Intro Class 3
38
description
optimization
Transcript of Intro Class 3
Mute ur call Plz add Engineering Optimization group of facebook (Rajesh Mishra)
3
Phases of an OR Study
Definition of the problem
Construction of the model
Solution of the model
Validation of the model
Implementation of the solution
Linear programming
Dynamic programming
Network programming
Integer programming
Nonlinear programming
Example: Product Mix Decision
• DJJ Enterprises makes automotive parts, Camshafts & Gears
• Unit Profit: Camshafts $25/unit, Gears $18/unit
• Resources needed: Steel, Labor, Machine Time. In total, 5000 lbs steel available, 1500 hours labor, and 1000 hours machine time.
• Camshafts need 5 lbs steel, 1 hour labor, 3 hours machine time.
• Gears need 8 lbs steel, 4 hours labor, 2 hours machine time.
• How many camshafts & gears to make in order to maximize profit?
Understanding the Problem
• Text-Based Formulation – Decision Variables:
Number of camshafts to make, number of gears to make
– Objective Function: Maximize profit
– Constraints: Don’t exceed amounts available of steel, labor, and machine time.
Algebraic Formulation • Decision Variables
– C = number of camshafts to make
– G = number of gears to make
• Objective Function – Maximize 25C + 18G (profit in $)
• Constraints – 5C + 8G <= 5000 (steel in lbs)
– 1C + 4G <= 1500 (labor in hours)
– 3C + 2G <= 1000 (machine time in hours)
– C >= 0, G >= 0 (non-negativity)
Important Concepts • Linear Program: The objective function and constraint
are linear functions of the decision variables. Therefore, this is a Linear Program.
• Feasibility – Feasible Solution. A solution is feasible for an LP if all
constraints are satisfied.
– Infeasible Solution. A solution is infeasible if one or more constraints is violated.
– Check the solutions C=75, G=200; and C=300, G=200 for feasibility.
• Optimal Solution. The optimal solution is the feasible solution with the largest (for a max problem) objective value (smallest for a min problem).
Solving Linear Programming Problems • Trial and error: possible for very small problems;
virtually impossible for large problems.
• Graphical approach: It is possible to solve a 2-variable problem graphically to find the optimal solution (not shown).
• Simplex Method. This is a mathematical approach developed by George Dantzig. Can solve small problems by hand.
• Computer Software. Most optimization software actually uses the Simplex Method to solve the problems. Excel’s Solver Add-In is an example of such software.
• Solver can solve LPs of up to 200 variables. Enhanced versions of Solver are available from Frontline Systems (http://www.solver.com).
Wild West produces two types of hats. Type I
hat requires twice as much labor as a Type II.
If all the available labor time is dedicated to
Type II alone, the company can produce a total
of 400 Type II hats in a day. The respective
market limits for the two types of hats are 150
and 200 hats per day. The profit is $8 per Type
I hat and $5 per Type II hat. Formulate the
problem as an LPP so as to maximize the
profit.
Solution: Assume that Wild West produces x1 Type I hats and x2 Type II hats per day.
8 x1 + 5 x2
Labour Time spent is (2 x1 + x2) c minutes
Per day Profit got =
Assume the time spent in producing one type II hat is c minutes.
The objective is to find x1, x2 so as to
maximise the profit z = 8 x1 + 5 x2
satisfying the constraints:
(2 x1 + x2 ) c ≤ 400 c
x1 ≤ 150
x2 ≤ 200
x1, x2 ≥ 0, integers
That is: The objective is to find x1, x2 so
as to maximise the profit z = 8 x1 + 5 x2
satisfying the constraints:
2 x1 + x2 ≤ 400
x1 ≤ 150
x2 ≤ 200
x1, x2 ≥ 0, integers
Trim Loss problem: A company has to manufacture the circular tops of cans. Two sizes, one of diameter 10 cm and the other of diameter 20 cm are required. They are to be cut from metal sheets of dimensions 20 cm by 50 cm. The requirement of smaller size is 20,000 and of larger size is 15,000. The problem is : how to cut the tops from the metal sheets so that the number of sheets used is a minimum. Formulate the problem as a LPP.
A sheet can be cut into one of the following three patterns:
Pattern I
Pattern II
Pattern III
10
20
20
10
10
10
20
10
Pattern I: cut into 10 pieces of size 10 by 10 so as to make 10 tops of size 1
Pattern II: cut into 2 pieces of size 20 by 20 and 2 pieces of size 10 by 10 so as to make
2 tops of size 2 and 2 tops of size 1
Pattern III: cut into 1 piece of size 20 by 20 and 6 pieces of size 10 by 10 so as to make 1 top of size 2 and 6 tops of size 1
So assume that x1 sheets are cut according to pattern I, x2 according to pattern II, x3 according to pattern III
The problem is to
Minimize z = x1 + x2 + x3
Subject to 10 x1 + 2 x2 + 6 x3 ≥ 20,000
2 x2 + x3 ≥ 15,000
x1, x2, x3 ≥ 0, integers
A Post Office requires different number of full-time employees on different days of the week. The number of employees required on each day is given in the table below. Union rules say that each full-time employee must receive two days off after working for five consecutive days. The Post Office wants to meet its requirements using only full-time employees. Formulate the above problem as a LPP so as to minimize the number of full-time employees hired.
Requirements of full-time employees day-wise
Day No. of full-time
employees required
1 - Monday 10
2 - Tuesday 6
3 - Wednesday 8
4 - Thursday 12
5 - Friday 7
6 - Saturday 9
7 - Sunday 4
Solution: Let xi be the number of full-time employees employed at the beginning of day i (i = 1, 2, …, 7). Thus our problem is to find xi so as to
Minimize
1 2 3 4 5 6 7z x x x x x x x
Subject to
1 4 5 6 7 10 (Mon)x x x x x
1 2 5 6 7 6 (Tue)x x x x x
1 2 3 6 7 8 (Wed)x x x x x
1 2 3 4 7 12 (Thu)x x x x x
1 2 3 4 5 7 (Fri)x x x x x
2 3 4 5 6 9 (Sat)x x x x x
3 4 5 6 7 4 (Sun)x x x x x
xi 0. integers
BITS wants to host a Seminar for five days. For the delegates there is an arrangement of dinner every day. The requirement of napkins during the 5 days is as follows:
Day 1 2 3 4 5
Napkins
Needed
80 50 100 80 150
Institute does not have any napkins in the beginning. After 5 days, the Institute has no more use of napkins. A new napkin costs Rs. 2.00. The washing charges for a used one are Rs. 0.50. A napkin given for washing after dinner is returned the third day before dinner. The Institute decides to accumulate the used napkins and send them for washing just in time to be used when they return. How shall the Institute meet the requirements so that the total cost is minimized ? Formulate as a LPP.
Solution Let xj be the number of napkins purchased on day j, j=1,2,..,5
Let yj be the number of napkins given for washing after dinner on day j, j=1,2,3
Thus we must have
Also we have y1 ≤ 80, y2 ≤ (80 – y1) + 50
y3 ≤ (80 – y1) + (50 – y2) + 100
x1 = 80, x2 = 50, x3 + y1 = 100, x4 + y2 = 80 x5 + y3 = 150
Thus we have to Minimize z = 2(x1+x2+x3+x4+x5)+0.5(y1+y2+y3)
Subject to
x1 = 80, x2 = 50, x3 +y1 =100,
x4 + y2 = 80, x5 + y3 = 150,
y1 ≤ 80, y1+y2 ≤ 130, y1+y2+y3 ≤ 230,
all variables ≥ 0, integers
Formulating Transportation Problems
Powerco has three electric power plants that supply the electric needs of four cities.
•The associated supply of each plant and demand of each city is given in the table 1.
•The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.
Transportation tableau
A transportation problem is specified by the supply, the demand, and the shipping costs. So the relevant data can be summarized in a transportation tableau. The transportation tableau implicitly expresses the supply and demand constraints and the shipping cost between each demand and supply point.
Table 1. Shipping costs, Supply, and Demand for Powerco Example
From To
City 1 City 2 City 3 City 4 Supply
(Million kwh)
Plant 1 $8 $6 $10 $9 35
Plant 2 $9 $12 $13 $7 50
Plant 3 $14 $9 $16 $5 40
Demand
(Million kwh)
45 20 30 30
Transportation Tableau
Solution
1. Decision Variable:
Since we have to determine how much electricity is sent from each plant to each city;
Xij = Amount of electricity produced at plant i and sent to city j
X14 = Amount of electricity produced at plant 1 and sent to city 4
2. Objective function
Since we want to minimize the total cost of shipping from plants to cities;
Minimize Z = 8X11+6X12+10X13+9X14
+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
3. Supply Constraints
Since each supply point has a limited production capacity;
X11+X12+X13+X14 <= 35
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40
4. Demand Constraints
Since each supply point has a limited production capacity;
X11+X21+X31 >= 45
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30
5. Sign Constraints
Since a negative amount of electricity can not be shipped all Xij’s must be non negative;
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
LP Formulation of Powerco’s Problem
Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
S.T.: X11+X12+X13+X14 <= 35 (Supply Constraints)
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40
X11+X21+X31 >= 45 (Demand Constraints)
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
Questions/Queries?
