1. **Problem Statement:** We need to determine the optimal number of Solar Home Kits to purchase to maximize expected gross profit after advertising in (i) one newspaper and (ii) two newspapers. Then calculate the expected net profit for both cases. 2. **Given Data:** - Selling price per kit $= 245$ - Cost price per kit $= 105$ - Salvage price per unsold kit $= 70$ Demand and probabilities: | Demand | $P_1$ (One Newspaper) | $P_2$ (Two Newspapers) | |--------|-----------------------|-----------------------| | 105 | 0.25 | 0.1 | | 140 | 0.25 | 0.2 | | 175 | 0.2 | 0.3 | | 210 | 0.1 | 0.2 | | 245 | 0.2 | 0.2 | 3. **Formula for Expected Gross Profit:** For a purchase quantity $Q$ and demand $D$: $$\text{Profit} = \begin{cases} (Q - D) \times 70 + D \times (245 - 105), & \text{if } Q > D \\ Q \times (245 - 105), & \text{if } Q \leq D \end{cases}$$ Explanation: If demand $D$ is less than $Q$, unsold kits are sold at salvage price 70. If demand $D$ is greater or equal to $Q$, all $Q$ kits are sold at profit margin $245 - 105 = 140$. 4. **Calculate expected profit for each possible purchase quantity $Q$ from the demand set:** Possible $Q$ values: 105, 140, 175, 210, 245 **(i) One Newspaper:** Calculate $E[Profit] = \sum P_1(D) \times Profit(Q,D)$ for each $Q$. - For $Q=105$: - If $D \geq 105$, profit $= 105 \times 140 = 14700$ - Expected profit $= \sum P_1(D) \times 14700 = 14700$ (since all $D \geq 105$) - For $Q=140$: - If $D < 140$ (i.e., 105): profit $= (140-105) \times 70 + 105 \times 140 = 35 \times 70 + 14700 = 2450 + 14700 = 17150$ - If $D \geq 140$: profit $= 140 \times 140 = 19600$ - Expected profit $= 0.25 \times 17150 + (0.25+0.2+0.1+0.2) \times 19600 = 4287.5 + 0.75 \times 19600 = 4287.5 + 14700 = 18987.5$ - For $Q=175$: - $D=105,140$: profit $= (175-D) \times 70 + D \times 140$ - $D=105$: $70 \times 70 + 105 \times 140 = 4900 + 14700 = 19600$ - $D=140$: $35 \times 70 + 140 \times 140 = 2450 + 19600 = 22050$ - $D \geq 175$: profit $= 175 \times 140 = 24500$ - Expected profit $= 0.25 \times 19600 + 0.25 \times 22050 + (0.2+0.1+0.2) \times 24500 = 4900 + 5512.5 + 0.5 \times 24500 = 4900 + 5512.5 + 12250 = 22662.5$ - For $Q=210$: - $D=105,140,175$: profit $= (210-D) \times 70 + D \times 140$ - $D=105$: $105 \times 70 + 105 \times 140 = 7350 + 14700 = 22050$ - $D=140$: $70 \times 70 + 140 \times 140 = 4900 + 19600 = 24500$ - $D=175$: $35 \times 70 + 175 \times 140 = 2450 + 24500 = 26950$ - $D \geq 210$: profit $= 210 \times 140 = 29400$ - Expected profit $= 0.25 \times 22050 + 0.25 \times 24500 + 0.2 \times 26950 + (0.1+0.2) \times 29400 = 5512.5 + 6125 + 5390 + 0.3 \times 29400 = 5512.5 + 6125 + 5390 + 8820 = 25847.5$ - For $Q=245$: - $D=105,140,175,210$: profit $= (245-D) \times 70 + D \times 140$ - $D=105$: $140 \times 70 + 105 \times 140 = 9800 + 14700 = 24500$ - $D=140$: $105 \times 70 + 140 \times 140 = 7350 + 19600 = 26950$ - $D=175$: $70 \times 70 + 175 \times 140 = 4900 + 24500 = 29400$ - $D=210$: $35 \times 70 + 210 \times 140 = 2450 + 29400 = 31850$ - $D=245$: profit $= 245 \times 140 = 34300$ - Expected profit $= 0.25 \times 24500 + 0.25 \times 26950 + 0.2 \times 29400 + 0.1 \times 31850 + 0.2 \times 34300 = 6125 + 6737.5 + 5880 + 3185 + 6860 = 28787.5$ **Maximum expected profit for one newspaper is $28787.5$ at $Q=245$ kits.** 5. **(ii) Two Newspapers:** Repeat the same calculations with $P_2$: - $Q=105$: - Profit $= 105 \times 140 = 14700$ - Expected profit $= 14700$ - $Q=140$: - $D=105$: profit $= 35 \times 70 + 105 \times 140 = 17150$ - $D \geq 140$: profit $= 140 \times 140 = 19600$ - Expected profit $= 0.1 \times 17150 + 0.9 \times 19600 = 1715 + 17640 = 19355$ - $Q=175$: - $D=105$: $70 \times 70 + 105 \times 140 = 19600$ - $D=140$: $35 \times 70 + 140 \times 140 = 22050$ - $D \geq 175$: $175 \times 140 = 24500$ - Expected profit $= 0.1 \times 19600 + 0.2 \times 22050 + 0.7 \times 24500 = 1960 + 4410 + 17150 = 23520$ - $Q=210$: - $D=105$: $105 \times 70 + 105 \times 140 = 22050$ - $D=140$: $70 \times 70 + 140 \times 140 = 24500$ - $D=175$: $35 \times 70 + 175 \times 140 = 26950$ - $D \geq 210$: $210 \times 140 = 29400$ - Expected profit $= 0.1 \times 22050 + 0.2 \times 24500 + 0.3 \times 26950 + 0.4 \times 29400 = 2205 + 4900 + 8085 + 11760 = 26950$ - $Q=245$: - $D=105$: $140 \times 70 + 105 \times 140 = 24500$ - $D=140$: $105 \times 70 + 140 \times 140 = 26950$ - $D=175$: $70 \times 70 + 175 \times 140 = 29400$ - $D=210$: $35 \times 70 + 210 \times 140 = 31850$ - $D=245$: $245 \times 140 = 34300$ - Expected profit $= 0.1 \times 24500 + 0.2 \times 26950 + 0.3 \times 29400 + 0.2 \times 31850 + 0.2 \times 34300 = 2450 + 5390 + 8820 + 6370 + 6860 = 29890$ **Maximum expected profit for two newspapers is $29890$ at $Q=245$ kits.** 6. **(b) Expected Net Profit:** Net profit is the expected gross profit minus total cost $Q \times 105$. - For one newspaper, $Q=245$: - Cost $= 245 \times 105 = 25725$ - Net profit $= 28787.5 - 25725 = 3062.5$ - For two newspapers, $Q=245$: - Cost $= 25725$ - Net profit $= 29890 - 25725 = 4165$ **Final answers:** - (a) Purchase 245 kits for both one and two newspapers to maximize expected gross profit. - (b) Expected net profit is 3062.5 for one newspaper and 4165 for two newspapers.