Operations Management, Chapter 6 Solutions

Chapter 6 Solutions 1. OT = 450 minutes a. Minimum cycle time = length of longest task, which is 2.4 minutes. Maximum cycle time =  task times = 18 minutes. b. Range of output: 450 = 187 .5 units 2.4 450 @18 min . : = 25 units 18 Dx  t 187 .5(18) N= = = 7.5, which rounds to 8 OT 450 @ 2.4 min . : c. OT 450 Solving for CT, CT = = 3.6 minutes per cycle CT 125 e. Potential output: OT 450 = = 50 units (1) CT = 9 min . : CT 9 d. Output = (2) CT = 15 min . : 450 = 30 units 15 2. 0.6 c 0.5 f 1.4 0.5 0.7 0.5 a b d h 1.0 g 0.8 e Desired output = 33.33 units per hour Operating time = 60 minutes per hour CT = Operating time 60 minutes per hour = = 1.80 minutes per unit Desired output 33 .33 units per hour Solutions (continued) a. Task A Number of following tasks 7 Positional Weight 6 B C D E F G H 6 2 2 2 1 1 0 4.6 1.6 2.2 2.3 1.0 1.5 0.5 Assembly Line Balancing Table (CT = 1.8) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 1.4 0.4 – II B 0.5 1.3 C, D, E E 0.8 0.5 – D 0.7 1.1 C C 0.6 0.5 F F 0.5 0 – G 1.0 0.8 H H 0.5 0.3 – III IV b. Assembly Line Balancing Table (CT = 1.8) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 1.4 0.4 – II B 0.5 1.3 C, D, E E 0.8 0.5 – D 0.7 1.1 C C 0.6 0.5 F F 0.5 0 – G 1.0 0.8 H H 0.5 0.3 – III IV Solutions (continued) c. Efficiency = Total time 6.0 = = 83 .3% CT x no. of stations 7.2 3. 3 2 4 a b c 7 4 9 5 d e h i 5 6 f g Desired output = 4 Operating time = 56 minutes Operating time 56 minutes per hour CT = = = 14 minutes per unit Desired output 4 units per hour Task A # of Following tasks 4 Positional Weight 23 B 3 20 C 2 18 D 3 25 E 2 18 F 4 29 G 3 24 H 1 14 I 0 5 Solutions (continued) a. First rule: most followers. Second rule: largest positional weight. Assembly Line Balancing Table (CT = 14) Work Station Task Task Time Time Remaining Feasible tasks Remaining I F 5 9 A,D,G A 3 6 B,G G 6 – – D 7 7 B, E B 2 5 C C 4 1 – E 4 10 H H 9 1 – I 5 9 – II III IV b. First rule: Largest positional weight. Assembly Line Balancing Table (CT = 14) Work Station Task Task Time Time Remaining Feasible tasks Remaining I F 5 9 A,D,G D 7 2 – G 6 8 A, E A 3 5 B,E B 2 3 – C 4 10 E E 4 6 – H 9 5 I I 5 – II III IV c. Efficiency = Solutions (continued) Total time 45 = = 80 .36 % CT x no. of stations 56 4. a, b a. l. c a d b h 1. e g f 1. 2. Minimum Ct = 1.3 minutes Task Following tasks a 4 b 3 c 3 d 2 e 3 f 2 g 1 h 0 Work Station Eligible Assign Time Remaining Idle Time I a A 1.1 b,c,e, (tie) B 0.7 C 0.4 E 0.3 0.3 0.0 II d D 0.0 III f,g F 0.5 G 0.2 0.2 H 0.1 0.1 IV h 0.6 3. Idle percent = 4. Output = (idle time) .6 = = 11 .54 percent N x CT 4(1.3) OT 420 min./day = = 323 .1 (rounds to 323 )units / day CT 1.3 min . / cycle Solutions (continued) b. 1. Total time = 4.6, CT = Total time 4.6 = = 2.3 minutes N 2 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time] Assign f, g, and h to station 2: 2.3 minutes OT 420 = = 182 .6 units / day 3. Output = CT 2.3 4. Maximum Ct is 4.6. Output = 5. 420 min./day = 91 .30 units / day 4.6 min . / cycle a. .2 a 1.2 .2 .4 b f c 1.0 1.2 .4 d g e b. The minimum cycle time = maximum task time =1.2 minutes The maximum cycle time = .2 +.4 +.2 +.4 +1.2 +1.2 + 1.0 = 4.6 minutes OT 480 min./day CT = = = 2 minutes (calculate d CT) output 240 units / day c. N =  t 4.6 = = 2.3 (rounds to 3) stations CT 2.0 d. Task A Number of following tasks 4 B 3 C 2 D 2 E 1 F 1 G 0 Solutions (continued) Assembly Line Balancing Table (CT = 2 minutes) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 0.2 1.8 B,D B 0.4 1.4 C, D D 0.4 1.0 C C 0.2 0.8 – II E 1.2 0.8 – III F 1.2 0.8 – IV G 1.0 1.0 – e. Idle percent = .8 + .8 + .8 + 1.0 3.4 = = 42 .5% (4)(2) 8.0 E= 6. 4.6 = 57 .5% (4)(2) a. (1,2) 0.1 0.2 0.9 0.6 a b c d (3.9) (3.8) (3.6) (2.7) Positional weights in parentheses 0.1 0.2 0.4 0.1 0.2 0.7 e f g h i j (2.2) (2.1) (1.9) (1.5) (1.4) (1.2) CT = 1.5 (0.5) k 0.3 (0.2) l 0.2 Solutions (continued) 6. b. Using both the greatest positional weight and the greatest number of following tasks rules result in the following balance. Assembly Line Balancing Table (CT = 1.5 minutes) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 0.1 1.4 B B 0.2 1.2 C C 0.9 0.3 E E 0.1 0.2 – D 0.6 0.9 F F 0.2 0.7 G G 0.4 0.3 H H 0.1 0.2 I I 0.2 – – J 0.7 0.8 K K 0.3 0.5 L L 0.2 0.3 – II III Total idle time = 0.2 + 0 + 0.3 = 0.5 c. For positional weights and greatest number of following tasks Idle rate = 7. 0.5 x100 = 11 .11 % 3(1.5) g a. a c b h d e f i j Solutions (continued) OT 7(60) = = .84 minutes = 50.4 seconds (maximum cycle time) b. CT = D 500 Minimum cycle time = maximum task time = 45 seconds (results in 560 units of production) c. N=  t 193 = = 3.83 or 4 stations CT 50 .4 d. Task A Number of followers 6 *PW 106 B 5 61 C 4 50 D 4 106 E 3 56 F 2 30 G 2 31 H 2 29 I 1 19 J 0 10 *Positional weight CT = 50 seconds Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 45 5 – III D 50 – – III B 11 39 C, E E 26 13 C, F C 9 4 – G 12 38 H, F F 11 27 H H 10 17 I I 9 8 – J 10 40 – IV V Solutions (continued) e. I = 1− 193 = 22 .8% (50 )(5) 8. b a b. c g h e d f 200 CT = = 2.0 minutes 400 j k i Station 1 Tasks a,b Time 1.9 Idle/Time .1 2 c,d 1.9 .1 3 e,f,i 2.0 0 4 g,h,j 1.5 .5 5 k,m 1.2 .8 1.5 c. Tasks a b Positional Weight 8.5 4.6 c 4.4 d 4.2 e f 3.2 3.5 g 1.9 h 1.5 i 2.5 j 2.0 k 1.2 m .3 m Solutions (continued) Work Station Task Task Time Time Remaining Feasible tasks Remaining I a .5 1.5 b, c, d b 1.4 .1 – c 1.2 .8 d, e d .7 .1 f 1.0 1.0 e, i e .5 .5 i, g i .5 0 – j .8 1.2 g g .4 .8 h h .3 .5 – k .9 1.1 m m .3 .8 – II III IV Total idle time = .1 + .1 + 0 + .5 + .8 = 1.5 minutes d. Balance delay: part b and c 1.5/10 = 15% 9. 1 4 3 2 5 6 10. 1 2 3 11. 1 5 4 4 3 8 7 5 6 2 6 A o o o A X o o o o A o o o o Solutions (continued) 12. 3 1 4 8 5 1 4 8 Or 5 2 7 6 5 1 7 8 3 2 7 6 3 1 7 8 Or 3 2 4 6 3 1 7 6 5 1 4 6 3 2 7 8 3 1 4 6 5 2 7 8 Or Or 5 2 4 6 5 1 7 6 3 2 4 8 Or 5 13. 14. 2 4 8 3 1 8 9 7 4 5 2 6 First rank or arrange the number of trips from high to low. Department Number of trips 2-4 90 1-4 80 3-4 55 2-3 40 From this we can see that departments 2 and 4 have the greatest interdepartmental work flow, so they should be close, perhaps locations C and B. Next, we can see that the work flows for 1 and 4, and 3 and 4 are high. Therefore department 4 has to be located at a central location (location B), while department 2 is in location C, department 1 is in location A, and department 3 is in location D. Distance * Number of trips matrix Department 1 2 1 – (10 x 80) = 800 3 (20x 70) = 1400 4 (80 x 40) = 3200 2 – – (40 x 60) = 2400 (90 x 40) = 3600 3 – – – (55 x 50) = 2750 – 4 – – – Total cost = $14,150 Solutions (continued) 15. No. of trips (two way) 1–2 Order of Assignment 10 1–3 5 A reasonable (intuitive) set of assignments is: 1–4 90 11 1–5 370 1 1–6 135 6 1–7 125 1–8 0 2–3 A #1 B #5 7 C #7 D #4 E #3 360 2 2–4 120 8 (tie) F #6 G #2 H #8 2–5 40 2–6 115 2–7 45 2–8 120 8 (tie) 3–4 350 3 3–5 110 10 3–6 40 3–7 20 3–8 200 No. of trips (two way) 4–5 190 4–6 70 4–7 50 4–8 190 5–6 10 5–7 40 5–8 10 6–7 50 6–8 20 7–8 20 9 4 Order of Assignment 5 (tie) 12 5 (tie) This set of assignments has a total cost of $143,650 per day. Slight variations would also be reasonable, as long as departments 2, 4 and 8 are close to 3, 4 is close to 5, and 5 is close to 1. Solutions (continued) 16. No. of trips (two way) 1–2 0 1–3 40 1–4 110 3–A, 5–B, 1–C, 4–D, 6–E, 2–F. 1–5 80 An equivalent solution is the reverse order: 1–6 50 2–F, 6–B, 4–C, 1–D, 5–E, 3–F. 2–3 0 2–4 50 2–5 40 2–6 120 3–4 10 3–5 250 3–6 10 4–5 40 4–6 90 5–6 20 A reasonable (intuitive) assignment is: (Ignore Reception since all locations are the same distance from it.) 17. Two-way trips can not be used here because of the one-way route restriction. Consequently, students are forced to develop a heuristic that will yield reasonable assignments. One possible heuristic is the following: Beginning with Department 1, identify the department which receives the greatest number of trips from that department (e.g., 40 to Department 2). Assign that department to the next location counter-clockwise. For that department (e.g., 2) identify the department which receives the greatest number of trips (e.g., 5) and assign it to the next position. Continue in this manner until all departments have been assigned. The resulting set of assignments for this problem is: A–1, 2–B, 5–C, 4–D, 9–E, 8–F, 6–G, 10–H, 7–I, 3–J. Students may raise the question about return trips to the original departments after delivery, which would seem to make all locations comparable. Three possible explanations are: 1. Return trips cost less because they are unloaded. 2. Unloaded trips may be permitted to move clockwise. 3. Material handlers (?) pick up new load at each new department and move it to the next department. (The last explanation seems to appeal most to students.) Solutions (continued) Work Station Task Task Time Time Remaining Feasible tasks Remaining I D 50 0.4 – II A 45 5.4 – III E 26 24 B ,F B 11 13 C, F F 11 C 9 41 G, H G 12 29 H H 10 19 I I 9 10 J J 10 IV 2.4 0.4 –