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Variance and Standard Daviation Part 2-Basic Mathematics-Assignment Solution, Exercises of Mathematics

This is solution to assignment of Basic Mathematics course. This was submitted to Karunashankar Sidhu at Institute of Mathematical Sciences. It includes: Average, Cost, Derivative, variance, Standard, deviation, Eliminating, Optimum

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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1) If in the Wilson’s Model, the set up cost instead of being fixed, is equal to
C2+BQ,where B is the set up cost per item produced, then show that there is no
change in the optimum order quantity produced due to change in the set up cost.
Solution:
So in this case;
Total Cost for one cycle = C2+BQ+(½) QtC3
Average Cost for one unit of time = C(Q) = C2/t + BQ/t +(½)QC3
Eliminating ‘t’, by using ‘t= Q/D’
C(Q) = C2D/Q +BD +(½)QC3
For minimum value, putting the 1st derivative of above w.r.t ‘Q’ equal to zero.
C’(Q) = - C2D/Q2 +(½)C3 = 0
C2D/Q2=(½)C3
Q2/ C2D =2/C3
Q2=2C2D/C3
2
3
2C D
QC
2) An item is produced at the rate of 50 items per day. The demand occurs at the rate
25 items per day. If the set up cost is Rs.100 per set up and the holding cost is
Rs.0.01 per unit of item per day, find the economic lot size for one run, assuming
that the shortage are not permitted.
Solution:
Here
R = 50/Day, D = 25/Day, C3 = Rs. 0.01/Day, C2 = Rs. 100/run
So
Economic Lot Size = Q* =
2
3
22 100 25 50 1000
0.01 50 25
CD Ritems
C R D



t* = Q/D=1000/25 = 40 Days
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1) If in the Wilson’s Model, the set up cost instead of being fixed, is equal to

C 2 + BQ ,where B is the set up cost per item produced, then show that there is no

change in the optimum order quantity produced due to change in the set up cost.

Solution:

So in this case;

Total Cost for one cycle = C 2 + BQ +(½) QtC 3

 Average Cost for one unit of time = C ( Q ) = C 2 / t + BQ / t +(½) QC 3

Eliminating ‘ t ’, by using ‘ t = Q /D’

 C ( Q ) = C 2 D / Q + BD +(½) QC 3

For minimum value, putting the 1 st derivative of above w.r.t ‘ Q ’ equal to zero.

C ’( Q ) = - C 2 D / Q^2 +(½) C 3 = 0

 C 2 D / Q^2 =(½) C 3

 Q^2 / C 2 D =2/ C 3

 Q^2 =2 C 2 D / C 3

^2

3

2C D

Q

C

2) An item is produced at the rate of 50 items per day. The demand occurs at the rate

25 items per day. If the set up cost is Rs.100 per set up and the holding cost is

Rs.0.01 per unit of item per day, find the economic lot size for one run, assuming

that the shortage are not permitted.

Solution:

Here

R = 50/Day, D = 25/Day, C 3 = Rs. 0.01/Day, C 2 = Rs. 100/run

So

Economic Lot Size = Q* =^2

3

C D R

items C R D

t* = Q / D= 1000/25 = 40 Days

3) The demand of an item is uniform at a rate of 25 units per month. The fixed cost

is Rs.15 each time a production is made. The production cost is Rs.1/ item, and

the inventory carrying cost is Rs. 0.30/item/month. If the shortage cost is

Rs.1.50/item/month, determine how often to make a production run and of what

size it should be?

Solution:

In this example the set up cost is variable i-e C 3 + BQ , where B is the production cost per

item. Such a cost is considered in Question-01. Here also it can be shown that the optimal

value of Q remains unchanged. Only the Total cost per unit of time will become

Here

C 1 = Rs. 0.30/item/month

C 2 = Rs. 1.50/item/month

C 3 = Rs.15/set up(fixed)

B =Rs. 1.00/item, R = 25/month

Considering one month as one unit of time optimal value,

  • (^2 3 )

3 4

C C C D

Q items C C

And the optimal time = Q */ D = 54/25 = 2.16 months.

4) Group the items given below into an ABC classification.

Solution:

Item no. Annual Usage Accumulated Usage in Rs.

Cumulative Usage Cumulative % of items catagory 9 95000 95000 28.96 10

20

3 75000 17000 51.82 A

B

C

Item no 3 4 2 1 7 8 6 5 9 10

Annual usage

in x10

3 Rs.