Question

(20 points) A component used in a manufacturing facility is ordered from an outside supplier. Because the component is used in a variety of end products, the demand is high. Estimated demand (in thousands) over the next 10 weeks is 6. Week Demand 10 22 34 3 12 8 44 5416 76 30 The components cost 65 cents each and the interest rate used to compute the holding cost is 0.5 percent per week. The fixed order cost is estimated to be $200. (Hint: express h as the holding cost per thousand units.) a. What ordering policy is recommended by the Silver-Meal heuristic? b. What ordering policy is recommended by the part period balancing heuristic? c. What ordering policy is recommended by the least unit cost heuristic? d. What ordering policy is recommended by the lowest-cost policy for this problem?

Answer 1

Answer K = 200

                  h = (.005)(.65) = .00325/unit

                  µ h = 3.25/thousand.

                  r = (22, 34, 32, 12, 8, 44, 54, 16, 76, 30)

            a)   Silver Meal Solution (S/M)

                  Start in period 1:

                  C(1) = 200

                  C(2) = [200 + ((3.25)(34)]/2 = 172.8

                  C(3) = [(155.25)(2) + (3.25)(2)(32)]/3 = 183.5     Stop.

                  y₁ = r₁ + r₂ = 56

                  Start in period 3:

                  C(1) = 200

                  C(2) = [200 + (3.25)(12)]/2 = 119.5

                  C(3) = [(2)(119.5) + (2)(3.25)(8)]/3 = 97

                  C(4) = [(3)(97) + (3)(3.25)(44)]/4 = 180     Stop.

                  y₃ = r₃ + r₄ + r₅ = 32 + 12 + 8 = 52

                        Start in period 6:

                  C(1) = 200

                  C(2) = [200 + (3.25)(54)]/2 = 187.75

                  C(3) = [(2)(187.75) + (2)(3.25)(16)]/3 = 159.83...

                  C(4) = [(3)(159.833..) + (3)(3.25)(76)]/4 = 305.125   Stop.

                  y₆ = r₆ + r₇ + r₈ = 44 + 54 + 16 = 114

                  Obtain y₉ = r₉ + r₁₀ = 106

                  Silver Meal Solution = (56, 0, 52, 0, 0, 114, 0, 0, 106, 0)

                  Cost = (4)(200) + (3.25)(34 + 20 + 8 + 70 + 16 + 30)

                           = $1378.50

Answer C:-   Least Unit Cost (LUC) method:

                  Start in period 1:

                  C(1) = 200/22 = 9.09

                  C(2) = [200 + (3.25)(34)]/(22 + 34) = 5.54

                  C(3) = [200+(3.25)(34+(2)(32))]/(22+34+32) = 5.89 Stop.

                  y₁ = r₁ + r₂ = 56

                  Start in period 3:

                  C(1) = (200)/(32) = 6.25

                  C(2) = [200 + (3.25)(12)]/(32 + 12) = 5.43

                  C(3) = [200 + (3.25)(12+(2)(8))]/(32+12 8) = 5.60   Stop.

                  y₃ = r₃ + r₄ = 44

                  Start in period 5:

                  C(1) = 200/8 = 25

                  C(2) = [200+(3.25)(44)]/(8 + 44) = 6.60

                  C(3) = [200+(3.25)(44+(54)(2))]/(8+44+54) = 6.55

                  C(4) = [200+(3.25)(44+(54)(2)+(16)(3))]/(8+44+54+16)= 7.97 Stop.

                  y₅ = r₅ + r₆ + r₇ = 106

                  Start in period 8:

                  C(1) = 12.5

                  C(2) = 4.9

                  C(3) = 5.3 Stop

                  y₈ = r₈ + r₉ = 92

                  y₁₀ = r₁₀ = 30

Answer d-

Summary of Results

                         Cost

                        S/M $1378.50

                        LUC $1890.50

           

*Note: None of these solutions is optimal. The true optimal solution is (56, 0, 52, 0, 0, 44, 70, 0, 106, 0) at a cost of $1351.00.   Silver Meal obviously comes the closest for this example.

                  LUC solution: (56, 0, 44, 0, 106, 0, 0, 92, 0, 30)

                  Cost = (200)(5)+(3.25)(34 + 12 + 98 + 54 + 76) = $1,890.50.