Chapter 5, Problem 44AP

a

Summary Introduction

Interpretation:Value of Q and R used to control the inventory of white dress shirts is to be determined.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 44AP

Order quantity (Q) is 240 units and Reorder point (R) is also 240 units.

Explanation of Solution

Given information:

Cost of each shirt = $6

Selling cost of each shirt = $15

Monthly demand = 120

Standard deviation of monthly demand = 32

Reorder point = Two months of demand,

Order quantity = two months of demand

Q refers to the order quantity and R refers to the reorder point.

Order quantity is two months of demand i.e. 120 units $\times$ 2 = 240 units.

The proprietor orders when stock falls below the two-month’s supply stock. The two month’s supply stock means 120 units $\times$ 2 = 240 units.

Order quantity (Q) is 240 units and Reorder point (R) is also 240 units.

b

Summary Introduction

Interpretation:Fill rate achieved with current policy is to be determined.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 44AP

The fill rate of 99.9% is being achieved in current policy.

Explanation of Solution

Given information:

Cost of each shirt = $6

Selling cost of each shirt = $15

Monthly demand = 120

Standard deviation of monthly demand = 32

Reorder point = Two months of demand,

Order quantity = two months of demand

Type 2 service:

  $\frac{n\left(R\right)}{Q}=1-\beta$

  $n\left(R\right)=EOQ\left(1-\beta \right)$

  $L\left(Z\right)=\left(1-\beta \right)Q/\sigma$

  $R=\sigma z+\mu$

  $\beta =F\left(R\right)$

  $\text{β = proportion}$

  $\text{μ = mean}$

  $\text{σ = standard}\text{deviation}$

Lead time given = 3 weeks = $\frac{\text{3}}{\text{4}}\text{month}$

Mean, $\mu =\frac{3}{4}\times 120=90$

Standard deviation, $\sigma =\sqrt{\frac{3}{4}}\times 32=27.71$

  $n\left(R\right)=\sigma L\left(z\right)$

  $z=\frac{R-\mu }{\sigma }=\frac{240-90}{27.71}=5.41$

  $n\left(R\right)=\left(27.71\right)\left(<0.00001\right)$

  $n\left(R\right)=0.000099$

Therefore, the fill rate of 99.9% is being achieved in current policy.

c

Summary Introduction

Interpretation:Optimal value of Q and R is to be determined based on the 99% fill rate criteria.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 44AP

Optimal value of Q and R is (Q, R) = (455,107)

Explanation of Solution

Given information:

Cost of each shirt = $6

Selling cost of each shirt = $15

Monthly demand = 120

Standard deviation of monthly demand = 32

Reorder point = Two months of demand,

Order quantity = two months of demand

  $\text{Mean,μ=90}$

  $\text{Standard}\text{deviation,σ=27}\text{.71}$

  $\text{Mean}\text{weekly}\text{demand,λ=120×12=1440}$

  $\text{Holding}\text{cost, h = \$0}\text{.20×6 =1}\text{.2}$

  $\text{Setup}\text{cost,K=80}$

Iteration 1:

  $Q_o=\sqrt{\frac{2\kappa \lambda }{h}}$

  $Q_o=\sqrt{\frac{2\times 80\times 1440}{1.2}}$

  $=438.17$

  $n\left(R_o\right)=\left(1-\beta \right)Q$

  $=\left(0.01\right)438.17$

  $=4.38$

  $L\left(z_o\right)=\frac{n\left(R_o\right)}{\sigma }=\frac{4.38}{27.71}=0.1580$

From $L\left(z_o\right)$ , calculate $z_o=0.64,F\left(z_o\right)=0.261$

Iteration 2:

  $Q_1=\frac{n\left(R_o\right)}{F\left(z_o\right)}+\sqrt{{\left(Q_o\right)}^2+{\left(\frac{n\left(R_o\right)}{F\left(z_o\right)}\right)}^2}$

  $Q_1=\frac{4.38}{0.261}+\sqrt{{\left(438\right)}^2+{\left(\frac{4.38}{0.261}\right)}^2}$

  $Q_1=455.10$

  $n\left(R_o\right)=\left(1-\beta \right)Q$

  $=\left(0.01\right)455.10$

  $=4.55$

  $L\left(z_o\right)=\frac{n\left(R_o\right)}{\sigma }=\frac{4.55}{27.71}=0.1642$

From $L\left(z_1\right)$ , calculate $z_1=0.61,F\left(z_o\right)=0.268$

  $R_1=\sigma z+\mu =\left(27.71\right)\left(0.61\right)+90=107$

Iteration 3:

  $Q_2=\frac{n\left(R_1\right)}{F\left(z_1\right)}+\sqrt{{\left(Q_1\right)}^2+{\left(\frac{n\left(R_1\right)}{F\left(z_1\right)}\right)}^2}$

  $Q_2=\frac{4.55}{0.268}+\sqrt{{\left(455\right)}^2+{\left(\frac{4.55}{0.268}\right)}^2}$

  $Q_2=455.10$

Since the Q value is repeating, we terminate further iteration

Hence, (Q, R) = (455,107)

d

Summary Introduction

Interpretation:Difference in average annual holding and set up costs is to be determined.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 44AP

Difference in average annual holding and set up costs is $257.

Explanation of Solution

Given information:

Cost of each shirt = $6

Selling cost of each shirt = $15

Monthly demand = 120

Standard deviation of monthly demand = 32

Reorder point = Two months of demand,

Order quantity = two months of demand

Considering (Q, R) from policy (b)

(Q, R) = (240,240)

  $\text{Annual}\text{holding}\text{and}\text{set-upcosts=h}\left[\frac{\text{Q}}{\text{2}}\text{+}\left(\text{R-μ}\right)\right]\text{+}\frac{\text{κλ}}{\text{Q}}$

  $=1.2\left[\frac{240}{2}+\left(240-90\right)\right]+\frac{80\times 1440}{240}$

  $=804$

Therefore, the annual holding cost and set up cost are $804.

Considering (Q, R) from policy (c)

(Q, R) = (455,107)

  $\text{Annual}\text{holding}\text{and}\text{set-upcosts=h}\left[\frac{\text{Q}}{\text{2}}\text{+}\left(\text{R-μ}\right)\right]\text{+}\frac{\text{κλ}}{\text{Q}}$

  $\begin{array}{l}=1.2\left[\frac{455}{2}+\left(107-90\right)\right]+\frac{80\times 1440}{455}\\ =546.6\end{array}$

Therefore, the annual holding cost and set up cost are $547.

Saving on total = $804-$547 = $257

e

Summary Introduction

Interpretation:Time required paying for $25000 inventory control system is to be calculated.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 44AP

Time required in paying $25,000 of inventory control is 4.86 yearsto the system.

Explanation of Solution

Given information:

Cost of each shirt = $6

Selling cost of each shirt = $15

Monthly demand = 120

Standard deviation of monthly demand = 32

Reorder point = Two months of demand,

Order quantity = two months of demand

Time required in paying $25,000 of inventory control is as shown below:

Assuming 20% of the annual interest rate to the estimate (Q, R)

Then the savings would be estimated as (20)($257) = $5,140

Here if the time value of money is ignored for the given inventory control of $25,000

Time required = 25,000/5140 = 4.86 years

Therefore, time required in paying $25,000 of inventory control is 4.86 yearsto the system.