ ON
Some Answers (in brief)
a)
Winco decides to raise the price of product 2 by 50 cents a unit. (Ex 3a, p 235)
This increase of 50 cents (0.50 dollars) less than the allowable increase for the coefficient of X2
( 0.66667). We CAN draw some conclusions:
With a 50 cent increase in the price of product 2, the optimal production schedule wont change.
The sales revenue will change. The new optimal value will be 4(0)+ 6.5(400) + 7(150) + 8(400)=
$6850. Alternatively, making this change will give us (0.5) times 400, or $200 additional revenue,
so the sales revenue would increase from $6650 to $6850.
b)
Winco decides to raise the price of product 1 by 60 cents a unit. (Ex 3b p 235)
This increase of 60 cents (0.60 dollars) is less than the allowable increase for the coefficient of X1
( 1.0000), so again we CAN draw some conclusions. The optimal production schedule will not change.
That schedule has X1 = 0, so the optimal Z-value wont change either.
c)
Winco decides to lower the price of product 3 by 60 cents a unit. (Ex 3c, p235)
The allowable decrease for the coefficient of X3 in the objective function is 0.50 (dollars), so this change
takes us outside the allowable range. The optimal production schedule WILL (probably) change, and we
cannot tell what it would be.
Can we say anything? The proposed change affects only the goal and not the constraints, so the present
optimal solution will still be feasible. That solution has X3 = 150, so the change wont reduce the sales
revenue by any more than (0.50) x150 = $75.00, i.e the new z-value will be at least $6650 - 75 = $ 6675
d)
Winco is required to produce a total of 980 units (not 950). (Ex 4a, p 237)
The current RHS for constraint 2 is 950, and the allowable range for that RHS is from 950-100=850 to
950+50 = 1000. The new requirement falls within this range, so we CAN say something.
What can we say? First, since the dual price (shadow price) for constraint 2 is +3.0000, we can say that
the sales revenue will improve BY (shadow price)x(increase in RHS) = (+3)(+30) = 90. The new optimal
sales revenue will be 6650+90 = 6740 dollars..
Can we (easily) say how to achieve this? No. But we can deduce a bit of information. Since the RHS of
this constraint is staying within the allowable range, the current basic SET (which has X2, X3, X4 and
S5 in it) will still give the optimal solution. Therefore, X1 will stay nonbasic (and so equal to zero) and
e3 will, too, which will force X4 = 400. It would take a some work to figure out what X2 and X3 would
be.
e)
(i) A check of inventory shows that only 4500 units of raw material are available, not the 4600 we
thought. (ii) Later we learn that water damage made an additional 100 units of raw material
unavailable use, for leaving only 4400 usable units. (Ex 4b, p 237)
The raw material availability appears in constraint 4. (i) The decrease here is 100 (from 4600 to 4500),
which is less than the allowable decrease of 150. Therefore we CAN SAY the following. First, the
shadow price for constraint 4 is +1, so the change in the optimal z-value will be (shadow price)x(increase
The Winco problem Interpreting the Lindo Output
page 6
in RHS) = (+1)(-100)= -100, and our optimal sales revenue will drop to $6550. While we cannot easily
say exactly what production schedule well want to use, we do know the current basic set will still give
the optimal solution, so (same reasons as in previous part) well have X1 = 0 and X4 = 400. (ii) In this
case the decrease in the RHS of constraint 4 is 200, which is more than the allowable decrease (150) .
Therefore there is little we can easily say with the information we have now.
f)
Returning to the original scenario, Winco has a chance to buy more raw material. Would it like to,
and if so, what would it be willing to pay? And how much would it want at that price?
At present, all the raw material is used up (S4 = 0 at the optimum). This leads us to suspect that yes, we
would like to have more. This suspicion is confirmed by the fact that the dual price for raw material is
+1.000 (positive). Therefore, each one-unit increase in RHS(4) would add +1.0000 dollars to the sales
revenue.
This is important! That increase of $ 1 revenue for each additional unit of raw material applies only if
additional raw material could be obtained under the same conditions as the present raw material was
obtained. Under the scenario given, the raw material was already there in a sense it was free, and
our goal was to maximize the sales revenue. Each additional unit of raw material obtained for free adds
one dollar to the sales revenue, so wed be willing to pay up to one dollar a unit for additional raw
material.
For example, if we could obtain some additional raw material at a cost of $0.75 per unit, we could
increase our sales revenue by $1.000 dollars per unit and therefore our bottom line profit would improve
by $1.00 minus $0.75 = $0.25 for each additional unit of raw material we got.
Important! We need to think very carefully of what the shadow price (dual price) for a constraint means:
The shadow price for a constraint is the amount that the objective will improve for each unit
increase in the RHS of that constraint by one unit, PROVIDED
that increase keeps us within the allowable range for RHS of that constraint, and
the increase in the RHS is attained under the same conditions as the original RHS value
was determined.
We will explore this idea more deeply below.
How much more raw material would Winco be willing to take if it could get it for no more than a dollar a
unit? The present RHS for constraint 4 is 4600 and the allowable increase in that RHS is 250. This tells
us wed be willing to take as much as 250 additional units of this resource if we could get it at the
right price.
If we increase the RHS of constraint 4 by more than 250, the present basic set will no longer be feasible.
A different basic set will give the optimal solution. In practical terms, once we get and use 250 more units
of this resource, we are probably bumping into another boundary of the feasible region (perhaps using up
all our labour) so having additional raw material beyond that 250 is of no use to us.
g)
Winco has a chance to hire more labour. Would it like to, and if so, what would it be willing to
pay? And how much would it want at that price?
The Winco problem Interpreting the Lindo Output
page 7
At the present optimal solution the value of the slack variable S5 is 250. This means we have 250 hours
of unused labour already. Additional labour is of no value to us this is why the dual price for
constraint 5 is zero.
Note that this constraint was
5)
3 X1 + 4 X2 + 5 X3 + 6 X4 <=
5000
and that at the optimal solution the slack variable equalled 250. Note also that the allowable DECREASE
for RHS(5) was also 250.
Is this an accident? No, it is not. With the present optimal solution, 3X1+4X2+5X3+6X4 =
3(0)+4(400)+5(150)+6(400) = 4750, which is why S5 worked out to 250. If the RHS(5) dropped by
(say) 115, to 4885, this current values of X1, X2, X3, and X4 would still meet that constraint and so
would still be feasible (and optimal). The optimal z-value would remain at 6650. In fact, the only change
would be that the value of S5 would become 4885 - 4750 = 135.
The discussion around Table 4 on pages 239-240 of the text discusses this point a bit further. Read it.
A more realistic example
Example 6 (page 246) provides us with a more realistic scenario for Winco. Here it is.
The company is planning a production run during which it will make units of four products. The
information we have is this.
Resources consumed in making one item of a
product line
Resource
Product
1
Product 2
Product 3
Product 4
Available
Cost
Raw material
2 units
3 units
4 units
7 units
4600
units
$ 4/unit
Hrs of labour
3
4
5
6
5000
$6 / hour
Sales price/unit
$ 30
$ 42
$53
$72
As before , the company plans to produce exactly 950 units in total, of which at least 400 had to be of
product 4. Our GOAL is the more realistic one of maximizing PROFIT.
Setting X1, ... X4 to be the numbers of units of products 1, .. 4 to be produced we see:
the sales revenue will be 30 X1 + 42 X2 +53 X3 + 72 X4 dollars;
the cost of raw material used will be 4(2 X1 + 3 X2 + 4 X3 + 7 X4) dollars;
the cost of labour used will be 6(3 X1 + 4 X2 + 5 X3 + 6 X4) dollars; and
the PROFIT will be (sales Revenue) - (raw material cost) - (labour cost), and this works out to
4 X1 + 6 X2 + 7 X3 + 8 X4.
Thus the mathematical problem is the same. So is the solution, except that we now interpret Z as being
profit (in dollars) rather than sales revenue. Under this revised scenario, lets ask the same questions we
did before. What is the effect on the optimal solution in these different situations?
a)
Winco decides to raise the price of product 2 by 50 cents a unit.
The Winco problem Interpreting the Lindo Output
page 8
Doing this would increase the profit for product 2 by 50 cents a unit. The analysis above is still
applicable, except that its the profit that would increase to $ 6850.
b)
c)
d)
e)
Winco decides to raise the price of product 1 by 60 cents a unit. (Ex 3b p 235)
Winco decides to lower the price of product 3 by 60 cents a unit. (Ex 3c, p235)
Winco is required to produce a total of 980 units (not 950). (Ex 4a, p 237)
(i) A check of inventory shows that only 4500 units of raw material are available, not the 4600 we
thought. (ii) Not only that, but water damage made an addition 100 units of raw material
unavailable for use, leaving only 4400 usable units. (Ex 4b, p 237)
The answers to these are the same as above, except that Z-values refer to profit, not sales revenue.
Heres the interesting one!!!
f)
Returning to the original scenario, Winco has a chance to buy more raw material. Would it like to,
and if so, what would it be willing to pay? And how much would it want at that price?
The shadow price for raw material is +1.0000 , and this means that Winco would be willing to pay $1.00
a unit more for additional raw material under current conditions . Those current conditions include
buying raw material at $4.00 per unit. Therefore, Winco would be willing to pay up to $5.00 per unit for
some additional raw material.
What would be the payoff if it did this? Lets say Winco did buy t additional units of raw material at a
price of 4.75 dollars per unit.. Then the new optimal solution would still have X2, X3, X4 and S5 as the
basic variables but their values would probably be different. (X1 would still be nonbasic and therefore
equal to zero; e3 would also be nonbasic and so equal to zero, making X4 stay at 400.) Whatever the
new optimal values of X1, ... X4 are, the optimal profit can be deduced this way. Imagine Winco pays
$4.75 for the extra material in two payments: first it pays the standard price of $4.00/unit, and later it
pays an additional payment of $0.75/unit.
After it has made the first payment, and produced and sold its output, the profit so far is
6650 + (shadow price for raw material)(amount of additional material bought) = 6650 + 1t
dollars.
Then it makes the additional payment, so the final profit figure is (6650+t) - 0.75 t = 6650 +0.25t
dollars.
NOTE: we are NOT saying it would be willing to pay up to $5.00 for all its raw material. If the cost
of all raw material was $5.00/unit, all the coefficients in the objective function would change; we would
have a quite different problem which most likely has a quite different optimal solution.
How much additional raw material would it be willing to buy (so long as this additional material costs no
more than $1.00/unit more than the regular price)? The answer, again, is 250 units, and the reasons for
this are exactly the same as outlined above.
g)
Winco has a chance to hire more labour. Would it like to, and if so, what would it be willing to
pay? And how much would it want at that price?
This, too must be considered carefully. First, Winco does not want any additional labour at the current
The Winco problem Interpreting the Lindo Output
page 9
price of $6.00 per hour. It is not using all of the labour it has available now at that price. Clearly, it would
be even less willing to buy any additional labour at a price of more than $6.00 per hour.
But what if some people came along and said Well work for less wages! Well provide up to 3000
hours of labour and only charge $4.75 an hour!? Would Winco want them? Here the answer depends on
how Winco might be able to use these workers.
If these workers could be used only after all the regular labour was used up, Winco would not
take up this offer. At the present optimal solution we are using up all the raw material but not all
the labour available. There would be no raw material left, so we could not make anything else.
On the other hand, if we could use this cheaper labour in place of some of the regular labour, it
seems clear that we would want to do this.
Here are the details. Constraint 5 deals with the labour used. Since we have 5000 hours of labour
available, and S5 = 250 in the optimal solution, we are using up 4750 hours of labour, and paying
$6.00 an hour for it. If we could replace 3000 of those hours with this cheaper labour, we would
gain an additional 3000 x (6.00 - 4.75) = $3750 in profit, bringing our total profit to 6650 + 3750
= $10,400.
Conclusion:
The discussion and examples above illustrate how the optimal solution to an LP problem provides just the
beginning and not the end of information that can be useful to those who needed the answer to that
problem in the first place. Sensitivity Analysis is an important aspect of using mathematical tools to aid
good decision making in the real world.