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Inventory Management Calculations

1) The document provides calculations to determine the order quantity, safety stock, and reorder point for an inventory system with known parameters. The optimal order quantity is 326 units and the reorder point is 289 units. 2) Different scenarios for an inventory system are analyzed to determine optimal order quantity, reorder point, and total annual inventory costs. Maintaining the current supplier arrangement is recommended based on having the lowest total cost. 3) Optimal order quantity calculations are shown for a product with known demand distribution and costs. Expected profit, in-stock probability, and fill rate are also determined for different order quantities. Ordering 8000 units instead of the optimal 6565 results in $44,868 lower expected profit.

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Oni Sunday
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55 views3 pages

Inventory Management Calculations

1) The document provides calculations to determine the order quantity, safety stock, and reorder point for an inventory system with known parameters. The optimal order quantity is 326 units and the reorder point is 289 units. 2) Different scenarios for an inventory system are analyzed to determine optimal order quantity, reorder point, and total annual inventory costs. Maintaining the current supplier arrangement is recommended based on having the lowest total cost. 3) Optimal order quantity calculations are shown for a product with known demand distribution and costs. Expected profit, in-stock probability, and fill rate are also determined for different order quantities. Ordering 8000 units instead of the optimal 6565 results in $44,868 lower expected profit.

Uploaded by

Oni Sunday
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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1(a) B = $300, D = 120 × 52 = 6240, H = hC = 0.26 × 43 = 11.

18, S = $95
EOQ = sqrt (2DS/H) = 325.65 or 326 = order quantity.
Standard deviation of demand during lead time = sqrt ((E[L] × σ²) + (E[d]² × σL²)) = 77.77
BD ÷ ((sqrt 2π)QH × 77.77) = 2.637
Therefore, the safety stock is k = max{sqrt(2ln(2.637), 0.50} = max{1.3926, 0.50}
In this case, the safety factor is 1.3926, which is more than the minimum value.
Therefore, ROP = E[d] × E[L] + k × 77.77 = 289
1(b) Average time between stockouts = six months = 0.5 year.
Now, Q ÷ (D. TBS) = 325.65 ÷ (6240 × 0.5) = 0.104 < 1
Therefore, k = NORM.S.INV(1 – 0.104) = 1.259 which is larger than the minimum allowable
safety factor.
Therefore, SS = 1.259 × 77.77 = 97.9 or 98
ROP = 120 × 1.5 + 1.259 × 77.77 = 278

2(a) E[DDLT) = E[d] × L = 16 × 3 = 48


Standard deviation of demand during lead time = σd × √L = 4 × √3 = 6.9282
If we are to assume that inventory policy is meant to minimize total annual inventory cost,
EOQ = sqrt (2DS/H) = 395
ROP = E[D] × L + Z × 6.9282 = 62.2305 or 63
2(b) Ordering Cost = SD/Q = (75 × 16 × 52)/395 = $157.97468
Holding Cost = H(Q/2 + R – E[DDLT]) = 0.2 × 4 (395/2 + 63 – 48) = $170
Total annual inventory cost = Holding Cost + Ordering Cost = $157.97 + $170 = $328
2(c) Case 1:
Unit Cost = $3.75
E(DDLT) = E[d] × E[L] = 3 × 16 = 48
Standard deviation of demand during lead time = sqrt(3 × 4² + 16² × 1²) = 17.4356
ROP = 16 × 3 + 2.054 × 17.4356 = 83.8127 or 84
EOQ = sqrt (2DS/H) = 408
Ordering Cost = (75 × 16 × 52)/408 = $152.94
Holding Cost = H(Q/2 + R – E[DDLT]) = $180
Total annual inventory cost = $152.94 + 180 = $332.94
Case 2: EOQ = sqrt (2DS/H) = 380.94 or 381
Standard deviation of demand during lead time = 4 × √2 = 5.6569
E(DDLT) = E[d] × L = 16 × 2 = 32
ROP = 16 × 2 + 2.054 × 5.6569 = 32 + 11.61927 = 43.62 or 44.
Ordering Cost = SD/Q = (75 × 16 × 52)/381 = $163.77
Holding Cost = 0.2 × 4.3(381/2 + 44 – 32) = $174.15
Total annual inventory cost = $174.15 + $163.77 = $337.92
It can be seen that total annual relevant cost in the cases surpass the total cost in the earlier
arrangement. Based on the lowest total relevant cost criterion, we would suggest that Bobbi
should stick with her current supplier.

3(a) Co = V – s
Cu = p – V
Where p = unit revenue and V = unit cost.
Co = 99 – 49 = $50 and Cu = 569 – 99 = $470
P(D≤Q) = Cu/(Co + Cu)
The critical ration = 470/(50+470) = 0.9038, Z = (6565 – 5000)/1200 = 1.3042
Therefore, Optimal quantity = Zσ + 5000 = 6564.8 or 6565
3b(i) E[Lost Sales] = Go(Z) × σ
Go(Z) = 0.0451
Therefore, E[Lost Sales] = 0.0451 × 1200 = 54.12
3b(ii) E[Sales] = E[Demand] – E[Lost Sales] = 5000 – 54.12 = 4945.88
3b(iii) E[Leftover] = Q* - E[Sales] = 6565 – 4945.88 = 1619.12
3b(iv) E[Profit] = Cu × E[Sales] – Co × E[Leftover] = 470 × 4945.88 – 50 × 1619.12 =
$2,243,607.60
3b(v) In-Stock Probability = F(Q) = F(6565) = Φ(1.3042) = 0.9039
3b(vi) Fill Rate = E[Sales]/E[Demand] = 4945.88/5000 = 0.9892
3(c) If Q = 8000, Z = 2.5
Go(Z) = 0.00202
E[Lost Sales] = 2.424
E[Sales] = 4997.576
E[Leftover] = 3002.424
E[Profit] = Cu × E[Sales] – Co × E[Leftover] = $2,198,739.52
Loss profit per each phone model’s selling period = $2243607.60 - $2198739.52 = $44,868.08

4(a) Lot Size, Q = 10,000, ROP = 6000, Average weekly demand = 2000, Average demand per
year = 52 × 2000 = 104,000
Standard deviation of demand = 500, unit cost = $40, H = 0.25 × 40 = 10
E[DDLT] = 2000 × 2 = 4000
Safety Inventory = ROP – LR = 6000 – 4000 = 2000.
4(b) Standard deviation of demand during lead time = 500√2 = 707.11
CSL = NORMDIST(6000;4000;707.11;1) = 0.9977
Cu = HQ/(1-CSL)D = 10 × 10000/(104000(1 – 0.9977)) = $418.06
4(c) If Cu = $80
CSL* = (DCu – HQ)/DCu = 1 – (HQ/DCu) = 1 – (10 × 10000/104000 × 80) = 1 – 0.0120 = 0.988
K = 2.257
Therefore, Safety Stock = K × standard deviation of demand during lead time = 2.257 × 707.11 =
1596.

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