Chapter 8.2, Problem 14P

a)

Summary Introduction

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.

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: $r_1=6-4=2,r_2=12,r_3=4,r_4=8,r_5=15,r_6=25,r_7=20,\mathrm{....},r_{12}=12+8=20$

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 $C\left(1\right)=K$

And $C\left(2\right)=\left(K+h{r}_2\right)/2$

  $C\left(3\right)=\left(K+h{r}_2+2h{r}_3\right)/3$

And general equation is $C\left(j\right)=\left(K+h{r}_2+2h{r}_3+\mathrm{...}+\left(j-1\right)h{r}_j\right)/j$

Once $C\left(j\right)\succ C\left(j-1\right),$ stop the function and set $y=r_1+r_2+r_3+\mathrm{....}{r}_{j-1}$

Now, calculate the order policy using the above formula as follows:

Starting in period 1:

  $C\left(1\right)=40$

  $C\left(2\right)=\left(40+12\right)/2=26$

  $C\left(3\right)=\frac{\left[40+12+\left(2\right)\left(4\right)\right]}{3}=20$

  $C\left(4\right)=\frac{\left[60+\left(3\right)\left(8\right)\right]}{4}=21$

Stop the process since $C\left(4\right)\succ C\left(3\right),$

  $y_1=r_1+r_2+r_3$

  $y_1=2+12+8=22$

Starting in period 4:

  $C\left(1\right)=40$

  $C\left(2\right)=\frac{\left(40+15\right)}{2}=27.5$

  $C\left(3\right)=\frac{\left[40+15+\left(2\right)\left(25\right)\right]}{3}=35$

Stop the process since $C\left(3\right)\succ C\left(2\right),$

  $y_3=r_4+r_5$

  $y_3=8+15=23$

Starting in period 6:

  $C\left(1\right)=40$

  $C\left(2\right)=\frac{\left(40+20\right)}{2}=30$

  $C\left(3\right)=\frac{\left[40+20+\left(2\right)\left(5\right)\right]}{3}=23.33$

  $C\left(4\right)=\frac{\left[40+30+\left(10\right)\left(3\right)\right]}{4}=25$

Stop the process since $C\left(4\right)\succ C\left(3\right),$

  $y_6=r_6+r_7+r_8$

  $y_6=25+20+5=50$

Starting in period 9:

  $C\left(1\right)=40$

  $C\left(2\right)=\frac{\left(40+20\right)}{2}=30$

  $C\left(3\right)=\frac{\left[40+20+\left(2\right)\left(5\right)\right]}{3}=23.33$

  $C\left(4\right)=\frac{\left[40+30+\left(12\right)\left(3\right)\right]}{4}=26.5$

Stop the process since $C\left(4\right)\succ C\left(3\right),$

  $y_9=r_9+r_{10}+r_{11}$

  $y_9=10+20+5=35$

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)

Summary Introduction

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.

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 $C\left(1\right)=K/r_1$

And $C\left(2\right)=\left(K+h{r}_2\right)+\left(r_1+r_2\right),$

And general equation is $C\left(j\right)=\left(K+h{r}_2+2h{r}_3+\mathrm{...}+\left(j-1\right)h{r}_j\right)/\left(r_1+r_2\mathrm{...}+r_j\right)$

Once $C\left(j\right)\succ C\left(j-1\right),$ stop the function and set $y=r_1+r_2+r_3+\mathrm{....}{r}_{j-1}$

Starting from period 1

  $C\left(1\right)=40/2=20$

  $C\left(2\right)=\left(40+12\right)/\left(2+12\right)=3.71$

  $C\left(3\right)=\frac{\left[40+12+\left(2\right)\left(4\right)\right]}{2+12+4}=3.33$

  $C\left(4\right)=\frac{\left[60+\left(3\right)\left(8\right)\right]}{2+12+4+8}=3.23$

  $C\left(5\right)=\frac{\left[84+\left(4\right)\left(15\right)\right]}{2+12+4+8+15}=3.51$

Stop the process since $C\left(5\right)\succ C\left(4\right),$

  $y_1=r_1+r_2+r_3+r_4$

  $y_1=2+12+4+8=26$

Starting in period 5:

  $C\left(1\right)=40/15=2.67$

  $C\left(2\right)=\frac{\left[\left(40+25\right)\right]}{\left(15+25\right)}=1.625$

  $C\left(3\right)=\frac{\left[40+25+\left(2\right)\left(20\right)\right]}{\left(15+25+20\right)}=1.75$

Stop the process since $C\left(3\right)\succ C\left(2\right),$

  $y_5=r_5+r_6$

  $y_5=15+25=140$

Starting in period 7:

  $C\left(1\right)=40/20=2$

  $C\left(2\right)=\frac{\left[\left(40+5\right)\right]}{\left(20+5\right)}=1.8$

  $C\left(3\right)=\frac{\left[40+5+\left(2\right)\left(10\right)\right]}{\left(20+5+10\right)}=1.85$

Stop the process since $C\left(3\right)\succ C\left(2\right),$

  $y_7=r_7+r_8$

  $y_7=20+5=25$

Starting in period 9:

  $C\left(1\right)=40/10=4$

  $C\left(2\right)=\frac{\left[\left(40+20\right)\right]}{\left(10+20\right)}=2$

  $C\left(3\right)=\frac{\left[40+20+\left(2\right)\left(5\right)\right]}{\left(10+20+5\right)}=2$

  $C\left(4\right)=\frac{\left[70+\left(3\right)\left(20\right)\right]}{\left(10+20+5+20\right)}=2.36$

Stop the process since $C\left(4\right)\succ C\left(3\right),$

  $y_9=r_9+r_{10}+r_{11}$

  $y_9=10+20+5=35$

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)

Summary Introduction

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.

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:

  $r=\left(2,12,4,8,15,25,20,5,10,20,5,20\right)$

Starting from Period 1

  

d)

Summary Introduction

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.

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