AAOC C222: OPTIMISATIONText Book: Operations Research: An Introduction By Hamdy A.Taha (Pearson Education) 8th Edition 7 August 2012 P K Sahoo, BITS-Hyderabad Campus 1 Reference Books: 1. Rao, S S: Engineering Optimization, New Age International 2. Pant, J.C: Optimization,Jain Brothers 3. Ravindran A, et al.: Operations Research, John Wiley & sons 7 August 2012 P K Sahoo, BITS-Hyderabad Campus 2 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, Heidelberg, Germany, First Edition, 2004. 7 August 2012 P K Sahoo, BITS-Hyderabad Campus 3 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) 7 August 2012 P K Sahoo, BITS-Hyderabad Campus 4 Today. OR is a dominant and indispensable decision making tool. Following the end of the war. 7 August 2012 P K Sahoo. the ideas advanced in military operations were adapted to improve efficiency and productivity in the civilian sector.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. BITS-Hyderabad Campus 5 . What are the decision alternatives? 2. Under what restrictions is the decision made? 3. What is an appropriate objective criterion for evaluating the alternatives? 7 August 2012 P K Sahoo.In decision making problem we have to answer three questions: 1. BITS-Hyderabad Campus 6 . sewing and packaging. The production process includes cutting.Example: The Cotton Gate garment company manufactures men's shirts and women’s blouses for Walmark Discount stores. 35 in the sewing department and 5 in the packaging 7 August 2012 P K Sahoo. Cotton Gate employs 25 workers in the cutting department. Walmark will accept all the production supplied by Cotton Gate. BITS-Hyderabad Campus 7 . The following table gives the time requirements and the profits per unit for the two garments: 7 August 2012 P K Sahoo. The factory works one 8hour shift. BITS-Hyderabad Campus 8 . 5 days a week.department. 00 Determine the optimal weekly production schedule for Cotton Gate. 7 August 2012 P K Sahoo.Minutes per unit Garment Cutting Sewing Packaging Unit profit($) Shirts Blouses 20 60 70 60 12 4 8.00 12. BITS-Hyderabad Campus 9 . BITS-Hyderabad Campus 10 .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 7 August 2012 P K Sahoo. integers 7 August 2012 P K Sahoo.The objective is to find x1. x2 ≥ 0. BITS-Hyderabad Campus 11 . 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. We are interested in finding the optimum feasible solution that gives the maximum profit while satisfying all the constraints. x2 that satisfy all the constraints of the model is called a feasible solution. 7 August 2012 P K Sahoo. Any values of x1. BITS-Hyderabad Campus 12 .This is a typical optimization problem. …. …. 7 August 2012 P K Sahoo. xn) satisfying the constraints gi (x1.More generally. x2. m). BITS-Hyderabad Campus 13 . an optimization problem looks as follows: Determine the decision variables x1. x2. 2. xn) ≤ bi (i=1. …. …. x2. xn so as to optimize an objective function f (x1. 2. namely xj ≥ 0 for all j=1. x1. …. BITS-Hyderabad Campus 14 . x2. Thus a typical LPP is of the form: 7 August 2012 P K Sahoo. n. We also include the “non-negativity restrictions”. xn.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. …. x2. BITS-Hyderabad Campus 15 .Optimize (i. 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 . ….e. xn 0 7 August 2012 P K Sahoo. . am1 x1 + am2 x2 + … + amn xn ≤ bm x1. . 7 August 2012 P K Sahoo.A LPP satisfies the three properties: Proportionality . 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 & certainty. BITS-Hyderabad Campus 16 . 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. 7 August 2012 P K Sahoo. Certainty means the objective and constraint coefficients of the LP model are known constants (deterministic). BITS-Hyderabad Campus 17 . BITS-Hyderabad Campus 18 . • 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 7 August 2012 P K Sahoo.• We shall first look at formulation of some LPPs. The profit is $8 per Type I hat and $5 per Type II hat. Type I hat requires twice as much labor as a Type II. The respective market limits for the two types of hats are 150 and 200 hats per day. 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 19 .Q. 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. Formulate the problem as an LPP so as to maximize the profit. Wild West produces two types of cowboy hats. Solution: Assume that Wild West produces x1 Type I hats and x2 Type II hats per day. BITS-Hyderabad Campus 20 . Labour Time spent is (2 x1 + x2) c minutes 7 August 2012 P K Sahoo. Per day Profit got = 8 x1 + 5 x2 Assume the time spent in producing one type II hat is c minutes. integers . x2 ≥ 0.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 x1 x2 7 August 2012 ≤ 150 ≤ 200 P K Sahoo. BITS-Hyderabad Campus 21 x1. BITS-Hyderabad Campus 22 x1. x2 ≥ 0. x2 so as to maximize the profit z = 8 x1 + 5 x2 satisfying the constraints: 2 x1 + x2 ≤ 400 x1 x2 7 August 2012 ≤ 150 ≤ 200 P K Sahoo.That is: The objective is to find x1. integers . in all. If there are. 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%. I and II to undertake quality control inspection.2B) A company has two grades of inspectors. find the 7 August 2012 P K Sahoo.Q4(2. BITS-Hyderabad Campus 23 . Any error made by an inspector costs $3 to the company. At least. 10 grade I inspector and 15 grade II inspector in the company. 1500 pieces must be inspected in an 8-hour day. The wages of two inspectors are $5 & $4 per hour. The company has to incur two types of costs: Wages paid to the inspectors and the cost of their inspection errors. Solution: Let x1 and x2 represent the number of these inspectors. BITS-Hyderabad Campus 24 .40 7 August 2012 P K Sahoo.04 × 20) = Rs. 7. The cost of grade I inspector/hour is Rs.optimal assignment of inspector that minimizes the daily inspection cost. (5 + 3 × 0. The objective is to minimize the daily cost of inspection. BITS-Hyderabad Campus 25 . (4 + 3 × 0.500.20 x1 + 58. of pieces to be inspected daily). x1.The cost of grade II inspector/hour is Rs.08 × 14) = Rs.40 x1 + 7. 7.36 x2) = 59.36 The objective function is Minimize Z = 8(7. (No.90 x2 subject to the constraints x1 ≤ 10 x2 ≤ 15 20 × 8 x1 + 14 × 8 x2≥ 1. x2 ≥ 0 7 August 2012 P K Sahoo. The requirement of smaller size is 20. one of diameter 10 cm and the other of diameter 20 cm are required. Formulate the problem as a LPP.Trim Loss problem: A company has to manufacture the circular tops of cans.000 and of larger size is 15. BITS-Hyderabad Campus 26 . The problem is : how to cut the tops from the metal sheets so that the number of sheets used is a minimum.000. They are to be cut from metal sheets of dimensions 20 cm by 50 cm. Two sizes. 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 27 10 10 .A sheet can be cut into one of the following three patterns: 10 10 Pattern I Pattern II 20 10 20 Pattern III 20 7 August 2012 P K Sahoo. 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 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 28 . x3 according to pattern III The problem is to Minimize z = x1 + x2 + x3 Subject to 10 x1 + 2 x2 + 6 x3 ≥ 20. x2. x2 according to pattern II.000 x1. x3 ≥ 0.000 2 x2 + x3 ≥ 15.So assume that x1 sheets are cut according to pattern I. integers 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 29 . 15. terminal docks include casual workers who are hired temporarily to account for peak loads. the minimum demand for casual workers during the seven days of the week (starting on Monday) is 20.Q2 (2. 10.3F) In an LTL (less-than-truckload) trucking company. 18. At the Omaha. Nebraska. 12 workers. dock. 14. 10. Each worker is contracted to work five consecutive days. Determine an optimal weekly hiring practice of casual workers for the company. BITS-Hyderabad Campus 30 . 7 August 2012 P K Sahoo. 2. BITS-Hyderabad Campus 31 . Thus our problem is to find xi so as to 7 August 2012 P K Sahoo.Solution: Let xi be the number of casuals required at the beginning of day i (i = 1. …. 7). BITS-Hyderabad Campus xi 0. integers 32 .Minimize z x1 x2 x3 x4 x5 x6 x7 Subject to x1 x4 x5 x6 x7 20 (Mon) x1 x2 x5 x6 x7 14 (Tue) x1 x2 x3 x6 x7 10 (Wed) x1 x2 x3 x4 x7 15(Thu) x1 x2 x3 x4 x5 18 (Fri) x2 x3 x4 x5 x6 10 (Sat) x3 x4 x5 x6 x7 12 (Sun) 7 August 2012 P K Sahoo. BITS wants to host a Seminar for five days. BITS-Hyderabad Campus 33 . The requirement of napkins during the 5 days is as follows: Day Napkins Needed 1 80 2 50 3 100 4 80 5 150 7 August 2012 P K Sahoo. For the delegates there is an arrangement of dinner every day.Q. Institute does not have any napkins in the beginning. 0. A new napkin costs Rs. the Institute has no more use of napkins.00. A napkin given for washing after dinner is returned the third day before dinner. The washing charges for a used one are Rs. How shall the Institute meet the requirements so that the total cost is minimized ? Formulate as a LPP. BITS-Hyderabad Campus 34 . 7 August 2012 P K Sahoo.50. After 5 days. 2. The Institute decides to accumulate the used napkins and send them for washing just in time to be used when they return. j=1.. x4 + y2 = 80 x5 + y3 = 150 Also we have y1 ≤ 80. j=1..3 Thus we must have x1 = 80.. BITS-Hyderabad Campus 35 .5 Let yj be the number of napkins given for washing after dinner on day j.Solution Let xj be the number of napkins purchased on day j.2.2. x2 = 50. y2 ≤ (80 – y1) + 50 y3 ≤ (80 – y1) + (50 – y2) + 100 7 August 2012 P K Sahoo. x3 + y1 = 100. all variables ≥ 0. x4 + y2 = 80. integers 7 August 2012 P K Sahoo.5(y1+y2+y3) Subject to x1 = 80. BITS-Hyderabad Campus 36 .Thus we have to Minimize z = 2(x1+x2+x3+x4+x5)+0. x3 +y1 =100. x2 = 50. y1 ≤ 80. y1+y2 ≤ 130. x5 + y3 = 150. y1+y2+y3 ≤ 230. The demands for each month are 100. $45. $55. $48. 220 and 110 units respectively. $52 and $50. 190. 140. Acme estimates the production cost per window over next 6 months to be $50.Inventory Control Ex: Acme manufacturing company has contracted to deliver home windows over the next 6 months. material & utilities. Production cost per window varies from month to month depending on the cost of labour. 250. 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 37 . To take an advantage of the fluctuations in manufacturing cost. Develop a linear program to determine the optimum production schedule. will incur storage costs at the rate of $8 per window per month assessed on end-ofmonth inventory. however. This. 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 38 . Acme may elect to produce more than is needed in a given month and hold the excess units for delivery in latter months. 2…. For i= 1..6 let xi= Number of units produced in month i Ii =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.Solution: The variables of the problem include the monthly production amount and the end-of-month inventory. BITS-Hyderabad Campus 39 . 7 August 2012 P K Sahoo. which means I0 =0. 7 August 2012 P K Sahoo.The system starts empty. 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. BITS-Hyderabad Campus 40 . Total production cost= 50x1 +45x2 +55x3 +48x4 +52x5 +50x6 Total inventory cost= 8(I1 +I2 +I3 +I4 +I5 +I6 ) The objective function is Min z= 50x1 +45x2 +55x3 +48x4 +52x5 +50x6 + 8(I1 +I2 +I3 +I4 +I5 +I6 ) The constraints can be determined from the figure. For each period we have the balance equation: 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 41 . Ii ≥ 0 for all i= 1.Beginning inventory + Production amount -Ending inventory = Demand For the individual month (Month 1) I0 +x1 –I1 = 100 (Month 2) I1 +x2 –I2 = 250 (Month 3) I2 +x3 –I3 = 190 (Month 4) I3 +x4 –I4 = 140 (Month 5) I4 +x5 –I5 = 220 (Month 6) I5 +x6 –I6 = 110 xi .…. BITS-Hyderabad Campus 42 .6 Put I0 =0 since the situation starts 7 August 2012 P K Sahoo.2. (2. 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.Q5. C2 and C3 are pulverized and mixed together to produce 50 tons per hour needed to power a plant for generating electricity.3G) Pollution Control Three types of coal. C1. The following table summarizes the data of the situation: 7 August 2012 P K Sahoo. BITS-Hyderabad Campus 43 . i= 1.3 Minimize Z=30x1+35x2+33x3 7 August 2012 P K Sahoo.2.C1 Sulfur (parts per million) 2500 Pulverizer capacity (ton/hr) 30 Cost per ton $30 C2 1500 30 $35 C3 1600 30 $33 Determine the optimal mix of the coals. BITS-Hyderabad Campus 44 . Solution: Let xi = tons of coal. x3 ≥ 0.Subject to the restrictions 2500x1+1500x2+1600x3≤ 2000(x1+x2+x3) x1 ≤30 x2 ≤30 x3 ≤30 x1+x2+x3≥ 50 x1. BITS-Hyderabad Campus 45 . integers 7 August 2012 P K Sahoo. x2. • There is also a commercial package “LINGO” 7 August 2012 P K Sahoo.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. BITS-Hyderabad Campus 46 . • There is also Microsoft’s Excel Solver. BITS-Hyderabad Campus 47 . 7 August 2012 P K Sahoo. J C Pant’s book contains in the end a C code for solving some of the LPP problems. • You may yourself develop programs to solve LPP problems.• Dr.