IE 361 HOMEWORK 1
In this homework, we have dealt with an inventory problem of an electronic store particularly their mobile phone stocks. We have worked on two alternative scenarios which selling with backordering strategy and without it.
Dt=Demand for mobile phones on week t.( Poisson Distribution) Dt={0,1….m}
Xt=Inventory level at the end of week t.
S=Maximum amount of inventory level. s=Minimum inventory level that does not required new order.(if inventory level strictly below of s , company gives order and complete their inventory level to S at the beginning of new week.) Xt <s company buys Ot=S-Xt Order of company(Ot) = Xt≥s company buys nothing. t=0,1……n CASE A In this case, company sell mobile phones with backordering strategy. i) State= the inventory level at the end of week t. S={(s-m),……………,S } At the beginning of the week the lowest inventory level can be s, so that the minimum inventory level can be s-m which m is the maximum number of demand. ii) Xt <s Xt+1=S-Dt+1 Xt≥s Xt+1= Xt -Dt+1

iii) Xt=i Xt+1=j So; i<s j=S-Dt+1 Dt+1=S-j i≥s j= i -Dt+1 Dt+1=i-j Due to the reason that the demand function is distributed with poisson distribution. We use the probability function of Poisson which is; P(X=a)= λa*eλ/a! Therefore, we use adjust this equation according to our demand function. P(Dt+1=S-j)= λ(S-j) *e-λ ÷ (S-j)! P(Dt+1=i-j)= λ(i-j) *e-λ ÷ (i-j)! Let X0=4, λ=2,m=6, s=2 and S=5 iv) According to the probability functions that are stated above we calculated the probabilities of state transitions and inserted these values in to the matrix cells which is given below. Table 1. case a first probability matrix. The we realized that sum of the values in a row doesn’t equal to 1 due to the approximations in calculator. However it should equal to one due to the function that is given below; P(Dt=d) = P(y=d) ÷ P(y≤m) So we divide the each value to the sum of the rows which is equal to 0.995 . Table2. The adjusted probability matrix of case A. By this way we can reach the exact results. v) We calculate the power of matrices by using MPower of Excel. Table3. P2 matrix. Table4. P5 matrix.

Table5. P10 matrix.

Table6. P100 matrix.

vi) X0=4 is given, first week there should not be any order then end of the second week the first order should be placed. Therefore; X0=4 , X1=a , X2=b Where 2≤a≤4 , Due to the reason that should not be lower than s=2 in order not to place order. Logically, it cannot be higher than 4 which is previous inventory level. Where -4≤b<2 , because first order should be placed at the end of week 2. P(X2=b,X1=a X0=4 )=P24ab=P4aPab = * = However we need the sum of these probabilities, and it is equal to 0.43999. vii) This markov chain is ergodic . Due to the reason that all the states communicate with each other, and they are aperiodic and recurrent. viii) Table7. steady state distributions of matrix. ix) In order to find long-run expected net inventory level ,we have multiplied the steady state probabilities with the number of related inventory level.*
= 1.890118 x) In order to find the expected number of week between two replenishments, we should separate the probabilities whether they are required replenishment or not. We called the probabilities which need order as R.

Table8. replenishment table.

Table9. steady state probability of matrix above.

We have found the mRR, which is equal to 1/πRR= 1/0.39854= 2.509151 xi) According to the value that we have found in the x part , the expected number of two successive replenishment is 2.509151. Therefore, in a year we have 52 weeks , if we divide 52 to 2.509151 we can easily find the number of orders which is equal to 20.72414135 . xii) In this part , we have found the backordering occasions per week for case a. Due to the reason that we are looking for occasions, we only sum the all possibilities which are required backorders.

Table10. the probabilities which are required backorders.
The expected number of backordering occasions are 0,103795096. xiii) In order to find this part, we have multiply with the probabilities with the number of cell phones which are required.
If the inventory level is below 0, we complete it to 5. For example for inventory level -4, we backorder 9 items which is equal to S-Xt . The total backorder is;
*= 0,157369

xiv) It is not necessary to compute this for case a.

xv) We have computed this part according to the demand probability function. Di is the probability of i unit products.

i=06(i*Di)=1.97484~2 xvi) Selling Price: 14* 1.97484=27.64776
Weekly Inventory Holding cost= Firstly we should calculate it by multiplying the inventory levels(0,1,2,3,4,5) with the steady state probabilities. i=05(i*Xi)= 2,047487

The inventory holding cost will be 2*2.047487=4.094974 .

Fixed cost of placing an order= We have found the number of order per year in part xi . The number of orders per week is 0.398541 .
Therefore, the fixed cost of placing an order is 0.398541*10= 3.985411. Variable cost of purchasing a phone= We have use the value in part xv. This variable cost is 4*1.97484=7.89936 . Fixed cost for backordering occasion= 5*0,103795=0.518975 Backordering cost = 3*0,157369=0.472107 The total profit = Total Sales- Total Cost The total profit =27.64776-16.970828= 10.676931 xvii)

CASE B In this case, company sells mobile phones without backordering strategy, accepts to face with possible lost sales. i) State= the inventory level at the end of week t. S={0,1,……………,S } There is no backordering in this case, so the minimum inventory level is “0”, and maximum is given “S”. ii) Xt <s Xt+1=max (S - Dt+1 , 0) Xt≥s Xt+1= max (Xt - Dt+1 , 0)

iii) Xt=i Xt+1=j So; i<s j=max (S - Dt+1 , 0) Dt+1=S-j i≥s j=max (i - Dt+1 , 0) Dt+1=i-j Due to the reason that the demand function is distributed with poisson distribution. We use the probability function of Poisson which is; P(X=a)= λa*eλ/a! 10.092925P(Dt+1=S-j)= λ(S-j) *e-λ ÷ (S-j)! P(Dt+1=i-j)= λ(i-j) *e-λ ÷ (i-j)! Let X0=4, λ=2,m=6, s=2 and S=5 iv) According to the probability functions that are stated above we calculated the probabilities of state transitions and inserted these values in to the matrix cells which is given below. Table 11. Case b first probability matrix. The we realized that sum of the values in a row doesn’t equal to 1 due to the approximations in calculator. However it should equal to one due to the function that is given below; P(Dt=d) = P(y=d) ÷ P(y≤m) So we divide the each value to the sum of the rows which is equal to 0.995 . Table12. The adjusted probability matrix of case B . By this way we can reach the exact results. v) We calculate the power of matrices by using MPower of Excel. Table13. P2 matrix Table14.P5 matrix

Table15.P10 matrix Table16.P100 matrix vi) X0=4 is given, first week there should not be any order then end of the second week the first order should be placed. Therefore; X0=4 , X1=a , X2=b Where 2≤a≤4 , Due to the reason that should not be lower than s=2 in order not to place order. Logically, it cannot be higher than 4 which is previous inventory level. Where 0≤ b<2 , because first order should be placed at the end of week 2. P(X2=b,X1=a X0=4 )=P24ab=P4aPab = * = However we need the sum of these probabilities, and it is equal to 0.439998. vii) This markov chain is ergodic . Due to the reason that all the states communicate with each other, and they are aperiodic and recurrent. viii) Table17. . Steady state distribution matrix. ix) In order to find long-run expected net inventory level ,we have multiplied the steady state probabilities with the number of related inventory level. * = 2,047487 x) In order to find the expected number of week between two replenishments, we should separate the probabilities whether they are required replenishment or not. We called the probabilities which need order as R. Table18. replenishment table.

Table19. . steady state probability of matrix above.

We have found the mRR, which is equal to 1/πRR= 1/0.39854= 2.509151 xi)
According to the value that we have found in the x part , the expected number of two successive replenishment is 2.509151. Therefore, in a year we have 52 weeks , if we divide 52 to 2.509151 we can easily find the number of orders which is equal to 20.72414135 .

xii) Logically, if we deal with lost sales in case B (sell without backordering), we can use the probability matrix that we found in case A. In this scenario company doesn’t meet the demand that exceeds inventory level. Lost sale occasions equals to backordering occasions found in case A.

Table20. the probabilities which are cause lost sale.
The expected number of lost sales occasions are sum of above probabilities; 0,103795096.

xiii) It is not necessary to compute this for case b.

xiv) We had calculated number of backorders in case A xiii which is 0.157369. In case B, there is no backordering, so the number of backorders in case A, will be the lost sales for company in case B. We can use the steady state matrix in case a.

* = 0.157369.

xv) Expected number of sold cell phones equals to expected demand minus lost sales.

Number of expected demand is i=06(i*Di)=1.97484 Number of expected total sales is 0.157369. (Found in part xiv )

Expected number of sold cell phones is 1.97484 – 0.157469 = 1.817371

xvi) Total sales : 14 * 1.817371 = 25.443194
Inventory cost: 2* 2,047487 = 4.094974
Fixed cost of placing order: 10 * ( 20.72414135 ÷ 52) = 3.985411
Variable cost: 1.817471 * 4 = 7.269884

Total profit = total sales – total cost = 10.092925

xvii) The main performance measure in company is profit.

In case A total profit per week is 10.676931 .

In case B total profit per week is 10.092925 .

It is seen that in case A total profit is larger than profit in case B for the company. So we can say that selling with backordering strategy is more profitable for company.