AAOC C222: OPTIM OPTIMISA ISATION TION Text Boo Book: k: Operations Research: An Introduction By Hamdy A.Taha (Pearson Education) 8th Edition
Reference Books: 1. Rao, Rao, S S: S: Eng Engin inee eeri ring ng Op Opti timi miza zati tion on,, New Age International 2. Pa Pant nt,, J.C: J.C: Opt Optim imiz izat atio ion, n,Ja Jain in Bro Broth ther erss 3. Ravi Ravind ndra ran n A, A, et et al. al.:: Ope Opera rati tion onss Research, John Wiley Wiley & sons
4. Hillier & Lieberman: Introduction to Operations Research, Tata McGraw-Hill
5. Winston, WL: Operations Research, Thomson Learning 6. GC Onwubolu and BV Babu, New Optimization Techniques in Engineering, Springer-Verlag, Springer-V erlag, Heidelberg, Germany Germany,, First Edition, 2004.
Section: 1 Room: G 201 (M W F) 5th Hour G 201 (W) 1st Hour Instructor: Dr. P. K. Sahoo COURSE HANDOUT IS AVAILABLE IN EDUCAN CHAMBER CONSULTATION HOUR THURSDAY 9th HOUR (C-317)
Introduction The formal activities of Operations Research (OR) were initiated in England during World War II when a team of British scientists set out to make decisions regarding the best utilization of war material. Following the end of the war, the ideas advanced in military operations were adapted to improve efficiency and productivity in the civilian sector. Today, OR is a dominant and indispensable decision making tool.
In decision making problem we have to answer three questions: 1. What are the decision alternatives? 2. Under what restrictions is the decision made? 3. What is an appropriate objective criterion for evaluating the alternatives?
Example: The Cotton Gate garment company manufactures men's shirts and women’s blouses for Walmark Discount stores. Walmark will accept all the production supplied by Cotton Gate. The production process includes cutting, sewing and packaging. Cotton Gate employs 25 workers in the cutting department, 35 in the sewing department and 5 in the packaging
department. The factory works one 8hour shift, 5 days a week. The following table gives the time requirements and the profits per unit for the two garments:
Minutes per unit Garment
Cutting
Sewing
Packaging
Unit profit($)
Shirts
20
70
12
8.00
Blouses
60
60
4
12.00
Determine the optimal weekly production schedule for Cotton Gate.
Solution: Assume that Cotton Gate produces x1 shirts and x2 blouses per week. Profit got = 8 x1 + 12 x2 Time spent on cutting = 20 x1 + 60 x2 mts Time spent on sewing = 70 x1 + 60 x2 mts Time spent on packaging =12 x1 + 4 x2 mts
The objective is to find x1, x2 so as to
maximize the profit z = 8 x1 + 12 x2 satisfying the constraints: 20 x1 + 60 x2 ≤ 25 40 60 70 x1 + 60 x2 ≤ 35 40
60
12 x1 + 4 x2 ≤ 5 40
60
x1, x2 ≥ 0, integers
This is a typical optimization problem. Any values of x1, x2 that satisfy all the constraints of the model is called a feasible solution. We are interested in finding the optimum feasible solution that gives the maximum profit while satisfying all the constraints.
More generally, an optimization problem looks as follows: Determine the decision variables x1, x2, …, xn so as to optimize an objective function f ( x1, x2, …, xn) satisfying the constraints gi ( x1, x2, …, xn) ≤ bi (i=1, 2, …, m).
Chapter-2 Linear Programming Problems(LPP) An optimization problem is called a Linear Programming Problem (LPP) when the objective function and all the constraints are linear functions of the decision variables, x1, x2, …, xn. We also include the “non-negativity restrictions”, namely x j ≥ 0 for all j=1, 2, …, n. Thus a typical LPP is of the form:
Optimize (i.e. Maximize or Minimize) z = c1 x1 + c2 x2+ …+ cn xn
subject to the constraints: a11 x1 + a12 x2 + … + a1n xn ≤ b1 a21 x1 + a22 x2 + … + a2n xn ≤ b2
. . . am1 x1 + am2 x2 + … + amn xn ≤ bm x1, x2, …, xn 0
A LPP satisfies the three properties: Proportionality , additivity & certainty.
Proportionality means the contributions of each decision variable in the objective function and its requirements in the constraints are directly proportional to the value of the variable.
Additivity stipulates that the total contributions of all the variables in the objective function and their requirements in the constraints are the direct sum of the individual contributions or requirements of each variable. Certainty means the objective and constraint coefficients of the LP model are known constants (deterministic).
• We shall first look at formulation of some LPPs, • Graphically solve some LPPs involving two decision variables • Study some mathematical preliminaries regarding the solutions of LPPs • Finally look at the Simplex method of solving a LPP
Q. Wild West produces two types of cowboy 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 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. Per day Profit got =
8 x1 + 5 x2
Assume the time spent in producing one type II hat is c minutes. Labour Time spent is (2 x1 + x2) c minutes
The objective is to find x1, x2 so as to
maximize the profit z = 8 x1 + 5 x2 satisfying the constraints:
(2 x1 + x2 ) c ≤ 400 c ≤ 150
x1 x2
≤ 200
x1, x2 ≥ 0, integers
That is: The objective is to find x1, x2 so as to maximize the profit z = 8 x1 + 5 x2 satisfying the constraints: 2 x1 + x2 ≤ 400
≤ 150
x1 x2
≤ 200
x1, x2 ≥ 0, integers
Q4(2.2B) A company has two grades of inspectors, I and II to undertake quality control inspection. At least, 1500 pieces must be inspected in an 8-hour day. Grade I inspector can check 20 pieces in an hour with an accuracy of 96% and grade II inspector can check 14 pieces in an hour with an accuracy of 92%. The wages of two inspectors are $5 & $4 per hour. Any error made by an inspector costs $3 to the company. If there are, in all, 10 grade I inspector and 15 grade II inspector in the company, find the
optimal assignment of inspector that minimizes the daily inspection cost. Solution: Let x1 and x2 represent the number of these inspectors. The objective is to minimize the daily cost of inspection. The company has to incur two types of costs: Wages paid to the inspectors and the cost of their inspection errors. The cost of grade I inspector/hour is Rs. (5 + 3 × 0.04 × 20) = Rs. 7.40
The cost of grade II inspector/hour is Rs. (4 + 3 × 0.08 × 14) = Rs. 7.36 The objective function is Minimize Z = 8(7.40 x1 + 7.36 x2) = 59.20 x1 + 58.90 x2 subject to the constraints x1 ≤ 10 x2 ≤ 15 20 × 8 x1 + 14 × 8 x2≥ 1,500. (No. of pieces to be inspected daily). x1, x2 ≥ 0
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: 10 Pattern I Pattern II
10
20
10 20
Pattern III
10 10
20
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
Q2 (2.3F) In an LTL (less-than-truckload) trucking company, terminal docks include casual workers who are hired temporarily to account for peak loads. At the Omaha, Nebraska, dock, the minimum demand for casual workers during the seven days of the week (starting on Monday) is 20, 14, 10, 15, 18, 10, 12 workers. Each worker is contracted to work five consecutive days. Determine an optimal weekly hiring practice of casual workers for the company.
Solution: Let xi be the number of casuals required at the beginning of day i (i = 1, 2, …, 7). Thus our problem is to find xi so as to
Minimize Subject to
z x1 x2 x3 x4 x5 x6 x 7
x1 x4 x5 x6 x 7
20 (Mon)
x1 x2 x5 x6 x 7
14 (Tue)
x1 x2 x3 x6 x 7
10 (Wed)
x1 x2 x3 x4 x 7
15(Thu)
x1 x2 x3 x4 x 5
18 (Fri)
x2 x3 x4 x5 x 6 x3 x4 x5 x6 x 7
10(Sat)
12 (Sun)
xi 0.
integers
Q.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 x j be the number of napkins purchased on day j, j=1,2,..,5 Let y j be the number of napkins given for washing after dinner on day j, j=1,2,3 Thus we must have x1 = 80, x2 = 50, x3 + y1 = 100, x4 + y2 = 80 x5 + y3 = 150 Also we have y1 ≤ 80, y2 ≤ (80 – y1) + 50 y3 ≤ (80 – y1) + (50 – y2) + 100
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
Inventory Control Ex: Acme manufacturing company has contracted to deliver home windows over the next 6 months. The demands for each month are 100, 250, 190, 140, 220 and 110 units respectively. Production cost per window varies from month to month depending on the cost of labour, material & utilities. Acme estimates the production cost per window over next 6 months to be $50, $45, $55, $48, $52 and $50.
To take an advantage of the fluctuations in manufacturing cost, Acme may elect to produce more than is needed in a given month and hold the excess units for delivery in latter months. This, however, will incur storage costs at the rate of $8 per window per month assessed on end-ofmonth inventory. Develop a linear program to determine the optimum production schedule.
Solution: The variables of the problem include the monthly production amount and the end-of-month inventory. For i= 1,2….,6 let xi= Number of units produced in month i I i =Inventory units left at the end of month i
The relationship between these variables and monthly demand over the six-month horizon is represented by the schematic diagram.
The system starts empty, which means I0 =0. x1 x2 x3 x4 x5 x6 I=0 I1 I2 I3 I4 I5 I 6
100
250
190
140
220
110
The objective function seeks to minimize the sum of the production and end-of-month inventory costs.
Total production cost= 50x1 +45x2 +55x3 +48x4 +52x5 +50x6
Total inventory cost= 8(I 1 +I 2 +I 3 +I 4 +I 5 +I 6 ) The objective function is Min z= 50x1 +45x2 +55x3 +48x4 +52x5 +50x6 + 8(I 1 +I 2 +I 3 +I 4 +I 5 +I 6 )
The constraints can be determined from the figure. For each period we have the balance equation:
Beginning inventory + Production amount -Ending inventory = Demand For the individual month (Month 1) I 0 +x1 – I 1 = 100 (Month 2) I 1 +x2 – I 2 = 250 (Month 3) I 2 +x3 – I 3 = 190
(Month 4) I 3 +x4 – I 4 = 140
(Month 5) I 4 +x5 – I 5 = 220
(Month 6) I 5 +x6 – I 6 = 110
xi , I i ≥ 0 for all i= 1,2,….6
Put I0 =0 since the situation starts
Q5. (2.3G) Pollution Control Three types of coal, C1, C2 and C3 are pulverized and mixed together to produce 50 tons per hour needed to power a plant for generating electricity. The burning of coal emits sulfur oxide (in parts per million) which must meet the Environment Protection Agency (EPA) specifications at most 2000 parts per million. The following table summarizes the data of the situation:
C1 Sulfur (parts per million) 2500 Pulverizer capacity (ton/hr) 30 Cost per ton $30
C2 1500 30 $35
Determine the optimal mix of the coals.
Solution: Let xi = tons of coal, i= 1,2,3 Minimize Z=30x1+35x2+33x3
C3 1600 30 $33
Subject to the restrictions 2500x1+1500x2+1600x3≤ 2000(x1+x2+x3) x1 ≤30
x2 ≤30
x1+x2+x3≥ 50 x1, x2, x3 ≥ 0, integers
x3 ≤30
There are many Software packages available to solve LPP and related problems. • Your book contains a CD having the package “TORA” probably developed by the author. • There is also Microsoft’s Excel Solver. • There is also a commercial package “LINGO”
