An interleaving crescent broadcasting protocol for near video-on-demand services

Abstract

This work presents a novel interleaving crescent broadcasting protocol for near video-on-demand service. The interleaving crescent broadcasting protocol is a trade-off among the subscriber’s access latency, maximum buffer requirement, needed subscriber’s bandwidth, and maximum disk I/O transfer rate. A longer subscriber’s access latency may cause a subscriber to leave. A lower maximum buffer requirement, a lower needed subscriber’s bandwidth, and a lower maximum disk I/O transfer rate reduce subscribers’ costs. The interleaving crescent broadcasting protocol not only makes access latency shorter, but also lowers the overall system’s cost. We prove the correctness of the interleaving crescent protocol; provide mathematical analyses to demonstrate its efficiency.

Appendix A. Proof of Lemma 5

By Theorem 4, we only consider the sum of the rates at which data is transferred from four consecutive channels to the subscriber’s disk. Let R(t) be the disk input transfer rate in the time slot t, $size\left( {i,x} \right)\,\left( { = d \cdot b/div(n)} \right)$ be the amount of data which can be transferred within one time slot in the subsubchannel $C\left[ n \right]\left( { = C_{{i,1}}^x } \right),\,D\left( {t,i,x} \right)$ be the amount of incoming data from subchannel $C_i^x$ where t is a time slot, $3 \le i \le k$ and x∈{A,B}. Let us consider 11 cases of the value R(t) in time slots $\beta_i^A, \gamma_{{i + 1}}^A, \beta_i^B, \gamma_{{i + 1}}^B, \alpha_{{i + 3}}^A \left( { = \alpha_{{i + 3}}^B } \right),\delta_i^B \left( { = \beta_{{i + 1}}^A - 1} \right)$ and the variation $R\left( {t + 1} \right) - R(t)$ for each time slot, $\beta_i^A \le t \le \beta_{{i + 1}}^A$, to show this lemma.

Case 1:

$t = \beta_i^A .$ In this case,$D\left( {t,i,A} \right) = D\left( {t,i,B} \right) = db/2$, $D\left( {t,l,x} \right) \ne 0$ for $l = i + 1,\,i + 2$, and $D\left( {t,i + 3,x} \right) = 0$ for x∈{A,B} because $\alpha_{{i + 3}}^A = \alpha_{{i + 3}}^B > \beta_i^A$. However, these data are evenly downloaded in time interval d. Therefore,

$$ \begin{array}{*{20}c} {R(t) = \left( {\sum\limits_{{\begin{array}{*{20}c} {l = i,i + 1,\,i + 2,} \hfill \\ {x \in \left\{ {A,B} \right\}} \hfill \\ \end{array} }} {D\left( {t,l,x} \right)} } \right)/d = b + \left( {\sum\limits_{{\begin{array}{*{20}c} {l = i + 1,\,i + 2} \hfill \\ {x \in \left\{ {A,B} \right\}} \hfill \\ \end{array} }} {D\left( {t,l,x} \right)} } \right)/d = b + \frac{1}{d} \cdot \left( {\sum\limits_{{x \in \left\{ {A,B} \right\}}} {D\left( {t,i + 1,x} \right)} + \sum\limits_{{x \in \left\{ {A,B} \right\}}} {D\left( {t,i + 2,x} \right)} } \right)} \\ { = b + \frac{1}{d} \cdot \left( {\left( {\beta_i^A - \alpha_{{i + 1}}^A + 1} \right)\left( {size\left( {i + 1,A} \right) + size\left( {i + 1,B} \right)} \right) + \left( {\beta_i^A - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {3 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {4 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {4 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + 2^{{i - 3}} \cdot \left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {15 \cdot db/24 + 5 \cdot db/48} \right) = 83b/48.} \\ \end{array} $$

Case 2:

$\beta_i^A \le t<\gamma_{{i + 1}}^A .$ In this case, we show R(t+1) is larger than R(t) by the following equation.

$$ \begin{array}{*{20}c} {R\left( {t + 1} \right) - R(t) = \left( { - size\left( {i,A} \right) + \left( {\left( {size\left( {i + 1,A} \right) + size\left( {i + 1,B} \right) + size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right.} \right)} \right)/d} \\ { = - b/\left( {2 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {4 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {4 \cdot 3 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right) = b/\left( {16 \cdot 2^{{i - 3}} } \right) > 0.} \\ \end{array} $$

Case 3:

$t = \gamma_{{i + 1}}^A .$ In this case, $D\left( {t,i,A} \right) \ne 0$, $D\left( {t,i,B} \right) = D\left( {t,i + 1,A} \right) = db/2$, $D\left( {t,i + 1,B} \right) \ne 0$, $D\left( {t,i + 2,x} \right) \ne 0$, and $D\left( {t,i + 3,x} \right) = 0$ for x∈{A,B} because $\alpha_{{i + 3}}^A = \alpha_{{i + 3}}^B > \gamma_{{i + 1}}^A$. Therefore,

$$ \begin{array}{*{20}c} {R(t) = \left( {\sum\nolimits_{{\begin{array}{*{20}c} {l = i,i + 1,i + 2,} \hfill \\ {x \in \left\{ {A,B} \right\}} \hfill \\ \end{array} }} {D\left( {t,l,x} \right)} } \right)/d = D\left( {t,i,A} \right)/d + b/2 + b/2 + D\left( {t,i + 1,B} \right)/d + \left( {\sum\nolimits_{{x \in \left\{ {A,B} \right\}}} {D\left( {t,i + 2,x} \right)} } \right)/d} \\ { = b\left( {1/d} \right) \cdot \left( {\left( {\delta_i^A - \gamma_{{i + 1}}^A + 1} \right)\left( {size\left( {i,A} \right)} \right) + \left( {\gamma_{{i + 1}}^A - \alpha_{{i + 1}}^B + 1} \right)\left( {size\left( {i + 1,B} \right)} \right) + \left( {\gamma_{{i + 1}}^A - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {2^{{i - 3}} \cdot \left( {db/\left( {4 \cdot 2^{{i - 3}} } \right)} \right) + 4 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {4 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + 2 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {db/4 + db/3 + 5db/24} \right) = 86b/48.} \\ \end{array} $$

Case 4:

$\gamma_{{i + 1}}^A \le t<\beta_i^B .$ In this case, we show R(t+1) is smaller than R(t) by the following equation.

$$ \begin{array}{*{20}c} {R\left( {t + 1} \right) - R(t) = \left( { - size\left( {i,A} \right) + \left( {size\left( {i + 1,B} \right) + size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right)/d} \\ { = \left\{ {\left[ { - size\left( {i,A} \right) + \left( {size\left( {i + 1,A} \right) + size\left( {i + 1,B} \right) + size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right] - size\left( {i + 1,A} \right)} \right\}/d.} \\ \end{array} $$

By Case 2: $\beta_i^A \le t<\gamma_{{i + 1}}^B, \;R\left( {t + 1} \right) - R(t) = b/\left( {16 \cdot 2^{{i - 3}} } \right) - size\left( {i + 1,A} \right)/d = b\left( {16 \cdot 2^{{i - 3}} } \right) - b/\left( {4 \cdot 2 \cdot 2^{{i - 3}} } \right) = - b/\left( {16 \cdot 2^{{i - 3}} } \right)<0.$

Case 5:

$t = \beta_i^B$. In this case, $D\left( {t,i,A} \right) = 0$, $D\left( {t,i,B} \right) = D\left( {t,i + 1,A} \right) = db/2$, $D\left( {t,i + 1,B} \right) \ne 0$, $D\left( {t,i + 2,x} \right) \ne 0$, and $D\left( {t,i + 3,x} \right) = 0$ for x∈{A,B} because $\alpha_{{i + 3}}^A = \alpha_{{i + 3}}^B > \beta_i^B$. Therefore,

$$ \begin{array}{*{20}c} {R(t) = b/2 + b/2 + D\left( {t,i + 1,B} \right)/d + \left( {\sum\nolimits_{{x \in \left\{ {A,B} \right\}}} {D\left( {t,i + 2,x} \right)} } \right)/d} \\ { = b + \left( {1/d} \right) \cdot \left( {\left( {\beta_i^B - \alpha_{{i + 1}}^B + 1} \right)\left( {size\left( {i + 1,B} \right)} \right) + \left( {\beta_i^B - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + \left( {size\left( {i + 2,B} \right)} \right.} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {5 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {4 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + 3 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {5db/12 + 15db/48} \right) = b\left( {1 + 5/12 + 15/48} \right) = 83b/48.} \\ \end{array} $$

Case 6:

$\beta_i^B \le t<\gamma_{{i + 1}}^B .$ In this case, we show R(t+1) is larger than R(t) by the following equation.

$$ \begin{array}{*{20}c} {R\left( {t + 1} \right) - T(t) = \left( { - size\left( {i,B} \right) + \left( {size\left( {i + 1,B} \right) + size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right)/d} \\ { = - b/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + b/\left( {4 \cdot 3 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right) = b/\left( {48 \cdot 2^{{i - 3}} } \right) > 0.} \\ \end{array} $$

Case 7:

$t = \gamma_{{i + 1}}^B$. In this case, $D\left( {t,i,A} \right) = 0$, $D\left( {t,i,B} \right) \ne 0$, $D\left( {t,i + 1,x} \right) = db/2$, $D\left( {t,i + 2,x} \right) \ne 0$, and $D\left( {t,i + 3,x} \right) = 0$ for x∈{A,B} because $\alpha_{{i + 3}}^A = \alpha_{{i + 3}}^B > \gamma_{{i + 1}}^B$. Therefore,

$$ \begin{array}{*{20}c} {R(t) = D\left( {t,i,B} \right)/d + b/2 + b/2 + \left( {\sum\nolimits_{{x \in \left\{ {A,B} \right\}}} {D\left( {t,i + 2,x} \right)} } \right)/d} \\ { = b + \left( {1/d} \right) \cdot \left( {\left( {\delta_i^B - \gamma_{{i + 1}}^B + 1} \right)\left( {size\left( {i,B} \right)} \right) + \left( {\gamma_{{i + 1}}^B - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {2 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + 4 \cdot 2^{{i - 3}} \cdot \left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \right)} \\ { = b + \left( {1/d} \right) \cdot \left( {db/3 + 5db/12} \right) = b + b/3 + 5b/12 = 84b/48.} \\ \end{array} $$

Case 8:

$\gamma_{{i + 1}}^B \left( { = 8 \cdot 2^{{i - 3}} - 1} \right) \le t<\alpha_{{i + 3}}^A \left( { = 8 \cdot 2^{{i - 3}} } \right)$. In this case, we show R(t+1) is smaller than R(t) by the following equation.

$$ \begin{array}{*{20}c} {R\left( {t + 1} \right) - R(t) = \left( { - size\left( {i,B} \right) + \left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right) + size\left( {i + 3,A} \right) + size\left( {i + 3,B} \right)} \right)} \right)/d} \\ { = - b/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right) + b/\left( {16 \cdot 2 \cdot 2^{{i - 3}} } \right) + b/\left( {16 \cdot 3 \cdot 2^{{i - 3}} } \right) = - b/\left( {48 \cdot 2 \cdot 2^{{i - 3}} } \right)<0.} \\ \end{array} $$

Case 9:

$t = \alpha_{{i + 3}}^A$. In this case, $D\left( {t,i,A} \right) = 0$, $D\left( {t,i,B} \right) \ne 0$, $D\left( {t,i + 1,x} \right) = db/2$, $D\left( {t,i + 2,x} \right) \ne 0$ and $D\left( {t,i + 3,x} \right) \ne 0$ for x∈{A,B}. Therefore,

$$ \begin{array}{*{20}c} {R(t) = D\left( {t,i,B} \right)/d + b/2 + b/2 + \left( {\sum\nolimits_{\begin{subarray}{l} x \in \left\{ {A,B} \right\} \\ l = i + 2,i + 3 \end{subarray} } {D\left( {t,l,x} \right)} } \right)/d} \\ { = b + \frac{1}{d} \cdot \left( {\left( {\delta_i^B - \alpha_{{i + 3}}^A + 1} \right)\left( {size\left( {i,B} \right)} \right) + \left( {\alpha_{{i + 3}}^A - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right) + size\left( {i + 3,A} \right) + size\left( {i + 3,B} \right)} \right)} \\ { = b + \frac{1}{d} \cdot \left( {\left( {2 \cdot 2^{{i - 3}} - 1} \right)\left( {db/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + \left( {4 \cdot 2^{{i - 3}} + 1} \right)\left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + db/\left( {16 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {16 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \\ { = b + \frac{1}{d} \cdot \left( {db/3 - db/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + 5db/12 + 5db/\left( {48 \cdot 2^{{i - 3}} } \right) + 5db/\left( {6 \cdot 16 \cdot 2^{{i - 3}} } \right)} \right)} \\ { = b + b/3 - b/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + 5b/12 + 5b/\left( {48 \cdot 2^{{i - 3}} } \right) + 5b/\left( {6 \cdot 16 \cdot 2^{{i - 3}} } \right) = 84b/48 - b/\left( {96 \cdot 2^{{i - 3}} } \right).} \\ \end{array} $$

Case 10:

$\alpha_{{i + 3}}^A \le t<\delta_i^B$. In this case, R(t+1) is smaller than R(t) since the equation

$$ R\left( {t + 1} \right) - R(t) = \left( { - size\left( {i,B} \right) + \left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right) + size\left( {i + 3,A} \right) + size\left( {i + 3,B} \right)} \right)} \right)/d - b/\left( {48 \cdot 2 \cdot 2^{{i - 3}} } \right)<0 $$

is the same as Case 8: $\gamma_{{i + 1}}^B \le t<\alpha_{{i + 3}}^A$.

Case 11:

$t = \delta_i^B$. In this case, $D\left( {t,i,A} \right) = 0$, $D\left( {t,i,B} \right) \ne 0$, $D\left( {t,i + 1,x} \right) = db/2$, $D\left( {t,i + 2,x} \right) \ne 0$, and $D\left( {t,i + 3,x} \right) \ne 0$ for x∈{A,B}. Therefore,

$$ \begin{array}{*{20}c} {R(t) = D\left( {t,i,B} \right)/d + b/2 + b/2 + \left( {\sum\nolimits_{{\begin{array}{*{20}c} {x \in \left\{ {A,B} \right\}} \hfill \\ {l = i + 2,i + 3} \hfill \\ \end{array} }} {D\left( {t,l,x} \right)} } \right)/d} \\ { = b + \frac{1}{d} \cdot \left( {size\left( {i,B} \right) + \left( {\delta_i^B - \alpha_{{i + 2}}^A + 1} \right)\left( {size\left( {i + 2,A} \right) + size\left( {i + 2,B} \right)} \right) + \left( {\delta_i^B - \alpha_{{i + 3}}^A + 1} \right)\left( {size\left( {i + 3,A} \right) + size\left( {i + 3,B} \right)} \right)} \right)} \\ { = b + \frac{1}{d} \cdot \left( {db/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + \left( {6 \cdot 2^{{i - 3}} - 1} \right)\left( {db/\left( {8 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {8 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right) + \left( {2.2^{{i - 3}} - 1} \right)\left( {db/\left( {16 \cdot 2 \cdot 2^{{i - 3}} } \right) + db/\left( {16 \cdot 3 \cdot 2^{{i - 3}} } \right)} \right)} \right)} \\ { = b + \frac{1}{d} \cdot \left( {db/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + 5db/8 - 5db/\left( {48 \cdot 2^{{i - 3}} } \right) + 5db/48 - 5db/\left( {96 \cdot 2^{{i - 3}} } \right)} \right)} \\ { = b + b/\left( {2 \cdot 3 \cdot 2^{{i - 3}} } \right) + \left( {5b/8 - 5b/\left( {48.2^{{i - 3}} } \right)} \right) + \left( {5b/48 - 5b/\left( {96.2^{{i - 3}} } \right)} \right) = 83b/48 + b/\left( {96.2^{{i - 3}} } \right).} \\ \end{array} $$

Up to now, we have calculated the values of R(t) in these time slots $\beta_i^A, \gamma_{{i + 1}}^A, \beta_i^B, \gamma_{{i + 1}}^B \alpha_{{i + 3}}^A \left( { = \alpha_{{i + 3}}^B } \right),\delta_i^B$ and the variations R(t+1)–R(t) for each time slot t, $\beta_i^A \le t<\beta_{{i + 1}}^A$ as summarized in Table 5. Thus $83b/48 \le R(t) \le 86b/48$ holds for any time slot t with $\beta_i^A \le t<\beta_{{i + 1}}^A$. □