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a)
Interpretation: Determine the order policy for the item based on silver-meal method.
Concept Introduction: Silver-Meal method is mainly used to determine the production quantities of the firm to be produced at the minimum cost.it is also provides the appropriate solutions to the time varying demand in production patterns.
a)
Answer to Problem 14P
The order policy according to silver meal method is 18 units in period-1,23 units in Period-4,50 units in Period-6,35 units in period-9 and 12 units in period-12.
Explanation of Solution
Given information: The anticipated demand for an inventory is as follows:
| Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Demand | 6 | 12 | 4 | 8 | 15 | 25 | 20 | 5 | 10 | 20 | 5 | 12 |
Current inventory is 4 and ending inventory is 8.
Holding cost (h) is $1 per period, set up cost (K) is $40
The objective of this method is to minimize the per-period-cost of ordering policy.
Net Requirements:
The order policy (lot-size) of the item under silver meal method can be calculated as follows:
According to silver meal method the average cost per period C (T) is a function of the average holding and set up cost per period for T number of Periods. The production in period 1 is equal to the demand in that period 1 to incur the order cost K.
Hence
And
And general equation is
Once
Now, calculate the order policy using the above formula as follows:
Starting in period 1:
Stop the process since
Starting in period 4:
Stop the process since
Starting in period 6:
Stop the process since
Starting in period 9:
Stop the process since
The same is explained with the help of a table shown below:
Table 1: Order Quantity using Silver-Meal Method:
| Months | No.of periods | Q | 11 | 12 | 13 | 14 | 15 | 16 | 17 | Total inventory | Holding Cost | Ordering Cost | Per Period Cost | Decision |
| 1 to 1 | 1 | 2 | 0 | =SUM(D3:J3) | =K3*1 | 40 | =SUM((L3:M3)/B3) | Continue | ||||||
| 1 to 2 | 2 | =2+12 | 12 | 0 | =SUM(D4:J4) | =K4*1 | 40 | =SUM((L4:M4)/B4) | Continue | |||||
| 1 to 3 | 3 | =14+4 | =C5-C3 | =C5-C4 | 0 | =SUM(D5:J5) | =K5*1 | 40 | =SUM((L5:M5)/B5) | Optimal | ||||
| 1 to 4 | 4 | =C5+8 | =C6-C3 | =C6-C4 | =C6-C5 | 0 | =SUM(D6:J6) | =K6*1 | 40 | =SUM((L6:M6)/B6) | Go Back | |||
| 4 to 4 | 1 | 8 | 0 | =SUM(D7:J7) | =K7*1 | 40 | =SUM((L7:M7)/B7) | Continue | ||||||
| 4 to 5 | 2 | =C7+15 | =C8-C7 | 0 | =SUM(D8:J8) | =K8*1 | 40 | =SUM((L8:M8)/B8) | Optimal | |||||
| 4 to 6 | 3 | =C8+25 | =C9-C7 | =C9-C8 | 0 | =SUM(D9:J9) | =K9*1 | 40 | =SUM((L9:M9)/B9) | Go Back | ||||
| 6 to 6 | 1 | 25 | 0 | =SUM(D10:J10) | =K10*1 | 40 | =SUM((L10:M10)/B10) | Continue | ||||||
| 6to 7 | 2 | =C10+20 | =C11-C10 | 0 | =SUM(D11:J11) | =K11*1 | 40 | =SUM((L11:M11)/B11) | Continue | |||||
| 6 to 8 | 3 | =C11+5 | =C12-C10 | =C12-C11 | 0 | =SUM(D12:J12) | =K12*1 | 40 | =SUM((L12:M12)/B12) | Optimal | ||||
| 6 to 9 | 4 | =C12+10 | =C13--C10 | =C13-C11 | =C13-C12 | =C13-C13 | =SUM(D13:J13) | =K13*1 | 40 | =SUM((L13:M13)/B13) | Go Back | |||
| 9 to 9 | 1 | 10 | 0 | =SUM(D14:J14) | =K14*1 | 40 | =SUM((L14:M14)/B14) | Continue | ||||||
| 9 to 10 | 2 | =C14+20 | =C15-C14 | 0 | =SUM(D15:J15) | =K15*1 | 40 | =SUM((L15:M15)/B15) | Continue | |||||
| 9 to 11 | 3 | =C15+5 | =C16-C14 | =C16-C15 | 0 | =SUM(D16:J16) | =K16*1 | 40 | =SUM((L16:M16)/B16) | Optimal | ||||
| 9 to 12 | 4 | =C16+12 | =C17--C14 | =C17-C15 | =C17-C16 | =C17-C17 | =SUM(D17:J17) | =K17*1 | 40 | =SUM((L17:M17)/B17) | Go Back | |||
| 12 to 12 | 1 | 12 | 0 | =SUM(D18:J18) | =K18*1 | 40 | =SUM((L18:M18)/B18) | Optimal |
b)
Interpretation: Determine the order policy for the item based on Least Unit Cost method.
Concept Introduction: Least unit cost produced the demand of the present periods based on the trial basis, yield the future periods. The method is calculated by adding the setup cost and carrying inventory cost and finally find the smallest cost per unit.
b)
Answer to Problem 14P
The order policy according to LUC method is 26 units in period-1, 40 units in Period-5, 25 units in Period-7, 35 units in period-9 and 12 units in period-12.
Explanation of Solution
Given information: The anticipated demand for a component VC is as follows:
| Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Demand | 42 | 42 | 32 | 12 | 26 | 112 | 45 | 14 | 76 | 38 |
Holding cost (h) is $0.60 per period, set up cost (K) is $132
The order policy (lot-size) of the item under Least Unit Cost(LUC)method can be calculated as follows:
LUC divides the average cost per period C (T) by the total number of units demanded. Hence
And
And general equation is
Once
Starting from period 1
Stop the process since
Starting in period 5:
Stop the process since
Starting in period 7:
Stop the process since
Starting in period 9:
Stop the process since
The same is explained with the help of a table shown below:
Table 2: Order Quantity using Least-Unit-Cost Method:
| Demand | Months | Q | 11 | 12 | 13 | 14 | 15 | Total inventory | Holding Cost | Ordering Cost | Total Cost | Per Period Cost | Decision |
| 2 | 1 to 1 | =A3 | 0 | =SUM(D3:H3) | =L3*1 | 40 | =K3+J3 | =L3/C3 | Continue | ||||
| 12 | 1 to 2 | =C3+A4 | =C4-C3 | 0 | =SUM(D4:H4) | =L4*1 | 40 | =K4+J4 | =L4/C4 | Continue | |||
| 4 | 1 to 3 | =C4+A5 | =C5-C3 | =C5-C4 | 0 | =SUM(D5:H5) | =L5*1 | 40 | =K5+J5 | =L5/C5 | Continue | ||
| 8 | 1 to 4 | =C5+A6 | =C6-C3 | =C6-C4 | =C6-C5 | =C6-C6 | =SUM(D6:H6) | =L6*1 | 40 | =K6+J6 | =L6/C6 | Optimal | |
| 15 | 1 to 5 | =C6+A7 | =C7-C3 | =C7-C4 | =C7-C5 | =C7-C6 | =C7-C7 | =SUM(D7:H7) | =L7*1 | =K7+J7 | =L7/C7 | Go Back | |
| 15 | 5 to 5 | =A8 | 0 | =SUM(D8:H8) | =L8*1 | 40 | =K8+J8 | =L8/C8 | Continue | ||||
| 25 | 5 to 6 | =C8+A9 | =C9-C8 | 0 | =SUM(D9:H9) | =L9*1 | 40 | =K9+J8 | =L9/C9 | Optimal | |||
| 20 | 5 to 7 | =C9+A11 | =C10-C8 | =C10-C9 | =C10-C10 | =SUM(D10:H10) | =L10*1 | 40 | =K10+J10 | =L10/C10 | Go Back | ||
| 20 | 7 to 7 | =A11 | 0 | =SUM(D11:H11) | =L11*1 | 40 | =K11+J11 | =L11/C11 | Continue | ||||
| 5 | 7 to 8 | =C11+A12 | =C12-C11 | 0 | =SUM(D12:H12) | =L12*1 | 40 | =K12+J12 | =L12/C12 | Optimal | |||
| 10 | 7 to 9 | =C12+A13 | =C13-C11 | =C13-C12 | =C13-C13 | =SUM(D13:H13) | =L13*1 | 40 | =K13+J13 | =L13/C13 | Go Back | ||
| 10 | 8 to 8 | =A14 | 0 | =SUM(D14:H14) | =L14*1 | 40 | =K14+J14 | =L14/C14 | Continue | ||||
| 20 | 8 to 9 | =C14+A15 | =C15-C14 | 0 | =SUM(D15:H15) | =L15*1 | 40 | =K15+J15 | =L15/C15 | Continue | |||
| 5 | 8 to 10 | =C15+A16 | =C16-C14 | =C16-C15 | 0 | =SUM(D16:H16) | =L16*1 | 40 | =K16+J16 | =L16/C16 | Optimal | ||
| 12 | 8 to 11 | =C16+A17 | =C17-C14 | =C17-C15 | -C17-C16 | =C17-C17 | =SUM(D17:H17) | =L17*1 | 40 | =K17+J17 | =L17/C17 | Go Back | |
| 12 | 12 to 12 | 12 | 0 | =SUM(D18:H18) | =L18*1 | 40 | =K18+J18 | ==L18/C18 | Optimal |
c)
Interpretation: Determine the order policy for the item based on Part Period Balancing method.
Concept Introduction: Part Period Balancing method is the lot-size method which use the starting and ending of the process function to consider the multiple periods to modifying the calculation based on the least total cost.
c)
Answer to Problem 14P
The order policy according to part period balancing method is 26 units in period-1, 60 units in Period-5, 35 units in Period-8, 17 units in period-11.
Explanation of Solution
Given information: The anticipated demand for a component VC is as follows:
| Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Demand | 42 | 42 | 32 | 12 | 26 | 112 | 45 | 14 | 76 | 38 |
Holding cost (h) is $0.60 per period, set up cost (K) is $132
The order policy (lot size) according to part period balancing method can be calculated as follows:
In this method the order horizon that equates holding and setup cost over that period has to be calculated as follows:
Starting from Period 1
d)
Interpretation: Determine the three lot-sizing method resulted in the lowest cost for the 12 periods.
Concept Introduction: Lot size is determined the quantity order during the production time. The size of the lot may be dynamic or fixed.ERP (Enterprise Resource Planning) is the inbuilt multiple heuristic methods to determine the size of the lot to the production unit.
d)
Answer to Problem 14P
The Silver-meal method is giving lowest cost.
Explanation of Solution
Given information: The anticipated demand for a component VC is as follows:
| Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Demand | 42 | 42 | 32 | 12 | 26 | 112 | 45 | 14 | 76 | 38 |
Holding cost (h) is $0.60 per period, set up cost (K) is $132
Calculate the total cost of the ordering for the three methods as shown below:
| Silver-meal | Least Unit Cost | Part Period Balancing | |
| Holding Cost | =20+15+30+30 | =44+25+5+30 | =44+65+50+12 |
| Setup Cost | =40*5 | =40*5 | =4*40 |
| Total Cost | =B3+B2 | =C3+C2 | =D3+D2 |
| Silver-meal | Least Unit Cost | Part Period Balancing | |
| Holding Cost | 95 | 104 | 171 |
| Setup Cost | 200 | 200 | 160 |
| Total Cost | 295 | 304 | 331 |
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Chapter 8 Solutions
EBK PRODUCTION AND OPERATIONS ANALYSIS
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