Practice Problems for End-Term Examination

Some Operations Management Practice Problems for exam

1. Practice for End Term

2. Centre of Gravity X 120, y 140 L-2000 X 180 y 40 L-1500 X 100 y 100 L-1800 X 80 y 200 L-2500 Location x co-ordinate= 153.45 Location y co-ordinate = 175.86

3. Weighted Factor Rating Method Wt A Wt Score A B Wt Score B C Wt Score C Cheap Land 0.1 60 75 30 Power 0.5 40 40 80 Labour 0.2 70 80 60 Raw Mat 0.15 30 60 50 Govt Policy 0.05 40 80 40 6 20 14 4.5 2 46.5 7.5 20 16 9 4 56.5 3 4 12 7.5 2 64.5

4. Forecast using Exponential Smoothing Smoothing constant α F t+1 = Forecast for next period D t = Actual demand for current period F t = Forecast demand for current period

5. Monday 85 Tuesday 88 Wednesday 92 Thursday 84 Friday 96 Saturday 102 Sunday α= 0.3 85 85 85.9 87.73 86.61 89.43 93.20

6. Two Machines N Parts Part P i  has two operations, of duration P i1 , P i2 , to be done on Machine M1, M2 in that sequence. Step 1. List A = { 1, 2, �, N }, List L1 = {}, List L2 = {}. Step 2. From all available operation durations, pick the minimum. If the minimum belongs to P k1 , Remove K from list A; Add K to end of List L1. If minimum belongs to P k2 , Remove K from list A; Add K to beginning of List L2. Step 3. Repeat Step 2 until List A is empty. Step 4. Join List L1, List L2. This is the optimum sequence.

7. Tie Breaking Rule This is applicable for regular 2 machine Johnson Algorithm If two equal smallest processing time, place the job on M1 at end of first list L1, and if it belongs to M2, put it at the beginning on second list L2 If both smallest processing time is for M1, select the job with lowest processing time for M2 at end of L1 If both smallest processing time is for M2, select the job with lowest processing time for M1 at beginning of L2

8. Jobs M1 M2 M3 P1 4 4 6 P2 7 6 8 P3 8 5 7 P4 3 1 9 P5 4 3 7 P6 1 2 10 P7 6 6 8 Does it fulfil condition for application of Johnson’s Rule 

9. Jobs M1 M2 M3 G1 G2 P1 4 4 6 P2 7 6 8 P3 8 5 7 P4 3 1 9 P5 4 3 7 P6 1 2 10 P7 6 6 8 8 10 13 14 13 12 4 10 7 10 3 12 12 14

10. Jobs G1 G2 P1 8 10 P2 13 14 P3 13 12 P4 4 10 P5 7 10 P6 3 12 P7 12 14 {1, 2, 3, 4, 5, 6, 7} L1= { } L2= { } 6 4 5 1 3 7 2 Optimal Sequence {6, 4, 5, 1, 7, 2, 3}

11. Finding out Idle Time Jobs M1 M2 M3 P1 4 4 6 P2 7 6 8 P3 8 5 7 P4 3 1 9 P5 4 3 7 P6 1 2 10 P7 6 6 8 Jobs M1 in M1 out M2 in M2 Out M3 in M3 out M2 Idle M3 Idle 6 4 5 1 7 2 3 1 1 4 4 8 8 12 12 18 18 25 25 33 1 3 4 5 8 11 12 16 18 24 25 31 33 38 1 1 3 1 2 1 2 11 3 13 13 22 22 29 29 35 35 43 4 3 51 51 58 3 3 Total Idle Time= 11+3 + 25+20 = 59

12. PERT - For Dealing With Uncertainty So far, times can be estimated with relative certainty, confidence For many situations this is not possible, e.g Research, development, new products and projects etc . Use 3 time estimates t m = most likely time estimate, mode. t o = optimistic time estimate, t p = pessimistic time estimate, and Expected Value ( t e ) = (t o + 4t m + t p ) /6 Variance (V) = ( ( t p – t o ) / 6 ) 2 Std Deviation (δ) = SQRT (V)

13. PERT Activity t o t m t p t e V 1-2 6 9 12 1-3 3 4 11 2-4 2 5 14 3-4 4 6 8 3-5 1 1.5 5 2-6 5 6 7 4-6 7 8 15 5-6 1 2 3 9 6 6 6 2 5 9 2 1.77 4 0.43 0.43 0.11 1.77 1 0.11

14. 9 5 6 6 2 6 9 2 9 5 7 15 24 24 15 22 9 9 Critical Path = 1-2-4-6 t e cp = 24 δ cp = SQRT ( 1 +4+1.77) = 2.61

15. Probability of completing in 26 days Calculate Z, where Z = (D-S) / δ D = Planned completion date S = Scheduled completion date (expected) Z= (26-24 )/ 2.61 = 0.766 Corresponding probability from Z table is about 0.7794 = 77.94%

16. If Purchase policy changes Item C minimum order 50 Item D multiple of 50

17. Jeevan Pharma - EOQ D=2000 , O= 100, I= 10% of unit cost = 1 EOQ= SQRT (2x2000x100)/1 = 632.45 Cost of Inventory = (2000/632.45)x100 + (632.45/2)x1 = 316.23+ 316.23 = 632.46 If Order quantity reduced by 10%, new order quantity would be 632.45x(1-10%)=569.21 New cost of inventory = (2000/569.21)x100 + (569.21/2)x1 = 351.36 + 284.61 = 635.97 Therefore additional cost = 635.97-632.46 = 3.51

18. Thank You Good Luck