UBS, GURGAON Executive MBA Program Course on Operations Management (EMB 108)

Transcription

1 UBS, GURGAON Executive MBA Program Course on Operations Management (EMB 108) Handout on Numerical Techniques: Facility Location and Layout (Prepared by: A. Ramachandran, Course Instructor) Single-facility location techniques 1. Factor rating method Example: A new medical facility is to be located in a town. The following table shows the various factors, and the corresponding weights and scores (1 = poor, 5 = excellent) on each factor, for two potential sites (A and B, say). The weights add up to 100 percent. Which of the two sites would you recommend? Factors for location (F i ) Weight Score (W i ) A B F1: Total patient miles per month F2: Facility utilization F3: Average time per emergency trip F4: Expressway accessibility F5: Land and construction costs F6: Employee preferences The total weighted score for each site is computed as shown below: Factors for location (F i ) Weight Weighted score (W i ) A B F1: Total patient miles per month F2: Facility utilization F3: Average time per emergency trip F4: Expressway accessibility F5: Land and construction costs F6: Employee preferences Total Note that for a given site (A, say), the weighted score for the i th factor (i = 1, say) is 25x4=100, and, similarly, for site B it is 25x5=125. The total weighted score values for the two sites are 340 and 360, respectively. Clearly, we would recommend site B for the location of the new medical facility. A. Ramachandran, Course Instructor) 1

2 2. Load-distance score method Example: Suppose the medical facility is meant to serve seven corporation wards in the town. The following table gives the coordinates of the center of each ward, along with projected population (in 000). Customers will travel from the seven ward centers to the new facility when they need healthcare. Two locations being considered for the new facility are at (5.5, 4.5) and (7,2), which are at the centers of wards C and F. If we use the population as the loads and use rectilinear distance, which location is better in terms of its total load-distance score? Ward Coordinates of ward center (x,y) Population ( 000) l A (2.5, 4.5) 2 B (2.5, 2.5) 5 C * (5.5, 4.5) 10 D (5, 2) 7 E (8, 5) 10 F * (7, 2) 20 G (9, 2.5) 14 * indicates that it is a candidate location. We compute the total load-distance score for each candidate location as shown below, using rectilinear distance: Ward Coordinates of ward center (x,y) Population ( 000) l Distance d if we locate at C (5.5, 4.5) ld for C Distance d if we locate at F (7, 2) ld for F A (2.5, 4.5) = = 7 14 B (2.5, 2.5) = = 5 25 C * (5.5, 4.5) = = 4 40 D (5, 2) = = 2 14 E (8, 5) = = 4 40 F * (7, 2) = = 0 0 G (9, 2.5) = = Total Note: Rectilinear distance between A and C is given by d AC, rect = abs ( ) + abs ( ) = = 3. Similarly, d AF, rect = abs (2.5 7) + abs (4.5 2) = = 7. [The abs function refers to the absolute value of the argument. E.g. abs (-10) = 10; abs (10) = 10.] Since load-distance score for F is less than that for C, location F is preferable to location C. A. Ramachandran, Course Instructor) 2

3 [Note: The Euclidean distance between A and C is given by d AC, Euc = sqrt (( ) 2 + ( ) 2 ) = sqrt (3 2 ) = 3. Similarly, d BC,Euc = sqrt (( ) 2 + ( ) 2 ) = sqrt ( ) = sqrt (13) = 3.6, approx. But d BC,rect = = 5. Hence, d PQ,Euc <= d PQ,rect for any point (P, Q).] 3. Center of gravity method Consider the medical facility location example above. The center of gravity method works as follows: Ward x i y i Popn. (l i ) l i x i l i y i A B C D E F G The center of gravity CG (x *, y * ) is given by x * = Σ (l i x i )/ Σ (l i ) = 453.5/68 = 6.67; y * = Σ (l i y i )/ Σ (l i ) = 205.5/68 = Series A. Ramachandran, Course Instructor) 3

4 4. Break-even analysis method Example: A firm has narrowed the search for a new facility location to three candidate locations. The annual fixed costs (in Rs. 000) and unit variable costs (in Rs.) of production for each candidate location are: Location Fixed cost per year Unit variable cost (Rs. 000) (Rs.) A B C a. Plot the total cost curves for all the three locations on the same graph. Identify on the graph the approximate range over which each location provides the lowest cost. (Assume a maximum output level of 20,000 units per year.) b. Using break-even analysis, compute the break-even quantities over the relevant ranges. c. If the expected demand is 15,000 units per year, identify which is the best location. (a) The three lines are: TC (A) = X TC (B) = X TC (C) = X, where X = Output level in thousand units; and TC stands for total cost, fixed cost being measured in thousand rupees. Taking output level Q on x-axis (up to 20), and total cost on y-axis, the three lines can be plotted. The graph (not to scale herein) may look like this: A. Ramachandran, Course Instructor) 4

5 TC TC (A) TC (B) TC (C) A best B best C best Q From the graph, from zero to 6.25 thousand units, location A is best (i.e., least total cost). From 6.25 to 14.3 thousand units, B is best. Beyond 14.3, C is best. (b) Point of intersection of TC (A) and TC (B) gives the break-even quantity for first range: i.e., 150, X = 300, X, yielding X = 6250 units. Similarly, TC (B) and TC (C) intersect at X= 14,286 units. (c) For an estimated demand = 15,000 units, location C is best, as it is more than 14,286. A (simple?!) facility layout problem The layout of a work center is being designed for the production of a single product. The center must house three sections, A, B, and C. Two options, viz. Op. I and Op. II, are being considered, as shown below: 12 Op. I A B C Op. II C B 12 A A. Ramachandran, Course Instructor) 5 15

6 (Note: the figures above are not to scale; and, the distances are in feet.) The workload flows between the three sections are as given below. It costs Rs.2 per load-foot to transport the product. Evaluate the two options for the layout on the basis of load-distance. Which option do you think is better? A B C A x B 20 x 40 C x The problem is solved as follows: Option I:- movements distance in feet AB BA AC CA BC CB Workload units Total load-feet = 12( ) + 15(40+50) = 12(90) + 15(90) = 27(90) = 2430 feet-units. Option II:- movements distance in feet CB BC CA AC AB BA Workload units Total load-feet = 12( ) + 15(30+20) = 12(130) + 15(50) = = 2410 feetunits. Obviously, Op. II is better! (And the cost per load foot information is irrelevant!) A. Ramachandran, Course Instructor) 6