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York University

Environmental Studies

ENVS 1000

Peter Timmerman

Fall

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SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
EXAMPLE 1:
Winco sells 4 types of products. The resources needed to produce one unit of each and
the sales prices are given below:
Table 1: Cost & Resource Requirements for Winco:
Product 1 Product 2 Product 3 Product 4
Raw material 2 3 4 7
Hours of labor 3 4 5 6
Sales price $4 $6 $7 $8
At present, 46000 units of raw material and 5000 labor hours are available. To meet
customer demands, exactly 950 total units must be produced. Customers also demand that
at least 400 units of product 4 be produced. Formulate an LP that can be used to
maximize Winco’s sales revenue.
Solution: Let x i number of units of product i produced by Winco
max z = 4x1 + 6x2 + 7x3 + 8x4
s.t. x1 + x2 + x3 + x4 = 950
x4 >= 400
2x1 + 3x2 + 4x3 + 7x4 =< 4600
3x1 + 4x2 + 5x3 + 6x4 =< 5000
x1, x2, x3, x4 >= 0
Excel output is given below:
Cost and Resource Requirements for Winco
Product 1 Product 2 Product 3 Product 4
Raw Material 2 3 4 7
Hours of Labor 3 4 5 6
Sales Price $4 $6 $7 $8
Total Demand Min # of Product 4 Available raw materiaAvailable labor hours
950 400 4600 5000
Total Number of Number of ProductAmount of raw materials Amount of labor hours
Units producedProduced used used
950 400 4600 4750
Product 1 Product 2 Product 3 Product 4
Units to be produced 0 400 150 400
Sales Revenue 6650
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 1 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
ANSWER REPORT:
Target Cell (Max)
Cell Name Original Value Final Value
$B$20 Sales Revenue Product 1 0 6650
Adjustable Cells
Cell Name Original Value Final Value
$B$19 Units to be produced Product 1 0 0
$C$19 Units to be produced Product 2 0 400
$D$19 Units to be produced Product 3 0 150
$E$19 Units to be produced Product 4 0 400
Constraints
Cell Name Cell Value Formula Status Slack
$B$15 Units produced 950 $B$15<=$B$11 Binding 0
$C$15 Produced 400 $C$15>=$C$11 Binding 0
$D$15 used 4600 $D$15<=$D$11 Binding 0
$E$15 used 4750 $E$15<=$E$11 Not Binding 250
SENSITIVITY REPORT:
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease
$B$19 Units to be produced Product 1 0 -1 4 1 1E+30
$C$19 Units to be produced Product 2 400 0 6 0.666666667 0.5
$D$19 Units to be produced Product 3 150 0 7 1 0.5
$E$19 Units to be produced Product 4 400 0 8 2 1E+30
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$15 Units produced 950 3 950 50 100
$C$15 Produced 400 -2 400 37.5 125
$D$15 used 4600 1 4600 250 150
$E$15 used 4750 0 5000 1E+30 250
Objective Function Changes:
In the sensitivity report, the columns called Allowable Increase (AI) indicates the amount
by which an objective function coefficient can be increased with the current basis (values
of the decision variables) remaining optimal. Similarly, the Allowable Decrease (AD)
column indicates the amount by which an objective function coefficient can be decreased
with the current basis remaining optimal. To illustrate these ideas, let c1 be the objective
function coefficient for x1. If c1 is changed, then the current basis remains optimal if
- ∞ = 4 - ∞ ≤ c1 ≤ 4 + 1 = 5
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 2 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
EXAMPLE 2:
a. Suppose Winco raises the price of product 2 by 50 cents per unit. What is the new
optimal solution to the LP?
b. Suppose the sales price of product 1 is increased by 60 cents by unit. What is the new
optimal solution to the LP?
c. Suppose the sales price of product 3 is decreased by 60 cents per unit. What is the new
optimal solution to the LP?
Solution:
a. Since AI for c2 is $0.666667, and we are increasing c2 by only 50 cents, the current
solution remains optimal. The optimal values of the decision variables remain unchanged
(x1 = 0, x2 = 400, x3 = 150, and x4 = 400 is optimal). The new optimal z-value may be
determined in two ways. First, we may simply substitute the optimal values of the
decision variables into the new objective function, yielding
New optimal z – value = 4(0) + 6.5(400) + 7(150) + 8(400) = $6850
Another way to see that the new optimal z – value is $6850 observe the only difference in
sales revenue: Each unit of product 2 brings in 50 cents more in revenue. Thus, the total
revenue should increase by 400(.50) = $200, so
New z-value = original z-value + 200 = $6850
b. The AI for c1 is 1, so the current basis remains optimal, and the optimal values of the
decision variables remain unchanged. Since the value of x1 in the optimal solution is 0,
the change in the sales price for product 1 will not change the optimal z-value – it will
remain $6650.
c. For c3, AD = .50, so the current basis is no longer optimal. Without resolving the
problem by hand or computer, we cannot determine the new optimal solution.
Reduced Cost and Sensitivity Analysis:
The REDUCED COST column of the Excel output gives us information about how
changing the objective function coefficient for a nonbasic variable will change the LP’s
optimal solution. If the objective function coefficient of a nonbasic variable is improved
by its reduced cost, then the LP will have alternative optimal solutions – at least one in
which the old nonbasic variable will be a basic variable. If the objective function
coefficient of a nonbasic variable is improved more than its reduced cost, then any
optimal solution to the LP will have this variable as a basic variable.
Keep in mind that: The reduced cost for a non-basic variable (in a max problem) is the
maximum amount by which the variable’s objective function coefficient can be
increased before the current basis becomes suboptimal and it becomes optimal for the
non-basic variable to enter the basis.
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 3 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
Right Hand Side Ranges:
In sensitivity report under the columns of Allowable Increase & Decrease, allowable
limits to the changes in the constraints are given. To illustrate, consider the first
constraint in Example 1. Currently, the RHS of this constraint is 950. The current basis
remains optimal if this value is decreased by up to 100 (Allowable Decrease) or increased
by up to 50 (Allowable Increase)
Shadow Prices:
th
Shadow price for the i constraint of an LP is the amount by which the optimal z-value is
imprthed - increased in a max problem and decreased in a min problem – if the RHS of
the i constrained is increased by 1.
EXAMPLE 3:
a. In Example 1, suppose that a total amount of 980 units must be produced.
Determine the new optimal z-value.
b. In Example 1, suppose that 4500 of raw materials are available. What is the new
optimal z-value? What if only 4400 units of raw material are available?
Solution:
a. Since the allowable increase is 50, the current basis remains optimal, and the
shadow price of $3 remains applicable. Then (1) yields
New optimal z-value = 6650 + 30(3) = $6740
b. Since the allowable decrease is -150, the shadow price of $1 remains valid. Then
(1) yields
New optimal z-value = 6650 + 100(1) = $6550
Let’s give an interpretation to the shadow price for each constraint in Example 1. Again
all discussions are assuming that we are within the allowable range where the current
basis remains optimal. The shadow price of $3 for Constraint 1 in Example 1 implies that
each one-unit increase in total demand will increase sales revenues by $3. The shadow
price of -$2 for Constraint 2 implies that each unit increase in the requirement for product
4 will decrease the revenue by $2. The shadow price of $1 for Constraint 3 implies that
an additional unit of raw material given to Winco (at no cost) increase total revenue by
$1. Finally, the shadow price of $0 for Constraint 4 implies that an additional unit of
labor given to Winco (at no cost) will not increase the total revenue. This is reasonable; at
present, 250 of the available 5000 labor hours are not being used, so why should we
expect additional labor to raise revenues?
Signs of Shadow Prices:
A ≥ constraint will always have a non-positive shadow price; a ≤ constraint will always
have a non-negative shadow price; and an equality constraint may have a positive, a
negative, or a zero shadow price. To see why this is true, observe that adding points to an
LP’s feasible region can only improve the optimal z value or leave it the same.
Eliminating points from an LP’s feasible region can only make the optimal value worse
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 4 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
or leave it the same. For example, let’s look at the shadow price of the raw material
constraint (a ≤ constraint) in Example 1. Why must this shadow price be non-negative?
The shadow price of the raw material constraint represents the improvement in the
optimal z-value if 4601 units (instead of 4600) of raw material are available. Since
having an additional unit of raw material available adds points to the feasible region –
points for which Winco uses > 4600 but ≤ 4601 units of raw material – we know that the
optimal z-value must increase or stay the same. Thus the shadow price of this constraint
must be non-negative.
Similarly, let’s consider the shadow price of the x4 ≥ 400 constraint in Example 1.
Increasing the right-hand side of this constraint to 401 eliminates points from the feasible
region. Thus, the optimal z-value must decrease or stay the same, implying that the
shadow price of this optimal z-value must decrease or stay the same, implying that the
shadow price of this constraint must be non-positive. Similar reasoning shows that for a
minimization problem, a ≥ constraint will have a non-negative shadow price, and a ≤ will
have a non-positive shadow price.
Sensitivity Analysis and Slack & Excess Variables:
It can be shown that for any inequality constraint, the product of the values of the
constraint’s slack or excess variable and the constraint’s shadow price must equal to 0.
This implies that any constraint whose slack or excess variable is > 0 will have a zero
shadow price. It also implies that any constraint with a non-zero shadow price must be
binding (have slack or excess equal to 0). To illustrate these ideas, consider the labor
constraint in Example 1. This constraint has positive slack, so its shadow price must be 0.
This is reasonable, because slack = 250 for this constraint indicates that 250 hours of
currently available labor are unused at present. Thus, an extra hour of labor would not
increase revenues. Now consider the raw material constraint of Example 1. Since this
constraint has a non-zero shadow price, it must have slack = 0. This is reasonable; the
non-zero shadow price means that additional raw material will increase revenue. This can
be the case only if all presently available raw materials are now being used.
Managerial Use of Shadow Prices:
EXAMPLE 4:
In Example 1, what is the most that Winco should be willing to pay for an additional unit
of raw material? How about an extra hour of labor ?
Solution:
Since the shadow price of the raw-material availability constraint is 1, an extra unit
would increase total revenue by $1. Thus, Winco would pay up to $1 for an extra unit of
raw material and be as well off as it is now. The labor availability constraint has a
shadow price of 0. This means that an extra hour of labor will not increase revenues, so
Winco should not be willing to pay anything for an extra hour of labor.
EXAMPLE 5:
Lets’ consider Example 1 with the following changes. Suppose up to 4600 units of raw
material are available, but they must be purchased at a cost of $4 per unit. Also, up to
5000 hours of labor are available, but they must be purchased at a cost of $6 per hour.
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 5 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
The per unit sales price of each product is as follows: product 1 - $30; product 2 - $42;
product 3 - $53; product 4 - $ 72. A total of 950 units must be produced, of which at least
400 must be product 4. Determine the maximum amount that the firm should be willing
to pay for an extra unit of raw material and an extra hour of labor.
Solution:
The contribution to profit from one unit of each product may be computed as follows:
Product 1: 30 – 4(2) – 6(3) = $4
Product 2: 42 – 4(3) – 6(4) = $6
Product 3: 53 – 4(4) – 6(5) = $7
Product 4: 72 – 4(7) – 6(6) = $8
Thus, Winco’s profit is 4x1 + 6x2 + 7x3 + 8x4. To maximize profit, Winco should solve
the same LP as in Example 1, and the relevant Excel Output is given in ANSWER
REPORT above. To determine the most Winco should be willing to pay for an extra unit
of raw material, note that the shadow price of the raw material constraint may be
interpreted as follows: If Winco has the right to buy one more unit of raw material (at $4
per unit), then the profits increase by $1. Thus, paying $4 + $1 = $5 for an extra unit of
raw materials will increase the profits by $1 - $1 = 0. SO Winco could pay up to $5 for an
extra unit of raw material and still be better off. For the raw material constraint, the
shadow price of $1 represents a premium above the current price Winco is willing to pay
for an extra unit of raw material.
The shadow price of the labor availability constraint is $0, which means that the
right to buy an extra hour of labor will not increase the profits. This tells us that at the
current price of $4 per hour, Winco should not buy extra labor.
EXAMPLE 6: (A DIET PROBLEM FOR A COLLEGE STUDENT)
A college student’s diet requires that all the food he eats come from one of the four
“basic food groups” (chocolate cake, ice cream, soda, and cheese cake). At the present,
the following four foods are available for consumption: brownies, chocolate ice cream,
cola, and pineapple cheesecake. Each brownie costs 50 cents, each scoop of chocolate ice
cream costs 20 cents, each bottle of cola costs 30 cents, and each piece of pineapple
cheesecake costs 80 cents. Each day, he must ingest at least 500 calories, 6 oz chocolate,
10 oz of sugar, and 8 oz of fat. The nutritional content per unit of each food is shown in
Table 2 below. Formulate a linear programming model that can be used to satisfy this
college student’s daily requirements at minimum cost.
Table 2: Nutritional values for Diet
Calories Chocolate (ounces) Sugar (ounces) Fat (ounces)
Brownie 400 3 2 2
Chocolate ice cream (1 scoop) 200 2 2 4
Cola (1 bottle) 150 0 4 1
Pineapple cheesecake (1 piece)500 0 4 5
Solution:
Decision variables:
* SOURCE: Winston, W.L. Operations Research, Applications & Algorithms 6 SENSITIVITY ANALYSIS* Murat Kristal
MGTS 2010 SECTIONS U, R, & V
X1 = number of brownies eaten daily
X2 = number of scoops of chocolate ice cream eaten

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