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OPERATIONS AND SUPPLY CHAIN MANAGEMENT 2022 ASSIGNMENT 6, Assignments of Operational Research

IIT Nptel OPERATIONS AND SUPPLY CHAIN MANAGEMENT 2022 ASSIGNMENT 6

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2021/2022

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OPERATIONS AND SUPPLY CHAIN MANAGEMENT 2022
ASSIGNMENT 6
Note: Answer all numerical questions up to three decimal places
The demand for 7 months is 500, 600, 800, 1200, 1400, 700 and 800. C0 = 100 and
Cc = 2/unit/year.
1. The ordering costs for 7 months using lot for lot heuristic is ------- (Ans: 700)
2. The inventory costs for 7 months using lot for lot heuristic is -------- (Ans: 500)
3. The total cost is ------- (Ans: Range 1200)
Solutions for questions 1,2,3:
1. Ordering cost = 7*100 = 700
2. Inventory cost = (2/12)*(6000/2) = 500
3. Total cost = 700 + 500 = 1200
The demand for 8 months is 600, 800, 1200, 1000, 1300, 900, 700 and 1200. C0 =
200 and Cc = 3/unit/year.
4. The first lot size using Part period balancing is ------------ (Ans: 600)
5. The second lot size using Part period balancing is ------------ (Ans: 800)
6. The first lot size using Silver-Meal heuristics is ------------ (Ans: 600)
7. The second lot size using Silver-Meal heuristics is ------------ (Ans: 800)
Solutions for questions 4,5,6,7:
4. Part period balancing:
When Q=600, avg inv = 300, CC = 300/4 = 75
When Q=1400, avg inv = 1500, CC = 375 (>200)
75 is close to 200. Hence first lot size = 600
5. When Q=800, avg inv = 400, CC = 400/4 = 100
When Q=2000, avg inv = 2200, CC = 550 (>200)
100 is close to 200. Hence second lot size = 800
6. Silver Meal heuristic:
When Q=600, avg inv = 300, CC = 300/4 = 75, TC = 275; per period cost = 275/1 = 275.
When Q=1400, avg inv = 1500, CC = 375, TC = 575; per period cost = 575/2 = 287.5
275 is lesser. Hence first lot size = 600
7. When Q=800, avg inv = 400, CC = 400/4 = 100, TC = 300; per period cost = 300/1 = 300.
When Q=2000, avg inv = 2200, CC = 550, TC = 750; per period cost = 750/2 = 375
300 is lesser. Hence second lot size = 800
Consider the four-item disaggregation problem with D1 = 200, D2 = 300, D3 = 200, D4 =
400 and P= 1500. Assume I1 = 100, I2 = 200, I3 = 150 and I4 = 250. Assume t 1, t2, t3, and t4
be the start times of production of items.
8. The maximum value of t1, t2, t3, t4should be respectively
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OPERATIONS AND SUPPLY CHAIN MANAGEMENT 2022

ASSIGNMENT 6

Note: Answer all numerical questions up to three decimal places The demand for 7 months is 500, 600, 800, 1200, 1400, 700 and 800. C0 = 100 and Cc = 2/unit/year.

  1. The ordering costs for 7 months using lot for lot heuristic is ------- (Ans: 700)
  2. The inventory costs for 7 months using lot for lot heuristic is -------- (Ans: 500)
  3. The total cost is ------- (Ans: Range 1200) Solutions for questions 1,2,3:
  4. Ordering cost = 7*100 = 700
  5. Inventory cost = (2/12)*(6000/2) = 500
  6. Total cost = 700 + 500 = 1200 The demand for 8 months is 600, 800, 1200, 1000, 1300, 900, 700 and 1200. C0 = 200 and Cc = 3/unit/year.
  7. The first lot size using Part period balancing is ------------ (Ans: 600)
  8. The second lot size using Part period balancing is ------------ (Ans: 800)
  9. The first lot size using Silver-Meal heuristics is ------------ (Ans: 600)
  10. The second lot size using Silver-Meal heuristics is ------------ (Ans: 800) Solutions for questions 4,5,6,7:
  11. Part period balancing: When Q=600, avg inv = 300, CC = 300/4 = 75 When Q=1400, avg inv = 1500, CC = 375 (>200) 75 is close to 200. Hence first lot size = 600
  12. When Q=800, avg inv = 400, CC = 400/4 = 100 When Q=2000, avg inv = 2200, CC = 550 (>200) 100 is close to 200. Hence second lot size = 800
  13. Silver Meal heuristic: When Q=600, avg inv = 300, CC = 300/4 = 75, TC = 275; per period cost = 275/1 = 275. When Q=1400, avg inv = 1500, CC = 375, TC = 575; per period cost = 575/2 = 287. 275 is lesser. Hence first lot size = 600
  14. When Q=800, avg inv = 400, CC = 400/4 = 100, TC = 300; per period cost = 300/1 = 300. When Q=2000, avg inv = 2200, CC = 550, TC = 750; per period cost = 750/2 = 375 300 is lesser. Hence second lot size = 800 Consider the four-item disaggregation problem with D1 = 200, D2 = 300, D3 = 200, D4 = 400 and P= 1500. Assume I1 = 100, I2 = 200, I3 = 150 and I4 = 250. Assume t 1 , t 2 , t 3 , and t 4 be the start times of production of items.
  15. The maximum value of t 1 , t 2 , t 3 , t 4 should be respectively

a. (0.5, 0.75, 0.4, 0.625) b. (0.5, 0.66, 0.75, 0.625) c. (0.25, 0.8, 0.75, 0.66)

  1. The order of the procedure is a. (1-2-3-4) b. (2-1-3-4) c. (1-4-2-3)
  2. If t 1 =0.04, t 2 =0.2, t 3 =0.15, t 4 =0.17 what is the inventory of item 1 at t 1 ---------- (Ans:

Solution for questions 8,9,10:

  1. r 1 = 100/200 = 0.5; r 2 =200/300 =0.66; r 3 =150/200 = 0.75; r 4 =250/400 = 0. Therefore, the constraints are t 1 ≤ 0.5; t 2 ≤ 0.66; t 3 ≤0.75; t 4 ≤0.
  2. r 1 ≤ r 4 ≤ r 2 ≤ r 3. Therefore the order is 1-4-2-3.
  3. inventory of 1 at t=0.04 is 100-(200*0.04) = 92
  4. In Disaggregation problem, the inventory is taken in a. Number of manhours b. Number of units c. Number of lots (lot size)
  5. Economic lot scheduling model can be applied for a machine that can produce multiple products. a. True b. False
  6. When the lead time and demand are deterministic, reorder level is: a. Sum of lead time and demand b. Demand during the lead time c. Lead time
  7. If inventory cost is less than the shortage cost: a. Reorder level is more than the expected value of lead time demand b. Reorder level is equal to the expected value of lead time demand c. Reorder level is less than the expected value of lead time demand d. Reorder level cannot be predicted
  8. The main objective in economic lot scheduling is a. Minimise total cost b. Minimise number of orders c. Minimise average inventory