Question

Exercise 2: The time phased net requirements for the base assembly are given as follows. Setup cost is $300, and the holding cost is $0.50 per assembly per week. Week 2 3 4 5 6 Req 250 150 140 400 200 200 a. Determine the lot sizes using Silver Meal heuristic. b. Determine the lot size using least unit cost heuristic.

please make sure the answer is correct 100%

Answer 1

(a). Lot Sizing based on Silver Meal heuristic:

The purpose of Silver-Meal Heuristic is to calculate production quantities to meet the requirement of operations at minimum cost.

Under this, the Forward method is used calculate the average cost per period.

Formula for average cost :

C(j) = [k+hr2+2hr3+...+(j-1)hrj]/j, where

C(j) = average holding cost and setup cost per period

k = order or setup cost

h = holding cost

r = demand

I am explaining the steps to be used as follows:

Step 1: Start the calculation from period 1 to next period :

C(1) = k
C(2) = (k+hr2)/2
C(3) = (k+hr2+2hr3)/3 and so on.
Step 2: Stop the calculation when C(j)>=C(j-1).

Step 3: Find Lot size (1) = r1+r2+....+rj-1

Step 4: Repeat steps 1-3 for finding a lot size for the next period starting from the period in which you stopped earlier

using Silver-Meal Heuristic:

Using the values given in the question, we have

k=300$

r= (300,200,150,400,200,200)

h = 0.50 $

Starting in week 1 :

C(1) = 300
C(2) = (300+0.5*200)/2= 200
C(3) = (300+0.5*200+2*0.5*150)/3 = 183.33
C(4) = (300+0.5*200+2*0.5*150+3*0.5*400)/4 = 287.5
No further calculation as C(4)>C(3).

So, Lot size(1) = r1+r2+r3 = 300+200+150 = 650 units

The next step is to start over in week 4 again and the calculations are :

C(1) = 300 (for period 4 )
C(2) = (300+0.5*200)/2= 200 (for period 5 )
C(3) = (300+0.5*200+2*0.5*200)/3 = 200 ( for period 6 ) [Stop as C(4)>= C(3)]
So, Lot size(4) = r4+r5 = 400+200 = 600 units

Because week 6 is the last period, there is no need to start the process again.

And hence Lot size(6) = r6 = 200 units

$\therefore$  the lot sizes are = (650,0,0,600,0,200)

(b). Lot Sizing based on Least unit Cost heuristic:

The purpose of Least unit Heuristic is to minimize cost per unit of demand.

Formula for average cost :

C(j) = [k+hr2+2hr3+...+(j-1)hrj]/(r1+r2+....+rj), where

C(j) = average holding cost and setup cost per period

k = order or setup cost

h = holding cost

r = demand

I am explaining the steps to be used as follows:

Step 1: Start the calculation from period 1 to next period :

C(1) = k/r1
C(2) = (k+hr2)/r1+r2
C(3) = (k+hr2+2hr3)/r1+r2+r3, and so on.
Step 2: Stop the calculation when C(j)>=C(j-1).

Step 3: Find Lot size (1) = r1+r2+....+rj-1

Step 4: Repeat steps 1-3 for finding a lot size for the next period starting from the period in which you stopped earlier

using Least Unit Cost Heuristic.

Using the values given in the question, we have

k=300$

r= (300,200,150,400,200,200)

h = 0.50 $

Starting in week 1 :

C(1) = 300/300 = 1
C(2) = (300+0.5*200)/(300+200)= 0.8
C(3) = (300+0.5*200+2*0.5*150)/(300+200+150) = 0.846
No further calculation as C(3)>C(2).

So, Lot size(1) = r1+r2 = 300+200 = 500 units

The next step is to start over in week 3 again and the calculations are :

C(1) = 300/150=2 (for period 3 )
C(2) = (300+0.5*400)/(150+400)= 0.91 (for period 4 )
C(3) = (300+0.5*400+2*0.5*200)/(150+400+200) = 0.93 ( for period 5 ) [Stop as C(3)> C(2)]
So, Lot size(3) = r3+r4 = 150+400 = 550 units

The next step is to start over in week 5 again and the calculations are :

C(1) = 300/200=1.5 (for period 5 )
C(2) = (300+0.5*200)/(200+200)= 1 (for period 6 )
So, Lot size(5) = r5+r6 = 200+200 = 400 units

Note: As week 6 is the last week, no need to repeat the steps.

Also, the goal under “least unit cost” is to minimize cost per unit, and hence, Lot size (5) is calculated by adding the demand for week 5 and week 6 covering the demand for both.

$\therefore$ the lot sizes are = (500,0,550,0,400,0)